Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Students can Download Chapter 1 The Living World Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examination

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Plus One The Living World One Mark Questions and Answers

Question 1.
As we go from species to kingdom in a taxonomic hierarchy, the number of common characteristics
(a) Will decrease
(b) Will increase
(c) Remain same
(d) May increase or decrease
Answer:
(a) Will decrease

Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Question 2.
Which of the following ‘suffixes’ used for units of classification in plants indicates a taxonomic category of ‘family’.
(a) -Ales
(b) -Onae
(c) -Aceae
(d) none of the above
Answer:
(c) -Aceae

Question 3.
The term ‘systematics’ refers to:
(a) Identification and classification of plants and animals
(b) Nomenclature and identification of plants and animals
(c) Diversity of kinds of organisms and their relationship
(d) Different kinds of organisms and their classification
answer:
(c) Diversity of kinds of organisms and their relationship

Question 4.
Genus represents
(a) An individual plant or animal
(b) A collection of plants or animals
(c) Group of closely related species of plants or animals
(d) None of these
Answer:
(c) Group of closely related species of plants or animals

Question 5.
The taxonomic unit ‘Phylum’ in the classification of animals is equivalent to which hierarchical level in classfication of plants.
(a) Class
(b) Order
(c) Division
(d) Family
Answer:
(c) Division

Question 6.
Correct and rewrite the following if there is any mistake.
Sativa Oriza, Tigris Panthera
Answer:
Oriza sativa, Panthera tigris

Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Question 7.
Kingdom – Carnivora
Phylum – Mammalia
Class – Felidsffe
Order – Chordata
Family – Tigris
Genus – Animalia
Sps – Panthera

Correct the mistakes of the flow chart to get the hierarchical arrangement of tiger in the ascending order.
Answer:
Kingdom – Animalia
Phylum – Chordata
Class – Mammalia
Order – Carnivora
Family – Felidae
Genus – Panthera
Sps – Tigris

Question 8.
Match the following:

  1. A Unit of classification – kingdom
  2. Lowest taxonomic category – Genus
  3. Panthera – Species
  4. Highest Taxonomic category – Taxon

Answer:

  1. A Unit of classification – Taxon
  2. Lowest taxonomic category – Species
  3. Panthera – Genus
  4. Highest Taxonomic category – kingdom

Question 9.
Expand the term ICBN and ICZN.
Answer:
1. ICBN: International Code for Botanical Nomenclature.
2. ICZN: International Code for Zoological Nomenclature.

Question 10.
Given below is the scientific name of Frog. Identify the correctly written name.
(a) Rana Tigrina
(b) Rana tigrina
Answer:
(b) Rana tigrina

Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Question 11.
Rearrange the order of classification.
Genus, Family, Phylum, Species, Class, Kingdom, Order
Answer:
Species, Genus, Family, Order, Class, Phylum, Kingdom

Question 12.
Arrange the given terms in their taxonomic hierarchy.
Primata, Homosapien, Chordata, Mammalia, Hominidae.
Answer:
Homosapiens, Hominidae, Primata, Mammalia, Chordata.

Plus One The Living World Two Mark Questions and Answers

Question 1.
You are provided with a stuffed rabbit and a dried leaf. As a student of biology which taxonomic aid will you choose to store them.
Answer:
1. Stuffed rabbit – Museum
2. Dried leaf – Herbarium.

Question 2.
Find the odd one out and give reason.
Herbarium, botanical garden, museum, aquarium
Answer:
Aquarium – Aquarium is not a taxonomical aid.
All other are taxonomical aids.

Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Question 3.
Once you visited a museum. There are different kinds of animals and plants are preserved. How these plants and animals all preserved in a museum.
Answer:
Plants and animals are preserved in the containers or jars in preservative solutions. These may also be preserved as dry specimen. Insects are preserved in insect box.

Larger animals like birds and mammals are usually stuffed and preserved. Museums often have collections of skeleton of animals too.

Question 4.
Zoological parks are different from museums. Give reason?
Answer:
Zoological parks are the places where wild animals are kept in protected environments, that the conditions similar to their natural habitats. Museums have collection of preserved plants and animal specimens for study and reference.

Question 5.
Raju collected a skull of an animal and a living rare animal during a study tour. Select the suitable location for each from the list given in the brackets. (Botanical garden, Zoological park, Herbarium, Museum)
Answer:
Skull of an animal – museum
Living rare animal – Zoological park

Question 6.
Distinguish between taxonomic category and taxonomic hierarchy.
Answer:
Classification involves hierarchy of steps in which each step represents a rank or category. Since category is a part of overall taxonomic arrangement, it is called taxonomic category. All the categories together constitute taxonomic hierarchy.

