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Students can Download Chapter 11 Dual Nature of Radiation and Matter Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter

Plus Two Physics Dual Nature of Radiation and Matter NCERT Text Book Questions and Answers

Question 1.
Find the

1. Maximum frequency, and
2. Minimum wavelength of X-rays produced by 30 kV electrons.

Answer:
Given V_o = 30 kV = 30 × 10³ V
v_max = ?
λ_max = ?
1. Since k_max = eV_o
So hv_max = eV_o
or

= 7.24 × 10¹⁸ Hz

2.

= 0.041 × 10^-9
or λ_min = 0.041 nm.

Question 2.
The work function of caesium metal is 2.14 eV. When light of frequency 6 × 10¹⁴Hz is incident on the metal surface, photoemission of electrons occurs. What is the

1. maximum kinetic energy of the emitted electrons,
2. maximum speed of the emitted photoelectrons?

Answer:
Given W₀ = 2.14 eV
= 2.14 × 1.6 × 10^-19 J
= 3.424 × 10^-19J
v = 6 × 10¹⁴ Hz

1. K_max = hv – W₀
= 6.62 × 10^-34 × 6 × 10¹⁴ – 3.424 × 10¹⁹
= 3.972 × 10^-19 – 3.424 × 10^-19
= 0.54 × 10^-19 J.

2. Since eV₀ =k_max

or v₀ = 0.34 V
Since \(\frac{1}{2}\) mV²_max = K_max

or v_max = 0.344 × 10⁶ ms-1 = 344 kms^-1

Question 3.
The photoelectronic cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Answer:
Given V₀ = 1.5 V
Since k_max = eV₀
= 1.5 eV
= 1.5 × 1.6 × 10^-19 J = 2.4 × 10^-19 J.

Question 4.
In an experiment on photoelectric effect, the slope of the cutoff voltage versus frequency of incident light is found to be 4.12 × 10^-15Vs. Calculate the value of Planck’s constant.
Answer:
Given, slope of graph = 4.12 × 10^-15 Vs
since, slope of graph = \(\frac{h}{e}\)
∴ h = e × slope of graph
= 1.6 × 10^-19 × 4.12 × 10^-15 = 6.59 × 10^-34 Js.

Question 5.
The threshold frequency fora certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.
Answer:
Given v₀ = 3.3 × 10¹⁴Hz
v = 8.2 × 10¹⁴Hz
Since eV₀ = hv – hv₀
So V₀ = \(\frac{h}{e}\) (v – v₀)
\(=\frac{6.62 \times 10^{-34}}{1.6 \times 10^{-19}}\) × (8.2 × 10¹⁴ – 3.3 × 10¹⁴)
= 4.14 × 10^-15 × 4.9 × 10¹⁴
= 2.02 V = 2.0 V

Question 6.
Light of frequency 7.21 × 10¹⁴Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer:
Given v = 7.21 × 10¹⁴Hz
u_max = 6.0 × 10¹⁴ms^-1
v₀ = ?
Since K_max = hv – hv₀
∴ v₀ = v – K_max

Plus Two Physics Dual Nature of Radiation and Matter One Mark Questions and Answers

Question 1.
Find out the wrong statement
(i) As frequency increases photo current increases
(ii) As frequency increase KE increases
(iii) As frequency increase velocity of electrons increases
(iv) As frequency increase stopping potential increases
(v) As frequency is below a certain value photo electrons are not emitted.
Answer:
(i) As frequency increases photo current increases.

Question 2.
Read the following statements and write whether true or false.

1. During photo electric effect photon share its energy with a group of electrons.
2. Intensity is directly proportional to square of amplitude

Answer:

1. False
2. True

Question 3.
In photoelectric emission the number of photoelectrons emitted per second depends on
(a) wavelength of incident light
(b) frequency of incident light
(c) intensity of incident light
(d) work function of the material
Answer:
(c) intensity of incident light

Question 4.
What is de-Broglie wave?
Answer:
The wave associated with material particle is called de-Broglie wave.

Question 5.
In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is
(a) 1.8 V
(b) 1.3 V
(c) 0.5 V
(d) 2.3 V
Answer:
(c) 0.5 V
Explanation:
The stopping potential V_s is related to the maximum kinetic energy of the emitted electrons K_max through the relation
K_max = eV_s
0.5 eV = eV_s or V_s = 0.5 V

Question 6.
Name the experiment, which establish the wave nature of moving electrons.
Answer:
Davisson Germer experiment.

Question 7.
Pick the odd one out of the following,
(a) Interference
(b) Diffraction
(c) Polarization
(d) Photoelectric effect
Answer:
(d) Photoelectric effect

Question 8.
Find out the wrong statement

1. If two particles have same momentum then they have same de-Broglie wave length.
2. If two particles have same KE the lighter particle has smallerwave length
3. As velocity of a given mass decreases wave-length increases

Answer:

1. True
2. False
3. True

Question 9.
Pick the odd one out from the following x-rays, visible light, matter waves, radio waves
Answer:
Matter waves.

Question 10.
If the electrons are accelerated by a potential of 50V, calculate the de-Broglie wavelength of electrons.
Answer:

Plus Two Physics Dual Nature of Radiation and Matter Two Mark Questions and Answers

Question 1.
Classify the following properties of the waves into de Broglie wave, em wave, and sound wave.

1. Associated with the moving particle.
2. Longitudinal wave
3. Electric field and magnetic field are perpendicular to each other.
4. Can produce photo electric effect.
5. Wave length is inversely proportional to mass of the moving particle.
6. Velocity in vacuum is 3×10⁸ m/s.

Answer:

1. de-Broglie wave – 1,5
2. Em wave – 3,4,6
3. Sound wave – 2

Question 2.
Table given below gives the work function of certain elements.

Identify the element in which photoelectric effect occurs easily. Justify your answer.
Answer:
Na – 2.70eV, work function is least.

Question 3.
“Louis De Broglie suggested existence of matter waves based on a hypothesis”

1. What do you mean by matter wave?
2. The objects in our daily life do no exhibit wave like properties. Why?

Answer:

1. The wave associated with material particle is called matterwave.
2. objects in ourdaily life have large mass. Hence λ is very small (In the order of 10^-34 cm) to be neglected.

Question 4.
A body of mass 1 Kg is moving with a velocity 1 m/s, a wave is associated with this body

1. Name the wave
2. Can you measure wave length of this wave. Explain?

Answer:

1. Matterwave
2. No. wave length of matterwave is very small.

Plus Two Physics Dual Nature of Radiation and Matter Three Mark Questions and Answers

Question 1.
In figure below represents the variation of current with potential fora metal

1. Identify the law governing it.
2. Even when the potential is zero, there is current. Explain.
3. Current is zero for a particular potential. How does this potential help in determining the velocity of electrons.

Answer:
1. Laws of photoelectric emission

• For a given frequency of radiation, number of photoelectrons emitted is proportional to the intensity of incident radiation.
• The K.E. of photoelectrons depends on the frequency of incident light but is independent of the light intensity.

2. When radiation falls on metal, photo electrons are emitted with certain velocity even if accelerating potential is zero.

3. At slopping potential (v₀), photocurrent is zero.
ie. 1/2 mv²_max = ev₀

Question 2.
A metal whose work function is 2 eV is illuminated by light of wavelength 3 × 10^-7 m. Calculate

1. threshold frequency
2. maximum energy of photoelectrons
3. the stopping potential.

Answer:
F₀ = h ν₀ – 2eV

1. F₀ = h ν₀

2. 1/2mv² = h(ν – ν₀)
= 6.63 × 10^-34 (10¹⁵ – 4.8 × 10¹⁴)
= 3.44 × 10^-19J

3. eV₀ = h(ν – ν₀)

Question 3.
The wave nature of electron was experimentally verified by diffraction of electron by Nickel crystal.

1. Name the experiment which establish wave nature of moving electron.
2. An electron and a proton have same kinetic energy which of these particles has shortest de-Broglie wave length?

Answer:
1. Davisson and Gemner experiment

2.

Mass of alpha particle is more than that of proton, hence it has shortest wavelength.

Question 4.
“Moving particles of matter shows wave like properties under suitable conditions’’

1. Who put forward this hypothesis?
2. A proton and an electron have been accelerated through same potential. Which one have higher matter wave length. Write the reason

Answer:
1. De broglie

2. \(\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}\)
The mass of proton is higher than electron. Hence wave length of proton is less than electron.

Question 5.
Three light beams of same frequency and different intensity I₁, I₂ and I₃ are incident on the same metal. I₁ >I₂ >I₃.

1. Which beam produce maximum photocurrent?
2. Which beam produce electrons of maximum speed and KE?
3. Draw a graph showing variation of photocurrent with intensity in same speed.

Answer:
1. I₃.

2. Frequency is same. Hence electron emitted from the metals will be same.

3.

Plus Two Physics Dual Nature of Radiation and Matter Four Mark Questions and Answers

Question 1.
The graph shows photoelectric current with anode potential.

1. The potential at ‘O’ is called

• accelerating potential
• retarding potential
• stopping potential
• saturation potential

2. Why current becomes constant in the region BC?

3.

Is the above graph possible? Justify your answer
Answer:

1. Stopping potential
2. The whole electrons emitted from the cathode will reach at anode. Hence current becomes saturation at BC.
3. This graph is not possible. Stopping potential is directly proportional to frequency of incident light, ie. when frequency of incident light increases, stopping potential also increases. Hence we expect a high stopping potential for v₂.

Question 2.
The magnification of an electron microscope is much larger than that of an optical microscope, because electron beams are used instead of light beams in an electron microscope.

1. Which property of electrons is used in the construction of the electron microscope?
2. Obtain expression for the wavelength of de-Droglie waves associated with an electron accelerated through a potential of V volts.

Answer:
1. Wave nature.

2. ev = 1/2mv²
mv² = 2 eV
m²v² = 2eVm

Question 3.
The wavelengths of violet and red ends of visible spectrum are 390nm and 760 nm respectively.
1. Evaluate the energy range of the photons of the visible light in electron volts.
2. The work function of four different materials is given in the table below.

Pick out the suitable metal/ metals for the construction of the photo cell which is to operate with visible light.
3. Calculate the threshold frequency of the selected metal/metals.
Answer:
1. 1.65ev to 3.1ev

2. Cs

3.

Question 4.
“To emit a free electron from a metal surface a minimum amount of energy must be supplied”.

1. It is called……..
2. Give three method to supply energy to a free electron
3. For metal A (tungsten) – work function is 4.52 eV for metal B (thoriated tungsten) it is 2.6 ev, for metallic (oxide coated tungsten) it is 1 eV. Which will you prefer as a good electron emitter and why?

Answer:

1. Threshold energy
2. Give light energy or heat energy
3. Work function for metallic oxide coated tungsten is small (1 ev.) Hence this material is good electron emitter.

Question 5.
Louis de Broglie argued that electron in circular orbit as proposed by Bohr, must be seen as a Particl wave.

1. From Bohr’s postulate of angular quantization momentum, arrive at an expression for wave length of an orbital electron. (2)
2. Comment on the above result (2)

Answer:
1.

2. Since λ = \(\frac{2 \pi r}{n}\), length of the first orbit is the de-Broglie wavelength of the orbit.

Plus Two Physics Dual Nature of Radiation and Matter Five Mark Questions and Answers

Question 1.
Schematic diagram of an experimental set up to study the wave nature of electron is shown below.

1. Identify the experiment.
2. In the experiment the intensity of electron beam is measured for different values of ‘q’. At 54V accelerating potential and q = 50°, a sharp diffraction maximum is obtained. What is the wave length associated with the electron.
3. A particle is moving three times as fast as electron. The ratio of debroglie wavelength of the particle to that electron is 1.813 × 10^-14. Calculate the mass of the particle.

Answer:
1. Davisson and German experiment

2. \(\lambda=\sqrt{\frac{150}{v}}=\sqrt{\frac{150}{54}}=1.66 \mathrm{A}^{0}\)

3.

Question 2.
An electron moves under a potential difference of 300V

1. The wave associated with electron is called……….
2. Derive an expression for its wave in terms of charge of particle.
3. Calculate the wavelength of above electron.

Answer:
1. matter wave

2.

3.

Question 3.
Figure below shows a version of Young’s Experiment performed by directing a beam of electrons on double slit. The screen reveals a pattern of bright and dark fringes similar to an interference as pattern produced when a beam of light is used.

1. Which property of electron is revealed in this observation?
2. If the electrons are accelerated by a p.d. of 54V, what is the value of wavelength associated with electrons.
3. In similar experiment if the electron beam is replaced by bullets fired form a gun, no interference pattern is observed. Give reason.

Answer:
1. Dual nature or wave nature.

2.

3. λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) Since the mass of the bullet is very large compared to the mass of electron, the de Broglie wavelength is not considerable.

Question 4.
A particles moving with KE 5Mev. Its mass is 1.6 × 10^-27 Kg

1. What is the energy of particle in joule?
2. Derive an equation of find De Broglie wave length in terms of KE
3. Calculate De-Broglie wave length of above particle.

Answer:
1. KE = 5 Mev = 5 × 10⁶ × 1.6 × 10^-19J

2.

3.

Question 5.
Einstein got Nobel Prize in 1921 for his explanation of photoelectric effect.

1. In order to start photoelectric emission, the minimum energy acquired by free electron in the metal is called as…… (1)
2. The minimum energy forthe emission of an electron from metallic surface is given below Na: 2.75eV K: 2.3eV Mo:4.17eV Ni:5.15eV Select the metal which is more photo sensitive. Why? (1)
3. Draw variation of photoelectric current with applied voltage for radiation of intensities I₁ and I₂ (I₁ > I₂). Comment on the relatiion between intensity of light and photoelectric current. (2)
4. Does Light from a bulb falling on an iron table emit photoelectron? Justify your answer. (1)

Answer:
1. Work function/ Threshold energy.

2. K is more photosensitive because it has less work function.

3.

As intensity increases photoelectric current also increases.

4. No. The work function of iron is very large.

Question 6.
Lenard and Hallwachs investigated the phenomenon of photoelectric effect in details during 1886-1902 through experiments
1. Write any two characteristic features observed in the above experiment? (2)
2. Explain with reason

• Green light emit electron from certain metal surface while yellow light does not
• When the wavelength of incident light is decreased, the velocity of emitted photo electrons increases (2)

3. Complete the following statement about photoelectric effect.
The radiations having minimum frequency called…….falls on a metallic surcace, electrons are emitted from it. The metal which emits photoelectrons are called………The kinetic energy of photoelectrons emitted by a metal depends on………of the radiations, while intensity of the incident radiations depends on……… (1)
Answer:
1. Any two statement of laws of photoelectric effect.

2. Explain with reason:

• Energy of incident photon is inversely proportional to its wavelength. Since λ of green light is less than that of yellow, it has larger energy. So it can emit photoelectrons
• As the wavelength decreases, frequency and hence energy of incident radiation increases and hence kinetic energy of photo electrons increases.

3. Threshold frequency, photosensitive frequency, number of photons.

Question 7.
Figure below shows variation of stopping potential (V₀) with frequency(?) of incident radiations for two different metals A and B.

1. Write down the values of work function A and B.

2. What is the significance of slope of the above graph? (1)

3. The value of stopping potential for A and B for a frequency γ₀₁ (which is greaterthan γ₀₂) of incident radiations are V₁ and V₀ respectively. Show that the slopes of the lines is equal to \(\frac{v_{1}-v_{2}}{\gamma_{01}-\gamma_{02}}\). (3)
Answer:
1. Work function of A, Φ₀₁ – hν₀₁
Work function of B, Φ₀₁ – hν₀₂.

2. The slope of the graph gives value of h/e.

3. For the metalA, hν₁ = hν₀₁ + eV₁………..(1)
For the metal B, hν₁ = hν₀₂ + eV₂…………..(2)
From equation (1) and (2)
hν₀₁ + eV₁ = hν₀₂ + eV₂
e(V₁ – V₂) = h(ν₀₂ – ν₀₁)
\(\frac{h}{e}=\frac{V_{1}-V_{2}}{v_{02}-v_{01}}\)

Question 8.
Albert Einstein proposed a radically new picture of electromagnetic radiation to explain photoelectric effect.
1. Identify Einstein’s photoelectric equation? (1)
2. With the help of Einstein’s photoelectric equation explain the following facts.

• Kinetic energy of photoelectrons is directly proportional to frequency not on intensity.
• Existence of threshold frequency for a given photosensitive material. (2)

3. A metal whose work function is 2 eV is illuminated by light of wavelength 3 × 10^-7m. Calculate

• threshold frequency
• maximum energy of photoelectrons
• the stopping potential. (3)

Answer:
1. hν = hν₀ + 1/2 mv²

2. Einstein’s photoelectric equation:
a. hν ∝ 1/2 mv²
1/2 mv² ∝ ν
Hence kinetic energy is proportional to frequency.

b. hν – hν₀ = 1/2 mv²
h(ν – ν₀) = 1/2 mv²
ν should be greater than ν₀ otherwise h(ν – ν₀) is negative and is not possible.

3.

a. Φ₀ = hν₀
ν₀ = \(\frac{2 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}\)
= 4.8 × 10¹⁴ Hz

b. 1/2 mv² = h(ν – ν₀)
= 6.63 × 10^-34 (10¹⁵ – 4.8 × 10¹⁴)
= 3.44 × 10^-19J

c. ev₀ = h(ν – ν₀)

Question 9.
The wave nature of electron was experimentally verified by diffraction of electron by Nickel crystal.

1. Name the experiment which establish wave nature of moving electron. (1)
2. With a neat diagram explain the existence of matter wave associated with an electron. (3)
3. An electron and a proton have same kinetic . energy which of these particles has shortest de-Broglie wave length? (2)

Answer:
1. Davisson and Germer experiment.

2.

Aim:
To confirm the wave nature of electron.
Experimental setup:
The Davisson and Germer Experiment consists of filament ‘F’, which is connected to a low tension battery. The Anode Plate (A) is used to accelerate the beam of electrons. A high voltage is applied in between A and C. ’N’ is a nickel crystal. D is an electron detector. It can be rotated on a circular scale. Detector produces current according to the intensity of incident beam.

Working:
The electron beam is produced by passing current through filament F. The electron beam is accelerated by applying a voltage in between A (anode) and C. The accelerated electron beam is made to fall on the nickel crystal.

The nickel crystal scatters the electron beam to different angles The crystal is fixed at an angle of Φ = 50° to the incident beam. The detector current for different values of the accelerating potential ‘V’ is measured. A graph between detector current and voltage (accelerating) is plotted. The shape of the graph is shown in figure.

Analysis of graph:

The graph shows that the detector current increases with accelerating voltage and attains maximum value at 54V and then decreases. The maximum value of current at 54 V is due to the constructive interference of scattered waves from nickel crystal (from different planes of crystal). Thus wave nature of electron is established.

Experimental wavelength of electron:
The wave length of the electron can be found from the formula
2d sinθ = n λ ………..(1)
From the figure, we get
θ + Φ + θ = 180°
2θ = 180 – Φ, 2θ = 180 – 50°
θ = 65°
for n = 1
equation (1) becomes
λ = 2d sinθ ………(2)
for Ni crystal, d = 0.91 A°
Substituting this in eq. (2), we get
wavelength λ = 1.65 A°

Theoretical wave length of electron:
The accelerating voltage is 54 V
Energy of electron E = 54 × 1.6 × 10^-19J
∴ Momentum of electron P = \(\sqrt{2 \mathrm{mE}}\)
P = \(\sqrt{2 \times 9.1 \times 10^{-31} \times 54 \times 1.6 \times 10^{-19}}\)
= 39.65 × 10^-25 Kg ms^-1
∴ De- Broglie wavelength λ = \(\frac{h}{P}\)

The experimentally measured wavelength is found in agreement with de-Broglie wave length. Thus wave nature of electron is confirmed.

3.

Mass of alpha particle is more than that of proton, hence it has shortest wavelength.

Question 10.
In Geiger-Marsden Scattering experiment alpha particles of 5.5 MeV is allowed to fall on a thin gold foil of thickness 2.1 × 10^-7m.
1. Draw Schematic diagram of above experimental arrangement.
2.

In the above graph nearly 10⁷ particles were detected when scattering angle is Zero. What do you understand by it?
3. Why gold foil is used in this experiment?
4. Does there exist any relation between impact parameter and scattering angle? If yes, explain your answer.
Answer:
1.

2. Most of the alpha particles get unscattered means that most of the space in an atom is empty.

3. Atomic number of gold 79, so number of protons is very high. Hence scattering between alpha and nucleons is larger. Gold foil can be made very thin so that the alpha particles suffer not more than one scattering.

4. Yes.
As impact Parameter increases, scattering angle decreases.

Question 11.
The study of emission line spectra of a material serve as a fingerprint for identification of the gas.

1. Name different series of lines observed in hydrogen spectrum. (1)
2. Draw energy level diagram of hydrogen atom? (2)
3. Write down the Balmer formula for wavelength of H_α line. (1)
4. Given Rydberg constant as 1.097 × 10⁷m^-1. Find the longest and shortest wavelength limit of Baler Series. (2)

Answer:
1. Lyman series, Balmer series, Paschen series, Bracket series, Pfund series

2.

3.

4.

Longest wavelength n₁ = 2 and n₂ = 3

Shortest Wavelength n₁ = 2 and n₂ = α

Question 12.
Bohr combined classical and early quantum concept and gave his theory in the form of three postulates.

1. State three postulates of Bohr Model of atom? (2)
2. The total energy of an electron in ground state of hydrogen atom is -13.6eV. What is the significance of negative sign? (1)
3. The radius of innermost electron orbit of hydrogen atom is 5.3 × 10¹¹m. What are the radii of n = 2 and n = 3 orbits? (2)

Answer:
1. Bohr postulates:
Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates.

• Electrons revolve round the positively charged nucleus in circular orbits.
• The electron which remains in a privileged path cannot radiate its energy.
• The orbital angular momentum of the electron is an integral multiple of h/2π.
• Emission or Absorption of energy takes place when an electron jumps from one orbit to an other.

2. Negative sign implies that the electrons are strongly bounded to the nucleus.

3.

• r_n = n²a₀ = 5.3 × 10^-11m
• r₁ = a₀ = 5.3 × 10^-11m
• r₂ = 4a₀ = 21.2 × 10^-11m
• r₃ = 9a₀ = 47.7 × 10^-11m.

Students can Download Chapter 5 Measures of Central Tendency Questions and Answers, Plus One Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency

Plus One Economics Measures of Central Tendency One Mark Questions and Answers

Question 1.
The midpoint of the class ‘5-10’ is:
(i) 5
(ii) 7.5
(iii) 10
(iv) 15
Answer:
(ii) 7.5

Question 2.
Mode is equal to:
(i) 3 median – 2 mean
(ii) 2 median – 3 mea
(iii) 3 median – 3 mean
(iv) 3 median -1 mean
Answer:
(i) 3 median – 2 mean

Question 3.
Which of the following is a positional average?
(i) Mean
(ii) Median
(iii) Mode
(iv) All the above
Answer:
(ii) Median

Question 4.
Which of the following divides the data into four equal parts?
(i) decile
(ii) percentile
(iii) quartiles
(iv) none of the above
Answer:
(iii) quartiles

Question 5.
Which of the following is the most commonly used average?
(i) Arithmetic Mean
(ii) Median
(iii) Mode
(iv) Percentile
Answer:
(i) Arithmetic Mean

Question 6.
Mode can be graphically located by means of
(i) bio diagram
(ii) pie diagram
(iii) histogram
(iv) ogive
Answer:
(iii) histogram

Question 7.
The most suitable average for qualitative measurement is
(i) Arithmetic mean
(ii) Median
(iii) Mode
(iv) Geometric mean
(v) None of the above
Answer:
(ii) Median

Question 8.
Which average is affected most by the presence of extreme items?
(i) median
(ii) Mode
(iii) Arithmetic mean
(iv) Geometric mean
(v) Harmonic mean
Answer:
(iii) Arithmetic mean

Question 9.
The algebraic sum of deviation of a set of n values from A.M. is
(i) n
(ii) 0
(iii) 1
(iv) None of the above
Answer:
(ii) 0

Question 10.
The average value of a given variable is known as ____
Answer:
A.M

Question 11.
Total of given variables can be depicted by _____
Answer:
Σx

Question 12.
Common factor is depicted by ____
Answer:
‘c’

Question 13.
A.M should be
(i) Simple
(ii) Based on all items
(iii) Rigidly defined
(iv) All the above
Answer:
(iv) All the above

Question 14.
Median is the _____ value in a series.
Answer:
Middle.

Question 15.
Q₃ represents ____ Quartile.
Answer:
Middle.

Question 16.
_____ is the division of the series into 100 equal parts.
Answer:
Percentiles.

Question 17.
\(\left(\frac{N+1}{10}\right)^{t h}\) is used to calculate _____.
Answer:
Deciles

Question 18.
Value of median is equal to _____Answer:
Answer:
II Quartile – 50^th percentile, 5^th Decile

Question 19.
Pick out the odd one out and Justify.
Arithmetic mean, Median, Standard deviation, Mode
Answer:
Standard deviation. Others are measures of central tendency.

Question 20.
Which average would be suitable in the following cases?

1. Average size of readymade garments.
2. Average intelligence of students in a class.
3. Average production in a factory per shift.
4. Average wages in an industrial concern.
5. When the sum of absolute deviations from average is least.
6. When quantities of the variable are in ratios.
7. In case of open-ended frequency distribution.

Answer:

1. Mode
2. Median
3. Mode or median
4. Mode or median
5. Mean
6. Mode or mean
7. Median

Plus One Economics Measures of Central Tendency Two Mark Questions and Answers

Question 1.
Name the types of positional averages.
Answer:
Positional averages are median and mode.

Question 2.
Explain weighted Arithmetic mean
Answer:
When calculating Arithmetic means it is important to assign weights to various items according to their importance. The arithmetic mean calculated with the relative importance to different items is known as weighted arithmetic mean.

Question 3.
Give the special features of arithmetic mean.
Answer:
It is interesting to know and useful for checking your calculation that the sum of deviations of items about arithmetic mean is always equal to zero. Symbolically, S (X-X) = 0.However, arithmetic mean is affected by extreme values. Any large value, on either end, can push it up or down.

Question 4.
If median and mean of distribution are respectively 18.8 and 20.2. What would be its made?
Answer:
Mode = 3 Median – 2 Mean
= 3 × 18.8 – 2 × 20.2
= 56.4 – 40.4 = 16

Question 5.
Mention two demerits of median.
Answer:

1. Median is not based on all observations
2. It cannot be given for further mathematical treatment

Question 6.
Give the formulae of median in all series
Answer:
Formulae of Median

Question 7.
If median is 15 and mean is 17, calculate mode?
Answer:
Mode = 3 median – 2 mean
= 3 × 15 – 2 × 17
= 45 – 34 = 11

Question 8.
Can there be a situation where mean, median and mode are equal?
Answer:
Yes. Mean, median and mode will be equal when all given variables are the same.

Question 9.
Mark the missing value of the following data. The mean marks are 10.05.
5, 6, 7, 8, 12, ?, 15, 17, 18, 3, 10 5, 10, 12, 15, 11, 13, 1
Answer:

Plus One Economics Measures of Central Tendency Three Mark Questions and Answers

Question 1.
What is the relative position of arithmetic mean, median and mode?
Answer:
Relative position of arithmetic mean, median and mode can be understood from the following narration. Suppose we express,
Arithmetic Mean = Me
Median = Mi
Mode = Mo
The relative magnitude of the three is
Me>Mi>Moor
Me<Mi<Mo
That is the median is always between the arithmetic mean and the mode.

Question 2.
Write down the advantages of median.
Answer:
Merits of median:

• It is easy to understand
• It is not affected by extreme values It can be graphically determined
• It is suitable in case of open-end classes
• It is suitable for qualitative measurement

Question 3.
Write the merits of mode.
Answer:
Merits of mode

• It is easy to understand and simple to calculate
• It is not affected by extreme values
• It can be graphically determined
• It is suitable in case of open-end classes.

Question 4.
Complete the following

1. ………………. divides the series into two equal parts
2. The central tendency based on all values is …………..
3. The average which can be determined through ogive is ………

Answer:

1. median
2. mean
3. median

Question 5.
The mean mark of 60 students in section A are 40 and mean mark if 40 student in section B is 35. Calculate the combined mean of all the students.
Answer:

Plus One Economics Measures of Central Tendency Four Mark Questions and Answers

Question 1.
Point out important features of a good average.
Answer:
The important features of a good average are given below.

1. It should be easy to understand
2. It should be simple to calculate
3. It should be rigidly defined
4. It should be based on all observations
5. It should not be affected by extreme values.
6. It should be capable for further statistical calculations.