Question 7.
Write the scientific name of following animals.

  1. Lion
  2. Frog
  3. Housefly
  4. Tiger

Answer:

  1. Pantheraleo
  2. Ranatigrina
  3. Musca domestica
  4. Panthera tigris

Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Question 8.
Fill in the blank spaces in the table given below.
Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World 1
Answer:

(a) Live
(b) Museum

Question 9.
Panthera jeo is the scientific name of Lion. List the rules you follow to write this scientific name.
Answer:

  1. Biological names are generally in Latin and written in italics.
  2. When handwritten, the words in a biological name are separately underlined.
  3. The first word in a biological name represents the genus and second word represents the species name.
  4. The first name (Genus) starts with capital letter and the second name (species) starts with small letter.

Question 10.
List the advantages of

  1. Taxonomical key
  2. Herbarium

Answer:
1. axonomical key is a taxonomical aid used for identification of plants and animals based on the similarities and dissimilarities. Key are generally analytical in nature.

2. Herbarium is a store house of collected plant specimens that are dried, pressed and preserved on sheets. These specimens, along with their descriptions on herbarium sheets, became a store house or repository for future use. Herbarium serves as quick reference system in taxonomical studies.

Question 11.
Give the terms.

  1. The actual account of habital and distribution of plants of a given area.
  2. Providing information for identification of names of species found in an area.
  3. Contain information on any one taxon.
  4. Identification of plants and animals based on the similarities and dissimilarities.

Answer:

  1. Flora
  2. Manuals
  3. Monographs
  4. Taxonomical key

Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Question 12.
Define a taxon. Give some examples of taxon at different hierarchical levels.
Answer:
Each category in the taxonomical hierarchy is considered as a taxonomic unit and is known as a taxon. The taxon used in the classification of animals are kingdom, phylum, class, order, family, genus and species.

Plus One The Living World NCERT Questions and Answers

Question 1.
What do we learn from identification of individuals and populations?
Answer:
In a diverse country like India can learn following things from identification of individuals and population:

  1. Native place
  2. MotherTongue
  3. Costumes
  4. Cuisine
  5. Religion
  6. Caste
  7. Socio-economic Background

Question 2.
Can you identify the correct sequence of taxonomical categories?
(a) Species → Order → Phylum → Kingdom
(b) Genus → Species → Order → Kingdom
(c) Species → Genus → Order → Phylum
Answer:
As clear from the table in previous answer, (a) and (c) are showing the correct order.

Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Question 3.
Why are the classification system changing every now and then? ,
Answer:
In any branch of science nothing is written in concrete. Theories keep on changing as more relevant and correct theories are being discovered. In case of living beings certain species become extinct and some new species is being formed in every era.

This process of addition and deletion of species necessitates the continuous change of the classification system.

Question 4.
Define a taxon. Give some examples of taxa at different hierarchical levels.
Answer:
A taxon is a particular level of hierarchy in the system of classification of living beings. The following figure gives taxa at different hierarchical levels:

Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World 2

Plus One The Living World Multiple Choice Questions and Answers

Question 1.
Binomial nomenclature is described in the book
(a) Genera Plantarum
(b) Historia Plantarum
(c) Systema Naturae
(d) Flora Japonica
Answer:
(c) Systema Naturae

Question 2.
Which of the following is not the main criteria for five kingdom system of classification?
(a) Cell structure and thallus organization
(b) Mode of nutrition and reproduction
(c) Phylogenetic relationship.
(d) Gram staining
Answer:
(d) Gram staining

Question 3.
Two plants can be conclusively said to belong to the same species if they
(a) Can reproduce freely with each other and form seeds
(b) have more than 90 percent similar genes
(c) look similar and possess identical secondary metabolites
(d) have same number of chromosomes
Answer:
(a) Can reproduce freely with each other and form seeds

Question 4.
ICZN stands for
(a) International code of Botanical Nomenclature
(b) International code of Zoological Nomenclature
(c) international code of Viral Nomenclature
(d) International code of Zoo Nomenclature
Answer:
(b) International code of Zoological Nomenclature

Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Question 5.
In Mangifera indica L. Generic epithet is
(a) Indica
(b) Mangifera
(c) Linnaeus
(d) None of these
Answer:
(b) Mangifera

Question 6.
ICBN stands for
(a) Indian Congress of Biological Names
(b) International Code of Botanical Nomenclature
(c) International Congress of Biological Names
(d) Indian Code of Botanical Nomenclature
Answer:
(b) International Code of Botanical Nomenclature

Question 7.
Taxon is the
(a) taxonomic group of any rank
(b) procedure to assign a scientific name
(c) process of classification
(d) process by which anything is grouped into convenient categories based on characters
Answer:
(a) taxonomic group of any rank

Question 8.
Which of the following is famous for stating that “Population increases much faster than its food supply”?
(a) Fredrick Losch
(b) R Vircow
(c) T R Malthus
(d) Karl Von Baer
Answer:
(c) T R Malthus

Question 9.
Reproduction is the characteristic feature of living organisms. Which of the following can not reproduce?
(a) Amoeba and Paramecium
(b) Fungi and filamentous algae
(c) Humans and Ayes
(d) Mules and worker bees
Answer:
(d) Mules and worker bees

Question 10.
Nicotiana sylvestris flowers only during long days and N.tobacurh flowers only during short days. If raised in the laboratory under different photoperiods, they can be induced to flower at the same time and can be cross fertilized to produce self-fertile offspring. What is the best reason for considering N. sylvestris and N. tobocum to be separate species?
(a) They are physiologically distinct
(b) They are morphologically distinct
(c) They cannot interbreed in nature
(d) They are reproductively distinct
Answer:
(c) They cannot interbreed in nature

Question 11.
Which of the following book is not associated with Carolus Linnaeus, the father of Taxonomy and Nomenclature?
(a) Systema Naturae
(b) Genera Plantarum
(c) Species Plantarurn
(d) Historia Generalis Plantarum
Answer:
(d) Historia Generalis Plantarum

Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Question 12.
Philosophie Zoologique, a book written by Jean Baptiste de Lamarck is based on
(a) Survival of the fittest
(b) Natural Selection
(c) Inheritance of acquired characters
(d) Biogenetic law
Answer:
(c) Inheritance of acquired characters

Question 13.
The book “Philosophic Zoologique” was written by
(a) Lamarck
(b) Mendel
(c) Haeckel
(d) Hugo deVries
Answer:
(a) Lamarck

Question 14.
Binomial nomenclature was introduced by
(a) Linnaeus
(b) Darwin
(c) Aristotle
(d) deCandoile
Answer:
(a) Linnaeus

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

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 Vo = 30 kV = 30 × 103 V
vmax = ?
λmax = ?
1. Since kmax = eVo
So hvmax = eVo
or
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 1
= 7.24 × 1018 Hz

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 2
= 0.041 × 10-9
or λmin = 0.041 nm.

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

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. Kmax = 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 =kmax
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 3
or v0 = 0.34 V
Since \(\frac{1}{2}\) mV2max = Kmax
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 4
or vmax = 0.344 × 106 ms-1 = 344 kms-1

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

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 kmax = 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.

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

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
umax = 6.0 × 1014ms-1
v0 = ?
Since Kmax = hv – hv0
∴ v0 = v – Kmax
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 5

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.

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

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

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

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 Vs is related to the maximum kinetic energy of the emitted electrons Kmax through the relation
Kmax = eVs
0.5 eV = eVs or Vs = 0.5 V

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

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

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

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 Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 6

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

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

Question 2.
Table given below gives the work function of certain elements.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 7
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.

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

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
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 8

  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 mv2max = ev0
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 9

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

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
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 43
1. F0 = h ν0
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 10

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

3. eV0 = h(ν – ν0)
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 11

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

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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 12
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.

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

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 Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 13

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

Question 1.
The graph shows photoelectric current with anode potential.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 14
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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 15
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.

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

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
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 16

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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 17
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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 18

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

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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 19

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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 20

  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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 21

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

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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 22

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 23

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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 24

  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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 25

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.

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

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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 26

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 27

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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 28
As intensity increases photoelectric current also increases.

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

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

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.

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

Question 7.
Figure below shows variation of stopping potential (V0) with frequency(?) of incident radiations for two different metals A and B.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 29
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)
01 + eV1 = hν02 + eV2
e(V1 – V2) = h(ν02 – ν01)
\(\frac{h}{e}=\frac{V_{1}-V_{2}}{v_{02}-v_{01}}\)

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

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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 30
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)
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 31

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

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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 32
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.

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

Analysis of graph:
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 33
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°

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

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 = \(\frac{h}{P}\)
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 34
The experimentally measured wavelength is found in agreement with de-Broglie wave length. Thus wave nature of electron is confirmed.

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 35
Mass of alpha particle is more than that of proton, hence it has shortest wavelength.

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

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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 36
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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 37

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.

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

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 Hα 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.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 38

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 39

4.
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 40
Longest wavelength n1 = 2 and n2 = 3
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 41
Shortest Wavelength n1 = 2 and n2 = α
Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter - 42

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

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.

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

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

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

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

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’

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

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 – 50th percentile, 5th 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.