Question 2.
Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.

Answer:

Question 3.
Calculate the median from the following data.

X	F
0-10	5
10-20	8
20-30	10
30-40	14
40-50	3

Answer:

Plus One Economics Measures of Central Tendency Five Mark Questions and Answers

Question 1.
Comment whether the following statements are true or false.

1. The sum of deviation of items from median is zero.
2. An average alone is not enough to compare series.
3. Arithmetic mean is a positional value.
4. The upper quartile is the lowest value of top 25% of items.
5. Median is unduly affected by extreme observations.

Answer:

1. False
2. True
3. False
4. True
5. False

Question 2.
There are three types of averages. Name them. Also, give appropriate definitions.
Answer:
There are several statistical measures of central tendency or “averages”. The three most commonly used averages are:

• Arithmetic Mean
• Median
• Mode

1. Arithmetic mean:
Arithmetic mean is the most commonly used measure of central tendency. It is defined as the sum of the values of all observations divided by the number of observations

2. Median:
Median is that positional value of the variable which divides the distribution into two equal parts, one part comprises all values greater than or equal to the median value and the other comprises all values less than or equal to it. The Median is the “middle” element when the data set is arranged in order of the magnitude.

3. Mode:
The word mode has been derived from the French word “la Mode” which signifies the most fashionable values of distribution because it is repeated the highest number of times in the series. Mode is the most frequently observed data value. It is denoted by Mo.

Question 3.
What are the merits and demerits of arithmetic mean?
Answer:
1. Merits

• It is simple to calculate
• It is regidly defined
• It is easy to understand
• It is based on all observations

2. Demerits

• It is affected by extreme values.
• It cannot be calculated in open-end series.
• It cannot be determined graphically.
• It may sometimes give misleading results.

Plus One Economics Measures of Central Tendency Eight Mark Questions and Answers

Question 1.
Prepare a list of peculiarities of median, quartiles, and percentiles.
Answer:
1. Median

• The arithmetic mean is affected by the presence of extreme values in the data.
• If you take a measure of central tendency which is based on middle position of the data, it is not affected by extreme items.
• Median is that positional value of the variable which divides the distribution into two equal parts, one part comprises all values greater than or equal to the median value and the other comprises all values less than or equal to it.
• The Median is the middle element when the data set is arranged in order of the magnitude.

2. Quartiles

• Quartiles are the measures that divide the data into four equal parts; each portion contains equal number of observations. Thus, there are three quartiles.
• The first Quartile (denoted by Q₁) or lower quartile has 25% of the items of the distribution below it and 75% of the items are greater than it.
• The second Quartile (denoted by Q₂) or median has 50% of items below it and 50% of the observations above it.
• The third Quartile (denoted by Q₃) or upper Quartile has75% of the items of the distribution below it and 25% of the items above it.
• Thus, Q₁ and Q₃ denote the two limits within which central 50% of the data lies.

3. Percentiles

• Percentiles divide the distribution into hundred equal parts, so you can get 99 dividing positions denoted by P₁ P₂, P_{3, ……….}, P₉₉.
• P50 is the median value.
• If you have secured 82 percentile in a management entrance examination, it means that your position is below 18 percent of total candidates appeared in the examination.

Question 2.
The daily sales of car of 20 distributing companies is given below. Calculate:

1. Median, upper quartile and lower quartile.
2. Interpret the result obtained.

 

Daily Sales	No. of Companies
0-20	1
20-40	3
40-60	9
60-80	5
80-100	2
	20

Answer:
1.



2. The median divides the values into two equal parts. Lower quartile (Q₁) divides the values into 1/4 and upper quartile (Q₃) divides the values into 3/4.

Students can Download Chapter 13 Probability Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

Plus Two Maths Probability Three Mark Questions and Answers

Question 1.
Determine P(E/F). A die is thrown three times, E: ‘4 appears on the third toss’, F: ‘6 and 5 appears respectively on the two tosses’.
Answer:
n(S) = 6³ = 216
E = {( 1, 1, 4), (1, 2, 4), (1, 3, 4)……….(1, 6, 4),
(2, 1, 4), (2, 2, 4), (2, 3, 4)……..(2, 6, 4),
(3, 1, 4), (3, 2, 4), (3, 3, 4)……..(3, 6, 4),
(4, 1, 4), (4, 2, 4), (4, 3, 4)…….(4, 6, 4),
(5, 1, 4), (5, 2, 4), (5, 3, 4)……..(5, 6, 4),
(6, 1, 4), (6, 2, 4), (6, 3, 4)……..(6, 6, 4)}
F = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
⇒ E ∩ F = {(6, 5, 4)}
P(F) = \(\frac{6}{216}\) and P(E ∩ F) = \(\frac{1}{216}\)
Then, P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}\).

Question 2.
Determine P(E/F). Mother, Father and son lineup at random for a photograph.
E: ‘Son on one end’, F: ‘ Father in middle.
Answer:
Let Mother-M, Father-F and Son-S.
n(S) = 3! = 6
E = {SMF, SFM, MFS, FMS},
F = {MFS, SFM}
⇒ E ∩ F = {SFM, MFS}
P(F) = \(\frac{2}{6}\) = \(\frac{1}{3}\) and P(E ∩ F) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Then, P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{3}}{\frac{1}{3}}=1\).

Question 3.
A black and a red dice are rolled

1. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
2. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Answer:
We have, n(S) = 36
1. E = Event of 5 on black die.
E = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
P(E) = \(\frac{6}{36}\) = \(\frac{1}{6}\)
F = Getting a sum greater than 9.
F = {(4, 6), (5, 5), (6, 4)(5, 6), (6, 5), (6, 6)}
⇒ E ∩ F = {(5,5), (5,6)}
P(E ∩ F) = \(\frac{2}{36}\)= \(\frac{1}{18}\)
Therefore the required probability
P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{18}}{\frac{1}{6}}=\frac{1}{3}\).

2. E = Event of a number less than 4 on red die.
E = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3),
(3, 1), (3, 2), (3, 3), (3, 1), (3, 2), (3, 3),
(5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}
P(E) = \(\frac{18}{36}\) = \(\frac{1}{2}\)
F = Getting a sum 8.
F = {(4, 4), (5, 3), (3, 5)(2, 6), (6, 2), (6, 6)}
⇒ E ∩ F = {(5, 3),(6, 2)}
P(E ∩ F) = \(\frac{2}{36}\)= \(\frac{1}{18}\)
Therefore the required probability
P(E/F) = \(\frac{P(E \cap F)}{P(E)}=\frac{\frac{1}{18}}{\frac{1}{2}}=\frac{1}{9}\).

Question 4.
An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult true/False questions, 500 easy multiple choice questions, and 400 difficult multiple choice questions. If a question is selected from the test question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Answer:
Describe the events as follows
E: ‘getting an easy question.’
F: ‘getting a multiple choice question.’
Total Questions = 300 + 200 + 500 + 400 = 1400
n(F) = 500 + 400 = 900, n(E ∩ F) = 500
P(F) = \(\frac{900}{1400}=\frac{9}{14}\), P(E ∩ F) = \(\frac{500}{1400}=\frac{5}{14}\)
Therefore the required probability
= P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{5}{14}}{\frac{9}{14}}=\frac{5}{9}\).

Question 5.
Two cards are drawn at random without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Answer:
Describe the events as follows
B1: ‘getting a black card in the first draw.’
B2: ‘getting a black card in the second draw.’
P(B1) = \(\frac{26}{52}\) = \(\frac{1}{2}\)
When the first event is executed and since no replacement is allowed, the remaining total number of cards become 51 and black cards become 25.
P(B2/B1) = \(\frac{25}{51}\)
Therefore the required probability
P(B1 ∩ B2) = P(Bl) × P(B2/B1) = \(\frac{1}{2} \times \frac{25}{51}=\frac{25}{102}\).

Question 6.
If P(A) = \(\frac{6}{11}\), P(B) = \(\frac{5}{11}\) and P(A ∪ B) = \(\frac{9}{11}\) Find

1. P(A ∩ B)
2. P(A/B)
3. P(B/A)

Answer:
1. P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

2. P(A/B)

3. P(B/A)

Question 7.
Events A and B are such that P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{12}\) and P(not A or not B) = \(\frac{1}{4}\) State whether A and B are independent.
Answer:
P(not A or not B) = \(\frac{1}{4}\) ⇒ \(P(\bar{A} \cup \bar{B})=\frac{1}{4}\)

We have, P(A) × P(B) = \(\frac{1}{2} \times \frac{7}{12}=\frac{7}{24}\)
Therefore, P(A ∩ B) ≠ P(A) × P(B)
Hence A and B are not independent.

Question 8.
Consider two events such that P(A) = \(\frac{1}{2}\), P(A ∪ B) = \(\frac{3}{5}\) and P(B) = p. Find p, if A and B are independent events.
Answer:
If A and B are independent then
P(A ∩ B) = P(A) × P(B)
We have,
P(A ∪ B) = P(A) + P(B) – P(A) × P(B)

Question 9.
One card is drawn at random from a well shuffled pack of 52 cards. In which of the following cases are the events E and F independent? (3 scores each)

1. E: ‘the card drawn is a spades.’
   F: ‘the card drawn is an ace.’
2. E: ‘the card drawn is a black.’
   F: ‘the card drawn is a king.’
3. E: ‘the card drawn is a king or a queen.’
   F: ‘the card drawn is queen or a jack.’

Answer:
1. P(E) = \(\frac{13}{52}=\frac{1}{4}\), P(F) = \(\frac{4}{52}=\frac{1}{13}\)
There is only one card which is an ace of spade.
P(E ∩ F) = \(\frac{1}{52}\)
We have,
P(E) × P(F) = \(\frac{1}{4} \times \frac{1}{13}=\frac{1}{52}\) = P(E ∩ F)
Hence E and F are independent events.

2. P(E) = \(\frac{26}{52}=\frac{1}{2}\), P(F) = \(\frac{4}{52}=\frac{1}{13}\)
There are two king of black.
P(E ∩ F) = \(\frac{2}{52}\) = \(\frac{1}{26}\)
We have,
P(E) × P(F) = \(\frac{1}{2} \times \frac{1}{13}=\frac{1}{26}\) = P(E ∩ F)
Hence E and F are independent events.

3. There are 4 king and 4 queen cards
P(E) = \(\frac{8}{52}\) = \(\frac{2}{13}\),
There are 4 queen and 4 jack cards.
P(F) = \(\frac{8}{52}\) = \(\frac{2}{13}\)
There 4 queen common for both.
P(E ∩ F) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
We have,
P(E) × P(F) = \(\frac{2}{13} \times \frac{2}{13}=\frac{4}{169}\) ≠ P(E ∩ F)
Hence E and F are not independent events.

Question 10.
A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.
Answer:
P(A) = \(\frac{1}{2}\) and P(B) = \(\frac{1}{6}\)
When a coin and die are tossed the sample space will be as follows.
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
Here head and 3 come only once.
⇒ P(A ∩ B) = \(\frac{1}{12}\)
P(A) × P(B) = \(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\) = P(A ∩ B)
Hence A and B are independent.

Question 11.
Rani and Joy appear in an interview for two vacancies in the same post. The probability of Rani’s selection is \(\frac{1}{7}\) and that of Joy’s selection is \(\frac{1}{5}\) .What is the probability that

1. Rani will not be selected? (1)
2. Both of them will be selected? (1)
3. None of them will be selected? (1)

Answer:
1. Let Rani’s selection be the event A and Joy’s selection be the event B.P(Rani will not be selected)

2. P(Both of them will be selected)
P(A ∩ B) = P(A).P(B) = \(\frac{1}{7} \cdot \frac{1}{5}=\frac{1}{35}\).

3. P(None selected)

Question 12.
Find the probability distribution of number heads in two tosses of a coin.
Answer:
S = {HH, HT, TH, TT}
Let X denotes the random variable of getting a head. Then X can take values 0, 1, 2.
P(X = 0) = P(no heads) = P({TT})
= P(T) × P(T) = \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
P(X = 1) = P(one heads)
= P({HT, TH})
= P(H) × P(T) + P(T) × P(H)
= \(\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2}=\frac{1}{2}\)
P(X = 2) = P(two heads) = P({HH})
= P(H) × P(H) = \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
The required Probability Distribution is

Question 13.
Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.
Answer:
Let X denotes the random variable of number of defective eggs in the 10 eggs drawn.
Clearly X has a Binomial Distribution with n = 10 and p = 10% = \(\frac{1}{10}\), q = 1 – p = \(\frac{9}{10}\)
⇒ P(X = x) = 10C_xq^10-xp^x = 10C_x\(\left[\frac{9}{10}\right]^{10-x}\left[\frac{1}{10}\right]^{x}\)
P(at least 1 defective egg) = P(X ≥ 1)
1 – P(X = 0) = 1 – 10C₀\(\left[\frac{9}{10}\right]^{10}=1-\frac{9^{10}}{10^{10}}\).

Plus Two Maths Probability Four Mark Questions and Answers

Question 1.
In a hostel 50 % of the girls like tea, 40 % like coffee and 20% like both tea and coffee. A girl is selected and random.

1. Find the probability that she likes neither tea nor coffee. (2)
2. If the girl likes tea, then find the probability that she likes coffee. (1)
3. If she likes coffee then find the probability she likes tea. (1)

Answer:
Let T denotes the set of girls who like tea and C denotes who like coffee.
1. P(T) = 50% = \(\frac{1}{2}\); P(C) = 40% = \(\frac{2}{5}\);
P(T ∩ C) = 20% = \(\frac{1}{5}\)
P(T ∪ C)^‘ = 1 – P(T ∪ C)
= 1 – {P(T) + P(C)-P(T ∩ C)}
\(=1-\left\{\frac{1}{2}+\frac{2}{5}-\frac{1}{5}\right\}=\frac{3}{10}\).

2.

3.

Question 2.
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Answer:
For the box to be approved all the three oranges should be selected from the 12 good ones. Since the events are executed without replacement the number for good oranges and total oranges reduce by one on each draw.
O1: ‘getting a good orange in the first draw.’
O2: ‘getting a good orange in the second draw.’
O3: ‘getting a good orange in the third draw.’
P(good orange in the first draw) =P(O1)= \(\frac{12}{15}\) = \(\frac{4}{5}\),
P(good orange in the second) = p(O2/O1) = \(\frac{11}{14}\),
P(good orange in the third)
= P(O3/(O1 ∩ O2)) = \(\frac{10}{13}\)
Therefore the required probability
= P(O1 ∩ O2 ∩ O3)
= P(O1)P(O2/O1)P(O3/(O1 ∩ O2))
\(=\frac{4}{5} \times \frac{11}{14} \times \frac{10}{13}=\frac{44}{91}\).

Question 3.
Let two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find

1. P(A and B)
2. P(A and not B)
3. P(A or B)
4. P (neither A nor B)

Answer:
1. P(A and B) = P(A ∩ B) = P(A) × P(B)
= 0.3 × 0.6 = 0.18.

2. P(A and not B) = P(A ∩ \(\bar{B}\)) = P(A) × P(\(\bar{B}\))
= 0.3 × 0.4 = 0.12.

3. P(A or B) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – P(A) × P(B)
= 0.3 + 0.6 – 0.3 × 0.6 = 0.72.

4. P(neither A nor B) = \(P(\bar{A} \cap \bar{B})\)
= P(\(\bar{A}\)) × P(\(\bar{B}\)) = 0.7 × 0.4 = 0.28.

Question 4.
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

1. both balls are red.
2. the first ball is a black and the second is red
3. one of them is black and the other red.

Answer:
Describe the events as follows.
Black-B and Red-R.n(S) = 18,
P(B) = \(\frac{10}{18}\) = \(\frac{5}{9}\) and P(R) = \(\frac{8}{18}\) = \(\frac{4}{9}\)
Since the event is executed with replacement, is independent.
1. P( both ball is red) = P(R) × P(R)
= \(\frac{4}{9} \times \frac{4}{9}=\frac{16}{81}\).

2. P( first black and second red) = P(B) × P(R)
= \(\frac{5}{9} \times \frac{4}{9}=\frac{20}{81}\).

3. P( one of them is a black and the other red)
= P(B) × P(R) + P(R) × P(B)
\(=\frac{5}{9} \times \frac{4}{9}+\frac{4}{9} \times \frac{5}{9}=\frac{40}{81}\).

Question 5.
Bag 1 contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the Bags and it is found to be red. Find the probability that it was from Bag II.
Answer:
Describe the events as follows.
A: ‘getting a defective ball’.
E₁: ‘choosing Bag I.’
E₂: ‘choosing Bag II.’
P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A/E₁) = P (drawing a red ball from Bag I) = \(\frac{3}{7}\)
P(A/E₂) = P (drawing a red ball from Bag II) = \(\frac{5}{11}\)
P (a ball from Bag II, being given that it is red)

Question 6.
A Bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One ball of the two Bag is drawn at random and the ball is drawn the Bag is found to be red. Find the probability that the ball is drawn from first Bag.
Answer:
Describe the events as follows.
A: ‘getting a red ball’.
E₁: ‘choosing Bag I.’
E₂: ‘choosing Bag II.’
P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A/E₁) = P (a red ball from Bag I) = \(\frac{4}{8}\) = \(\frac{1}{2}\)
P(A/E₂) = P (a red ball from Bag II) = \(\frac{2}{8}\) = \(\frac{1}{4}\)
P (a ball from Bag II, being given that it is red)

Question 7.
In a factory which manufactures blots, machines A, B, and C manufacture respectively 25%, 35%and 40% of the bolts. Of their outputs 5%, 4% and 2%are defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?
Answer:
Describe the events as follows.
A: ‘getting a defective bolt’.
E₁: ‘choosing machine A.’
E₂: ‘choosing machine B.’
E₃: ‘choosing machine C.’
P(E₁) = 0.25, P(E₂) = 0.35, P(E₃) = 0.4
P(A/E₁) = P
(a defective bolt from machine A)
= 5% = 0.05
P(A/E₂) = P
(a defective bolt from machine B)
= 4% = 0.04
P(A/E₃) = P
(a defective bolt from machine C)
= 2% = 0.02
P (a bolt from machine B, being given that it is defective)
= P(E₂/A)

Question 8.
Suppose 5% of men and 0.25% of women have hair grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume there are equal number males and females.
Answer:
Describe the events as follows.
A: ‘person is grey haired’.
E₁: ‘choosing man.’
E₂: ‘choosing woman.’
P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A/E₁) = P (a grey haired person from men)
= 5% = 0.05
P(A/E₂) = P (a grey haired person from women) = 0.25% = 0.0025
P(selecting a male, being given that it is grey haired)

Question 9.
An Insurance company insured 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers. The probabilities of an accident are .01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Answer:
Describe the events as follows.
A: ‘accident happens’.
E₁: ‘choosing Scooter driver.’
E₂: ‘choosing Car driver.’
E₃: ‘choosing Truck driver.’
Total drivers = 2000 + 4000 + 6000 = 12000

P(A/E₁) = P
(accident of a Scooter driver) = 0.01
P(A/E₂) = P
(accident of a Car driver) = 0.03
P(A/E₃) = P
(accident of a Truck driver) = 0.15
P (accident happens, given that it is a Scooter driver).
= P(E₁/A)

Question 10.
A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
Answer:
Describe the events as follows.
A: ‘getting a defective item.’
E₁: ‘choosing machine A.’
E₂: ‘choosing machine B.’
P(E₁) = 60% = 0.6, P(E₂) = 40% = 0.4
P(A/E₁) = P (a defective from machine A) = 2% = 0.02
P(A/E₂) = P (a defective from machine B) = 1% = 0.01
P (a defective from machine B)

Question 11.
Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4 she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Answer:
Describe the events as follows.
A: ‘getting exactly one head.’
E₁: ‘she getting 5 or 6.’
E₂: ‘she getting 1, 2, 3 or 4.’
When a die is thrown the sample space is{1, 2, 3, 4, 5, 6}
P(E₁) = \(\frac{2}{6}\) = \(\frac{1}{3}\), P(E₂) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
When she gets 5 or 6, throws a coin 3 times. Then sample space is {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
P(A/E₁) = P (one head given that 5 or 6 happened)
\(\frac{3}{8}\)
When she gets 1, 2, 3 or 4, throws a coin once. Then sample space is {H, T}
P(A/E₂) = P
(one head given that 1, 2, 3 or 4 happened)
\(\frac{1}{2}\)
P (She gets exactly one head threw 1, 2, 3 or 4)

Question 12.
Vineetha and Reshma are competing for the post of school leader. The probability Vineetha to be elected is 0.6 and that of Reshma is 0.4 Further if Vineetha is elected the probability of introducing a new pattern of election is 0.7 and the corresponding probability is 0.3 if Resma is elected. Find the probability that the new pattern of election is introduced by Reshma.
Answer:
Let E₁ and E₂ be the respectively probability that Vineetha and Reshma will be elected. Let Abe the probability that a new pattern of election is introduced.
P(E₁) = 0.6; P(E₂) = 0.4
P(A|E₁) = 0.7; P(A|E₂) = 0.3

Question 13.
Find the probability of number of doublets in three throws of a pair of dice.
Answer:
Let X denotes the random variable of getting a Doublet. Possible doublets are (1, 1),(2, 2),(3, 3),(4, 4),(5, 5),(6, 6).
Then X can take values 0, 1, 2, 3.
P(getting a doublet) = \(\frac{6}{36}\) = \(\frac{1}{6}\)
P(not getting a doublet) = \(\frac{30}{36}\) = \(\frac{5}{6}\)
P(X = 0) = P(no doublet) = \(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{125}{216}\)
P(X = 1) = P(one doublet and 2 non-doublets)

P(X = 2) = P(2 doublet and 1 non-doublets)

P(X = 3) = P(3 doublet) = \(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}=\frac{1}{216}\)
The required Probability Distribution is

Question 14.
Find the probability distribution of the number of white balls drawn when three balls are drawn one by one without replacement from a bag containing 4 white and 6 red balls.
Answer:
Let X denotes the random variable of number of white balls. Clearly X can take values 0, 1, 2, 3. Describe the events as follows.
W: ‘getting white ball.’
R: ‘getting red ball.’
P(X=0) = P(no white balls)
= P(RRR)= \(\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}=\frac{5}{30}\)
P(X=1) = P(1white, 2red balls)
= P(WRR) + P(RWR) + P(RRW)

P(X=2) = P(2white, 1 red balls)
= P(WWR) + P(RWW) + P(WRW)

P(X=3) = P(3white)
P(WWW) = \(\frac{4}{10} \times \frac{3}{9} \times \frac{2}{8}=\frac{1}{30}\)
The required Probability Distribution is

Question 15.
Two dice are thrown simultaneously. If X denotes the number of sixes, find expectation of X. Also find the variance.
Answer:
Let X denotes the random variable of getting a 6. Clearly X can take values 0, 1, 2.
P(X = 0) = P(non-six, non-six) = \(\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}\)
P(X = 1) = P((six, non-six),(non-six, six))

P(X = 2) = P(six, six) = \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\)
The required Probability Distribution is

Variance = σ² = E(X²) – [E(X)]²

Question 16.
A random variable X has the following probability distribution

Determine

1. k
2. P(X < 3)
3. P(X > 6)
4. P(0 < X < 3)

Answer:
1. We know that sum of the probabilities is = 1
0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
10k² + 9k – 1 = 0
(k + 1)(10k – 1) = 0; k= -1 or k = \(\frac{1}{10}\)
(negative value cannot be accepted).

2. P(X < 3) = P(0) + P( 1) + P(2)
0 + k + 2k + 2k = 3k = \(\frac{3}{10}\).

3. P(X < 3) = P(7) = 7k² + k

4. P(0 < X < 3) = P(1) + P(2) = k + 2k = 3k = \(\frac{3}{10}\). Question

Question 17.
(i) P(A) = \(\frac{7}{13}\); P(B) = \(\frac{9}{13}\); (A ∩ B) = \(\frac{4}{13}\), then P(A/B) is
(a)  \(\frac{9}{4}\)
(b)  \(\frac{16}{13}\)
(c)  \(\frac{4}{9}\)
(d)  \(\frac{11}{13}\)
(ii) Probability of solving a specific problem independently by A and B are \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively. If both try to solve the problem independently, then
(a) Find the probability that the problem is solved. (2)
(b) Find the probability that exactly one of them solve the problem. (1)
Answer:
(i) (c) \(\frac{4}{9}\)

(ii)

P(Problem is solved)
= P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

(iii) P(Exactly one of them solve)

Plus Two Maths Probability Six Mark Questions and Answers Question

Question 1.
(i) A and B are two events such that P(A) = \(\frac{1}{5}\) and P(A ∪ B) = \(\frac{2}{5}\) Find P(B) if they are mutually exclusive
(a) \(\frac{1}{5}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{3}{5}\)
(d) \(\frac{4}{5}\)
(ii) A box contains 3 red and 4 blue balls. Two balls are drawn one by one without replacement. Find the probability of getting both balls red.
(iii) Three cards are drawn successively without replacement from a pack of 52 cards. What is the probability that first two cards are queen and the third is king.
Answer:
(i) (a) \(\frac{1}{5}\).

(ii) Let A be the event that the first ball drawn is red and B be the event of drawing red ball in the second draw
P(A) = \(\frac{3}{7}\)
Probability of getting one red ball in the second draw = P(B/A) = \(\frac{2}{6}\) = \(\frac{1}{3}\). P(A ∩ B) = P(A).P(B/A)
\(=\frac{3}{7} \times \frac{1}{3}=\frac{1}{7}\).

(iii) Let Q denote the event that the card drawn is Queen and K denote the event of drawing a King
P(Q) = \(\frac{4}{52}\), P(Q/Q) = \(\frac{3}{51}\)
P(K/QQ) is the probability of drawing the third card is a king
P(K/QQ) = \(\frac{4}{50}\)
P(QQk) = P(Q)P(Q/Q)P(K/QQ)
\(=\frac{4}{52} \cdot \frac{3}{51} \cdot \frac{4}{50}=\frac{2}{5525}\).

Question 2.
60 shirts of different colours are on sale. If one shirt is chosen at random.

1. What is the probability that it is red? (1)
2. What is the probability that it is plain and extra-large? (1)
3. What is the probability that it is small, given that it is blue? (2)
4. If A is the event ‘the shirt is medium’ and B is the event ‘the shirt is blue’. Are the events A and B independent? (2)

Answer:
1. P(Red) = \(\frac{8+8+2+4}{60}=\frac{11}{30}\).

2. P (Plain and extra-large) = \(\frac{4+5}{60}=\frac{3}{20}\).

3. P(small/blue)

4.

∴ Not independent.

Question 3.
From a box containing balls numbered from, 1 to 100, one ball is drawn at random. The events X and Y are as follows. X: A perfect square is drawn. Y: An even number is drawn.

1. Find P(X) and P(Y). (2)
2. Compute P (X/Y). (2)
3. Are X and V independent? Justify. (2)

Answer:
Perfect square numbered balls are
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
Therefore, there are 10 perfect square numbered balls, 50 even numbered balls and 5 perfect square even numbered balls.
1. P(X) = \(\frac{10}{100}=\frac{1}{10}\), P(Y) = \(\frac{50}{100}=\frac{1}{2}\)
P (Even perfect square number) = P (X ∩ Y) \(\frac{50}{100}=\frac{1}{20}\).

2. P(X/Y)
= P(Drawing a perfect square numbers from even numbers) = \(\frac{5}{50}=\frac{1}{10}\).

3. We have, P (X ∩ Y) = \(\frac{1}{20}=\frac{1}{10} \times \frac{1}{2}\)
= P(X).P(Y).
Therefore X and Y are independent events.

Question 4.
(i) The probability of three mutually exclusive events A, B, and C are given by 2/3, 1/4, 1/6 respectively. Is this statement ________ (1)
(a) true?
(b) false?
(c) cannot be said?
(d) data not sufficient?
A husband and wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is 1/7 and that of wife’s selection is 1/5. What is the probability that
(ii) Only one of them will be selected? (3)
(iii) None will be selected? (2)
Answer:
(i) (b) Probability should be less than or equal to 1.
Here A, B, C are mutually exclusive.
Then,
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = \(\frac{2}{3}+\frac{1}{4}+\frac{1}{6}=\frac{26}{24}>1\)
∴ statement is false.

(ii) H – Event of husband selected,
W – Event of wife selected

(iii) P (None of them will be selected)

Question 5.
(i) Find P(A∩B) if A and B are independent events with P(A) = \(\frac{1}{5}\) and P(B) = \(\frac{5}{8}\)

(ii) An unbiased die is thrown twice. Let the event A be getting prime number in the first throw and B be the event of getting an even number in the second throw. Check the independence of the events A and B. (3)
(iii) The probability of solving a problem independently by A and B are \(\frac{1}{3}\) and \(\frac{1}{4}\) respectively. Find the probability that exactly one of them solves the problem. (2)
Answer:
(i) (c) \(\frac{1}{8}\).