Plus One Economics Chapter Wise Questions and Answers Chapter 5 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

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

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
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 1

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 Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 2

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

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

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 Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 3

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.
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 4
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 5

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

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 Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 6

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.

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

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, P3, ………., 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.

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

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.
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 7
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 8
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 9
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.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 9 Small Business

Students can Download Chapter 9 Small Business Questions and Answers, Plus One Business Studies Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examination

Kerala Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 9 Small Business

Plus One Small Business Two Mark Questions and Answers

Question 1.
Define small scale industries.
Answer:
Small scale industry:
A small scale industrial undertaking is defined as one in which the investment in fixed assets of plant and machinery does not exceed rupees one crore.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 9 Small Business

Question 2.
State the features of cottage industries.
Answer:
Cottage industries:
cottage industries are characterised by certain features:

  1. These are organised by individuals, with private resources.
  2. Normally use family labour and locally available talent.
  3. The equipment used is simple.
  4. Capital investment is small.
  5. Produce simple products, normally in their own premises.
  6. Production of goods using indigenous technology.

Plus One Small Business Three Mark Questions and Answers

Question 1.
Discuss the problems faced by small scale industries. (3)
Answer:
Problems of small business:
Small businesses are faced with the following problems:

  1. Small scale industries find it difficult to get adequate finance from banks and other financial institutions.
  2. They are not able to get quality raw materials at reasonable prices.
  3. Small business is generally operated by people who may not have all the managerial skills required to run the business.
  4. Small business firms cannot afford to pay higher salaries to the employees. So productivity per employee is relatively low and employee turnover is generally high.
  5. Small business depends excessively on middlemen for marketing the products. Middlemen exploit them by paying low price and delayed payments.
  6. Small business organization uses outdated technology to produce products. So they cannot compete with global enterprises with low quality products.
  7. Due to lack of marketing skills or lack of demand, many small business firms have to operate below full capacity.
  8. Use of outdated technology results low productivity and uneconomical production.
  9. Small-scale units find it very difficult to compete .with the product of multinational companies which are comparatively very cheap and of better quality
  10. Other important problems are poor project planning, inefficient management, transportation problems, lack of power, and lack of adequate warehousing, etc.

Plus One Small Business Four Mark Questions and Answers

Question 1.
Write short note on

  1. NSIC
  2. DIC (4)

Answer:
1. National Small Industries Corporation (NSIC):
This was set up in 1955 to promote, aid and foster the growth of small scale units in India.
Functions of NSIC :

  • It supplies imported machines and raw materials to small scale industries on easy hire-purchase schemes.
  • It exports the products of small units.
  • It provides technology to small scale Industries.
  • Helps in up-gradation of technology
  • Provides advisory service
  • Developing software technology parks and technology transfer centres.

2. The District Industries Centers (DICs):
DICs were established in May 1978. District Industries Centers is the institution at the district level which provides all the services and support facilities to the entrepreneurs for setting up small and village industries. Identification of suitable schemes, preparation of feasibility reports, arranging for credit, machinery and equipment, provision of raw materials and other extension services are the main activities undertaken by these centers.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 9 Small Business

Question 2.
What measures has the government taken to solve the problem of finance and marketing in the small scale sector? (4)
Answer:
Government assistance to small Industries and Small Business:
Institutional Support:
1. National Bank for Agriculture and Rural Development (NABARD):
NABARD was set up in 1982 to promote integrated rural development.
Functions of NABARD:

  • It provides financial support to small Industries, cottage and village industries and agriculture.
  • It provides counseling and consultancy services.
  • It also organises training and development programme for rural entrepreneurs.

2. The Rural Small Business Development Centre (RSBDC):
It aims at providing management and technical support to current and prospective micro and small entrepreneurs in rural areas. RSBDC has organized several programmes on rural entrepreneurship, skill upgradation workshops, training programmes, etc.

3. National small Industries Corporation (NSIC):
This was set up in 1955 to promote, aid and foster the growth of small scale units in India.
Functions of NSIC :

  • It supplies imported machines and raw materials to small scale industries on easy hire-purchase schemes.
  • It exports the products of small units.

4. Small Industries Development Bank of India (SIDBI):
SIDBI was set up in 1980. SIDBI is the main financial Institution for financing and development of small Business in India.
Functions of SIDBI:

  • Helps SSI unit for modernization and technology upgradation by providing loan
  • Meet working capital requirements of SSI.

5. The National Commission for Enterprises in the Unorganised Sector (NCEUS):
The NCEUS was constituted in September, 2004.
The objectives of NCEUS:

  • To recommend measures for improving the productivity of small enterprises
  • To generate more employment opportunities in rural areas

6. Rural and Women Entrepreneurship Development (RWED):
RWE provides the following services:

  • Creating a business environment that encourages initiatives of rural and women entrepreneurs.
  • Enhancing the human and institutional capacities required to foster entrepreneurial dynamism and enhance productivity.
  • Providing training manuals for women entrepreneurs and training them.
  • Rendering all types of advisory services.