(ii) P(A) = \(\frac{18}{36}\) = \(\frac{1}{2}\)
P(B) = \(\frac{18}{36}\) = \(\frac{1}{2}\)
P(A∩B) = P( prime number in first throw and even number in the second throw)

∴ A and B are independent events.

(iii)

Probability of exactly one of them solves the problem = P(A)P(B’) + P(B)P(A’)

Question 6.
(i) A set of events E₁ + E₂,…….E_n are said to be a partition of the Sample Space, then which of the following conditions is always not true (1)
(a) E₁∪ E₂ ∪……..∪ E_n = S,
(b) E₁ ∩ E_n = Φ,
(c) P(E₁) > 0,
(d) P(E₁) ≥ P(E_n)
(ii) A person has undertaken a business. The probabilities are 0.80 that there will be a crisis, 0.85 that the business will be completed on time if there is no crisis and 0.35 that the business will be completed on time if there is a crisis. Determine the probability that the business will be completed on time. (2)
(iii) A box contains 5 red and 10 black balls. A ball is drawn at random, its colour is noted and is returned to the box. More over 2 additional balls of the colour drawn are put in the box and then a ball is drawn. What is the probability that the second ball is red? (3)
Answer:
(i) P(E₁) ≥ P(E_n).

(ii) Let A be the event that the business will be completed on time and B be the event that there will be a crisis
P(B) = 0.80
P(no crisis) = P(B’) = 1 – P(B) = 0.20
P(A/B) = 0.35 P(A/B’) = 0.85
By theorem on total probability
P(A) = P(B)P(A/B) + P(B’)P(A/B’)
= 0.8 × 0.35 + 0.2 × 0.85 = 0.45.

(iii) Let a red ball be drawn in the first attempt P(drawing a red ball) = \(\frac{5}{15}=\frac{1}{3}\)
If two red balls are added to the box, then the box contains 7 red balls and 10 black balls
P(drawing a red ball) = \(\frac{7}{17}\)
Let a black ball be drawn in the first attempt P(drawing a black ball) = \(\frac{10}{15}=\frac{2}{3}\)
If two black balls are added to the box, then the box contains 5 red and 12 black balls
P(drawing a red ball) = \(\frac{5}{17}\)
Probability of drawing the second ball red is

Question 7.

1. Bag I contains 5 red and 6 black balls. Bag II contains 7 red and 5 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag I. (3)
2. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being diamond. (3)

Answer:
1. Let E₁ be the event of choosing Bag I and E₂ be the event of choosing Bag II
A be the event of drawing a red ball

2. Let E₁ be the event of choosing a diamond and E₂ be the event of choosing a non diamond card
A be the event that a card is lost

When a diamond card is lost, there are 12 diamond cards in 52 cards. Then

When a non diamond card is lost, there are 13 diamond cards in 51 cards. Then

Question 8.
(i) If X denotes number of heads obtained in tossing two coins. Then which of the following is false (1)
(a) X(HH) = 2
(b) X(HT) = 1
(c) X(TH)= 0
(d) X(TT) = 0
(ii) Find the probability distribution of the number of tails in the simultaneous toss of two coins. (2)
(iii) A coin is tossed so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails. (3)
Answer:
(i) (c) X(TH)= 0.

(ii) Sample space is S = {HH, HT, TH, TT}
Let X denote the number of tails, then
X(HH) = 2, X(HT) = 1, X(TH) = 1, X(TT) = 0
Therefore X can take the values 0, 1 or 2
P(HH) = P(HT) = P(TH) = P(TT) = \(\frac{1}{4}\)
P(X = 0) = \(\frac{1}{4}\) P(X = 1) = \(\frac{1}{2}\) P(X = 2) = \(\frac{1}{4}\)
Then the Probability distribution is

(iii) Let the probability of getting a tail in the biased coin be x.
P(T) = x P(H) = 3x
P(T) + P(H) = 1 ⇒ x + 3x = 1 x = \(\frac{1}{4}\)
P(T) = \(\frac{1}{4}\) P(H) = \(\frac{3}{4}\)
Let X denote the random variable representing the number of tails
P(X = 0) = P(HH) = P(H).P(H) = \(\frac{9}{16}\)
P(X = 1) = P(HT) + P(TH)

P(X = 2) = P(TT) = P(T).P(T) = \(\frac{1}{16}\)
Then the Probability distribution is

Question 9.
If a fair coin is tossed 10 times, find the probability of

1. Exactly 6 heads. (2)
2. At least 6 heads. (2)
3. At most 6 heads. (2)

Answer:
Let X denotes the random variable of number of heads in an experiment of 10 trials.
Clearly X has a Binomial Distribution with
n = 10 and p = \(\frac{1}{2}\) ⇒ P(x) = nC_xq^n-x P^x

1. P(x = 6)

2. P(at least 6 heads) = P(X ≥ 6)
= P(X = 6) + P(X = 1)+P(X = 8)

P(at most 6 heads) = P(X ≤ 6)
= P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)

Question 10.
Five cards are drawn successively with a replacement from a pack of 52 cards. What is the probability that

1. All the 5 cards are spades? (2)
2. only 3 cards are spade? (2)
3. none is a spade? (2)

Answer:
Let X denotes the random variable of number of spades cards in an experiment of 5 trials. Clearly X has a Binomial Distribution with n = 5

1. P(all 5 are spades) = P(X = 5)

2. P(3 are spades) = P(X = 3)

3. P(non-spade) = P(X = 0) =

Question 11.
Find the probability distribution, Mean and Variance of the number of success in two tosses of a die, where a success is defined as

1. number greater than 4. (3)
2. 6 appears on at least on die. (3)

Answer:
1. Let X denotes the random variable of getting a 5, 6. Clearly X can take values 0, 1, 2.
When number 1, 2, 3, 4 appears in both die. Number of such cases = 4 × 4 = 16
P(X=0) = P(no success) = \(\frac{16}{36}=\frac{4}{9}\)
When 5, 6 in one die and other with 1, 2, 3, 4 and visa versa.
Number of such cases is 2 × 4 + 4 × 2 = 16.
P(X= 1) = P(1 success and 1 no success) = \(\frac{16}{36}=\frac{4}{9}\)
When number 5, 6 appears in Jjoth die. Number Hof such cases = 2 × 2 = 4.
P(success) = \(\frac{4}{36}=\frac{1}{9}\)
The required Probability Distribution is

2. Let X denotes the random variable of getting at least 6 on one die. Clearly X can take values 0, 1.
No success means 1, 2, 3, 4, 5 appears on both die. Number of such cases is 5 × 5 = 25.
P(X=0) = P(no success) = \(\frac{25}{36}\)
When 6 in one die and other with 1, 2, 3, 4, 5 and visa versa. Number of such cases is 1 × 5 + 5 × 1 = 10
When both the die is 6. Number of such case is 1. Therefore total cases is 1 + 10 = 11
P(X=1) = P(1 success) = P(at least 1 six) = \(\frac{11}{36}\)
The required Probability Distribution is

Question 12.
(i) If A and B are two events such that A ⊂ B and P(A) ≠ 0 then P(A/B) is (1)

(ii) There are two identical bags. Bag I contains 3 red and 4 black balls while Bag II contains 5 red and 4 black balls. One ball is drawn at random from one of the bags.
(a) Find the probability that all the ball drawn are red. (3)
(b) If the balls drawn is red what is the probability that it was drawn from Bag I? (2)
Answer:
(i) Describe the events as follows.
A: ‘getting a red ball’.
E₁: ‘choosing Bag I.’
E₂: ‘choosing Bag II.’
P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A/E₁) = P
(drawing a red ball from Bag I) = \(\frac{3}{7}\)
P(A/E₂) = P
(drawing a red ball from Bag II) = \(\frac{5}{9}\)
P (All the balls drawn is red)
= P(A) = P(E₁)P(A/E₁) + P(E₂)P(A/E₂)

Question 13.
Consider the following probability distribution of a random, variable X.

1. Find the value of k. (2)
2. Determine the Mean and Variance of X. (4)

Answer:
(i) We have; Σp_i = 1

(ii)

Students can Download Chapter 5 Market Equilibrium Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium

Plus Two Economics Market Equilibrium One Mark Questions and Answers

Question 1.
Complete the statement given below. Free entry and of firms imply that the market price will always be equal to ……………
Answer:
Minimum average cost (P = Min. AC)

Question 2.
Choose the correct answer. The imposition of price ceiling below the equilibrium price leads to ……….
Answer:
Excess demand

Question 3.
Market equilibrium of a commodity shows,
(a) excess demand
(b) quantity demanded greater than quantity supplied
(c) quantity demanded equals quantity supplied
(d) excess supply
Answer:
(c) quantity demanded equals quantity supplied

Question 4.
When there is increase in demand, the demand curve.
(a) shifts reight ward
(b) shifts leftward
(c) shifts downward
(d) remains constant
Answer:
(a) shifts reight ward

Question 5.
The government imposing upper limit on the price of a good or service is called:
(a) price floor
(b) price ceiling
(c) equilibrium price
(d) fair price
Answer:
(b) price ceiling

Plus Two Economics Market Equilibrium Two Mark Questions and Answers

Question 1.
At what price – higher or lower than the equilibrium price, there will be excess demand?
Answer:
When the market price is lower than the equilibrium price, there will be excess demand.

Question 2.
Make pairs.
Price floor, below equilibrium price, above equilibrium price, price ceiling
Answer:

• Price floor – above equilibrium price.
• Price ceiling – below equilibrium price.

Question 3.
Point out the consequences of price ceiling.
Answer:

1. Black marketing
2. Malpractices by fair price shops
3. Sale of inferior quality goods.

Question 4.
What do you mean by control price?
Answer:
Fixation of price of a commodity at a lower level than equilibrium price is called control price. Control price is determined to protect the interest of the consumers.

Question 5.
Market equilibrium of apple is given below in the diagram below.

1. Define market equilibrium
2. Find out the market price and equilibrium quantity

Answer:

1. market equilibrium is a situation where quantity demanded is exactly equal to the quantity supplied.
2. The price of apple is ₹40 and the equilibrium quantity is 30kg.

Plus Two Economics Market Equilibrium Three Mark Questions and Answers

Question 1.
Match the following.

A	B
Price lower than equilibrium price	Excess demand
Equilibrium price	Excess supply
Price higher than equilibrium price	Demand = Supply

Answer:

A	B
Price lowerthan equilibrium price	Excess supply
Equilibrium price	Demand = Supply
Price higher than equilibrium price	Excess demand

Question 2.
The demand function and supply function of a product are given as q^D = 60 – P for 0 = P = 60 q^S = 30 + P for P > 10 Calculate equilibrium price.
Answer:
The demand and supply functions are given as
q^D= 60 – P for 0 = P = 60
q^S = 30 + P for P > 10
Equilibrium price is considered as the price at which quantity demanded is exactly equal to quantity supplied. Therefore we get.
60 – P = 30 + P
60-30 = 2 P
30 = 2P
P = 30/2 = 15
Therefore equilibrium price is ₹15.

Question 3.
Complete the following statements.

1. Long-run price under perfect competition will be equal to ……….
2. Minimum price fixed by government for a product is known as ……….
3. Maximum price fixed by government for a product is known as ………

Answer:

1. average cost
2. floor price
3. price ceiling

Question 4.
Demand curve for labour is downward sloping. Explain
Answer:
Demand curve for labour is downward sloping indicating that more and more labour is demanded at lower wages. This is due to the operation of declining marginal productivity of labour. Marginal productivity of labodr declines due to the operation of diminishing returns. That is why the demand curve for labour slopes downward.

Question 5.
The market demand function and market supply functions are given as, Find the equilibrium price and equilibrium quantity.
q^D = 200 – P for 0 = P = 200
q^S = 120 + P for P > 10
Answer:
We find equilibrium price by equating market demand function and market supply functions as shown below.
q^D = q^S
200-P = 120 + P
2P = 80
P = 80/2 = 40
Therefore equilibrium price is ₹40. Equilibrium quantity is obtained by substituting the equilibrium price into either the demand or supply function equations. Applying the value of price ₹40 in demand equation we have.
q^D= 200 – P
q^D =200 – 40 = 160
Therefore equilibrium price is ₹40 and equilibrium quantity is 160.

Question 6.
Prepare a note on market equilibrium.
Answer:
Equilibrium is defined as a situation where the plans of all consumers and firms in the market match and the market dears. In equilibrium, the aggregate quantity that all firms wish to sell equals the quantity that all the consumers in the market wish to buy; in other words, market supply equals market demand. The price at which equilibrium is reached is called equilibrium price and the quantity bought and sold at this price is called equilibrium quantity.

Therefore, q^D (P*) = q^S (P*) where P* denotes the equilibrium price and q^D (P*) and q^S (P*) denote the market demand and market supply of the commodity respectively at price P*

Question 7.
Suppose the demand and supply functions of commodity X are given by, Qd = 500 + 3P and Qs = 700 – P Qd = 500 + 3P Qs = 700 – P Find out the equilibrium price and quantity demanded and supplied.
Answer:
Equilibrium price and quantity can be determined by equating the demand and supply functions
Qd = Qs
500 + 3P = 700 – P
4P = 200
\(P=\frac{200}{4}=50\)
Equilibrium price is ₹50. Applying the price in the demand function, we get
500 + 3 × 50
500 + 150 =650
Therefore, equilibrium price is ₹50 and quantity is 650.
Qd = Qs
500 + 3P = 700 – P
4P = 200
\(P=\frac{200}{4}=50\)
Qd = 500 + 3 × 50
= 500 + 150 = 650

Question 8.
The diagram below illustrates the supply and demand for television sets. The original demand curve is D₂

Using the diagram, state the new demand curve (D₁ or D₃) which will apply after each of the following changes taken place. (The same answer may be used more than once)

1. A successful advertising campaign for television sets occurs
2. Income decreases
3. An increase in the population

Answer:

1. Demand increases (D₂ curve shifts right to D₃)
2. Demand decreases (D₂ curve shifts left to D₁)
3. Demand increases (D₁ curve shifts right to D₃)

Question 9.
Calculate equilibrium price and quantity based on the following information.
q^d= 400 – P (1)
q^s= 240 + 3 (p – 4) (2)
Answer:
At equilibrium,
q^d = q^s
Putting the values,
400 – p = 240 + 3(p – 4)
400 – p = 240 + 3p – 12
400-240 + 12 = 3p + p
172 = 4p
\(p=\frac{172}{4}\)
p =43
Putting p = 43 in the first equation, we get,
q^d =400 – 43 = 357 Therefore, equilibrium price = 43 and
equilibrium quantity is = 357 units

Question 10.
The diagram shows relationship between two commodities A and B.

1. Identify the commodities A and B
2. Explain what happens to the price and quantity demanded of A when the price of A falls.

Answer:

1. A and B are substitutes.
2. When the price of A falls people will demand more A. So the demand for B will fall. That will result in a decrease in the price of B.

Question 11.
The diagram below shows one of the government intervention programmes in the market.

1. Identify the programme and calculate the excess supply.
2. Explain how the government is monitoring the higher price fixed.

Answer:
1. Minimum price/floor price, 40 unit excess supply.

2. The government announces the minimum price above the market price. As a result of this intervention there occurs excess supply in the market. The government has to remove the excess from the market to maintain the price. So the government store the excess supply in the warehouses and redistribute it at the time of shortage.

Question 12.
Under perfect competition, a market for a good is in equilibrium. There is simultaneous “decrease” both in demand and supply, but there is no change in market price. Explain with the help of a diagram how it is possible.
Answer:
The decrease in demand and supply is the same, and hence the price remain the same. Shows this by drawing appropriate diagram

Question 13.
The diagrams below indicate four possible shifts in demand or in supply that could happen in particular markets. Relate each of the events described below to one of them. Also, give reason for the shift.

1. How does the lorry strike in Karnataka and Tamil Nadu affect the market for vegetables in Kerala?
2. People become aware of the fact that Birds Eye Chilly is very much helpful to prevent Cholesterol. What happens to the market for Birds Eye Chilly?
3. How do you think the rising income affect the market for fish?
4. A new technique is discovered for manufacturing computer that greatly lowers their production cost. What happens to the market for computers?

Answer:

1. figure C, supply falls and price rises.
2. figure A, demand increases and price rises.
3. figure B, demand increases and price rises.
4. figure D, supply increases and price falls.

Plus Two Economics Market Equilibrium Five Mark Questions and Answers

Question 1.
Mention the impact of the following.

1. Imposition of price ceiling below equilibrium price
2. Imposition of price floor above the equilibrium price

Answer:

1. Imposition of price ceiling below equilibrium price leads to excess demand.
2. Imposition of price floor above the equilibrium price leads to an excess supply

Question 2.
Complete the following table to show the impact of simultaneous shifts of demand and supply on equilibrium price and quantity.

Answer:

Question 3.
What will happen if the price prevailing in the market is?

1. above the equilibrium price
2. below the equilibrium price.

Answer:
1. If the price prevailing in the market is above equilibrium price, supply will exceed demand. Under such a situation some firms will not be able to sell their desired quantity; so they will lower their price. All other things remaining constant as price falls quantity demanded rises quantity supplied falls, and finally equilibrium price P* will be restored. At P* quantity demanded will be equal to quantity supplied.

2. If the price prevailing in the market is above equilibrium price, demand will exceed supply. Under such a situation some consumers will be ready to pay more prices to get the commodity. This will tend to increase the price. All other things remaining constant as price rises quantity demanded falls, quantity supplied rises, and finally, equilibrium price P* will be restored. At P* quantity demanded will be equal to quantity supplied.

Question 4.
Draw distinction between floor pricing and price ceiling.
Answer:
Floor price means minimum price. Floor price is fixed to protect producers like farmers from price crashes. It ensures a remunerative price to producers. In India, floor prices are fixed for a variety of agricultural commodities like paddy, wheat, coconut, rubber etc.

On the other hand, price ceiling mean maximum price. It is the maximum price fixed by the government. The aim of price ceiling is to protect consumers. Government fixes price ceiling for essential products and medicines to protect the interests of the consumers.

Question 5.

1. Identify the situations depicted in the following figures in panel A and B.
2. Why do such policies are followed and explain the impact of such policies?

Answer:
1. PANEL A – Price ceiling PANEL B –Price floor.

2. Price ceiling is fixed below equilibrium price. Imposition of price ceiling at ‘Pg’ gives rise to excess demand in the market price floor is fixed above equilibrium price. Imposition of floor price at ‘pg’ gives rise to excess supply.

Question 6.
Mention the factors that cause shift in the supply curve.
Answer:
The factors that cause shift in the supply curves are:

1. The change in the number of firms
2. The change in the price of factor inputs
3. Change in production technology
4. Change in the prices of related goods
5. Change in production tax.

Question 7.
How will a change in price of coffee affect the equilibrium price of tea? Explain the effect on equilibrium quantity through a diagram.
Answer:
Coffee and tea are substitutes. If prices of coffee are increased then its demand will decrease and demand tea would increase.lt will shift the demand curve of tea upwards. The equilibrium price and quantity will increase. This is shown in the following diagram.

In the diagram, when the price of coffee increased then the demand for tea increases. This has resulted in equilibrium price and quantity of tea.

Question 8.
How is price determined in labour market?
Answer:
The price of labour is determined by the forces of demand and supply of labour. The households are the suppliers of labour and demand for labour comes from firms. Labour means the hours of work provided by labourers. The wage rate is determined at the intersection of the demand and supply curve of labour.

The firm being a profit maximiser will always employ labour up to the point where the extra cost it incurs for the last labour is equal to the additional benefit he earns from employment that labour. The extra cost of hiring one more labour is the wage rate. For each extra unit of labour, he gets a benefit equal to marginal revenue product of labour.

Thus firm employs labour up to a point where: W = MRPL Where MRPL = MR x MPL As long as MRPL is greater than the wage rate the firm will earn more profit by hiring one more labour and if at any level of labour employment MRPL is less than the wage rate the firm can increase here profit by reducing labour employed.

Question 9.
Prepare a table showing differences between price ceiling and price floor.
Answer:

Price ceiling	Price floor
Upper limit set by the government for some commodities	Lower limit set by the government for some commodities
Imposed on essential goods such as wheat, rice etc.	Agricultural goods, workers etc. are benefitted
To maintain price ceiling, fair price shops may be opened	To maintain price floor, government needs to buy the excess quantity supplied
Price lower than the equilibrium price	Price higher than the equilibrium price
Creation of excess demand	Creation of excess supply

Question 10.

1. With the help of a diagram show how the wage rate is determind in a free market.
2. Analyse the impact of an increased entrance of foreign migrant labourers into the labour market.

Answer:
1. The diagram below shows how the wage rate of labour in a free market is determined. DL is the demand for labour and SL is the supply of labour, ‘e’ is the point of equilibrium, ‘ow’ is the wage rate and ‘oq’ is the quantity of labours.

2. When the foreign migrant labour enters into the labour market, the supply of labour will shift rightward and the wage rate will come down as shown in the figure. ‘ow’ is the original wage rate and ‘OQ’ is the original quantity of labour. ow1 is the new wage rate and OQ, is the new quantity of labourers.

Question 11.
Suppose we have two equations, one for demand and other for supply.
Demand equation: Qx^d = 100 – 10P_x
Supply equation : Qx^s = 60 + 10P_x

1. Calculate equilibrium price and quantity using the equations.
2. Construct demand and supply schedules by assigning various prices. Obtain equilibrium price and quantity graphically.

Answer:
1. Equilibrium price =2, equilibrium quantity=80
2. Demand & supply schedules

Question 12.
Let us take market of commodity ‘X’, which is in equilibrium. Suppose demand for the commodity increases. Explain the chain of effects of this change till the market again reaches equilibrium. Use diagram.
Answer:
Increase in demand leads to disequilibrium-price in-creases – super profit – new firms enter the industry – or existing firms expand production – increase in output – supply increases – supply curve shifts – the process continue until price returns the to the equilibrium level. Draws the diagram, and explains the process.

Plus Two Economics Market Equilibrium Eight Mark Questions and Answers

Question 1.
Observe the following table.

1. Find equilibrium price.
2. Fill the fourth column
3. Why ₹35 and ₹40 are not equilibrium prices?
4. Product surplus drives prices up and shortage drives them down. Do you agree?
5. Draw a diagram of the above table showing the equilibrium price determination

Answer:
1. The equilibrium price is ₹37. At this price both demand and supply are equal.
2.

3. At ₹35, demand exceeds the supply causing a shortage in the market. At ₹40, supply exceeds demand causing surplus. Therefore, these prices are equilibrium prices.
4. No. I do not agree with this argument.
5.

Question 2.
Discuss the impact of the factors mentioned below on equilibrium price and quantity.

1. shift in demand to right
2. shift in demand to left
3. shift in supply to right
4. shift in supply to left.

Answer:
1. When demand curve shifts to right (increase in demand), there will be increase in equilibrium price and increase in equilibrium quantity. This change is shown in the diagram.

2. When demand curve shift to left (decrease in demand), both equilibrium quantity and equilibrium price falls. This is shown in the diagram.

3. When supply curve shift to right (increase in supply), the equilibrium price deceases and the equilibrium quantity increases. This is given in the following diagram.

4. When supply curve shifts to left (decrease in supply), the equilibrium price increases and the equilibrium quantity decreases. This is given in the following diagram.

Question 3.
Suppose the demand and supply curves of salt are given by. q^D = 1000 – P q^S = 700 + 2P

1. Find the equilibrium price and quantity
2. Suppose that the price of input used to produce salt has increased so that the supply curve is q^S = 400 + 2P How does the equilibrium price and quantity change? Does the change conform to your expectation?
3. Suppose the government has imposed a tax of 3 per unit of salt. How does it affect the equilibrium price and quantity?

Answer:
1. equilibrium price and quantity
q^D = 1000 – P
q^S = 700 + 2P
For equilibrium
q^D = q^S
1000 – P = 700 + 2P
1000 – 700 = 3 P
3P = 300
P = 300/3 = 100
Put the value of P in supply equation
q^S = 700 + 2P
q^S = 700 + 2×100
q^S =700 + 200 = 900
Therefore the equilibrium price = ? 100 and the equilibrium quantity is = 900 units

2. For equilibrium
q^D = q^S
1000 – P = 400 + 2P
1000 – 400 = 3P
600 = 3 P
P = 600 / 3 = 200
Put the value of P in demand equation
Q^D= 1000 – P
Q^D = 1000  – 200 = 800
Therefore the equilibrium price = ₹200 and the equilibrium quantity is = 800 units This change confirms to our expectations, i.e., rise in input prices raises prices and lowers supply.

3. q^D= 1000 – P
q^S= 700 + 2P
When ₹3 as tax is imposed on sale of salt the new demand and supply function will change
q^D= 1000 – (P + 3)
q^S= 700 + (2P +3)
In part A equilibrium price was ₹100 which goes up to ₹103 with imposition of tax
q^D = 1000 – (100 + 3) = 1000 – 103 =897
q^S = 700 + (2P + 3)
= 700 + 2(100 + 3)
= 700 + 2×103
= 700 + 206 = 906
q^D < q^S
Therefore, new price and quantity has to be adjusted.

Question 4.
The diagram below shows how the price of wheat is determined in a free market.
a. Show in a seperate diagram the changes on price and quantity demanded of wheat due to the following factors

1. The price of fertilizers increases.
2. The price of rice a substitute of wheat increases.

b. Assess the impact of an increased demand for wheat and an increase in its production.

Answer:
a. price and quantity demanded of wheat.
1. When the price of fertilizers increases the supply of wheat decreases. Its price increases and the quantity falls.

2. When the price of rice, a substitute of wheat increases people may switch to consume wheat, this will increase the demand for wheat. Its price will increase and quantity also will increase.

b. When the demand for heat increases its demand curve will shift rightward. When production increases its supply curve will shift rightward as shown in the diagram below.

Due to these shifts, the quantity will increase anyhow. But the effect on the price will be in different forms. The price may fall, will be constant or even may increase. Whether the price will increase, decrease or remain constant is determined by the respective shifts in demand and supply.

If both demand and supply shift in the same magnitude the price will be the same. If the shift in the demand is more than the supply the price will increase. And if the shift in the supply is more than the shift in the demand the price will fall.

Read More:

GLENMARK Pivot Point Calculator

Students can Download Chapter 2 National Income Accounting Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in their examinations

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting

Plus Two Economics National Income Accounting One Mark Questions and Answers

Question 1.
GNP – depreciation is called
(a) GDP
(b) NNP
(c) PCI
(d) PI
Answer:
(b) NNP

Question 2.
The GDP deflator is equal to
i) Real GDP-Nominal GDP

Answer:
iii) \(\frac{{ No minal GDP }}{\text { Real GDP }} \times 100\)

Question 3.
NFIA is included in:
(a) NNP_FC
(b) NDP_FC
(c) GDP_FC
(d) All the above
Answer:
(a) NNP_FC

Question 4.
Which among the following in a flow concept?
(a) export
(b) wealth
(c) capital
(d) foreign exchange reserve
Answer:
(a) export

Question 5.
When does net factor income from abroad become negative?
(a) NDP < NNP
(b) NNP < NDP
(c) NDP = NNP
(d) none of the above
Answer:
(b) NNP < NDP

Question 6.
When does GDP and GNP of an economy become equal?
(a) When net factor income from abroad is positive
(b) When net factor income from abroad is negative
(c) When net factor income from abroad is zero
(d) None ofthe above.
Answer:
(c) When net factor income from abroad is zero

Plus Two Economics National Income Accounting Two Mark Questions and Answers

Question 1.
Same job is done by a servant and housewife, whose service is included in the national income calculation? Why?
Answer:
Service of a servant is included in the national income calculation, whereas, the service of housewife is not included in the national income. This is because the housewife is not paid for the service she does.

Question 2.
From the following, classify the material into final goods and intermediary goods. Wheat, Bench, Bread, Wood, Rubber, Tyre.
Answer:

Final Goods	Intermediary goods
Bench	Wheat
Bread	Wood
Tyre	Rubber

Question 3.
Distinguish between real flow and money flow?
Answer:
Flow of goods and services from firms to households is called real flow. Factors of production receive reward for their services in the form of money. Households use this money to buy goods and services produced by firms. This flow of money from firms to households and back to firms is called money flow.