7. World Association for Small and Medium Enterprises (WASME):
It is the only International Non-Governmental Organisation of micro, small and medium enterprises based in India, which set up an International Committee for Rural Industrialisation. Its aim is to develop an action plan model for sustained growth of rural enterprises.

There are several schemes to promote the non-farm sector initiated by the Government of India, i.e. IRDP, PMRY, TRYSEM, JRY, Development of Women and Children in Rural Areas (DWCRA) etc.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 9 Small Business

8. Scheme of Fund for Regeneration of Traditional Industries (SFURTI):
To make the traditional industries more productive and competitive, the Central Government set up a fund with₹100 crores.
The main objectives of the scheme are as follows:

  • To develop clusters of traditional industries in various parts of the country
  • To make traditional industries competitive, profitable and sustainable
  • To create employment opportunities in traditional industries.

9. The District Industries Centers (DICs):
DICs were established in May 1978. District Industries Centers is the institution at the district level which provides all the services and support facilities to the entrepreneurs for setting up small and village industries.

Identification of suitable schemes, preparation of feasibility reports, arranging for credit, machinery and equipment, provision of raw materials and other extension services are the main activities undertaken by these centers.

Plus One Small Business Five Mark Questions and Answers

Question 1.
What are the incentives provided by the Government for industries in backward and hilly areas? (5)
Answer:
Incentives:
The incentives offered by the government to develop backward areas are:

  1. Availability of land at concessional rate.
  2. Supply power at a concessional rate of 50% or exempt such units from payment in the initial years.
  3. Water is supplied on a no-profit, no-loss basis or with 50 percent concession or exemption from water charges for a period of 5 years.
  4. All union territories, industries are exempted from sales tax,
  5. Most states have abolished octroi.
  6. Units located in backward areas get scarce raw materials at concessional rates.
  7. Subsidy of 10-15% for building capital asset. Loans are offered at concessional rates.
  8. Some states encourage setting up of industrial estates in backward areas.
  9. Exemption from paying taxes for 5 or 10 years is given to industries established in backward, hilly and tribal areas.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 9 Small Business

Question 2.
Small scale industrial sector has an important role in Indian economy. To meet the financial and developmental needs of small business in India a separate bank had been set up. (5)

  1. Give the name of the bank.
  2. What are its major functions?

Answer:
1. SIDBI: Small Industries Development Bank of India (SIDBI). SIDBI was set up in 1980. SIDBI is the main financial Institution for financing and development of small Business in India.

2. Functions of SIDBI:

  • Helps SSI unit for modernisation and technology upgradation by providing loan
  • Meet working capital requirements of SSI.
  • Provides means forthe rehabilitation of sick units
  • Provides service like leasing, hire purchase, venture capital financing etc.
  • Provide equity support to small entrepreneurs.

Plus One Small Business Six Mark Questions and Answers

Question 1.
Describe the role of small business in rural India. (6)
Answer:
Role of small business in Rural India:

  1. Cottage and rural industries provide employment opportunities in the rural areas especially for the traditional artisans and the weaker sections of society.
  2. It prevents migration of rural population to urban areas in search of employment.
  3. Small business helps to eradicate poverty, income inequalities, etc in rural area.
  4. Small scale industries are powerful instrument for the accelerated industrial growth and creating productive employment in rural and backward areas.

Plus One Small Business Eight Mark Questions and Answers

Question 1.
How do small scale industries contribute to the socio-economic development of India? (8)
Answer:
Role of small business in India:

  1. Small industries are labour intensive and less capital intensive. They generate more number of employment opportunities.
  2. The share of product from small industries is 45% of total export from India. So it earn valuable foreign exchange.
  3. Small scale Industries produces a wide variety of goods e.g. readymade garments, stationery, soaps, leather goods, plastic and rubber goods.
  4. The contribution of small industries to the balanced regional development of the country is very significant.
  5. Small industries provide ample opportunity for entrepreneurship.
  6. It enjoys the advantage of low cost of production because they used local resources in their product.
  7. Due to the small size of the organisation, quick and timely decisions can be taken without consulting many people.
  8. Small industries are best suited for the products which are designed according to the taste, needs and preferences of the customers.
  9. Small industries maintain good personal relations with both customers and employees.

Plus One Economics Chapter Wise Questions and Answers Chapter 1 Statistics for Economics Introduction

Students can Download Chapter 1 Statistics for Economics Introduction 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 1 Statistics for Economics Introduction

Plus One Economics Statistics for Economics Introduction One Mark Questions and Answers

Question 1.
The word statistics is used as ……………
(a) Singular
(b) Plural
(c) Both (a) and (b)
(d) None of (a) and (b)
Answer:
(c) Both (a) and (b)

Question 2.
Statistics deals with …………..
(a) Qualitative data
(b) Quantitative data
(c) Both (a) and (b)
(d) None of (a) and (b)
Answer:
(b) Quantitative data

Question 3.
The statistical method to analyze the data maybe ……….
(a) Explanatory
(b) Non-explanatory
(c) Both (a) and (b)
(d) None of (a) and (b)
Answer:
(a) Explanatory

Plus One Economics Chapter Wise Questions and Answers Chapter 1 Statistics for Economics Introduction

Question 4.
Who is known as the father of statistics?
Answer:
Grottfried Achenwale is known as the father of statistics.

Plus One Economics Statistics for Economics Introduction Two Mark Questions and Answers

Question 1.
Distinguish the singular and plural senses of statistics.
Answer:

  • singular sense statistics is the science of collecting, classifying and using statistics or a statistical fact.
  • In plural sense, it means numerical facts systematically collected or it is data.

Question 2.
Classify the following data into qualitative and quantitative.
Height, weight, marks scored in an exam, truth, honesty, beauty, credibility, income.
Answer:

  • Quantitative data: Income, height, weight, marks scored in an exam.
  • Qualitative data: Truth, honesty, beauty, credibility.

Question 3.
Mention a few examples each for quantitative and qualitative data.
Answer:
1. Quantitative data

  • height
  • weight
  • marks
  • price
  • income

Qualitative data

  • colour
  • intelligence
  • attitude
  • honesty
  • beauty

Plus One Economics Chapter Wise Questions and Answers Chapter 1 Statistics for Economics Introduction

Question 4.
Write some examples for distrust of statistics.
Answer:
Distrust of statistics means lack of confidence in statistical statements and statistical methods. Examples of such distrust are:

  • People keep blind faith in statistics. So they are not careful while accepting statistical figures.
  • People manipulate statistical data to present a wrong idea of their interest.

Plus One Economics Statistics for Economics Introduction Three Mark Questions and Answers

Question 1.
Give the limitations of statistics.
Answer:
The limitations of statistics are given below.

  • Statistical laws are true only on average.
  • Statistics can be misused.
  • Statistics deals only with quantitative data.
  • Statistical results lack mathematical accuracy.
  • Statistical facts are collected for a predetermined purpose.

Question 2.
Identify the data from the following statement.

  1. Indian economy grew at an average rate of 7% after the reforms.
  2. RBI increased the bank rate to 6.5.
  3. The percentage of people below poverty line in India is 26.

Answer:

  1. Average growth rate – 7%
  2. Bank rate – 6.5
  3. Percentage of people below poverty line – 26%

Question 3.
Mark the following statements as true or false

  1. statistics can only deal with quantitative data
  2. statistics solves economic problems
  3. statistics is of no use to economics without data.

Answer:

  1. True
  2. True
  3. True

Plus One Economics Chapter Wise Questions and Answers Chapter 1 Statistics for Economics Introduction

Question 4.
Explain how statistics is helpful in the formulation of economic planning.
Answer:
Statistics is the most important tool in economic planning. Economic planning is the best use of national resources, both in planning, without statistics is a leap in the dark. Every phase in planning, drawing a plan, execution, and review is based on statistics. The success of plan depends upon sufficient and accurate statistical data available at all the stages.

Question 5.
Make a list of economic activities found in your daily life.
Answer:
Some of the economic activities found in our daily life are:

  • Teachers teaching in the school.
  • People working in a factory.
  • Doctor treating a patient.
  • Milkman working in a dairy farm.

Question 6.
‘Statistics plays a vital role in policymaking’. Illustrate with examples.
Answer:
The government and the policymakers use statistical data to formulate suitable policies of economic development. No analysis of a problem would be possible without the availability of data on various factors underlying an economic problem. For example, if the government wants to make policy to solve the problem of unemployment and poverty, reliable data are required for it.

The number of persons who are waiting for a job, unemployed, the number of illiterate people, those living below the poverty line, etc are required. If the government wants to help the people affected by the tsunami earthquakes, the bind flu dangers then the data should be collected and put together about disaster’s cost systematically and correctly.

Modern economists include learning the basic skills involved in making useful studies for measuring poverty, how incomes are distributed, how earning opportunities are related to education, how environmental disasters affect our lives and so on.