Question 4.
Some variables are given below. Classify them into Stock and Flow

1. Wealth
2. Income of a household
3. Consumption
4. Capital
5. Money Supply
6. Capital formation
7. Inventories
8. Saving of a household

Answer:
a. Stock

• Wealth
• Inventories
• Capital
• Money supply

b. Flow

• Income of a household
• Consumption
• Capital formation
• Saving of a household

Question 5.
GDP = C + I + G + (X – M) = C + S + T Derive the Budget Deficit and Trade Deficit equations from the above identity.
Answer:
GDP = C + I + G + (X – M) = C + S + T
Budget deficit = G – T
Trade deficit = M – X

Plus Two Economics National Income Accounting Three Mark Questions and Answers

Question 1.
“Transfer payments are not included in the national income calculation”. Do you agree? Justify your answer.
Answer:
Yes. Transfer payments like pension, old age pension, etc. are not included in the national income. This is because they are transfer earnings not generated by any economic activity. These payments are usually made by the government out of tax revenue collected from the public. Since these generated incomes are already included in national income calculation there is no need to include transfer payment in the national income calculation again.

Question 2.
State whether the following are included or excluded in the national income.

1. purchase of second hand goods
2. operating surplus
3. production for self-consumption
4. interest
5. windfall gains and loses

Answer:

1. Purchase of second hand goods – excluded
2. operating surplus – included
3. old age pension – excluded
4. Production for self consumption – excluded
5. interest – included
6. windfall gains and loses – excluded

Question 3.
Provide appropriate term.

Answer:

1. Value added
2. GNP
3. NNP
4. NNP_FC

Question 4.
Point out any 3 uses of national income accounting.
Answer:
The uses of national income accounting are given below.

1. It shows the distribution of national income among the various factors of production.
2. National income statistics indicate the contribution of different sectors in the economy.
3. Structural changes in the economy can be assessed by the national income accounting.

Question 5.
Classify the following under proper heads.
Flow of teacher services, Flow of subsidies and taxes, Flow of factor rewards, flow of finished goods, Flow of consumption expenditure, Flow of import goods.

Answer:

Real Flow	Money Flow
Flow of teacher services	Flow of subsidies and taxes
Flow of finished goods	Flow of factor rewards
Flow of import goods	Flow of consumption

Question 6.

• Does not includes prices of imported goods
• Weights are different
• It includes all goods and services
• Includes prices of imported goods
• Weights are constant
• Does not include all goods and services

Answer:
a. Consumer price index

• Includes prices of imported goods
• Weights are constant
• Does not include all goods and services

b. GDP deflator

• Does not include prices of imported goods
• Weights are different
• It includes all goods and services

Question 7.
Assume that there are three goods produced in an economy and they are sold at different prices in dif-ferent years. Calculate GDP Deflator.

Answer:

Question 8.
Calculate Depreciation, Net Indirect Tax and NNPFC from the below data.
GDP_MP = 11300
NDP_MP = 10300
NDP_FC = 10000
NFIA = 1500
Answer:
1. Depreciation = GDP_MP – NDP_MP
= 11300 – 10300
= 1000

2. Net Indirect tax = NDP_MP – NDP_FC
= 10300 – 10000 = 300

3. NNP_FC = NDP_FC + NFIA
= 10000 + 1500
= 11500

Plus Two Economics National Income Accounting Five Mark Questions and Answers

Question 1.
Find the odd one out. Justify your answer.

1. GNP, NNP, CSO, GDP
2. Salary, bonus, GPF, free housing, saving
3. Smuggling, production of wheat, sale of second-hand goods, services of housewives
4. Services of teacher, services of engineer, services of lawyer, services of housewife
5. Unemployment allowances, scholarships, old age pension, support price.

Answer:

1. C.S.O. Others are national income concepts.
2. Saving. Others come under compensation to employees
3. Production of wheat. Others are excluded from national income
4. Services of housewife. Others are included in the national income calculation.
5. Support price. Others are transfer payments.

Question 2.
Match the following.

A	B
NNP	GDP – net factor income from abroad
GNP	Personal income – direct taxes
Value added	GNP-depreciation
GDP at market prices	value of output – intermediate consumption
Disposable income	GDP at factor cost – net indirect tax

Answer:

A	B
NNP	GNP – depreciation
GNP	GDP – net factor income from abroad
Value added	Value of output- intermediate consumption
GDP at market prices	GDP at factor cost – net indirect tax
Disposable income	Personal income – direct taxes

Question 3.
Categorize the following into stocks and flows, wealth, salary, food grain stock, foreign exchange reserves, export, gross domestic saving, capital, change in money supply, quantity of money, capital formation.
Answer:

Stock	Flow
Wealth	Export
Foreign exchange reserves	Salary
Food grain stock	Gross domestic saving
Capital	Change in money supply
Quantity of money	Capital formation

Question 4.
The phase of circular flow of income in a two sector economy is given below.

1. Complete the diagram.
2. Explain the process of circular flow

Answer:

1.

2. Circular flow of income:
The concept that the aggregate value of goods and services produced in an economy is going around in a circular way. Either as factor payments, or as expenditures on goods and services, or as the value of aggregate production.

Question 5.
Suppose that in a two sector economy the value of finished goods is equal to ₹100 crore and the income generated as factor rewards is also equal to ₹100 crore. The households spend only ₹80 crore.

1. What will happen to the circular flow?
2. Which system can be introduced to correct the circular flow?
3. Name the leakages and injections.

Answer:

1. There will be a mismatch between the real flow and money flow in the circular flow. In other words, the flow will be broken.
2. As a corrective measure, the financial system can be introduced.
3. The leakages is the difference between the income generates and household spending.

This is saving. The injection are the savings that the households, firms and the government take from the financial institutions as borrowings.

Question 6.
1. Estimate the NI of India and Pakistan from the data given below.

2. Which method is used here?
3. What are the other methods of measuring national income?
Answer:

1. National income of India = ₹2885 crore
   National income of Pakistan = ₹1860 crore
2. The method used here is the product method or value added method.
3. Income method and expenditure method are the other two method of measuring national income.

Question 7.
What do you mean by GDP deflator? How far GDP deflator differs from Consumer Price Index?
Answer:
The ratio of nominal to real GDP is a well known index of prices. This is called GDP Deflator. GDP deflator differs from Consumer Price Index. The major points of difference are given below.

1. The goods purchased by consumers do not represent all the goods which are produced in a country. GDP deflator takes into account all such goods and services.

2. CPI includes prices of goods consumed by the representative consumer; hence it includes prices of imported goods. GDP deflator does not include prices of imported goods.

3. The weights are constant in CPI – but they differ according to production level of each good in GDP deflator.

Question 8.
Write down some of the limitations of using GDP as an index of welfare of a country.
Answer:
GDP is the sum total of value of goods and services created within the geographical boundary of a country in a particular year. It gets distributed among the people as incomes. So we may be tempted to treat higher level of GDP of a country as an index of greater well-being of the people of that country. But there are at least three reasons why this may not be correct. They are discussed below.

1. Distribution of GDP – how uniform is it:
If the GDP of the country is rising, the welfare may not rise as a consequence. This is because the rise in GDP may be concentrated in the hands of very few individuals or firms. For the rest, the income may, in fact, have fallen.

In such a case the welfare of the entire country cannot be said to have increased. If we relate welfare improvement in the country to the percentage of people who are better off, then surely GDP is not a good index.

2. Non-monetary exchanges:
Many activities in an economy are not evaluated in monetary terms. For example, the domestic services women perform at home are not paid for. The exchanges which take place in the informal sector without the help of money are called barter exchanges.

This is a case of underestimation of GDP. Hence GDP calculated in the standard manner may not give us a clear indication of the productive activity and well-being of a country.

3. Externalities:
Externalities refer to the benefits (or harms) a firm or an individual causes to another for which they are not paid (or penalized). Externalities do not have any market in which they can be bought and sold. Therefore, if we take GDP as a measure of welfare of the economy we shall be overestimating the actual welfare.

This was an example of negative externality. There can be cases of positive externalities as well. In such cases, GDP will underestimate the actual welfare of the economy.

Question 9.
Assume that GDP in the year 2007 was ₹1,200 which rose to ₹1,800 in 2008. Calculate GDP deflator.
Answer:
GDP deflator = Current year GDP / Base year GDP x 100
= 1800/1200 × 100
= 1.5 × 100
= 1.5 (in percentage terms 150)

Question 10.
Relate and complete the identities/equations in column A with column B.

Answer:

Question 11.
Estimate the Gross National Product at market price and GNP at factor cost through the expenditure method.

Item	Amount (in Crores)
Inventory investment	15
Net factor income from abroad	10
Personal consumption expenditure	475
Gross residential construction investment	48
Exports	25
Government purchase of goods and services	175
Gross public investment	15
Gross business fixed investment	38
Imports	12
Net indirect tax	8

Answer:
GNP_MP = private consumption expenditure + govt, final consumption expenditure( gross fixed capital formation + change in stock or inventory investment) + net export + net factor income from abroad
= 475 + 175 + 101 (i.e., 48 + 15 + 38) + 15 + 13
= ₹779 crores.
GN_PC = GNPUD – net indirect taxes
= 779 – 8 = ₹771 crores

Question 12.
Suppose that in a two sector economy, the value o finished goods is equal to ₹200 crore and the income generated as factor rewards is equal to ₹200 crore. The households spend only ₹180 crore. The remaing 20 crore economy saved then.

1. Is ₹20 (saving) included in the circular flow?
2. Which system can be introduced to correct the circular flow?
3. Is saving leakage or injection.

Answer:

1. No, saving (₹20) is excluded in the circular flow.
2. Financial system can be introduced to correct the circular flow.
3. Yes, saving is a leakage.

Question 13.
Fill in the blanks

1. GNP_MP – ……….. = NNP_MP
2. NNP_MP – ………… = NNP_FC
3. GDP_FC+ – ………… = GDP_MP
4. GDP + -………….. = GNP

Answer:

1. GNP_MP – depreciation = NNP_MP
2. NNP_MP – net indirect tax = NNP_FC
3. GDP_FC + net indirect tax = GDP_MP
4. GDP + net factor income from aborad = GNP

Question 14.
Write down the 3 identities of calculating the GDP of a country by the 3 methods. Also briefly explain why each of those should give us the same value of GDP.
Answer:
Gross National Product (GNP) equals Gross National Income equals Gross National Expenditure, i.e.
GNP = GNI = GNE
These are equal because national income is a circular flow of income. Aggregate expenditure is equal to aggregate output which in turn, is equal to aggregate income. However each method has some different items, yet they show exactly identical results.

Their identity can be shown in the following manner:
Reconciling Three Methods of Measuring Gross

Question 15.
The economic recession of 2008 affected the market economics in general and the US in particular. Thou-sands of Indians working abroad lost their job especially in IT and banking sectors and they returned to India. Evaluate its consequences on Indian economy with regard to the following macro variables.

1. The value of GNP
2. Gneral unemployment level
3. Foreign exchange rate

Answer:

1. The value of GNP decreases due to reduction in NFIA.
2. General unemployment level increases.
3. Foreign exchange rate increases.

Plus Two Economics National Income Accounting Eight Mark Questions and Answers

Question 1.
Given below some macro economic indicators. Derive the equations of the following terms:

1. GNP
2. NNP
3. NNP at factor cost
4. Personal income
5. Personal disposable income
6. Private Income
7. National Disposable Income

Answer:
1. GNP = GDP + Factor income earned by the domestic factors of production employed in the rest of the world – Factor income earned by the factors of production of the rest of the world employed in the domestic economy

2. NNP = GNP – Depreciation

3. NNP at factor cost = National Income (NI) = NNP at market prices – (Indirect taxes – Subsidies)

4. Personal income (PI) = NI – Undistributed profits – Net interest payments made by households – Corporate tax + Transfer payments to the households from the government and firms.

5. Personal Disposable Income (PDI) = PI – Personal tax payments – Non-tax payments.

6. Private Income = Factor income from net domestic product accruing to the private sector + National debt interest + Net factor income from abroad + Current transfers from government + Other net transfers from the rest of the world

7. National Disposable Income = Net National Product at market prices + other current transfers from the rest of the world

Question 2.
Prepare a seminar report on the topic ‘Measurement of National Income’.
Answer:
Measurement of National Income Respected teachers and dear friends,
The topic of my seminar paper is ‘measurement of national income or the methods of measuring national income’. The concept of national income occupies an important place in economic theory.

National income is the aggregate money value of all goods and services produced in a country during an accounting year. In this seminar paper, I would like to present various methods of measuring national income.

Content:
National income can be measured in different ways. Generally there are three methods for measuring national income. They are

1. Value-added method
2. Expenditure method
3. Income method

1. Value-added method:
The term that is used to denote the net contribution made by a firm is called its value-added. We have seen that the raw materials that a firm buys from another firm which are completely used up in the process of production are called ‘intermediate goods’.

Therefore the value-added of a firm is the value of production of the firm – value of intermediate goods used by the firm. The value-added of a firm is distributed among its four factors of production, namely, labor, capital, entrepreneurship, and land.

Therefore wages, interest, profits, and rents paid out by the firm must add up to the value-added of the firm. Value-added is a flow variable.

2. Expenditure Method:
An alternative way to calculate the GDP is by looking at the demand side of the products. This method is referred to as the expenditure method. The aggregate value of the output in the economy by expenditure method will be calculated.

In this method we add the final expenditures that each firm makes. Final expenditure is that part of expenditure which is undertaken not for intermediate purposes.

3. Income Method:
As we mentioned in the beginning, the sum of final expenditures in the economy must be equal to the incomes received by all the factors of production taken together (final expenditure is the spending on final goods, it does not include spending on intermediate goods).

This follows from the simple idea that the revenues earned by all the firms put together must be distributed among the factors of production as salaries, wages, profits, interest earnings, and rents.
That is GDP = W + P + In + R

Conclusion:
Thus it can be concluded that there are three methods for measuring national income. These methods are value-added method, income method and expenditure method. Usually in estimating national income, different methods are employed for different sectors and sub sectors.

Question 3.
From the following data, calculate personal income and personal disposable income (₹in Crores).

1. NDPFC – 8,000
2. net factor income from abroad – 200
3. Undistributed profit – 1,000
4. Corporate tax – 500
5. Interest received by households – 1,500
6. Interest paid by households – 1,200
7. Transfer income – 300
8. Personal Tax – 500

Answer:
Personal income = NDPfc + Net factor income from abroad – undistributed profits – corporate taxes + transfer payments + net interest received from households.
= 8000 + 200-1000 – 500 + 300 (1500 -1200)
= 7,300 crores
Personal disposable income = Personal income – personal tax
= 7,300 – 500 = 6,800 crores

Question 4.
Production generates income. Prove this statement with the help of a simple two sector model of circular flow of income.
Answer:
circular flow of income:
It is a pictorial representation of interdependence or interrelationship between the various sectors of the economy. It is a concept associated with income earning and spending. The circular flow of income in a simple economy works on the basis of certain assumptions.
They are as follows:

1. Households and firms are the only two sectors in an economy (2 sector model)
2. Households supply factor services to firms.
3. Firms hire factor services households
4. Household spends their entire income on consumption and thereby no savings are left with them.
5. Firms sell their entire products to the households
6. There is no government in the economy.
7. The economy is not related to any other economies or the economy is a ‘closed’ system. As a result, there is no export or imports from the economy.

In such an economy, there would be two types of markets.
They are:

1. product-market for goods and services
2. factor markets for buying and selling various factor services.

The relationship between the sectors of an economy can be explained with the help of a diagram.

The households own the factors of production such as land, labour, capital, and organization. The households sell these factors of production to the firms for producing goods and services are known as real flow. The rewards for factors of production are rent to land, interest to capital, wage to the labour and profit to the entrepreneur is known as the money flow.

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Students can Download Chapter 3 Functions Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions

Plus Two Computer Application Functions One Mark Questions and Answers

Question 1.
To read a single character for gender i.e. ’rri’ or ’f’.___function is used.
(a) getch()
(b) getchar()
(c) gets()
(d) getline()
Answer:
(b) getchar()

Question 2.
To use getchar(), putchar(), gets() and puts(), which header file is used?
(a) iostream
(b) cstdio
(c) input
(d) output
Answer:
(b) cstdio

Question 3.
To use cin and cout, which header file is needed?
(a) iostream
(b) cstdio
(c) input
(d) output
Answer:
(a) iostrem

Question 4.
Predict the output of the following code snippet
#include<cstdio>
int main()
{
char name[ ] = “ADELINE”;
for(int i=0; name[i]!=’\0′;i++)
putchar(name[i]);
}
Answer:
The output is “ADELINE”.

Question 5.
From the following which is equivalent to the function getc(stdin)
(a) putchar()
(b) gets()
(c) getchar()
(d) puts()
Answer:
(c) getchar()

Question 6.
From the following which is equivalent to the function putc(ch,stdout)
(a) putchar(ch)
(b) ch=gets()
(c) ch=getchar()
(d) puts(ch)
Answer:
(a) putchar(ch)

Question 7.
To print a single character at a time which function is used?
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) putchar()

Question 8.
To read a string____function is used.
(?) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) gets()

Question 9.
To print a string_____function is used.
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) puts()

Question 10.
Consider the following code snippet
main()
{
char str[80];
gets(str);
for(int i=0,len=0;str[i]!-\0′;i++,len++);
cout<<“The length of the string is” <<len;
getch();
}
Select the equivalent for the under lined statement from the following
(a) int len= strlen(str)
(b) int len=strcmp(str)
(c) int len = strcount(str)
(d) None of these
Answer:
(a) int len= strlen(str)

Question 11.
Arjun wants to read a string with spaces from the following which is suitable
(a) cin>>
(b) cin.getline(str,80)
(c) str=getc(stdin)
(d) none of these
Answer:
(b) cin.getline(str,80)

Question 12.
State whether the following statement is true or false. The'<<‘ insertion operator stops reading a string when it encounters a space.
Answer:
True.

Question 13.
The process of dividing big programs into small programs are called____.
Answer:
Modularization.

Question 14.
The big programs are divided into smaller programs. This smaller programs are called_____.
Answer:
Functions.

Question 15.
The execution of the program begins at____function.
Answer:
main function.

Question 16.
One of the following is not involved in the creation and usage of a user defined function
(a) Define a function
(b) Declare a function
(c) invoke a function
(d) None of these
Answer:
(d) None of these

Question 17.
The default data type returned by a function is_____.
(a) float
(b) double
(c) int
(d) char
Answer:
(c) int

Question 18.
After the execution of a function, it is returned back to the main function by executing____keyword.
Answer:
return.

Question 19.
Supplying data to a function from the called function by using______.
Answer:
parameters (arguments).

Question 20.
_____keyword is used to give a value back to the called function.
Answer:
return.

Question 21.
____key word is used to specify a function returns nothing.
Answer:
void

Question 22.
One of the following is not necessary in the function declaration. What is it?
(a) name of the function
(b) return type
(c) number and type of arguments
(d) name of the parameters
Answer:
(d) name of the parameters.

Question 23.
A function declaration is also called_____.
Answer:
prototype.

Question 24.
Considerthe following declaration
int sum(int a , int b)
{
return a+b;
}
From the following which is the valid function call.
(a) n=sum(10)
(b) n=sum(10, 20)
(c) n=sum(10, 20, 30)
(d) n=sum()
Answer:
(b) n=sum(10, 20)

Question 25.
The ability to access a variable or a function from some where in a program is called_____.
Answer:
scope.

Question 26.
A variable ora function declared within a function is have_____scope.
Answer:
local.

Question 27.
A variable or a function declared out side of all the functions is have_____scope.
Answer:
global.

Question 28.
State True/False

1. A local variable exist till the end of the program
2. A global variable destroyed when the sub function terminates

Answer:

1. False
2. False

Question 29.
consider the following declaration
int x;
int main()
{
–
}
Here x is a____variable.
Answer:
global.

Question 30.
consider the following declaration
int main()
{
int x;
–
}
Here x is a_____variable.
Answer:
local.

Question 31.
______parameter is used when the function call does not supply a value for parameters.
Answer:
default.

Question 32.
Consider the following function declaration with optional (default) arguments and state legal or illegal and give the reasons

1. int sum(int x=10, int y, int z)
2. int sum(int x=10, int y=20, int z)
3. int sum(int x=10, int y=20, int z=30)
4. int sum(int x, int y=20, int z)
5. int sum(int x, int y=20, int z=30)
6. int sum(int x, int y, int z=30)
7. int sum(int x=10, int y, int z=30)

Answer:
There is a rule to make an argument as default argument,i.e., to set an argument with a value that must be in the order from right to left. All the arguments in the right side of an argument must be set first to make an argument as a default argument.

1. illegal, because y and z are not have values
2. illegal, because z has no value
3. legal
4. illegal, because z has no value
5. legal
6. legal
7. illegal, because x has a value but y has no value.

Question 33.
The parameter used to call a function is called_____.
Answer:
actual parameter.

Question 34.
The parameters appear in a function definition are_____.
Answer:
formal parameters.

Question 35.
After the distribution of answer scripts, the teacher gives the Photostat copy of the mark list to the students to check the marks. If the students make any change that do not affect the original mark list. There is a similar situation to pass the arguments to a function. What is this method?
(a) call by value
(b) call by reference
(c) call by address
(d) none of these
Answer:
(a) call by value

Question 36.
Your class teacher gives you the original mark list to check the mark. If you make any change it will affect the original mark list. There is a similar situation to pass the arguments to a function. What is this method?
(a) call by value
(b) call by reference
(c) call by function
(d) none of these
Answer:
(b) call by reference

Question 37.
Consider the following function declaration
int sum(int a, int b)
{
Body
}
Here the arguments are passed by______.
Answer:
call by value method.

Question 38.
Consider the following function declaration
int sum(int &a, int &b)
{
Body
}
Here the arguments are passed by______.
Answer:
call by reference method.

Qn. 39
A function calls it self is known as______
Answer:
recursive function.

Question 40.
Varun wants to copy a string by using strcpy() function. From the following which header file is used for this?
(a) cstdio
(b) cmath
(c) cstring
(d) cctype
Answer:
(c) cstring

Question 41.
____is a named group of statements to perform a job /task and returns a value.
Answer:
Function.

Question 42.
To use the function setw(), from the following which header file is used.
Answer:
iomanip.h

Question 43.
In his C++ program Ajith wants to accept a lengthy text of more than one line. Which function in C++ can be used in this situation.
Answer:
gets() function can be used to accept a lengthy text.

Question 44.
Choose the C++ function which can print a string.
(a) getche()
(b) putchar()
(c) getline()
(d) puts()
Answer:
(d) puts()

Question 45.
Which of the following is a console function?
(a) getline()
(b) write()
(c) put()
(d) getchar()
Answer:
(d) getchar()

Question 46.
Pick the odd one out and give reason.
(a) abs()
(b) strlen()
(c) strcmp()
(d) strcpy()
Answer:
(a) abs() – it is a mathematical function. All others are string functions.

Question 47.
Consider the following C++ statement and answer the following question:
char Word[10]=”GOOD DAY”;
Identify the correct output statement to display the string
(a) write (word);
(b) cout.write(word);
(c) cout (word);
(d) cout.write (word, 10);
Answer:
(d) cout.write(wond, 10);

Question 48.
When the number -7 is given as the argument of a predefined function in C++, it returns the value 7. Identify the function.
Answer:
abs(): This function returns the absolute value.

Question 49.
Pick out the correct statement for prototype declaration from the following and also explain the various information it contains.
(a) product (int a, int b);
(b) int product (a,b);
(c) int product (int, int);
(d) product (int, int);
Answer:
(c) int product(int, int);
This prototype specifies the return type, name of function, number, and type of arguments.

Question 50.
One among the following function prototypes is wrongly written. Identify it. Also given reason.
(a) float test (float);
(b) float test (float, int);
(c) test (float);
(d) int test (int);
Answer:
(c) test(float);
Here the prototype contains no return type.

Question 51.
A user defined function definition is given below. Choose the most appropriate function call statement from the options.
float calc(int x, float y)
{
return (x+y) / 2.0;
}
(a) calc (2, 4)
(b) calc (2.5, 4)
(c) calc (2.5, 4.5)
(d) calc (2, 4.5)
Answer:
(d) calc(2, 4.5);

Question 52.
Which of the following statements are FALSE about a local function?
(a) Declared inside a function
(b) Accessible only within the function it is declared
(c) Accessible from anywhere in the program
(d) Declared outside all other functions
Answer:
c and d are false to a local function.

Plus Two Computer Application Functions Two Mark Questions and Answers

Question 1.
In a C++ program, you forgot to include the header file iostream.h. What are the possible errors occur in that Program? Explain ?
Answer:
Proto type error. To use cin and cout the header file iostream is a must.

Question 2.
Pick the odd one out from the following and give reason.

• gets()
• getline()
• getch()
• getchar()

Answer:
getline() – It is a stream function where as the others are console functions.

Question 3.
Consider the following code snippet.
using namespace std;
int main()
{
int n;
cout<<“Enter a number”; cin>>n;
cout<<‘The number is “<<n;
}
Write down the names of the header files that must be included in this program
Answer:
Here cin and cout are used so the header file iostream must be included.

Question 4.
How does C++ support modularity in programming
Answer:
The process of converting big programs into smaller programs is known as modularisation. This small programs are called modules or sub programs or functions. C++ supports modularity in programming called functions.

Question 5.
The following assignment statement will generate a compilation error.
char str[20]; str=”Computer”
Write a correct C++ statement to perform the same task
Answer:
char str[20] = “Computer”;

OR

char str[20];
strcpy(str,”Computer”); (The header file should be included).

Question 6.
float area(const float pi=3.1415, const float r)
{
r=10;
return pi*r;
}
Is there any problem? If yes what is it?
Answer:
There is an error. The error is , Y is a constant T must be initialised and cannot be changed during execution.

Question 7.
What are the jobs of a return statement in a program.
Answer:
In the case of a sub function a return statement helps to terminate the sub function and return back to the main function or called function. But in the case of a main function it terminates the program.

Question 8.
Match the following

(a) strcmp()	(1) cctype
(b) tolower()	(2) cstring
(c) sqrt()	(3) cstdlib
(d) abort ()	(4) cmath

Answer:
(a) 2
(b) 1
(c) 4
(d) 3

Question 9.
How to invoke a function in C++ program.
Answer:
A function can be called or invoked by providing the name of the function followed by the arguments in parenthesis Eg. sum(m,n);

Question 10.
Briefly explain constant arguments Constant arguments.
Answer:
By using the key word const we can make argument (parameter) of a function as a constant argument.
The value of the const argument cannot be modified within the function.

Question 11.
void initialise()

1. Identify the error in the, above code and explain its reasons.
2. Correct the errors.

Answer:

1. K is a local variable in the function initialize() – It is not accessible in main()
2. Making the variable K as global we can correct the error.

Question 12.
List down the advantages of modular programming.
Answer:
Merits of modular programming

• It reduces the size of the program
• Less chance of error occurrence
• Reduces programming complexity
• Improves reusability

Question 13.
Some statements are given below. Analyse these statements and predict the output:
char str1 (15], str2[15];
str1[15]=” DATA”;
str2[15]=” STORAGE”;
strcat (str2, str1);
cout<<str2;
Answer:
The output is STORAGE DATA. The strings str2 and str1 are concatenated.

Question 14.
If char name [ ] = “Rajeev Kumar”; then what will be output of the following statement? cout<<strlen(name);
Answer:
The length(number of characters) is 12 including space.

Question 15.
Choose the value of ‘n’ after executing the following statements in C++.
char s1[ ]=”KIRAN”; char s2[ ]=”kiran”;
int n = strcmp (s1,s2);
cout<<n;
(a) 0
(b) >0
(c) <0
(d) None of these
Answer:
(c) <0. Here string 2, i.e. s2 is greaterthan string1 i.e. s1.
strcmp()- It is used to compare two strings and returns an integer.
Syntax: strcmp(string1 ,string2)

• if it is 0 both strings are equal.
• if it is greater than 0(i.e. +ve) string1 is greater than string2
• if it is less than 0(i.e. -ve) string2 is greater than string1

Question 16.
C++ has a built-in function with which we get the result of 4².

1. Identify the name of the function.
2. Identify the header file for the above function.

Answer:

1. pow(4, 2);
2. The header file used is cmath.

Question 17.
Consider the following C++ statements and predict the output.
int p=isalpha(‘5’);
cout<<p;
Answer:
0.
isalpha() – To check whether a character is an alphabet or not. If the character is an alphabet it returns a value 1 otherwise it returns 0.

Question 18.
Predict the output of the following C++ statements:

1. cout<<toupper(‘a’);
2. cout<<(char) toupper(‘a’);

Answer:

1. It prints 65
2. It prints A

Question 19.
Differentiate formal arguments and actual arguments.
Answer:
The parameter used to call a function is called actual parameter. The parameters appear in a function definition are formal parameters.

Plus Two Computer Application Functions Three Mark Questions and Answers

Question 1.
Suresh wants to print his name and native place using a C++ program. The program should accept name and native place first
Name is: Suresh Kumar
Address is: Alappuzha
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()

Question 2.
“Programming is Fun”. Write a C++ program to read a string like this in lowercase and print it in UPPER CASE. Without using toupper() library function.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()

Question 3.
An assignment, Kumar has written a C++ program which reads a line of text and print the number of vowels in it. What will be his program code?
Answer:
# include<iostream>
# include<cstdio>
# include<cctype>
using namespace std;
int main()

Question 4.
Write a program to display the following output
A
BB
CCC
Answer:
# include<iostream>
using namespace std;
int main()

Question 5.
Distinguish getchar and gets.
Answer:
getchar is a character function but gets is a string function. The header file cstdio must be included. It reads a character from the keyboard.
Eg.
char ch;
ch=getchar();
cout<<ch;
gets is used to read a string from the key board. It reads the characters upto enter key. The header file
cstdio must be included.
char str[80];
cout<<” Enter a string”;
gets(str);

Question 6.
Write a program to check whether a string is palindrome or not. (A string is said to be palindrome if it is the same as the string constituted by reversing the characters of the original string. Eg. “MALAYALAM”, “MADAM”, “ARORA”, “DAD”, etc.
Answer:
# include<iostream>
using namespace std;
int main()

Question 7.
Explain multi character function.
Answer:
getline() and write() functions are multi character functions.
1. getline():
It reads a line of text that ends with a newline character. It reads white spaces also.
Eg.
char line[80];
cin.getline(line,80);

2. write():
It is used to display a string.
Eg.
charline[80];
dn.getline(line, 80);
cout.write(line, 80);

Question 8.
Distinguish between get() and put() functions.
Answer:
1. get() function:
get() is an input function. It is used to read a single character and it does not ignore the white spaces and newline character.
Syntax is cin.get(variable);
Eg. char ch;
cin.get(ch);

2. put() function:
put() is an output function. It is used to print a character.
Syntax is cout.put(variable);
Eg. char ch;
cin.get(ch);
cout.put(ch);

Question 9.
Write a program to read a string and print the number of consonents
Answer:
# include<iostream>
# include<cstdio>
# include<cctype>
using namespace std;
int main()

Question 10.
Write a program to read a string and print the num¬ber of spaces.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()

Question 11.
Write a program to count the number of words in a sentence
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()

Question 12.
Write a program to input a string and display its reversed string using console I / O functions only. For example if the input is “AND” the output should be “DNA”.
Answer:
# include<iostream>
using namespace std;
int main()

Question 13.
Write a program to input a word(say COMPUTER) and create a triangle as follows.

Answer:
# include<iostream>
# include<cstring>
using namespace std;
int main()

Question 14.
Write a program to input a line of text and display the first characters of each word. Use only console I /O functions. For example, if the input is “Save Water, Save Nature”, the output should be “SWSN”.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()

Question 15.
Consider the following code snippet
char ch;
cout<< “Enter an alphabet”; cin>>ch;
cout<<toupper(ch);
What is the output of the above code? Give a sample output. If the above code is used in a computer that has no cctype file, how will you modify the code to get the same output?
Answer:
It reads a character and convert it into uppercase.
Eg:
Enter an alphabet: a
The output is A.
If a computer has no cctype header file the code is as follows.
char ch;
cout<< “Enter an alphabet”; cin>>ch;
if (ch>=97 && ch<<122)
cout<

Question 16.
Read the following program
# include<iostream.h>
int main()
{
cout<<sum(2, 3);
}
int sum(int x, int y)
{return (x + y);}
On compilation on the program, an error will be dis-played. Identify and explain the reason. How can you rectify the problem
Answer:
The compilation of the program starts from the first line and next line and so on( i.e. line by line). While compiling the line cout<<sum(2, 3); The compiler does not understand the word sum(2, 3) because it is not declared yet hence the error prototype required. To rectify this problem there are two methods
First method
Give the function definition just before the main function as follows.
# include<iostream>
using namespace std;
int sum(int x, int y)
{return (x+y);}
int main()
{
cout<<sum(2, 3);
}

Second Method
Give the function declaration(prototype only) in the main function as follows.
# include<iostream>
using namespace std;
int main()
{
int sum(int, int);
cout<<sum(2, 3);
}
int sum(int x, int y)
{return (x+y);}

Question 17.
Considering the following function definition;

The expected, desired output is 5! = 120
What will be the actual output of the program? It is not the same as above, why? What modification are required in the program to get the desired output.
Answer:
The output is 0! = 120
Because the address of variable ‘a’ is given to the variable ‘n’ of the function fact(call by reference method). So the function changes its value (i.e. n- -) to 0. Hence the result.

To get the desired result call the function as call by value method in this method the copy of the value of the variable ‘a’ is given to the function. So the actual value of ‘a’ will not changed. So instead of int fact(int &n) just write int fact(int n), i.e., no need of & symbol.

Question 18.
A function is defined as follows
int sum (int a, int b=2)
{return (a+b);}
Check whether each of the following function calls is correct or wrong, Justify your answer

1. cout<<sum(2, 3);
2. cout<<sum( 2);
3. cout<<sum();

Answer:
Here the function is declared with one optional argument. So the function call with minimum one argument is compulsory.

1. 0 It is valid. Here a becomes 2 and b becomes 3.
2. It is also valid . Here a becomes 2 and b takes the default value 2.
3. It is not a valid call. One argument is compulsory.

Question 19.
How do two functions exchange data between them? Compare the two methods of data transfer from calling function to called function.
Answer:
There are two methods they are call by value and call by reference
1. call by value:
In call by value method, a copy of the actual parameters are passed to the formal parameters. If the function makes any change it will not affect the original value.

2. call by reference:
In call by reference method, the reference of the actual parameters are passed to the formal parameters. If the function makes any change it will affect the original value.

Question 20.
Write down the operation performed by the following statements

1. int l=strten(“Computer Program”);
2. charch [] = tolower(“My School”);
3. cout<<(strcmp(“High”, “Low”)>0 ?

toupper(“High”):tolower(“Low”));
Answer:

1. The built in function strlen find the length of the string i.e. 16 and assigns it to the variable I.
2. This is an error because tolower is a character function.
3. This is also an error because tolower and toupper are character functions.

Question 21.
A line of given length with a particular character is to be displayed. For example, ********** is a line with ten asterisks (*). Define a function to achieve this output
Answer:
#include<iostream>
using namespace std;
void line (char ch, int n)

Question 22.
Read the following function
int fib(int n)
{
if (n<3)
return 1;
else
return (fib(n-1) + fib(n-2));
}

1. What is the speciality of this function
2. How does it work ?
3. What will be the output of the following code?

for (int i=1; i<5; i++)
cout<<fib(i)<<‘\t’;
Answer:
1. This function is a recursive function. That means the function calls itself.

2. It works as follows
if i = 1, The function fib calls with value 1. i.e. fib(1) returns 1
if i = 2, The function fib calls with value 2. i.e. fib(2) returns 1
if i = 3, The function fib calls with value 3. i.e. fib(3) returns fib(2) + fib(1) i.e. it calls the function again. So the result is 1 + 1 = 2
if i = 4, The function fib calls with value 4. i.e. fib(4) returns fib(3) + fib(2) i.e. it calls the function again. So the result is 2 + 1 = 3

3. The output will be as follows
1 1 2 3.

Question 23.
Explain scope rules of functions and variables in a C++ program
Answer:
1. Local variable or function:
A variable or function declared inside a function is called local variable or function. This cannot be accessed by the outside of the function.
Eg.
main()
{
int k;//local variable ,
cout<<sum(a,b); // local function
}

2. Global variable or function:
A variable or function declared out side of a function is called global variable or function. This can be accessed by any statements.
Eg.
int k; // global variable
int sum(inta, int b); //global function
main()
{
}

Question 24.
Briefly explain default arguments.
Answer:
A default value can be set for a parameter(argument) of a function. When the user does not give a value the function will take the default value. An important thing remember is an argument cannot have a default value unless all arguments on its right side must have default value.

Functions with valid default arguments are given below

• float area(int x, int y, int z=30);
• float area(int x, int y=20, int z=30);
• float area(int x=10, int y=20, int z=30);

Functions with invalid default arguments are given below

• float area(int x=10, int y, int z);
• float area(intx, inty=20, int z);
• float area(int x=10, int y=20, int z);

Question 25.
Write a program to read a character and check whether it is alphabet or not. If it is an alphabet check whether it is upper case or lower case?
Answer:
#include<iostream>
using namespace std;
int main()

Question 26.
Write a program to read 2 strings and join them
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()

Question 27.
Write a program to read 2 strings and compare it.
Answer:
# include<iostream>
# include<cstdio>
# include<cstring>
using namespace std;
int main()

Question 28.
Write a program to read a string and display the number of alphabets and digits and special characters.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()

Question 29.
A. void change(int&);
int main()

B. void change(int);
int main()

1. Predict the output of both programs.
2. Justify your predictions.

Answer:
1. A. Output
value = 40
B. output
value = 0

2. In the first case (A) the argument x is passed by reference method. So the changes made in the function reflects in main()

In the second case (B) the argument x is passed by value method. So the changes made in the function will not reflect in main().

Question 30.
Write a program to read 2 strings and join them using string function
Answer:
# include<iostream>
# include<cstring>
using namespace std;
int main()

Question 31.
Differentiate the outputs of the folloiwng C++ statements and also give reason

1. cout< <strcmp(“world”, “WORLD”);
2. cout<<strcmpi(“world”, “WORLD”);

Answer:

1. >0 Here first string “world” is greater than “WORLD”.
2. It prints 0. Because strcmpi is same as strcmp() but it is not case sensitive. That means uppercase and lowercase are treated as same.

Question 32.
Match the following.

1. strcmp()	a. To combine two strings
2. strcpy ()	b. To get 5 from 25
3. strcat ()	c. To get 10 from -10
4. sqrt ()	d. To change a to A
5. abs()	e. To compare two strings
6. toupper()	f. To copy one string another

Answer:
1-e, 2-f, 3-a, 4-b, 5-c, 6-d.

Question 33.
Write a C++ program to find the sum of first ‘N’ natural numbers using a user defined function.
Answer:
# include<iostream>
using namespace std;
int sum(int n)

Question 34.
Write the need for function prototype in a C++ program.
Answer:
When the function is defined after the main function then there is an error called “function should have a prototype”. This is because of the function is defined after the main function. To resolve this a prototype should be declared inside the main function.

Question 35.
Write suitable function prototype after reading the following cases.

• Case I : The function Volume() takes two arguments, one is float the other is int and it returns its volume.
• Case II : A function Big() has no arguments and no return type.
• Case III: A function PrintO takes two floating point type arguments and nothing is returned.

Answer:

• Case I: float Volume(float,int);
• Case II: void BigO; or void Big(void);
• Case III : void Print(float,float) or void Print(double,double);

Question 36.
Find the error in the following C++ program and rectify it.
#include<iostream>
using namespace std;
int main ()

Answer:
Error 1: Here function prototype is missing.
Error 2: no need for variable z.
Error 3: no need of the statement z = x*y;
#include<iostream>
using namespace std;
int multi(int, int);
int main()

Question 37.
Consider the following function definition.
int add (int a, int b=2, int c=5)
{
int s = a + b + c;
cout<<“Sum is :”<<s;
}
Predict the output of the above code forthe following function calls:

1. add (5, 8, 10);
2. add (5, 8);
3. add (5);

Answer:

1. add(5, 8, 10). Here a = 5, b = 8 and c = 10. Then It prints 23(5 + 8 + 10)
2. add(5, 8). Here a = 5, b = 8 and no value for c then c will take the default value 5. Hence it prints 18(5 + 8 + 5(default value for c)).
3. add(5). Here a = 5 and no values for b and c, then b and c will take the default values 2 and 5 respectively. Hence it prints 12(5 + 2 + 5(default values for b and c)).

Question 38.
Consider the following C++ program, predict the output and justify it.
#include<iostream>
using namespace std;
intsqr(int&);
int main()

Answer:
The output are 25 and 6. Here the function uses call by reference method. The function call sqr(a) passes the original value to the function sqr. The function changes the value of b(here a and b are same) to 6. That means a also becomes 6.

Question 39.
Differentiate local variable and global variable in C++ program.
Answer:
1. Local scope:
A variable declared inside a block can be used only in the block. It cannot be used any other block.
Eg: int sum(int n1,int n2)
{
int s;
s=n1+n2;
return(s);
}
Here the variable s is declared inside the function sum and has local scope;

2. Global scope:
A variable declared outside of all blocks can be used anywhere in the program.
Eg:
int s;
intsum(int n1,int n2)
{
s=n1+n2;
return(s);
}
Here the variable s is declared out side of all functions and we can use variable s any where in the program.

Question 40.
Consider the following C++ code fragment and identify the local function and global function. Also justify your selection.
int main()

Answer:
Here the function print() is declared inside a function hence it is a local function but the function sum() is declared outside of all functions hence it is called global function.

Question 41.
Read the following C++ program and identify the error and give reason.
# include<iostream>
using namespace std;
void disp(int);
int main()
int x=10; disp (x); return 0;

Answer:
The variable ‘x’ will not be printed because it is declared in the main() function. That is x is a local variable.

Plus Two Computer Application Functions Five Mark Questions and Answers

Question 1.
What will be the output of the following code if the userenterthe value “GOOD MORNING”

1. char String [80];
   gets(string);
   cout<<string;
2. char String [80]; cin>>string;
   cout<<string;
3. char ch;
   ch=getchar();
   cout<<ch;
4. char String [80];
   cin.getline(string,9);
   cout<<string;

Answer:

1. GOOD MORNING
2. GOOD
3. G
4. GOOD MORN

Question 2.
Read a string and print the number of vowels.
Answer:
# include<iostream>
# include<cstdio>
# include<cctype>
using namespace std;
int main()

Question 3.
Describe in detail about the unformatted console i/o functions.
Answer:

1. Single character functions: This function is used to read or print a character at a time.
   • getchar(): It reads a character from the keyboard and store it in a character variable.
     Eg. char ch;
     ch=getchar();
   • putchar(): This function is used to print a character on the screen.
     Eg. char ch;
     ch=getcharO;
     putchar(ch);
2. String functions: This function is used to read or print a string.
   • gets(): This function is used to read a string from the keyboard and store it in a character variable.
     Eg. char str[80];
     gets(str);
   • puts(): This function is used to display a string on the screen.
     Eg. char str[80];
     gets(str);
     puts(str);

Question 4.
Write a program to input a string and find the number of uppercase letters, lowercase letters, digits, special characters, and white spaces.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()

Question 5.
Write a program to input a string and replace all lowercase vowels by the corresponding uppercase letters.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()

Question 6.
A list of C++ built in functions are given. Classify them based on the usage and prepare a table with proper group names.

• strcmp()
• sin()
• getch()
• isalpha()
• pow()
• puts()
• strcat()
• tolower()
• getchar()
• isalnum()
• sqrt()
• exp()
• write()

Answer:

Question 7.
The factorial of a number, say N is the product of first N natural numbers. Thus, factorial of 5 can be obtained by taking the product of 5 and factorial of 4. Similarly factorial of 4 be found out by taking the product of 4 and factorial of 3. At last the factorial of 1 is 1 itself. Which technique is applicable to find the factorial of a number in this fashion? Write a C++ function to implement this technique. Also explain the working of the function by giving the number 5 as input.
Answer:
A function calls itself is known as recursion.
# include<iostream>
using namespace std;
int fac(int);
int main()

The working of this program is as follows. If the value of n is 5 then it calls the function as fac(5). The function returns value 5*fac(4). That means this function calls the function again and returns 5*4*fac(3). This process continues until the value n=1. So the result is 5*4*3*2*1 = 120.

Question 8.
Read the following program
# include<iostream>
using namespace std;
int a = 0;
int main ()

Write down the value displayed by the output of the above program with suitable explanation. What are the inferences drawn regarding the scope of variables?
Answer:
The output is 061.
1. Global variable: A variable declared out side of all functions it is known as global variable.

2. Local variable: A variable declared inside of a function it is known as local variable.

If a variable declared inside a function(main or other) with the same name of a global variable. The function uses the value of local variable and does not use the value of the global variable.

Here int a=0 is a global variable. In the main function the global variable ‘a’ is used. There is no local vari-able so the value of ‘a’, 0 is displayed. The statement ‘a++’ makes the value of ‘a’ is 1. It calls the function showval with argument ‘a=1’.

The argument ‘x’ will get this value i.e. ‘x=T. But in the function showval there is a local variable ‘a’ its value is 5 is used. So this function returns 6 and it will be displayed. After this the value 1 of the global variable ‘a’ will be displayed. Hence the result 061.

Question 9.
The following are function calling statements. Some of them will be executed, while some other generate compilation error. Write down your opinion on each of them with proper justification
Answer:

1. char ch=getch();
2. sqrt(25);
3. strcat (“Computer”, “Program”);
4. double num = pow(2, 3, 5)
5. put char(getchar());

Answer:

1. getch get a character from the console(key board) but does not echo to the screen. So we can’t read a character from the console.
2. It returns the square root of 25.
3. It concatenates Program to computer, i.e. we will get a string “computer program”
4. The function pow should contains only two arguments. But here it contains 3 arguments so it is an error. We can write this function as follows Double num = pow(pow(2, 3, 5)
5. It reads a character from the console and display it on the screen.

Question 10.
Short notes about character functions and string functions
Answer:
a. Character functions:
1. isalnum(): It is used to check whether a character is alphabet or digit. It returns a non zero value if it is an alphabet or digit otherwise it returns zero.

2. isalpha(): It is used to check whether a character is alphabet or not. It returns a non zero value if it is an alphabet otherwise it returns zero.

3. isdigit(): It is used to check whether a character is digit or not. It returns a non zero value if it is digit otherwise it returns zero.

4. islower(): It is used to check whether a character is lower case alphabet or not. It returns a non zero value if it is a lower case alphabet otherwise it returns zero.

5. isupper(): It is used to check whether a character is upper case alphabet or not. It returns a non zero value if it is an upper case alphabet otherwise it returns zero.

6. tolower(): It is used to convert the alphabet into lowercase.

7. toupper(): It is used to convert the alphabet into upper case.

b. String functions:
1. strcpy(): This function is used to copy one string into another.

2. strcat(): This function is used to concatenate(join) second string into first string.

3. strlen(): This function is used to find the length of a string.

4. strcmp(): This function is used to compare 2 strings. If the first string is less than second string then it returns a negative value. If the first string is equal to the second string then it returns a value zero and if the first string is greater than the second string then it returns a positive value.

Question 11.
Write a program to perform the following operations on a string

1. Length of a string
2. Search a character
3. Display the string

Answer:
# include<iostream>
# include<cstdio>
using namespace std;
void len()

Question 12.
Write functions to perform the following operations.

1. sqrt()
2. power of 2 numbers
3. sin
4. cos

Answer:
# include<iostream>
# include<cmath>
using namespace std;
void sqroot()

Question 13.
Read the following C++ programs and answer the questions:
Case I
# include<iostream>
using namespace std;

Case II
# include<iostream>
using namespace std;

1. Identify the type of function call in each case.
2. How do they differ?

Answer:

1. In case I the method used is Call by value and in Case II is Call by reference.
2. There are two methods they are call by value and call by reference

a. call by value:
In call by value method, a copy of the actual parameters are passed to the formal parameters. If the function makes any change it will not affect the original value.

b. call by reference:
In call by reference method, the reference of the actual parameters are passed to the formal parameters. If the function makes any change it will affect the original value.

Students can Download Chapter 10 Enterprise Resource Planning Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 10 Enterprise Resource Planning

Plus Two Computer Application Enterprise Resource Planning One Mark Questions and Answers

Question 1.
______is a group of people and other resources working together for a common goal.
Answer:
An enterprise

Question 2.
ERP stands for_______.
Answer:
Enterprise Resource Planning

Question 3.
The 4 M’s related to resources are
Answer:
Man, Material, Money, and Machine.

Question 4.
_____consists of single database and a collection of programs to handle the database hence handle the enterprise efficiently and hence enhance the productivity.
Answer:
ERP

Question 5.
state True or False
An ERP package consists of only one module
Answer:
False. It consists of many modules.

Question 6.
_____module is the core of ERP package
Answer:
Financial

Question 7.
ERP stands for_____.
(a) Entertain Resource Package
(b) Enterprise Retain Plan
(c) Enterprise Resource Planning
(d) None of these
Answer:
(c) Enterprise Resource Planning

Question 8.
_____module of ERP contains rules to manage production process.
Answer:
Manufacturing Module

Question 9.
_____module of ERP contains rules to plan the production process.
Answer:
Production Planning Module

Question 10.
_____module of ERP focus on human resource and human capital.
Answer:
HR module

Question 11.
______module of ERP is used to get the raw materials in the right time and right price.
Answer:
Purchasing Module

Question 12.
______module of ERP is used for monitoring and tracking customer oriented activities.
Answer:
Marketing Module

Question 13.
______module of ERP deals with important parts of sales cycle.
Answer:
Sales and distribution module

Question 14.
______module of ERP is used for managing the quality.
Answer:
Quality management module

Question 15.
BPR stands for ______.
Answer:
Business Process Re-engineering (BPR)

Question 16.
In general_____is the series of activities such as rethinking and redesign of the business process to enhance the enterprise’s performance such as reducing the cost (expenses), improve the quality, prompt and speed (time bound) service.
Answer:
BPR (Business Process Re-engineering)

Question 17.
Odd one out
(a) Inputs
(b) Processing
(c) Outcome
(d) SAP
Answer:
(d) SAP is a ERP package the others are 3 elements of business process.

Question 18.
Odd one out
(a) Oracle
(b) SAP
(c) Odoo
(d) C++
Answer:
(d) C++ is a general purpose programming language. The others are ERP packages.

Question 19.
Odd one out
(а) html
(b) C++
(c) JavaScript
(d) Tally ERP
Answer:
(d) Tally ERP is a ERP package.

Question 20.
From the following select the open source ERP.
(a) Microsoft Dynamics
(b) Tally ERP
(c) Odoo
(d) SAP
Answer:
(c) Odoo

Question 21.
Odd one out. Give the reason.
(a) Microsoft Dynamics
(b) Tally ERP
(c) Odoo
(d) SAP
Answer:
(c) Odoo is an open source ERP. The others are not.

Question 22.
PLM stands for_____.
Answer:
Product Life cycle Management

Question 23.
CRM stands for______.
Answer:
Customer Relationship Management

Question 24.
MIS stands for______.
Answer:
Management Information System

Question 25.
SCM stands for______.
Answer:
Supply Chain Management

Question 26.
DSS stands for_____.
(a) Decision Signal System
(b) Decision Support System
(c) Decision Support Scheme
(d) Design Support System
Answer:
(b) Decision Support System

Question 27.
Pick the odd one out and justify.
(a) SAP
(b) Oracle
(c) C++
(d) Tally
Answer:
(c) C++. This is a programming Language. The others are ERP packages

Question 28.
SAP stands for______.
Answer:
Systems, Applications, and Products for data processing

Question 29.
Pick the odd one out from the following list and justify the selection.
CRM, MIS, SCM, SAP
Answer:
SAP others are ERP related technologies. SAP is an ERP package.

Question 30.
Consider the following two statements.
Statement 1: “The number of functional modules in an ERP vary with nature of enterprise” Statement 2: “There is no connection between BPR and ERP”
Then choose the correct one from the following.
(i) Both statements are true
(ii) Both statements are false
(iii) Statement 1 is true and statement 2 is false
(iv) Statement 1 is false and statement 2 is true
Answer:
(iii) Statement 1 is true and Statement 2 is false.

Question 31.
Choose the correct answer from the following: Implementation of ERP in an enterprise ______.
(a) minimize planning risks.
(b) integrates different functional units of an enter-prise.
(c) uses centralized data base.
(d) All the above.
Answer:
(d) All the above.

Question 32.
State True or False

1. Every ERP package can manage all the functional units of an enterprise.
2. In ERP, a centralized database is used for integrating functional units.

Answer:

1. False
2. True

Question 33.
ERP stands for_____.
(a) Enterprise Resource Project
(b) Enterprise Resource Processing
(c) Enterprise Resource Planning
(d) Enterprise Requirement Planning
Answer:
(c) Enterprise Resource Planning

Plus Two Computer Application Enterprise Resource Planning Two Mark Questions and Answers

Question 1.
Define an Enterprise.
Answer:
An Enterprise is a group of people and other resources working together for a common goal. It is also an example for System.

Question 2.
Define Enterprise Resource Planning ERP.
Answer:
An enterprise(organization) is considered as a system (A system is an orderly grouping of interdependent components linked together to achieve an objective, according to a plan. Human body is an example for System).

All the departments of an enterprise are connected to a centralized data base. ERP consists of single database and a collection of programs to handle the database hence handle the enterprise efficiently and hence enhance the productivity.

Question 3.
Give the significance of HR module in an ERP package.
Answer:
This model ensures the effective use of Human resources and Human capital and enhance the productivity of the Enterprise hence increase the profit.

Question 4.
Give the relation between business process reengineering (BPR) and enterprise resource planning.
Answer:
ERP and BPR will not make much change if they are in stand alone. To improve the efficiency of an enterprise integrate both ERP and BPR because they are the two sides of a coin.

For the better results conducting BPR before implementing ERP, this will help an enterprise to avoid unnecessary modules from the software.

Question 5.
“The key concept of ERP is a centralised database management system”. Justify.
Answer:
ERP is an integrated business management system which uses a single database to store and communicate information of various departments of an enterprise.

Question 6.
Match the following.

(i) Financial module	(a) Supply chain
(ii) Oracle	(b) Functional unit of ERP
(iii) BPR	(c) ERP package
(iv) SCM	(d) Reengineering

Answer:
(i) – (b), (ii) – (c), (iii) – (d), (iv) – (a)

Question 7.
Briefly explain the importance of Business Process Re-engineering (BPR) in the implementation of ERP in an enterprise.
Answer:
Business Process Reengineering: In general BPR is the series of activities such as rethinking and redesign of the business process to enhance the enterprise’s performance such as reducing the cost(expences), improve the quality, prompt and speed(time bound) service.

Plus Two Computer Application Enterprise Resource Planning Three Mark Questions and Answers

Question 1.
Write short notes about BPR.
Answer:
In this world, tight competition is based on price, quality, wide variety of selection and quick service. To increase the business and hence increase the profit of a Business firm various activities are involved.

IT and Re-engineering plays major roles to increase the productivity. In general BPR is the series of activities such as rethinking and redesign of the business process to enhance the enterprise’s performance such as reducing the cost(expenses), improve the quality, prompt and speed(time bound) service. BPR enhances the productivity and profit of an enterprise.

Question 2.
Give an example for an enterprise from your real life and identify different departments/functional units in it.
Answer:
Indian Oil Corporation is an example for an enterprise. The activities involved are planning, purchasing raw material, production, storing finished goods (warehouse), sales, finance, etc. These activities are performed by different departments and their duties are interlinked.

Question 3.
The first five phases of ERP implementation are listed below. Rearrange them in correct order.
Package selection, BPR, Gap analysis, pre evaluation screening, project planning.
Answer:
The correct order is as follows

1. Pre evaluation screening
2. Package selection
3. Project planning
4. Gap analysis
5. Business Process Reengineering

Question 4.
Write a short note about the following terms:

1. DSS
2. MIS

Answer:
1. Decision Support System (DSS):
It is a computer based system that takes inputs as business data and after processing it produces good decisions as output that will make the business easier. Management Information.

System (MIS):
Management is the decision and policy makers. A good management can take good decision and that will help to do the business well. The good relationship between Management and employees is a key to success.

MIS will collect relevant data from inside and outside of a company. Based upon this information produce reports and take appropriate decisions.

Plus Two Computer Application Enterprise Resource Planning Five Mark Questions and Answers

Question 1.
Explain functional units of ERP in detail.
Answer:
Different modules are given below:

1. Financial Module:
It is the core. This is used to generate financial report such as balance sheet, general ledger, trial balance, financial statement, etc.

2. Manufacturing Module:
It provides information for the production and capable to change the methods in manufacturing sector.

3. Production planning Module:
This module ensures the effective use of resources and helps the enterprise to enhance the productive hence increase the profit.

4. HR (Human Resource) Module:
This model ensures the effective use of Human resources and Human capital.

5. Inventory control Module:
This model is useful to maintain the appropriate level of stock(includes raw material, work in progress and finished goods)

6. Purchasing Module:
This module is useful to make available the required raw materials in good condition and in the right time and price.