Question 7.
Write the important functions of statistics.
Answer:
The important functions of statistics are the following:
1. It enables an economist to present economic facts in a precise and definite form that helps in proper comprehension of what is stated.

2. Statistics also helps in condensing the mass of data into a few numerical measures (such as mean, variance, etc., about which you will learn later). These numerical measures help summarize data.

3. Statistics is used in finding relationships between different economic factors. Statistics find economic relationships using data and verifies them.

4. Statistical tools are used in the prediction of future trends. The economist might be interested in predicting the changes in one economic factor due to the changes in another factor.

5. Statistical methods help formulate appropriate economic policies that solve economic problems.

Plus One Economics Chapter Wise Questions and Answers Chapter 1 Statistics for Economics Introduction

Question 8.
The knowledge of statistics is helpful in several ways. List out the beneficiaries of statistics with examples for each.
Answer:
1. The applications of statistics are so numerous and ever interesting. It can be applied to any field of study whether it be trade, sociology, medicine, etc. Its application in economics, business, research, etc are discussed below.

2. In economics variables like income, consumption, input, saving, etc. are systematically and scientifically analyzed and studied with the tools of statistics. Therefore, it is highly useful in economics.

3. With the help of statistical methods abundant quantitative information can be obtained which can help immensely in formulating suitable policies. Statistics are so important to the state that government in most countries is the biggest collector and user of statistical data.

4. There is hardly any research work today that one can complete without statistical data and statistical methods. The statistical tools are highly needed for research.

5. Thus in this respect, the knowledge of statistics is helpful to society in different ways.

Plus One Economics Statistics for Economics Introduction Four Mark Questions and Answers

Question 1.
List out examples for quantitative data and qualitative data.
Answer:
A. Some examples of quantitative data in daily life are:

  1. family income
  2. family expenditure
  3. monthly saving of a family
  4. expenditure on education
  5. expenditure on medical treatment
  6. marks scored in examination

B. Some examples of qualitative data in daily life are:

  1. nationality
  2. sex
  3. religion
  4. caste
  5. colour
  6. educational qualification

Plus One Economics Chapter Wise Questions and Answers Chapter 1 Statistics for Economics Introduction

Question 2.
Distinguish between quantitative and qualitative data.
Answer:
1. Quantitative data are those data which can be measured in numerical terms.
Eg: height, weight, etc.

2. On the other hand, qualitative data are those facts that cannot be numerically measured.
Eg: intelligence, beauty, etc.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 6 Social Responsibilities of Business and Business Ethics

Students can Download Chapter 6 Social Responsibilities of Business and Business Ethics Questions and Answers, Plus One Business Studies Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examination

Kerala Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 6 Social Responsibilities of Business and Business Ethics

Plus One Social Responsibilities of Business and Business Ethics One Mark Questions and Answers

Question 1.
Identify which of the following is not a social responsibility of business _________
1. Offering goods and services at reasonable price.
2. Keeping environment pollution free
3. Giving reasonable pay to workers
4. Help in nation building activities
Answer:
3. Giving reasonable pay to workers

Question 2.
State Bank of India gives scholarship to bright students each year. This is a part of their __________
1. Sales promotion
2. Social responsibility
3. Charity work
4. Business ethics
Answer:
2. Social responsibility

Question 3.
Application of general ethics and principles in business practice is ____________
Answer:
Business ethics

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 6 Social Responsibilities of Business and Business Ethics

Question 4.
It is the belief in what is right, proper and just. Identify it.
Answer:
Business ethics

Question 5.
______________ is the code of conduct followed and performed by businessmen.
Answer:
Business ethics

Question 6.
Which among the following is not a case for an ethical activity.
1. child labour
2. equal pay for equal work
3. dumbing of waste in river
4. high usage of preservatives in products.
Answer:
2. Equal pay for equal work

Plus One Social Responsibilities of Business and Business Ethics Two Mark Questions and Answers

Question 1.
Roy and Rejitha joined in a factory as peons on the same day. Both of them are doing 8 hrs work a day. Roy is paid ₹100 while Rejitha ₹80 per day. Do you think the owner of the factory is fair to Rejitha? If not, give your opinion. (2)
Answer:
No. The owner of the factory is not fair to Rejitha. It is the social and moral responsibility of the business to offer equal pay for equal work without considering any caste, sex, etc.

Question 2.
Explain the need for pollution control. (2)
Answer:
Need for Pollution Control:

1. Reduction of health hazard:
Pollution control measures can check diseases like cancer, heart attack & lung complications and support a healthy life on earth.

2. Reduced Risk of Liability:
It is a sound business policy to install pollution control devices in its premises to reduce the risk of liability of paying compensation to the affected people.