7. Marketing Module:
It is used for handle the orders of customers.

8. Sales and distribution Module:
Existence of a company is based on the income from sales. This module will help to handle the sales enquiries, order placement, and scheduling, dispatching and invoicing.

9. Quality (Ql & QC) management module:
Quality of a product or service is very much important to a company. This module helps to maintain the quality of the product. Quality planning, inspection, and control are the main activities involved in this module.

Question 2.
Explain the different phases of ERP implementation.
Answer:
The different phases of ERP implementation are given below

1. Pre evaluation screening:
Many ERP packages are available in the markets. At most care should be taken before implementing a ERP. Select a few from from the available ERP packages.

2. Package selection:
Selection of right ERP to our enterprise is a laborious task and it needs huge investment. Various factors should be keep in mind before you purchase an ERP that should meet our complete needs.

3. Project planning:
A good planning is essential to implement an ERP. From the beginning to the end activities are depicted in this phase.

4. Gap analysis:
A cent percent(100%) problem solving ERP is not available in the market. Most of them solve a maximum of 70% to 80% problems. The rest (30% to 20%) of the problems and their solutions are mentioned here.

5. Business Process Reengineering:
In general BPR is the series of activities such as rethinking and redesign of the business process to enhance the enterprise’s performance such as reducing the cost(expences), improve the quality, prompt and speed(time bound) service. BPR enhances the productivity and profit of an enterprise.

6. Installation and configuration:
In this phase the new system are installing, before implementing the whole system a miniature of the actual system is going to be implemented as a test dose. Then check the reactions if it is good it is the time to install the whole system completely.

7. Implementation team training:
In this phase the company trains its employees to implement and run the system.

8. Testing:
This phase is very important. It determines whether the system produces proper result. Errors in design and logic are identified.

9. Going live:
Here a change over is taken place to new system from old system. It is not an easy process without the support and service from the ERP vendors.

10. End user training:
This phase will start familiarising the users with the procedures to be used in the new system. It is very important.

11. Post implementation:
Once the system is implemented maintenance and review begin. In this phase repairing or correcting previously ill defined problems and upgrade or adjust the performance according to the company needs.

Question 3.
Write short notes regarding ERP package Companies.
Answer:
Popular ERP packages are given below
1. Oracle:
American based company famous in database(Oracle 9i-SQL) packages situated in Redwood shores, California. Their ERP packages is a solution for finance and accounting problems. Their other products are

• Customer Relationship Management(CRM)
• Supply Chain Management (SCM)Software SAP

2. SAP:
stands for Systems, Applications, and Products for data processing. It is a German MNC in Walldorf and founded in 1972. Earlier they developed ERP packages for large MNC. But nowadays they developed for small scale industries also.
The other software products they developed are

• Customer Relationship Management(CRM)
• Supply Chain Management(SCM)
• Product Life cycle Management(PLM)

3. Odoo:
Formerly known as OpenERP.
It is an open source code ERP. Unlike other companies their source code is available and can be modified as and when need arises.

4. Microsoft Dynamics:

• American MNC in Redmond, Washington
• ERP for midsized companies.
• This ERP is more user friendly
• Other s/w is Customer Relationship Management(CRM)

5. Tally ERP:

• Indian company situated in Bangalore.
• This ERP provides total solution for accounting, inventory, and Payroll.

Question 4.
Explain the merits and demerits of ERP packages.
Answer:
ERP packages have a lot of advantages as well as many drawbacks also.
Benefits of ERP system:

1. Improved resource utilization:
Resources such as Men, Money, Material, and Machine are utilized maximum hence increase the productivity and profit.

2. Better customer satisfaction:
Without spending more money and time all the customer’s needs are considered well. Because customer is the king of the market. Nowadays a customer can track the status of an order by using the docket number through Internet.

3. Provides accurate information:
Right information at the right time will help the company to plan and manage the future cunningly. Acompany can increase or reduce the production based upon the right information hence increase the productivity and profit.

4. Decision making capability:
Right information at the right time will help the company to take good decision.

5. Increased flexibility:
A good ERP will help the company to adopt good things as well as avoid bad things rapidly. It denotes the flexibility.

6. Information integrity: A good ERP integrates various departments into a single unit. Hence reduce the redundancy, inconsistency, etc.

Risks of ERP implementation:

1. High cost:
Very huge investment is required to purchase and configure an ERP. Moreover, it requires up gradation or. replacement of hardware(Man, computer or machine) is an additional investment. So small scale enterprise cannot afford this.

2. Time consuming:
The full fledge implementation of ERP package needs one or two years. That is highly time consuming.

3. Requirement of additional trained staff:
The existing staffs may not capable to work with ERP. To overcome this give proper training to them otherwise appoint trained and experienced employees to Cop up.

4. Operational and maintenance issues:
The first major problem is that the resistance from the existing employees. To overcome this give awareness to the existing employees. The second problem is that ERP package is a cyclic process oriented package. It is a continuous process and should be maintained well otherwise the correct output will not available.

Question 5.
Explain in detail the ERP packages and related technologies.
Answer:
It is an all in one system. It integrates various functions such as raw material purchase, production planning, marketing, financial, etc., into a single application.

1. Product Life Cycle Management (PLM):
It manages the entire life cycle of a product. PLM consists of programs to increase the quality and reduce the price by the efficient use of resources.

2. Customer Relationship Management (CRM):
As we know that customer is the king of the market. The existence of a company mainly the customers. CRM consists of programs to enhance the customer’s relationship with the company.

3. Management Information System (MIS):
Management is the decision and policy makers. A good management can take good decision and that will help to do the business well. The good relationship between Management and employees is a key to success. MIS will collect relevant data from inside and outside of a company. Based upon this information produce reports and take appropriate decisions.

4. Supply Chain Management (SCM):
This is deals with moving raw materials from suppliers to the company as well as finished goods from company to customers. The activities includes are inventory(raw materials, work in progress and finished goods) management, warehouse management, transportation management, etc.

5. Decision Support System (DSS):
It is a computer based system that takes inputs as business data and after processing it produces good decisions as output that will make the business easier.

Question 6.
“The number and functioning of modules vary with the nature of enterprise and the type of ERP package.” List any six common modules of an enterprise.
Answer:
Different modules are given below

1. Financial Module:
It is the core. This is used to generate financial report such as balance sheet, general ledger, trial balance, financial statement, etc.

2. Manufacturing Module:
It provides information for the production and capable to change the methods in manufacturing sector.

3. Production planning Module:
This module ensures the effective use of resources and helps the enterprise to enhance the productive hence increase the profit.

4. HR (Human Resource) Module:
This model ensures the effective use of Human resources and Human capital.

5. Inventory control Module:
This model is useful to maintain the appropriate level of stock(includes raw material, work in progress and finished goods)

6. Purchasing Module:
This module is useful to make available the required raw materials in good condition and in the right time and price.

7. Marketing Module:
It is used for handle the orders of customers.

8. Sales and distribution Module:
Existence of a company is based on the income from sales. This module will help to handle the sales enquiries, order placement, and scheduling, dispatching and invoicing.

9. Quality (Ql & QC) management module:
Quality of a product or service is very much important to a company. This module helps to maintain the quality of the product. Quality planning, inspection, and control are the main activities involved in this module.

Question 7.
Mr. Suresh uses separate software for managing different functional units of an enterprise and Mr. Saleem uses an integrated software package formanaging overall functioning of the enterprise. Compare the benefits and risks of above two methods of an enterprise management.
Answer:
Benefits of ERP system:

1. Improved resource utilization:
Resources such as Men, Money, Material, and Machine are utilized maximum hence increase the productivity and profit.

2. Better customer satisfaction:
Without spending more money and time all the customer’s needs are considered well. Because customer is the king of the market. Nowadays a customer can track the status of an order by using the docket number through Internet.

3. Provides accurate information:
Right information at the right time will help the company to plan and manage the future cunningly. A company can increase or reduce the production based upon the right information hence increase the productivity and profit.

4. Decision making/apability:
Right information at the right time will help the company to take good decision.

5. Increased flexibility:
A good ERP will help the company to adopt good things as well as avoid bad things rapidly. It denotes the flexibility.

6. Information integrity:
A good ERP integrates various departments into a single unit. Hence reduce the redundancy, inconsistency, etc.

Risks of ERP implementation:

1. High cost:
Very huge investment is required to purchase and configure an ERP. Moreover, it requires up gradation or replacement of hardware(Man, computer or machine) is an additional investment. So small scale enterprise cannot afford this.

2. Time consuming:
The full fledge implementation of ERP package needs one or two years. That is highly time consuming.

3. Requirement of additional trained staff:
The existing staffs may not capable to work with ERP. To overcome this give proper training to them otherwise appoint trained and experienced employees to cop up.

4. Operational and maintenance issues:
The first major problem is that the resistance from the existing employees. To overcome this give awareness to the existing employees. The second problem is that ERP package is a cyclic process oriented package. It is a continuous process and should be maintained well otherwise the correct output will not available.

Question 8.
“Implementation of an ERP system in an enterprise is not a single step action”. Justify this statement by listing all the phases of ERP implementation in correct order.
Answer:
The different phases of ERP implementation are given below:

1. Pre evaluation screening:
Many ERP packages are available in the marketfs. At most care should be taken before implementing a ERP. Select a few from from the available ERP packages.

2. Package selection:
Selection of right ERP to our enterprise is a laborious task and it needs huge investment. Various factors should be keep in mind before you purchase an ERP that should meet our complete needs.

3. Project planning:
A good planning is essential to implement an ERP. From the beginning to the end activities are depicted in this phase.

4. Gap analysis:
A cent percents 00%) problem solving ERP is not available in the market. Most of them solve a maximum of 70% to 80% problems. The rest (30% to 20%) of the problems and their solutions are mentioned here.

5. Business Process Reengineering:
In general BPR is the series of activities such as rethinking and redesign of the business process to enhance the enterprise’s performance such as reducing the cost(expences), improve the quality, prompt and speed(time bound) service.

BPR enhances the productivity and profit of an enterprise:

1. Installation and configuration:
In this phase the new system are installing, before implementing the whole system a miniature of the actual system is going to be implemented as a test dose. Then check the reactions if it is good it is the time to install the whole system completely.

2. Implementation team training:
In this phase the company trains its employees to implement and run the system.

3. Testing:
This phase is very important. It determines whether the system produces proper result. Errors in design and logic are identified.

4. Going live:
Here a change over is taken place to new system from old system. It is not an easy process without the support and service from the ERP vendors.

5. End user training:
This phase will start familiarising the users with the procedures to be used in the new system. It is very important.

6. Post implementation:
Once the system is implemented maintenance and review begin. In this phase repairing or correcting previously ill defined problems and upgrade or adjust the performance according to the company needs.

Question 9.
“Selection of ERP package is one of the important phases of ERP implementation”.
Write a short note about any of the ERP packages.
Answer:
Popular ERP packages are given below:

1. Oracle:
American based company famous in database (Oracle 9i-SQL) packages situated in Redwood shores, California. Their ERP packages is a solution for finance and accounting problems. Their other products are

• Customer Relationship Management(CRM)
• Supply Chain Management (SCM)Software

2. SAP:
SAP stands for Systems, Applications, and Products for data processing. It is a German MNC in Walldorf and founded in 1972. Earlier they developed ERP packages for large MNC. But nowadays they developed for small scale industries also.
The other software products they developed are

• Customer Relationship Management(CRM)
• Supply Chain Management(SCM)
• Product Life cycle Management(PLM)

3. Odoo:
Formerly known as OpenERP.
It is an open source code ERP. Unlike other companies their source code is available and can be modified as and when need arises.

4. Microsoft Dynamics:

• American MNC in Redmond, Washington
• ERP for midsized companies.
• This ERP is more user friendly
• Other s/w is Customer Relationship Management(CRM)

5. Tally ERP:

• Indian company situated in’Bangalore.
• This ERP provides total solution for accounting, inventory, and Payroll.

Students can Download Chapter 7 The p Block Elements Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements

Plus Two Chemistry The p Block Elements One Mark Questions and Answers

Question 1.
The gas obtained by the thermal decomposition of barium azide is _________.
Answer:
Nitrogen

Question 2.
The number of replaceable hydrogen atoms in ortho phosphorous acid (H₃PO₃) is
(a) 3
(b) 2
(c) 1
(d) 0
Answer:
(b) 2

Question 3.
The stable allotrope of sulphur at room temperature is ___________ .
Answer:
Rhombic sulphur

Question 4.
Which element would readily replace oxygen from an oxide?
(a) N
(b) S
(c) F
(d) Cl
(Answer:
(c) F

Question 5.
The geometry of XeF₄ is.
Answer:
Square planar

Question 6.
The shape and hybridisation of XeF₄ molecule
Answer:
Square planar and sp³d²

Question 7.
The least stable hydride of 15^th group elements is.
Answer:
(BiH₃)

Question 8.
Choose the weak monobasic acid among the following.
(a) H₃BO₃
(b) H₃PO₂
(c) H₃PO₄
(d) HNO₃
Answer:
(a) H₃BO₃

Question 9.
The bond enthalpy is highest for ________
Answer:
H₂

Question 10.
Magnetic moment of an atom with atomic no. 24 in aqeous solution is ____________.
Answer:
4.90 B.M.

Plus Two Chemistry The p Block Elements Two Mark Questions and Answers

Question 1.
In the class, a student argued that “heterogenous” catalysis and Le-Chatlier’s principles are applied in contact process.”

1. Do you agree with this statement?
2. Justify.

Answer:
1. Yes

The reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are the favourable conditions for maximum yield according to Le Chatelier’s principle.

2. Contact process is an example of heterogeneous catalysis since the catalyst V₂O₅ is a solid while the reactants are gases.

Question 2.
ClF₃ exists, but FCl₃ doesn’t, why?
Answer:
In the valence shell of Cl vacant 3d orbitals are available. Hence it can form ClF₃. But, F has no vacant d orbital to show higher oxidation state and FCl₃ is not possible.

Question 3.
In a classroom discussion, a student argued that carbon has the maximum catenation property.

1. Do you agree?
2. What do you understand by the term catenation?
3. List out some other elements which can show catenation?

Answer:

1. Yes.
2. Catenation is the self linking property of an atom. It is maximum for carbon, because C-C bond is very strong.
3. Sulphur and Phosphorus.

Question 4.
Some elements and their ores are given in table. Arrange them correctly.

A	B
a) Calcium	Galena
b) Potassium	Magnetite
c) Lead	Fluorapatite
d) Tin	Gypsum
e) Phosphorus	Cassiterite
	Carnallite

Answer:

A	B
a) Calcium	Gypsum
b) Potassium	Carnallite
c) Lead	Galena
d) Tin	Cassiterite
e) Phosphorus	Fluorapatite

Question 5.
Analyse the statement: ‘Ionisation enthalpy of halogens decreases with a decrease in atomic size’. Is it true? Justify your answer.
Answer:
No. On moving from top to bottom in a given group, the size of the atom increases and hence ionisation enthalpy decreases.

Question 6.
Hydrides of group 15 are Lewis bases.

1. Give reason.
2. Arrange the group 15 hydrides in the decreasing order of basic strength.

Answer:

1. Due to the presence of lone pair on the central atom which is available for donation.
2. NH₃ > PH₃ > ASH₃ > SbH₃ ≥ BiH₃.

Question 7.
According to VSEPR theory assign structure to XeOF₄.
Answer:
XeOF₄ possess square pyramidal shape.

Question 8.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:

1. Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.
2. Nitrogen does not contain vacant d-orbitals in its valence shell whereas phosphorus contains vacant d-orbitals in its valence shell.
3. Nitrogen has the ability to form triple bond ( N ≡ N) as a result of which its bond enthalpy (941.4 kJ mol^-1) is very high making it less reactive.

Question 9.
Why does NH₃ form hydrogen bond but PH₃ does not?
Answer:
In NH₃, the nitrogen atom forms hydrogen bond because of the following reasons:

• Small size of nitrogen
• High electronegativity (3.0) of nitrogen

N-H bond is polar forming hydrogen bond.
P-H bond is almost purely covalent due to larger size and lesser electronegativity.

Question 10.
The HNH angle is higher than HPH, HAsH and HSbH angles. Why?
Answer:
Because in NH₃ is sp³ hybridised. Due to lone pair of electrons the bond angle contracts from 109° 28’ to 106.5°. The sp³ hybridisation becomes less and less distinct with increasing size of the central atom. Thus, the bond angle of the hydrides of group 15 decreases as.

Question 11.
Can PCl₃ act as an oxidising as well as a reducing agent? Justify.
Answer:
This is because in PCl₃ phosphorus is in the intermediate oxidising state of -3.
1. As reducing agent:
The following reactions support the reducing behaviour of PCl₃.

• PCl₃ + SO₂Cl₂ → PCl₅ + SO₂
• PCl₃ + SO₃ → POCl₃ + SO₂

2. As an oxidising agent:
It oxidises metals to their respective chlorides.

• 12Ag + 4PCl₃ → 12AgCl + P₄
• 6Na + PCl₃ → 3NaCl + Na₃P

Question 12.
Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.
Answer:
Due to its larger size (99 pm) as compared to oxygen (66 pm).

Question 13.
Ozone is a strong oxidising agent. Why?
Answer:
O₃ undergoes dissociation to form nascent oxygen. This nascent oxygen is responsible for oxidising property of ozone.

O₃ → O₂ + [O]

Question 14.
White phosphorus is more reactive than other solid phases of phosphorus. Give reason.
Answer:
This is beacuse of angular strain in the P₄ molecule where the angles are only 60°.

Question 15.
What happens when

1. Concentrated H₂SO₄ is added to CaF₂?
2. SO₃ is passed through water?

Answer:
1. It form hydrogen fluoride.
CaF₂ + H₂SO₄ → CaSO₄ + 2HF

2. It dissolves SO₃ to give H₂SO₄.
SO₃ + H₂O → H₂SO₄

Question 16.
Give two important fluorides of Xenone. Predict their structure.
Answer:

• XeF₂ – linear
• XeF₄ – Square planar
• XeF₆ – Distorted octahedral

Question 17.
Why is N₂ less reactive at room temperature?
Answer:
Due to the presence of triple bond between N atoms N₂ has high bond dissociation energy and is less reactive.

Question 18.
Arrange the following in the increasing order of thermal stability.
ASH₃, NH₃, PH₃
Answer:
Thermal stability of hydrides, decreases from nitrogen to bismuth: NH₃ > PH₃ > AsH₃.

Question 19.
The 3 allotropic forms of Pare white, red and black.

1. Name the thermodynamically most stable allotrope.
2. Which allotrope of P is stored in water?

Answer:

1. Black phosphorus
2. White phosphorus

Question 20.
Nitrogen and Phosphorus are in the same group. PCI₅ is known but NCI₅ is not known. Why?
Answer:
In phosphorus vacant d-orbitals are present. It can form pentahalides. But due to absence of d-orbitals nitrogen cannot form NCI₅.

Question 21.
Why is helium used in diving apparatus?
Answer:
Helium is used in diving apparatus because of its very low solubility in blood.

Question 22.
Why has it been difficult to study the chemistry of radon?
Answer:
Radon is radioactive with very short half-life which makes the study difficult.

Question 23.
Account for the following:
SO₃ is more covalent than SO₂.
Answer:
Due to high charge and small size of sulphur in +6 oxidation state in SO₃ it is more covalent than SO₂, in which sulphur is in +4 oxidation state.

Question 24.
I₂ is more soluble in KI than in water.
Answer:
I₂ combines with KI to form the soluble complex, KI₃.
KI + I₂ → KI₃

Question 25.
H₃PO₃ is dibasic while H₃PO₄ is tribasic.
Answer:
There are three ‘P-OH’ bonds that are ionisable in H₃PO₄. So, it is tribasic. In H₃PO₃ there are only 2 ionisable ‘P-OH’ bonds. Hence it is dibasic.

Plus Two Chemistry The p Block Elements Three Mark Questions and Answers

Question 1.
Consider the given reaction:
2A + 6SiO₂ → 6CaSiO₃ + B
B + 10C → D + 10CO

1. Identify A, B and D.
2. D is stored underwater. Why?
3. Give two examples of oxoacids of D and compounds of D.

Answer:
1. A – Ca₃(PO₄)₂
B – P₄O₁₀
D – P₄

2. Because P₄ readily catches fire in air.

3. Two examples of oxoacids of D and compounds of D

• Oxoacids of D:
  Oxoacids – Phosphorous acid(H₃PO₃), Phosphoric acid (H₃PO₄).
• Compounds of D:
  phosphine (PH₃), PCI₅, PCI₄, P₄O₆

Question 2.

1. Suggest a method for the preparation of PCI₃.
2. Why does PCI₃ fume in moist air?

Answer:
1. By passing dry chlorine overheated white phosphorus.
P₄ + 6CI₂ → 4PCI₃

2. PCI₃ hydrolyses in the presence of moisture giving fumes of HCI.
PCI₃ + 3H₂O → H₃PO₃ + 3HCI

Question 3.
A student argued that electronegativity of p-block elements decrease along period and increases down the group.

1. Do you agree with this? Explain.
2. Write about the metallic character of p-block elements.
3. Arrange the following of p-block elements in the order of decreasing oxidising power.

F₂/F-( E° =+2.85 V), Br/Br^–( E° =+1.07 V), CI₂/CI^– (E° =+1.36V), I₂/I^–( E° =+0.57V).
Answer:

1. No. electronegativity of p-block elements increases along a period and decreases down a group.
2. Most of the p-block elements are non-metallic in nature. Among p-block elements the metallic character decreases along a period and increases down a group.
3. F₂ > CI₂ > Br₂ > I₂, because the standard electrode potential values decreases in the same order.

Question 4.
The shape and hybridisation of some interhalogen compounds are given below in wrong order. Match them correctly.

Answer:

Question 5.
Account for the following:

1. H₂O is a liquid while H₂S is a gas.
2. H₂S is more acidic than H₂O
3. SF₆ is known while SH₆ is not known.

Answer:

1. There is intermolecular hydrogen bonding in H₂O molecule but there is no hydrogen bonding in H₂S .
2. The S-H bond is weaker than O-H bond because size of S atom is greater than that of O atom. Hence H₂S can dissociate to give H⁺ ions in aqueous solution.
3. In the higher oxidation state, S can combine only with highly electronegative elements like F.

Question 6.
Sulphur forms many allotropes such as a – sulphur, β -sulphur etc.

1. What do you mean by allotropy?
2. How can you convert a – sulphur to b-sulphur?

Answer:

1. Certain elements can exist in different forms with different physical property and same chemical properties.
2. b -sulphur is prepared by melting a -sulphur in a dish and cooling, till crust is obtained. Holes are then pierced into the crust, and the liquid is taken out. On removing the crust, needle shaped β -sulphur is obtained.

Question 7.
Account for the following:

1. PH₃ has lower boiling point than NH₃.
2. Pentahalides are more covalent than trihalides.
3. ICI is more reactive than I₂.

Answer:

1. NH₃, PH₃ molecules are not associated through intermolecular hydrogen bonding in liquid state.
2. Higher the positive oxidation state of central atom, more will be its polarising power which, in turn, increases the covalent character.
3. Bond energy of I-CI bond is less than that fo I-I bond.

Plus Two Chemistry The p Block Elements Four Mark Questions and Answers

Question 1.
My name is ‘X’. I am a poisonous colourless gas with the smell of rotten fish.

1. Identify ‘X’.
2. Explain the laboratory preparation of X.

Answer:

1. Phosphine(PH₃).
2. By heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO₂.

Question 2.

1. How is bleaching powder prepared?
2. Give the composition of bleaching powder.

Answer:
1. By treating Cl₂ with dry slaked lime.
2Ca(OH)₂ + 2CI₂ → Ca(OCI)₂ + CaCI₂ + 2H₂O

2. Ca(OCI)₂ CaCI₂.Ca(OH)₂.2H₂O.

Question 3.

1. Why do noble gases form compounds with fluorine and oxygen only?
2. Does the hydrolysis of XeF₆ lead to a redox reaction?

Answer:
1. Fluorine and oxygen are small atoms with high value of electronegativity. Fluorine is also highly reactive in nature. It is for this reason they form compounds with noble gases.

2. No, the products of hydrolysis are XeOF₄ and XeO₂F₂ where the oxidation states of all the elements remain the same as it was in the reacting state.

Question 4.

1. Interhalogen compounds are more reactive than halogens. Why?
2. Is there any exception to the above generalisation. Explain.

Answer:

1. This is because the bond in the interhalogen (X – X’) is weaker than X – X bond in the halogens.
2. Yes. F₂ is more reactive than interhalogen compounds. This is because the F – F bond is weaker than X – X’ bond in interhalogen compounds.

Question 5.
Nitrogen cannot extend its covalency beyond four but it forms wide variety of oxides and also forms oxoacids.

1. Name the two oxoacids of nitrogen.
2. Action of nitric acid with metals depends on many factors. Justify.

Answer:
1. HNO₂, HNO₃

2. The product of oxidation depends upon the concentration of the acid, temperature and the nature of the material undergoing oxidation, e.g.

a. 3Cu + 8HNO₃(dilute) → 3Cu(NO₃)₂ + 2NO + 4H₂O
Cu + 4HNO₃(conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

b. Zinc reacts with dilute nitric acid to give N₂O and with concentrated acid to give NO₂.
4Zn + 4HNO₃(dilute) → 4Zn(NO₃)₂ + 5H₂O + N₂O
Zn + 10HNO₃(conc.) → Zn(NO₃)₂ + 2H₂0 + 2NO₂

Some metals (e.g., Cr, Al) do not dissolve in concentrated nitric acid due to the formation of a passive film of oxide on the surface.

Question 6.

1. Arrange hydrides of group 16 in the order of increasing acidic strength.
2. Draw the structures of any three oxoacids of phosphorus and find out their basicity.

Answer:
1. H₂O < H₂S < H₂Se < H₂Te
2.

Question 7.

1. Noble gases have very low boiling points. Why?
2. XeF₂ and XeF₄ are important Xe compounds. How can we prepare them and what is the action of water on them?

Answer:
1. Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.

2. XeF₂ and XeF₄ are formed by the direct reaction of elements under the specific conditions.

They are readly hydrolysed even by traces of water,
e.g. 2XeF₂(s) + 2H₂O(l) → 2Xe(g) + 4HF(aq) + O₂(g)

Question 8.
It is greenish yellow gas with an offensive smell used in water purification. It partially dissolves in water to give a solution which turns blue litmus to red. When it is passed through NaBr solution bromine is formed.

• Identify the gas.
• Identify the group to which it belongs.
• Write the electronic configuration.
• Write the equation showing its reaction with water.

Answer:

• Chlorine
• Halogen family (17^th group)
• 1s² 2s² 2p⁶ 3s² 3p⁵
• CI₂ + H₂O → HCI + HOCI

Question 9.
Basic character of hydrides of group 15 is due to the presence of lone pair on the central atom.

1. NH₃ is strongly basic while BiH₃ is weakly basic. Why?
2. Differentiate between allotropic forms of phosphorus.

Answer:
1. Due to the presence of lone pair of electrons on nitrogen atom of NH₃. But in BiH₃, due to the large size of Bi the electron density decreases and hence is less basic.

2. The allotropic forms of P are White, Red and Black phosphorus. White phosphorus consists of tetrahedral P₄ molecules. Red phosphorus is polymeric in structure consisting of chains of P₄ tetrahedra linked together. Black phosphorus has a layer type structure and has α and β forms.

Question 10.

1. Give a method for preparation of XeO₃.
2. Deduce the molecular shape of BrF3 on the basis of VSEPR theory.

Answer:
1. XeO₃ is prepared by hydrolysis of xenon hexa fluride in presence of water.
XeF₆ + 3H₂0 → XeO₃ + 6HF

2. In BrF₃ the central atom Br is sp³d hybridised with 3 bond pairs and 2 lone pairs. According to VSEPR theory the expected geometry is trigonal bipyramidal. But due to strong Ip – Ip and Ip – bp repulsions compared to weak bp – bp repulsion it Assumes a bend T – shape as shown below

Question 11.
Account for the following:

1. BiH₃ is the strongest reducing agent among all the hydrides of group 15 elements.
2. Bleaching action of chlorine.

Answer:

1. It is because BiH₃ is least stable among the hydrides of group 15 elements.
2. Bleaching action of chlorine is due to oxidation. The nacent oxygen ([O]) produced is responsible for the bleaching action.

CI₂ + H₂O → 2HCI + [O]
Coloured substance + [O] → Colourless substance

Plus Two Chemistry The p Block Elements NCERT Questions and Answers

Question 1.

1. Noble gases have very low boiling points. Why?
2. XeF₂ and XeF₄ are important Xe compounds. How can we prepare them and what is the action of water on them?

Answer:
1. Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.

2. XeF₂ and XeF₄ are formed by the direct reaction of elements under the specific conditions.

They are readly hydrolysed even by traces of water,
e.g. 2XeF₂(s) + 2H₂O(l) → 2Xe(g) + 4HF(aq) + O₂(g)

Question 2.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:

1. Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.
2. Nitrogen does not contain vacant d-orbitals in its valence shell whereas phosphorus contains vacant d-orbitals in its valence shell.
3. Nitrogen has the ability to form triple bond ( N ≡ N) as a result of which its bond enthalpy (941.4 kJ mol^-1) is very high making it less reactive.

Question 3.
Why does NH₃ form hydrogen bond but PH₃ does not?
Answer:
In NH₃, the nitrogen atom forms hydrogen bond because of the following reasons:

• Small size of nitrogen
• High electronegativity (3.0) of nitrogen

Due to more difference of electronegativity between N and H atom the N-H bond is polar forming hydrogen bond. On the contrary, in PH₃ the P-H bond is almost purely covalent due to larger size and lesser electronegativity (2.11) of phosphorus and hence does not form hydrogen bond.

Question 4.
The HNH angle is higher than HPH, HAsH and HSbH angles. Why?
Answer:
Because in NH₃ is sp³ hybridised. Due to lone pair of electrons the bond angle contracts from 109° 28’ to 106.5°. The decreased bond angle in other hydrides is because of the fact that the sp³ hybridisation becomes less and less distinct with increasing size of the central atom i.e., pure p-orbitals are utilised in M-H bonding or in simple words the s- orbital of H atom overlaps with orbital having almost pure p-character. Thus, the bond angle of the hydrides of group 15 decreases as

Question 5.
Can PCI₃ act as an oxidising as well as a reducing agent? Justify.
Answer:
Yes, This is because in PCI₃ phosphorus is in the intermediate oxidising state of -3.
1. As an reducing agent:
The following reactions support the reducing behaviour of PCI₃.

• PCl₃ + SO₂Cl₂ → PCl₅ + SO₂
• PCl₃ + SO₃ → POCl₃ + SO₂

2. As an oxidising agent:
It oxidises metals to their respective chlorides.

• 12Ag + 4PCl₃ → 12AgCl + P₄
• 6Na + PCl₃ → 3NaCl + Na₃P

Question 6.
Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Answer:
Oxygen atom can form hydrogen bond whereas chlorine does not. The tendency for hydrogen bonding depends upon

1. Small size and
2. High electronegativity values

Although the electronegativities of O and Cl are nearly the same yet chlorine does not form hydrogen bond due to its larger size (99 pm) as compared to oxygen (66 pm).

Students can Download Chapter 8 Environmental Issues Questions and Answers, Plus Two Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Botany Chapter Wise Questions and Answers Chapter 8 Environmental Issues

Plus Two Botany Environmental Issues One Mark Questions and Answers

Question 1.
CPCB stand for
(a) Central pollution consistent board
(b) Central particulate collection & analysis board
(c) Central pollution Control Board
(d) Central phenomenon coexisting Board (pollution)
Answer:
(c) Central pollution Control Board

Question 2.
Catalytic converters may contain expensive metals like……….as catalytic in automobiles
(a) Platinum
(b) Palladium
(c) Rhodium
(d) All of these
Answer:
(d) All of these

Question 3.
The participation of local communities helps to conserve the forest in sustainable manner. Name the recently started project of Govt, of India.
Answer:
In 1980s Government of India has introduced the concept of Joint Forest Management (JFM).

Question 4.
According to the Central Pollution Control Board, particles that are responsible for causing great harm to human health are of diameter:
(a) 2.50 micrometers
(b) 5.00 micrometers
(c) 10.00 micrometers
(d) 7.5 micrometers
Answer:
(a) 2.50 micrometers

Question 5.
Match the items in column I and column II and choose the correct option:

Column I	Column II
A. UV	i)  Biomagnification
B. Biodegradable Organic matter	ii)  Eutrophication
C. DDT	iii) Snow blindness
D. Phosphates	iv) BOD

(a) A-ii, B-i, C-iv, D-iii
(b) A-iii, B-ii, C-iv, D-i
(c) A-iii, B-iv, C-i, D-ii
(d) A-iii, B-i, C-iv, D-i
Answer:
(c) A-iii, B-iv, C-i, D-ii

Question 6.
Three-mile Island, Chernobyl incidents are typical example for………pollution
(a) Sound
(b) Radioactivity
(c) Water
(d) Physical
Answer:
(b) Radioactivity

Question 7.
Match correctly the following and choose the correct option

i) EnvironmeriPProtection Act	A. 1974
ii) Air Prevention and Control of Pollution Act	B. 1987
iii) Water Act	C.  1986
iv) Amendment of Air Act to include noise	D. 1981

The correct matches is:
(a) i-C, ii-D, iii-A, iv-B
(b) i-A, ii-C, iii-B, iv-D
(c) i-D, ii-A, iii-B, iv-C
(d) i-C, ii-D, iii-B. iv-A
Answer:
(a) i-C, ii-D, iii-A, iv-B

Question 8.
Catalytic converters are fitted into automobiles to reduce emission of harmful gases. Catalytic converters change unbumt hydrocarbons into:
(a) carbon dioxide and water
(b) carbon mono oxide
(c) methane
(d) carbon dioxide and methane
Answer:
(a) carbon dioxide and water

Question 9.
In the modern world there are different types of substance that may cause pollution. A prominent waste in the modern computer world is “e-waste” The “e-waste” stand for
(a) Eliminated waste
(b) Eroded waste
(c) Electronic waste
(d) Enriched waste
Answer:
(c) Electronic waste

Question 10.
Humans, as well as other animals, have been dumping their wastes into the environment for thousands of years. What is the reason that this appears to be such a problem today?
Answer:
The human population is increasing rapidly, produc-ing more wastes, and a significant amount of the wastes are non-biodegradable.

Question 11.
Air pollution is maximally caused by
(a) Household detergents and pesticides
(b) Automobile exhausts and chemicals from industry
(c) Sewage and pesticides
(d) Sewage and industrial effluents
Answer:
(b) Automobile exhausts and chemicals from industry

Question 12.
Observe the figure given below. Identify the equipment and write its use.

Answer:
Scrubber: A Scrubber can remove gases like sulphur dioxide.

Question 13.
Government of India has instituted an award for individuals or communities from rural areas who have shown extraordinary courage and dedication in protecting wild life. Name that award.
Answer:
Amrita Devi Bishnoi Wildlife Protection Award.

Question 14.
Ecological sanitation is a sustainable system of handling human excreta, using dry composting toilets. Write the advantages of this method.
Answer:
It is Hygienic .efficient, cost effective and human excreta can be recycled into a natural fertilizer.

Question 15.
Choose the correct answer. Substances that cause biomagnification are
(a) Mercury and DDT
(b) Mercury and Phosphorus
(c) DDT and Phosphorus
(d) Phosphorus and nitrogen
Answer:
(a) Mercury and DDT

Question 16.
Observe the relationship between the first two terms and fill in the blank. Green house effect: carbon dioxide; ozone depletion:…………..
Answer:
Chlonofluorocarbons.

Question 17.
Choose the correct statement regarding biomagnification in an aquatic ecosystem.
(a) Decrease in concentration of the toxicant in successive tropic levels.
(b) Higher concentration of toxicant is seen in the first trophic level.
(c) Bird population is least affected.
(d) Causes decline in bird population.
Answer:
(d) Causes decline in bird population.

Question 18.
In 1986 Govt, of India passed Environment Act. Write the purpose of its implementation.
Answer:
To protect and improve the quality of our environment (air, water, and soil).

Question 19.
The participation of local communities helps to conserve the forest in sustainable manner. Name the recently started project of Govt, of India.
Answer:
In 1980s Government of India has introduced the concept of Joint Forest Management (JFM).

Question 20.
Give the term for reradiating of heat radiations by atmosphere dust, water vapours, CO-₂ CH-₄, O₂, etc.
Answer:
Green house effect

Question 21.
How CFCS cause the depletion of Ozone shield?
Answer:
By producing active chlorine which breaks ozone.

Plus Two Botany Environmental Issues Two Mark Questions and Answers

Question 1.
Expand CPCB – Write its significance.
Answer:
CPCB – Central Pollution Control Board.
CPCB – Aims for controlling particulate matter in the air. According to CPCB the particulate size 2.5 micrometers or less than in diameter (PM 2.5) are responsible for causing greatest harm to human health.

Question 2.
Working condition of electrostatic precipitator is given. By analysing the figure explain the process.

Answer:
At high voltage the electrons produced are attached to dust particles giving them a net negative charge. These charged dust particles are attracted by collecting plates. Then reducing the velocity of air between the plates which help the dust to fall.

Question 3.
Pollutant free Metropolitan cities can be raised by the use of CNG.

1. Expand CNG?
2. What are the?

Answer:
1. CNG is compressed natural gas.

2. advantages of CNG:

• CNG is better than diesel
• It burns most efficiently
• CNG is cheaper than petrol or diesel
• It cannot be adulterated like petrol or diesel.

Question 4.
Montreal protocol is a procedure signed at Montreal (Canada) in 1987.

1. Identify the environmental issue related to this.
2. Which is the chemical causing the environmental problem?

Answer:

1. Ozone hole
2. Chloro fluro carbons release active chlorine atoms which results in degradation of the ozone in the stratosphere.

Question 5.
“Pollutant free Metroes” – Can this be achieved in future?

OR

Pollutant free Metropolitan cities can be raised by the use of CNG. What does CNG implies? Can you suggest the priorities and limitations of CNG?

Answer:
CNG is compressed natural gas. It burns more efficiently than Petrol and diesel, thus brings down the amount of pollutants from automobiles (unburnt hydrocarbons).

CNG is cheaper than petrol or diesel. It cannot be siphoned out by thieves and adultered like petrol or diesel.

The problem in using CNG as fuel is the difficulty of laying down pipelines to deliver CNG through distribution points.

Question 6.
If motor vehicles are equipped with catalytic converter the pollution rate can be reduced.

1. Give reason.
2. Catalytic converters have its limitations. Mention the limitations.

Answer:
1. Catalytic converters are used in motor vehicles to reduce pollutants. Catalytic converters convert unburnt hydrocarbons into CO₂ and water. The catalytic converter uses expensive metals like Platinum, Palladium or rhodium as catalysts.

2. These catalysts are inactivated if leaded petrol is used.

Question 7.
Particulate and gaseous pollutants along with harmless gases are released from the thermal power plants.

1. Name any two harmless gases released.
2. Name the most widely used device of removing particulate pollutants from the air. Explain how the device is used.

Answer:
1. NO₂ & O₂.

2. Electrostatic precipitator
It has electrodes which are maintained at several 1000 watts which releases electrons. These electron attach to dust particle giving them a net -ve charge. The collecting plates attract the charged dust particle. The velocity of air between the plates are lowered to allow dust to fall.

Question 8.
Higher BOD could lower the D.O.

1. Do you agree with the statement.
2. Give reason.

Answer:
1. Yes.

2. Higher BOD lower the D.O. content of water. D.O. is the dissolved 02 content of water. When organic content is increased in water the D.O. is consumed for its oxidation by bacteria which is referred as BOD (Biochemical Oxygen demand) when B.O.D. is greater the D.O. becomes low.

Question 9.
Pesticide DDT was used to control insect such as mosquitoes and agricultural pests. But DDT persisted in the environment and transported from one trophic level to higher trophic levels.

1. Name the phenomenon and explain it.
2. What happens to the concentration of DDT in each trophic level?

Answer:
1. Biomagnification.
Biomagnification refers to increase in concentration of the toxicant at successive trophic levels.

2. Non biodegradable toxic substance accumulated by an organism cannot be metabolised or excreted and it is passed to the next higher trophic level.

Question 10.
Sewage from common hospitals if disposed to water bodies may cause the outbreak of serious water-borne diseases.

1. Mention 2 water-borne diseases.
2. Which is the popular method used to remove hospital waste?

Answer:

1. Dysentry, typhoid, jaundice, etc.
2. Incineration

Question 11.
Match the following.

Column I	Column II
a) Catalytic converter	(i) Solid wastes
b) Polyblend	(ii) Agrochemicals
c) Organic farming	(iii) Plastic wastes
d) Sanitary hand fills	(iv) Air pollution

Answer:

1. Catalytic converters – Air pollution
2. Polyblend – Plastic wastes
3. Organic farming – Agro chemicals
4. Sanitary land fills – Solid wastes

Question 12.
Solid waste can create a major problems in metro cities.

1. Which is the best way for the safe disposal of the solid waste
2. Classify solid waste into 3 groups, which is the alternative remedy for the use of plastic

Answer:
1. Sanitary land fills.

2. The Solid wastes can be categorised into 3 types

• bio degradable
• Recyclable
• non-biodegradable.

The biodegradable waste material can be disposed in deep pits for natural break down. Recycling of material is made easy by the separation done by rag pickers.

Non-biodegradable waste should reduced by using eco-friendly packaging & natural fiber carry bags. State government can play important role by applying laws for reducing the use of plastics.

Question 13.
Integrated organic farming is a zero waste procedure.

1. Do you agree with the statement.
2. Give reason.

Answer:
1. Yes.

2. Integrated organic farming is a cyclical zero waste procedure, where waste products from one process are cycled and act as nutrients for other processes. This allows the maximum utilization of resources and increases the efficiency of production.

For example if dung is used as manure for the crop, the crop waste can be used to create compost which in turn can be used as fertilizer after generating energy needs of the farm.

Question 14.
Analyse the situation given below and comment on it. “Fertilizers from a paddy field flows to nearby pond.’’
Answer:
Presence of large-amount of nutrients caused by sewage disposal causes excessive growth of planktonic algae which imparts distinct color & decreases quality of water & increases mortality of fishes.

Question 15.
Rag pickers do a great job for our nation. Justify this statement.
Answer:
The Solid wastes can be categories into 3 types

• biodegradable
• Recyclable
• non-biodegradable.

The biodegradable waste material can be disposed in deep pits for natural break down. Recycling of material is made easy by the separation done by rag pickers.

Question 16.
Ecologist argued that organic fertilizers are better for the soil than the chemical fertilizers, (hint – organic fertilizers:- green manures, cattle dung, biofertilizers). Write your opinions.
Answer:
Organic fertilizers are better for the soil as they do not cause pollution of soil and water. Chemical fertilizers which drained into the water bodies will cause eutrophication and algal blooms.

They also cause increase in BOD and lead to the destruction aquatic flora & fauna. Organic fertilizers are cheaper they tend to remain in the soil for long period.

Question 17.
Ahmed khan in Bangalore proved himself a good citizen in using his talents in solving a social issue. Comment on his contribution.
Answer:
Ahamed khan in Bangalore who is a plastic sac manufacturer effectively use poly blend – a fine powder produced from recycled plastic & mixed it with bitumen that is used to lay roads.

By doing so he could enhance the water repellent property of bitumen & which in turn help to road life by three times. It also reduced pollution caused by plasticis. It is an example for the effective way for recycling non biodegradable – plastic.

Question 18.
The concentration of DDT like toxins are higher in man and eagle.

1. Name the process.
2. Give the reason

Answer:
1. Biomagnification

2. Non bio degradable chemical which accumulate in the body of organism and is passed on to the organisms belonging to the next trophic level. Human beings & Eagle belong to the last trophic level of the food chain. Hence the bio magnification threaten human life.

Question 19.

1. Identify the process
2. Find out the reason for these.

Answer:
1. Biomagnification.

2. The concentration of DDT is increased at successive trophic levels. It starts at 0.003 ppm in water, it can ultimately can reach 25 ppm in fish-eating birds, through biomagnification.

High concentrations of DDT disturb calcium metabolism in birds, which causes the thinning of eggshell and their premature breaking, causing decrease in bird populations.

Question 20.
Integrated organic farming is a zero waste procedure. Do you agree with the statement. Justify.
Answer:
Yes. Integrated organic farming is a cyclical zero waste procedure, where waste products from one process are cycled and act as nutrients for other processes. This allows the maximum utilization of resources and increases the efficiency of production.

Question 21.
Solid waste can create a major problems in metro cities.

1. Which is the best way for the safe disposal of the solid waste
2. Classify solid waste into 3 groups, which is the alternative remedy for the use of plastic

Answer:
1. Sanitary land fills

2. The Solid wastes can be categorised into 3 types

• biodegradable
• Recyclable
• non-biodegradable.

Non-biodegradable waste should reduced by using eco-friendly packaging & natural fiber carry bags.

Question 22.
Ahmed khan in Bangalore proved himself a good citizen in using his talents in solving a social issue. Comment on his contribution.
Answer:

• Ahmed Khan in Bangalore has found a solution for problem of accumulating plastic waste.
• Polyblend, a fine powder of recycled modified plastic, is mixed with the bitumen that is used to lay roads.
• Bitumen is a water repellant substance helps to increase road life.
• The raw material for creating Polyblend is any plastic film waste that brought about by rag pickers.

Question 23.
In north- eastern state of India major hectors of Forest land is degraded by a mode of cultivation.

1. Name the cultivation method
2. Explain the procedure of this cultivation

Answer:
1. Jhum cultivation.

2. In Jhum cultivation or slash & burn cultivation, forest land is converted to agricultural land by cutting and burning trees, ash obtained is used as fertilizer for crop cultivation and other land is used for grazing cattle. The process repeated and it leads to deforestation.

Question 24.
What are Euro II norms?
Answer:

1. Sulphur be controlled at 350 ppm in diesel and 150 ppm in the petrol.
2. Aromatic hydrocarbons are to be contained at 42% of concerned fuel.

Question 25.
It has been recorded that the temperature of the earth’s atmosphere has increased by 0.6°C.

1. What has caused this increase?
2. Explain its consequences.

Answer:
1. Increase in the level of green house gases causes global warming.

2. Its consequences:

• Polar ice caps and glaciers will melt
• rise in sea level

Question 26.
Air pollution causes serious health problems in men, in industries ECP is a fitted in place of exhaust coming from smokestack to control pollution.

1. Which is the device used to control Sulphur Dioxide pollution?
2. Name the agency of Government of India monitor air pollution.

Answer:

1. Scrubber
2. CPCB (Central pollution control board)

Question 27.
Higher concentration of DDT is found in third trophic level than first trophic level

1. Name the phenomenon associated with the accumulation of DDT
2. Bird population is decreased by the accumulation of DDT. How is it occurs?

Answer:

1. biomagnification
2. DDT affects the calcium metabolism of birds, causes the thinning of egg shell and premature breaking that it result the reduction in bird population.

Question 28.
Many low lying land become submerged in coming centuries due to the increasing temperature of atmosphere

1. Name two greenhouse gases causes global warming
2. Write down the measures to reduce the emission of greenhouse gases

Answer:
1. Carbon dioxide and chioro fluoro carbon

2. Measures to reduce the emission of greenhouse gases:

• Cutting down the use of fossil fuels
• Reduce deforestation activities.

Question 29.
Plastic film waste that is used to made modified recycle plastic by Ajmal Khan in collaboration with RV College of Engineering

1. Name the product formed after the recycling of film waste
2. Name the water repellent substances is mixed with modified a plastic to lay down Road in Bangalore city Corporation

Answer:

1. Polyblend
2. Bitumen

Question 30.
There is a balance between production and degeneration of ozone in the stratosphere.

1. Name the gas that disturbs this balance.
2. Explain how this balance is disrupted?

Answer:

1. Chlorofluorocarbons (CFCs)
2. In stratosphere ,UV rays act on CFCs and release active Cl atoms, that degrades ozone releasing molecular oxygen.

Question 31.
The amount of biodegradable organic matter in sewage water can be estimated by measuring BOD.

1. Expand BOD
2. Write the impact of the discharge of sewage into a river

Answer:

1. BOD – Biochemical Oxygen Demand
2. Polluted water contains microorganisms they consume lot of oxygen from water for the biodegradation of organic material that Causes sharp decline in dissolved oxygen.

Question 32.
Radiation by nuclear waste is extremely dangerous to organisms.

1. Write any two dangers.
2. Write the recommendations for the storage of nuclear waste.

Answer:

1. Causes mutation at a very high rate At high doses nuclear radiation is lethal
2. Nuclear waste must be stored after sufficient pre-treatment. It must be done in suitable shielded containers buried within the rocks, about 500m deep below the earth’s surface.

Question 33.
There is a balance between production and degeneration of ozone in the stratosphere.

1. Name the gas that disturbs this balance.
2. Explain how this balance is disrupted?

Answer:

1. Chlorofluorocarbons (CFCs)
2. In stratosphere, UV rays act on CFCs and release active Cl atoms, that degrades ozone releasing molecular oxygen.

Question 34.
Algal bloom and eutrophication are the two effects of water pollution. Write the differences between the two.
Answer:
1. Algal bloom:
Excessive growth of planktonic algae due to the presence of large amount of nutrients in water.

2. Eutrophication: Natural aging of a lake by nutrient enrichment of its water.

Question 35.
Match the items of column A with B.

Answer:

Question 36.
A number of human activities contribute to deforestation.

1. Write any one such activity.
2. Write the consequences of deforestation.

Answer:
1. Cutting trees for timber and firewood.

2. consequences of deforestation:

• Increase carbon dioxide concentration in the atmosphere
• Loss of biodiversity due to habitat destruction
• Disturbs hydrologic cycle
• Causes soil erosion

Question 37.
In 1980s the Government of India has introduced JFM to conserve forests.

1. Expand JFM.
2. Write its significance.

Answer:

1. Joint Forest Management.
2. Participation by local communities for protecting and managing forest and these communities get benefit of various forest products.

Question 38.
Catalytic converters are fitted into automobiles for reducing emission of poisonous gases into the atmosphere. Write its working in reducing pollution.
Answer:
As the exhaust from the automobiles passes through the catalytic converter, hydrocarbons are converted into harmless carbon dioxide and water. Carbon monoxide is changed into carbon dioxide.

Plus Two Botany Environmental Issues Three Mark Questions and Answers

Question 1.
As a part of project work, a group of students collected water samples from different locations. Water samples and its BOD contents are given below.

1. Analyse the data given below and find out the sample which is highly polluted.
2. Give reason. Expand BOD.

Sample	BOD
A	20%
B	12%
C	8.5%
D	8%

Answer:

1. Sample. A is highly polluted.
2. BOD (Biochemical Oxygen demand).- refers to the amount of O₂ that is consumed when organic matter is oxidised by bacteria. If greater the BOD of water, the rate of pollution is high.

Question 2.
Global warming leads to so many hazards in the environment.

1. What are the green house gases?
2. How does it leads to global warming
3. Suggest any three methods to control it.

Answer:
1. Green house gases- CFC, N₂O, CH₄, CO.

2. In the atmosphere green house gases causes re radiation of infrared radiation & results in the over heating of earth. This is called global warming. Global warming causes changes in the climate Eg. EL Nino effect, Increased melting of polar ice caps rise in sea levels & submerging of coastal areas.

3. Suggest any three methods to control it:

• Cutting down the use of fossil fuel,
• Improving efficiency of energy usage,
• Reducing deforestation, planting trees

Question 3.
Ramu is comparing the graph given in the magazine and recognized that the content of aromatic hydrocarbons, CO, and NOx were decreased from 1997 to 2006 period. Give the reason of your findings.
Answer:
The new auto fuel policy introduced by the government of India is to reduce the sulphur,carbon monoxide and aromatics content in petrol and diesel fuels. The above mentioned things can be put in practice by different engine norms.

1. Bharat Stage II (equivalent to Euro-llnorms)
2. Euro III emission specifications
3. Euro-IV

Question 4.
A factory drains its waste water into the nearby lake. It has caused algal bloom.

1. How was the algal bloom caused?
2. What would be the consequences?
3. Name the phenomenon that caused it?

Answer:

1. Nutrients of waste water causes extensive growth of planktonic algae or free floating algae (algal bloom).
2. Algae use oxygen .biochemical oxygen demand go high, high fish mortality, deterioration of water quality.
3. Eutrophication.

Question 5.
The existence of human beings is under great threat due to the uncontrolled emission of green house gases. Can you suggest any 2 remedial measures to reduce this effect?
Answer:
The emission of green house gases can be reduced by cutting down use of fossil fuel, improving the efficiency of energy usage, reducing deforestation planting trees & slows down the growth of human population.

Question 6.
The balance between the formation and degradation of ozone disrupt due to the overuse of chloro fluoro carbons.

1. Name the degraded product of chlorofluorocarbon causes of ozone depletion
2. Which is the unit used to measure ozone thickness in stratosphere
3. Give two diseases affects men due to ozone depletion

Answer:
1. Active chlorine
2. Dobson (Du)
3. Two diseases affects men due to ozone depletion:

• snow blindness
• cataract

Question 7.
In Kerala, the major source of atmospheric pollution comes from automobile sector.

1. Expand CNG
2. State some novel practical steps to minimize this type of pollution.
3. Which is the govt, agency monitor pollution?

Answer:
1. CNG -Compressed Natural Gas.

2. Air pollution can be controlled by using lead free petrol & by using catalytic converters. The catalytic converters convert unburned hydro carbons into CO₂, H₂O, CO & nitric oxide are converted to CO₂ & N₂ gas.

3. CPCB (Central pollution control board).

Question 8.
Ozone balance has been disrupted due to enhancement of ozone degradation by CFC’s.

1. Enumerate the role of Cl in O₃ depletion.
2. Give a note on Montreal protocol.

Answer:
1. Ozone is found in the stratosphere It acts as shield by absorbing UV radiation from the sun. UV rays are injurious to living organisms since DNA & proteins of living organism are affected by UV rays. The refrigerators use CFC.

The CFC’s discharged into the atmosphere it reaches the stratosphere. In stratosphere UV rays act on them releasing chlorine atom. Chlorine degrade ozone and releasing molecular oxygen. As the chlorine atom are not consumed in this reaction they remain in the stratosphere & will have permanent & continuing effect on ozone level.

2. Recognizing the deleterious effects of ozone depletion an international treaty known as Montreal Protocol was signed at Montreal(canada) in 1987 & is effective from 1989. This protocol emphasizes to control the emission of ozone depleting substances.

Question 9.
Night blindness is caused by the deficiency of Vitamin. What is the cause of snow blindness.
Answer:
Due to ozone hole the UVB rays reach the earth and its high dose causes inflamation of cornea called snow blindness, cataract, etc. Such an exposure permanently damage the cornea.

Question 10.
Discuss the role of women and communities in protection and conservation of forests.
Answer:
Amrita devi bishnoi wild life protection award (in the name of lady who sacrificed her life for protecting trees), chipko movement( women belonging to garhwal showed courage by hugging trees to protect them from axe of contractors).

Question 11.
If the ESP is not fitted across the way of exhaust come out from thermal power plant. How does it affect humans.
Answer:
The particulate matter present in exhaust spread the air and causes respiratory problems like asthma, chronic bronchitis, hay fever, etc.

Question 12.
Two years ago, the lake was pure, clean and rich in variety of organisms. While now it is colourless, rich in algae and have a dirty smell. Name the pollutants and the phenomenon.
Answer:
The pollutans are phosphatic and nitrogenous in organic fertilisers. The phenomenon is called eutrophication.

Question 13.
Hundred metre surroundings of hospitals and schools are considered as silent zone. Justify it.
Answer:
It is considered as silent zone because in hospitals sound intensity increases the heart beat altering respiratory patterns of patients etc.but in schools it affects the teaching and learning process.

Question 14.
In 1990, the ozone thickness in a particular area was 190 Du while in 2001 and 2005 it was decreased to 150 Du and 135 Du respectively.

1. What is the causes of fall of Du level.
2. Which is the good ozone situated and named so?

Answer:
1. Itis due to the increased emission of chlorofluro carbons. It causes the upset of balance between formation and degradation of ozone and results in ozone depletion or thinning of ozone layer.

2. Ozone found in stratosphere filter uv rays coming into the earth.

Question 15.
CO and SO₂ are pollutants but they are different from, DDT. BHC etc. Do you agree. Give justifications of your answer.
Answer:
Yes. The former pollutants are gaseous, biodegradable and inorganic but latter are organic and non biodegradable.

Question 16.
What you meant by the auto fuel policy of Govt, of India? Name the latest reformed engine norm.
Answer:
The aim of auto fuel policy is to decrease the emission of sulphur, carbon monoxide and aromatics content in petrol and diesel fuels. Euro-IVnorms- April 1, 2010.

Question 17.
Now a days the acuumulation of plastic waste is the major problem because such wastes are non biodegradable. In some places it is used for beneficial purpose

1. Name the mixtures used by Ahmed Khan to lay down roads in Bangalore.
2. How was it minimized the accumulation of solid wastes?

Answer:

1. Polyblend and bitumen.
2. The raw material for creating Polyblend is any plastic film waste. So the accumulation of plastic wastes can be controlled.

Question 18.
Thermal Power plants are inevitable in an industrial and densely populated country like our-what harm they do to the environment? Also mention any precaution that could be taken to save our environment.
Answer:
Hot effluents of thermal power plants are generally released in the nearby water bodies. This causes rise in temperature of such water bodies which causes deoxygenation of water leading to decomposition of organic wastes and killing of aquatic animals. Thermal water pollution can be checked by employing dry or wet cooling towers.

Question 19.
Accumulation of solid waste (e-waste) is the major problem in developed countries.

1. What do you mean by e-waste?
2. Give its ill effects.

Answer:

1. It is formed of irreparable computers and other electronic goods.
2. During manual recycling of such waters in the developing countries the workers are exposed to toxic compounds present in them.

Question 20.
It is said that the use of refrigerators cause ozone depletion.

1. How is it possible?
2. Explain the process of ozone depletion.
3. Name any one treaty signed in the international level to control this phenomenon.

Answer:
1. The refrigerators emit CFC (Chloro flora carbon).

2. The CFC’s discharged into the atmosphere reaches the stratosphere. In stratosphere UV rays act on them releasing chlorine atom, chlorine Degrade ozone releasing molecular oxygen. As the chlorine atom are not consumed in this reaction they remain in the stratosphere and have permanent and continuing effect on ozone level.

3. Montreal protocol and Kyoto protocol.

Question 21.
Mention the two major environmental issues of global nature.
Answer:

1. Increasing greenhouse effect.
2. Depletion of ozone in the stratosphere.

Question 22.
Thermal power plants are inevitable in an industrial and densely populated country like ours. What harm do they do to the environment? Also mention any precaution that could be taken to save our environment.
Answer:

1. They release particulate and gaseous air pollutants, hot thermal waste kills organisms sensitive to high temperature, indigenous flora ad fauna lose.
2. Use of electrostatic precipitator.

Question 23.
1. On seeing bad state of roads in your locality, as a student, you have recommended to the Municipal corporation to use polyblend.

• What is poly blend? Point out its raw material.
• How will it be advantageous?

2. What are e-wastes? Explain the method of their disposal.
Answer:
1.

• It is a fine powder of recycled modified plastic. This mixture is mixed with bitumen used to lay roads. Raw material – Plastic film waste
• Blends of polyblend and bitumen, when used to lay roads, enhances the bitumen’s water repellent properties and helps to increase road life.

2. Irreparable computers and other electronic goods are known as e-wastes. Burried in landfills or incinerated.

Question 24.
Eutrophication is the main problem in of rivers due to the adding of more and more nitrates and phosphates.

1. How does it affect aquatic water body
2. What is algal bloom?
3. How does it related to D.O. in water bodies.

Answer:

1. It change the colour and quality of water and causes the mortality of fish
2. Presence of large amounts of nutrients in waters also causes excessive growth of planktonic(free-floating) algae, called an algal bloom
3. This results in the decrease of D.O. in water.

Question 25.
DDT content in the water of a lake, that supplies drinking water to the nearby villages, is found to 0.0Q3ppm. The kingfishers of that area are reported to have2ppm DDT.

1. Why has the concentration increased in these birds?
2. What harm will this cause to bird population?
3. Name the Phenomenon.

Answer:
1. Increase in concentration of DDT (non biodegradable) from the water bodies (0.003ppm) to the final consumers like kingfishers (2ppm) is due to increase in cone, of DDT in successive tropic levels (eg. Phytoplankton → Zooplanktons → Fishes → King Fishers) of food chain.

2. It affects ca metabolism and finally premature breaking of egg and declining bird population.

3. Biomagnification.

Question 26.
Observe the figure below.

1. Identify the greenhouse gases 1,2 and write their percentage of contribution to total global warming.
2. Write any four measures to control global warming.

Answer:
1. greenhouse gases 1,2 and write their percentage of contribution to total global warming:

• CO₂ – 60%
• CFCs 14%

2. four measures to control global warming:

• Cutting down use of fossil fuel
• Improving efficiency of energy usage
• Reducing deforestation
• Planting trees

Question 27.
Presence of large amounts of nutrients in water cause excessive growth of plankton.

1. Name this process.
2. Write the dangers caused by this process.

Answer:
1. Algal bloom.

2. Dangers caused by this process:

• Imparts a distinct colour to the water bodies
• Deterioration of the water quality
• Fish mortality
• Toxic to human beings and animals.

Question 28.
Pollutants from man’s activities like effluents from the industries and homes accelerate the aging process of lakes.

1. Name this phenomenon.
2. Write any four harmful effects of this process.

Answer:
1. Cultural or Accelerated eutrophication.

2. Any four harmful effects of this process:

• Over stimulate the growth of algae
• Cause unpleasant odor
• depletes dissolved oxygen
• Death of fish

Question 29.
CNG is better than diesel. Substantiate this statement writing any four advantage of CNG. Expand CNG.
Answer:
1.

• CNG bums most efficiently
• Cheaper than petrol or diesel
• Cannot be siphoned off by thieves
• cannot be adulterated like petrol or diesel

2. Compressed Natural Gas

Question 30.
Observe the flow chart given below.

1. Fill in the blanks 1 and 2.
2. Identify the process illustrated in the flow chart.

Answer:
1.

• Small fish (DDT 0.5 ppm)
• Large fish (DDT 2 ppm)

2. Biomagnification

Question 31.
People in the town of Areata created an integrated waste water treatment within a natural system. Explain the method.
Answer:
The cleaning of waste water is conducted in two stages. In the first stage conventional sedimentation, filtering, and chlorine treatment are given.

Through this step all materials present in water can be removed except dangerous pollutants like dissolved heavy metals.

In the second stage this heavy metal containing water is passed through a series of six connected marshes over 60 hectares of marshland.

In marshy area, plants, algae, fungi, and bacteria were seeded, they can neutralise, absorb and assimilate the pollutants as the water flows through the marshes, after this process water gets purified naturally.

Question 32.
Match the items of column A with B and C.

Answer:

Plus Two Botany Environmental Issues NCERT Questions and Answers

Question 1.
Match the items given in column A and B.

Column A	Column B
a) Catalytic inverter	i) Particulate Matter
b) Electrostatic precipitator	ii) carbon monoxide and Nitrogen oxides
c) Earmuffs	iii) High noise level
d) Landfills	iv) Solid wastes.

Answer:
(a) – (ii)
(b) – (i)
(c) – (iii)
(d) – (iv)

Question 2.
Write critical notes on following:

1. Eutrophication
2. Biological magnification
3. Ground-water depletion and ways for its replenishment.

Answer:
1. Natural aging of a lake by nutrient enrichment of its water.

2. Biomagnification, also known as bio amplification or biological magnification, is any concentration of a toxin, such as pesticides, in the tissues of tolerant organisms at successively higher levels in a food chain.

3. The level of ground water is falling day by day due to high demand in urban areas as well as in agriculture. Due to over use of surface water, people rely on ground water for irrigation, drinking, and industrial use.

About 85% of rural water supply and more than 50% of urban and industrial supply is mined. This results in depletion of ground water.

Ground water can be replenished by-

1. Rain water harvesting
2. Reduction in consumption and waste.

Question 3.
Why ozone hole forms over Antarctica? How will enhanced ultraviolet radiation affect us?
Answer:
Decline in the thickness of ozone layer over a restricted area is called ozone hole. It was first discovered over Antarctica. Antarctica air is completely isolated from the rest of the world by natural circulation of wind called as Polar Vertex.

CFCs released in the atmosphere slowly enters the stratosphere and wind push them towards the poles. Environmental conditions prevailing in Antarctica during winter months are conductive for the formation of ozone hole.

During winter months, lack of sunlight and low temperature facilitates the formation of ice clouds, which provide the catalytic surface for the reaction of chlorine.

Effect of Ultraviolet Rays

1. UV-B radiations are very harmful.Coronea of eye absorbs these radiations and becomes inflamed. This disorder is known as ‘snow blindness’ cataract and leads to diminishing of eye sight.
2. UV-B radiations damage skin cells and cause skin cancer.

Question 4.
Discuss the role of women and communities in protection and conservation of forests.
Answer:
Women and tribal communities have played a significant role in protection and conservation of forests. Amritha devi Devi Bishnoi Wild Lide Protection Award, has instituted by government of India for individuals and rural communities for their contribution in protection of wild life.

In 1731, a woman Amritha Devi showed exemplary courage by hugging a tree to prevent its cutting. Her three daughters and hundreds of other Bishnoi followed her. They were killed by soldiers of Jodhpur.

Question 5.
What measures, as ah individual, you would like to reduce environmental pollution?
Answer:

1. I will use those articles which are either disposable or can be recycled.
2. I would help in tree plantation in my school and surroundings.
3. I will minimize the use of fossil fuels.

Plus Two Botany Environmental Issues Multiple Choice Questions and Answers

Question 1.
The maximum biological magnification of DDT through food web is seen in.
(a) Bacteria
(b) Algae
(c) Man
(d) Higher plants
Answer:
(c) Man

Question 2.
Eutrophication leads to death of fish due to
(a) Increased O₂ content
(b) Increased algae content
(c) Decreased algae content
(d) Decreased O₂ content
Answer:
(d) Decreased O₂ content

Question 3.
Match the following:-

A	B
1. The insecticide Act	A. 1968
2. The water (prevention and control of pollution) Act	B. 1974
3. The air (prevention and control of pollution)Act	C. 1981
4. The environment (prevention)Act	D. 1986

(a) 1 – d, 2 – c,  3 – b, 4 – a
(b) 1 – a, 2 – d,  3 – c,  4 – b
(c) 1 – a, 2 – b,  3 – c,  4 – d
(d) 1 – c, 2 – d,  3 – a, 4 – b
Answer:
(c) 1 – a, 2 – b,  3 – c,  4 – d

Question 4.
Nitrogen oxides produced from the emission of automobiles and power plants are the source of fine air borne particles which lead to
(a) photochemical smog
(b) dry acid deposition
(c) industrial smog
(d) wet acid deposition
Answer:
(d) wet acid deposition

Question 5.
Drinking of mineral water with very low level of pesticides (about 0.02 ppm) For long periods may
(a) produce immunity against mosquito
(b) cause leukemia (blood cancer) in
(c) cause cancer of the intestine
(d) lead to accumulation of pesticide residues in body fat
Answer:
(d) lead to accumulation of pesticide residues in body fat

Question 6.
The soil pollutants that affect the food chain and food web by killing microorganisms and plants are
(a) nitrogen oxides
(b) pathogens chemical fertilizers
(c) agricultural waste
(d) pesticides
Answer:
(d) pesticides

Question 7.
In a coal-fired power plant, electrostatic precipitators are installed to control emission of
(a) SO
(b) NOX
(c) SPM
(d) CO
Answer:
(c) SPM

Question 8.
Green-house effect is due to the presence of
(a) ozone layer in the atmosphere
(b) moisture layer in the atmosphere
(c) CO₂ layer in the atmosphere
(d) infrared light reaching the earth
Answer:
(c) CO₂ layer in the atmosphere

Question 9.
Which one of the following pairs pf organisms are exotic species introduced in India?
(a) Ficusreligiosa, Lantana camara
(b) Lantana camara, water hyacinth
(c) water hyacinth, Prosopis cineraria
(d) Nile perch, Ficusreligiosa
Answer:
(b) Lantana camara, water hyacinth

Question 10.
A lake with an inflow of domestic sewage rich in organic waste may result in
(a) drying of the lake very soon due to algal bloom
(b) an increased production of fish due to lot of nutrients
(c) death of fish due to lake of oxygen
(d) increased population of aquatic food web organisms
Answer:
(c) death of fish due to lake of oxygen

Question 11.
A substantial fall in C02 and S02 level has been found in Delhi between 1997 and 2005 due to
(a) use of purified, unleaded petrol
(b) use of purified, unleaded diesel
(c) use of Compressed Natural Gas (CNG) in public transports
(d) use of LiquifiedPetroleam Gas (LPG) in public transports
Answer:
(c) use of Compressed Natural Gas (CNG) in public transports

Question 12.
Electrostatic precipitator is used to remove
(a) CO
(b) particulate matter
(c) lead
(d) secondary pollutant
Answer:
(b) particulate matter

Question 13.
CNG is used to control
(a) water pollution
(b) air pollution
(c) radioactive pollution
(d) thermal pollution
Answer:
(b) air pollution

Question 14.
Major pollutant present in automobile exhaust is
(a) lead
(b) CO
(c) CO₂
(d) hydrocarbon
Answer:
(b) CO

Question 15.
If there is no green house gases present in an atmosphere, the average temperature of earth surface is
(a) 100c
(b) 1000 c
(c) 00 c
(d) -150 c
Answer:
(d) -150 c

Question 16.
Deforestation activities and forest fires leads to the rise in global temperature, it is due to
(a) increase in CO level
(b) decrease in CO level
(c) increase in CO₂ level
(d) decrease in CO₂ level
Answer:
(c) increase in CO₂ level

Question 17.
A geing of lake is mainly caused by the accumulation of nutrients such as
(a) N₂ & P
(b) N₂ & P
(c) P & Ca
(d) Ca & N₂
Answer:
(b) N₂ & P

Question 18.
Appearance of hole in strato spheric ozone layer is caused by the emission of green gases, it is due to the reaction between
(a) O₃ and Cl
(b) O₃ and F
(c) O₂ and F
(d) O₂ and Cl
Answer:
(a) O₃ and Cl

Question 19.
The pollutant that seriously affect the marble monuments mainly contains the chemical have the nature is
(a) acidic
(b) basic
(c) neutral
(d) none of the above
Answer:
(a) acidic

Question 20.
Chlorinated hydrocarbons enters through one trophic level and its concentration is increased as ppm in successive trophic levels, its nature is
(a) biodegradable
(b) non biodegradable
(c) degradable on combustion
(d) not degradable on combustion
Answer:
(b) non biodegradable

Question 21.
Increase in noise level above particular dB affect the hearing abilities, that is
(a) 20 dB
(b) 60 dB
(c) 80 dB
(d) 150dB
Answer:
(d) 150dB

Students can Download Chapter 13 Nuclei Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 13 Nuclei

Plus Two Physics Nuclei NCERT Text Book Questions and Answers

Question 1.
Obtain the binding energy of a nitrogen nucleus (\(_{ 7 }^{ 14 }{ N }\)) Given m \(_{ 7 }^{ 14 }{ N }\) = 14.00307 u.
Answer:
Here Z = 7 and A = 14, A – Z = 14 – 7 = 7
∴ Mass defect
= [Z m_H + (A – Z)m_n – M_n] u
= (7 × 1.00783 + 7 × 1.00867 – 14.00307) u
= (7.05481 + 7.06069-14.00307) u = 0.11243 u
Since I u = 931.5 MeV
∴ B.E. of 14N = 0.11243 × 931 Mev
= 104.7 MeV.

Question 2.
Obtain approximately the ratio of the nuclear radii of the gold isotope \(_{ 79 }^{ 197 }\)Au and the silver isotope \(_{ 47 }^{ 107 }\)Ag
Answer:
Here A₁ = 197 and A₂ = 107

Question 3.
How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction as.

Answer:
Number of deuterium atoms is 2 kg
= \(\frac{6.023 \times 10^{23}}{2}\) × 2000 = 6.023 × 10²⁶
Energy released when 6.023 × 10²³ nuclei of deuterium fuse together

= 15.42 × 10¹³J = 15.42Ws
Power of lamp = 100 W
If the lamp glows fortime t, then electric energy consumed = 100 t
∴ 100 t = 15.42 × 10³³
∴ t = 0.1542 × 10¹³s
\(=\frac{0.1542 \times 10^{13}}{365 \times 86400} y\)
or t = 4.0 × 10⁴y.

Question 4.
From the relation R = R₀ A^{\(\frac{1}{3}\)}, Where R₀ is a constant and A is the mass number of a nucleus show that the nuclear matter density is nearly constant (i.e. Independent of A).
Answer:
Density of nucleus(p) It is defined as the nuclear mass per unit volume.

Thus the nuclear density is of the order of 10¹⁷kgm^-3 and is independent of its mass number. Therefore, all nuclei have the same approximate density.

Plus Two Physics Nuclei One Mark Questions and Answers

Question 1.
Fusion reaction takes place at high temperature because
(a) nuclei break up at high temperature
(b) atoms get ionized at high temperature
(c) kinetic energy is high enough to overcome the coulomb repulsion between nuclei
(d) molecules break up at high temperature
Answer:
(c) kinetic energy is high enough to overcome the coulomb repulsion between nuclei.

Question 2.
Give the relation between half-life and mean life
Answer:
t_1/2 = 0.693τ.

Question 3.
Write down the expression for nuclear radius
Answer:
R = R₀A_1/3.

Question 4.
Complete the given nuclear reaction

Answer:

Plus Two Physics Nuclei Two Mark Questions and Answers

Question 1.
A beam of radio active radiation is unaffected in a combined electric and magnetic field in mutually perpendicular direction. A boy argues that, it is essentially a gamma ray. Do you agree with him. Justify your answer.
Answer:
It need not be gama ray. It can be α or β ray as the electric and magnetic field are in mutually perpendicular direction.

Question 2.
Pick the odd one out of the following.
a.

1. Curie
2. Roentgen
3. Becquerel
4. Rutherford

b.

1. γ – decay
2. β – decay
3. β⁺ – decay
4. α – decay

Answer:
a. odd one:
2. Roentgen.

b. odd one:
1. γ – decay

Question 3.
Figure below represents radiation coming out from a radioactive element.

1. Identify the radiation B. Give reason.
2. If the electric field is replaced by magnetic field perpendicular and into the plane of the paper, identify the particle deflecting towards right (C).

Answer:

1. Gamma-ray. It has no change
2. C which is β ray

Question 4.
Two radioactive substances P and Q have half-life 6 months and 3 months respectively. Find the ratio of the activity of these two materials after one year.
Answer:
Activity R = λN
R₁ = λ₁N₁
R₂ = λ₂N₂

R₁: R₂ = 1: 1

Question 5.
Match the following.

Answer:

Question 6.
Suppose you are a health physicist and you are being consulted about a spill occurred in a radio chemistry lab. The isotope spilled out in the lab was 500 micro cure of ¹³¹Ba which has half-life of 12 – days.

1. What is the decay constant of ¹³¹Ba
2. Your recommendation is to clear the lab until the radiation level is down to 1 micro curie. How length will the lab have to be closed.

Answer:
1.

2. 9 × 12 = 108 days.

Question 7.
Classify the following statement in to nuclear fission and nuclear fusion

1. Products are radio active
2. Can be controlled
3. Reaction is spontaneous
4. Can’t proceed as a chain reaction

Answer:

1. fission
2. fission
3. fission
4. fusion

Question 8.

Answer:

Plus Two Physics Nuclei Three Mark Questions and Answers

Question 1.
1. Nuclear fusion can liberate more energy than nuclear fission. But nuclear fusion is not commonly used in energy production. Why?
2. Match the following.

A	B
1. Nuclear fission	β – decay
2. Nuclear fusion	Hydrogen spectrum
3. Transition between atomic energy level	Nuclei with low atomic numbers
4. Electron emission from nucleus	Generally possible for nuclei with high atomic number
	Photoelectric emission

Answer:
1. Nuclear fusion is a thermonuclear reaction. It occurs at high temperature (10⁶k) and it is difficult to attain such a high temperature. Hence fusion is not commonly used in energy production.

2.

A	B
1. Nuclear fission	Generally possible for nuclei with high atomic number
2. Nuclear fusion	Nuclei with low atomic numbers
3. Transition between atomic energy level	Hydrogen spectrum
4. Electron emission from nucleus	β – decay

Question 2.
β – Particles does not exist inside a nucleus. But it is emitted from the nucleus!

1. What is β Particle
2. What happens to nucleus of a atom, when a particle is emitted? Explain
3. Why β particle is emitted from the nucleus?

Answer:

1. It is electron
2. When α is emitted, mass number decreases to 4 and atomic decreases by 2.
3. To get stability, β is emitted from the nucleus.

Question 3.
Nuclear radius depends on the mass number of the element.

1. Write down the expression for nuclear radius.
2. Prove that the density of the nucleus is independent of mass number A.

Answer:
1. R = R₀ A^1/3.

2. Density of nucleus = (mass of the nucleus)/ (volume of the nucleus)

This shows that nuclear density is independent of mass number.

Question 4.
Match the following.

Answer:

Question 5.
The fission of one nucleus of ₉₂U²³⁹ release 200 Mev of energy.

1. What is meant by fission
2. Express 200 Mev energy in joule
3. How many fission of ₉₂U²³⁹ should occur per send for producing a power of 1 Mev

Answer:
1. The splitting of heavy nucleus into two nucleus is called fission.

2. E = 200Mev
= 200 × 10⁶ev
E = 200 × 10⁶ × 1.6 × 10^-19J
= 3.2 × 10^-11J.

3. Number of fission

Question 6.
Match the following.

Answer:

Question 7.
Size of the nucleus increases with number of nucleons. As size increases, volume increase (ie mass number increases).

1. Based on the above facts find an expression for radius of nucleus.
2. Calculate radius of ₁₃Al²⁷ nucleus. The constant R₀ = 1.2 fermi.

Answer:

1. R = R₀A^1/3
2. R = 1.2 × (27)^1/3 fermi = 3.6 fermi.

Question 8.
Binding energy curve shows the variation of Binding energy per nucleon of nuclei with mass number.
1. Binding energy per nucleon is maximum for mass number……. (1)
2. The figure shows dis integration of Deuteron. (2)

What should be the frequency of the incident photon to break Deuteron into proton and neutron?
Mass of proton m_p = 1,007276u.
Mass of neutron m_n = 1,008665u
Mass of deuteron = 2.013553u
Answer:
1. 56

2. Mass defect = (1.007276 + 1.008665) – 2.013553 = 0.002388u
Binding Energy = 0.002388 × 931 MeV = 2.223
MeV Energy supplied to the photon
= 2.223 × 10⁶ × 1.6 × 10^-19
= 3.56 × 10^-13J
Frequency of photon

Question 9.
The figure shows the potential energy of a pair of nuclear particles and their distance of separation in Fermi (fm).

1. Fill in the blanks.

2. What conclusion do you obtain about the nature of nuclear force from the graph.
Answer:
1.

2. The nuclear force is a short range force.

Plus Two Physics Nuclei Four Mark Questions and Answers

Question 1.
Atomic mass of \(_{ 8 }^{ 16 }{ O }\) is found to be 16.0000u

1. what is the mass of \(_{ 8 }^{ 16 }{ O }\) nucleus
   (Hint: mass of an electron = 0.00055u)
2. determine the total mass of the constituents particles of the \(_{ 8 }^{ 16 }{ O }\) nuclei.
   (Hint : Mass of neutron = 1.00864 u, Mass of proton = 1.007274 u)
3. Give a general expression for mass defect and explain what is binding energy?
4. Binding energy per nucleon is lower for both very light nuclei (Z ≤ 10) and very heavy nuclei (Z ≥ 70) Justify a nuclear fission and fusion

Answer:
1. Mass of nucleus = (8 × 1.00864 + 8 × 1.007274) amu.

2. (8 × 1.00864 + 8 × 1.007214 + 8 × 0.0055)amu

3. ∆m = [ZM_p + (A-Z) M_n – M)
The energy equivalent to mass defect is called binding energy.

4. When two light nuclei are combined to form a heavy nucleus, the binding per nucleon increases. Hence the stability of atom increases. When heavy nucleus split into two light nuclei; B.E. for nucleon also increases.

Question 2.
Classify the following into alpha, beta, and gama

1. Similar to fast moving electron
2. It is an electromagnetic wave
3. Similar to helium nucleus
4. Travel with 1/10^th the velocity of light
5. Travel with 99/100^th the velocity of light
6. Travel with the velocity of light
7. Positively charged
8. Negatively charged

Answer:

1. beta particle – 1, 3, 8
2. Gama ray – 2, 6
3. alpha particle – 3, 4, 7

Question 3.
a. Two protons and two neutrons may bound to form a single particle it is called

1. α particle
2. β particle
3. deuteron
4. triton

b. If such a particle is emitted what changes will occur in the nucleus
c. The penetrating power of the particle is very small in air. Why?
d. If the particle is projected upward in a uniform magnetic field with direction perpendicular to plane in ward, towards which plate (A or B) it is deflected? To find this which law is applied?

Answer:
a.
1. α particle.

b. Mass number decreases to 4 and atomic number decreases to 2

c. penetrating power of the particle is very small in air:
(i) Mass of α particle is 4. Hence penetrating power of the particle is small.

d. Towards A (left) Flemings left hand rule is used.

Question 4.
Nuclear radius depends on the mass number of the element.

1. Write down the expression for nuclear radius. (1)
2. Prove that the density of the nucleus is independent of mass number A. (2)
3. Read the following statement and choose the correct option.

“Electric dipole moment is zero for nuclei in stationary state.” (1)
Assertion:
All nuclei have spherical symmetry about the centre of mass.
Reason:
The zero dipole moment for stationary nuclei is due to the symmetry about the centre of mass.

• Assertion and reason are true.
• Assertion is false reason is true
• Assertion is true reason is false
• Assertion and reason are false.

Answer:
1. R = R₀A^1/3.

2. Density of nucleus = (mass of the nucleus)/(volume of the nucleus)

This shows that nuclear density is independent of mass number.

3.
(ii) Assertion is false reason is true.

Question 5.

1. Which process is represented by the equation?
2. What happens to the parent nucleus after the process?
3. After this process the nucleus will be in a higher energy state. How will it come to ground state?
4. If a proton splits in a nucleus what are the changes. Represent it with an equation.

Answer:

1. β – emission.
2. Atomic increases to one unit. But mass number remains constant
3. The parent atom comes to ground state by emitting gamma ray.
4. P → n + e⁺ + ν
   Atomic number decreases by one unit. But mass number remains constant.

Plus Two Physics Nuclei Five Mark Questions and Answers

Question 1.
Rutherford and Soddy’s laws of radioactivity explain the rate of decay of radioactive material.

1. Arrive at the expression for the number of radio active atoms of a radioactive material remaining after an interval of time. (2)
2. Draw the curve showing the variation of log\(\left(\frac{N}{N_{0}}\right)\) with time. (1)
3. Two radioactive substances P and Q have half life 6 months and 3 months respectively. Find the ratio of the activity of these two materials after one year. (2)

Answer:
1. According to Law of Radioactive decay,

Integrating
InN = -λt + C ………(1)
C is the constant of integration. To get value of C, let us assume that initially (t=0) the number of nuclei be N₀.
∴ C = In N₀
Substituting for C in equation (1) we get,
In N – In N₀ = -λt

\(\frac{\mathrm{N}}{\mathrm{N}_{0}}\) e^-λt
N = N₀e^-λt

2.

3. Activity R = λN
R₁ = λ₁N₁
R₂ = λ₂N₂

Question 2.
a. In the given figure a radioactive source is placed inside a lead block. Identify the rays incident on the photographic plates.

b. Which of the following statement is correct.

1. Gamma rays consist of high energy neutrons.
2. Alpha rays are equivalent to singly ionized He atoms.
3. Protons and neutrons have exactly the same mass.
4. Beta rays are same as cathode rays.

c. How many alpha and beta particles are emitted in the following reaction.

Answer:
a.

• 1 – Alpha
• 2-Gamma
• 3 – Beta

b.
4. Beta rays are same as cathode rays

c.

Difference in mass number = 32
Mass number of one alpha particle = 4
Hence number of alpha particles = 8
Change in atomic number = 10
Change in atomic number due to alpha particles = 16
Charge of beta particles = -1
Hence number of beta particles emitted = 6

Question 3.
The figure shows a nuclear reactor based on thermal neutron fission.

1. The energy of thermal neutrons is……..ev. (1)
2. Name the parts labelled X and Y in the figure. (1)
3. Write the function of X and Y (1)
4. The multiplication factor has great significance in nuclear reactor. Give reason. (2)

Answer:
1. 0.025eV.

2. Control rods.

3. Control rods are used in nuclear reactors to control the fission rate of Uranium and Plutonium.

4. The ratio, K, of number of fission produced by a given generation of neutrons to the number of fission of the preceding generation is called the multiplication factor, it is the measure of the growth rate of the neutrons in the reactor.

For K = 1, the operation of the reactor is said to be critical. Unless the factor K is brought down very close to unity, the reactor will become supercritical and can even explode.