3. Cost Saving:
An effective pollution control programme is needed to save costs of operating business.

4. Improved Public Image:
A firm that adopts pollution control measures enjoys a good reputation as a socially responsible enterprise.

5. Other social benefits:
Pollution control results in many other benefits like clearer visibility, cleaner buildings, better quality of life, and the availability of natural products in a purer form.

Plus One Social Responsibilities of Business and Business Ethics Three Mark Questions and Answers

Question 1.
George is working in a manufacturing industry. The manager compels him to do 10hrs work continuously without giving any additional benefits. Do you agree with the manager? Give your suggestions. (3)
Answer:
No. We cannot agree with the decision of the manager. It is against the social responsibility of the business. George has to work only 8 hours a day. Compelling to do 10 (without any additional benefits) hours is a violation of human right. Managers cannot ask a worker to do work continuously without giving remuneration. It is also a violation of human rights.

Plus One Social Responsibilities of Business and Business Ethics Four Mark Questions and Answers

Question 1.
Match column A with columns B and C
Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 6 Social Responsibilities of Business and Business Ethics 1
Answer:

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 6 Social Responsibilities of Business and Business Ethics 2

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 6 Social Responsibilities of Business and Business Ethics

Question 2.
Jaleel, owner of a fish stall, usually sells damaged and decayed fish by pouring chemicals, ice, along with sea sand for creating an impression in the minds of consumers that it is newly caught. Is it justifiable from the point of view of business ethics? How will it affect the reputation of the business? (4)
Answer:
Business ethics refer to the socially determined moral principles which should govern business activities.
A few examples of business ethics are

  • charging fair prices from customers
  • providing quality products
  • using fair weights for measurement of commodities
  • giving fair treatment to workers & earning only reasonable profits.

Ethical business behaviour improves public image, earns people’s confidence and trust, and leads to greater success. Selling damaged and decayed fish is unethical.

It will affect the reputation of the business and finally it leads to business loss. He must not goods of low quality for personal gain. No business can make progress without adopting ethical values.

Plus One Social Responsibilities of Business and Business Ethics Five Mark Questions and Answers

Question 1.
Anchu, a stationery dealer distributed good quality products at low price by suffering huge losses, in order to eliminate the competition from Chinju, another stationery dealer, in the same locality. When Chinju stopped her business, Anchu started to supply low quality products at high price. (5)

  1. Can you justify Anchu’s action from point of view of the society?
  2. Give a few undesirable practices which a business follow in a real life situation.

Answer:
1. No. We cannot justify Anchu’s action. It is against business ethics. Business ethics refers to moral values which should be followed in business activities.

2. Examples of unethical practices are

  • Providing low quality products at high prices.
  • Making misleading advertisements.
  • Creating artificial scarcity to increase demand.
  • Keeping false accounts to avoid paying taxes.
  • Acceptance of bribe for personal benefits.
  • Manufacture and sale of adulterated goods.
  • Polluting the environment.
  • Exploiting workers.

Plus One Social Responsibilities of Business and Business Ethics Six Mark Questions and Answers

Question 1.
What steps can an enterprise take to protect the environment from the dangers of pollution? (6)
Answer:
Role of Business in Environmental Protection:

1. A definite commitment by top management to create a work culture for environmental protection

2. Ensuring that commitment of environmental protection is shared throughout the enterprise by all divisions and employees.

3. Developing clear cut policies and programmes for purchasing good quality raw materials, introducing superior technology, using scientific techniques of disposal of waste and developing employee skills for pollution control

4. Complying with the laws and regulations enacted by the Government for prevention of pollution.

5. Participation in government programs relating to management of hazardous substances, cleaning up of polluted rivers, plantation of trees, and checking deforestation.

6. Periodical assessment of pollution control programmes in terms of costs and benefits with a view to improve them.

7. Arranging educational workshops and training materials to share technical information with everyone involved in pollution control.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 6 Social Responsibilities of Business and Business Ethics

Question 2.
Explain the elements of business ethics. (6)
Answer:
Business Ethics:
Ethics is concerned with what is right and what is wrong in human behavior. Business ethics refer to the socially determined moral principles which should govern business activities.

Business ethics is the code of conduct followed and performed by every business. Ethical business behavior improves public image earn’s public confidence and leads to greater success.
Examples of Business Ethics:

  1. Charging fair prices from customers
  2. Using fair weights for measurement of commodities
  3. Giving fair treatment to workers
  4. Earning reasonable profits.
  5. Avoiding adulteration, hoarding etc.
  6. Using environmentally friendly products

Elements of Business Ethics:

1. Top management commitment:
The Chief Executive Officer and higher level managers must give continuous leadership for developing and upholding the moral values of the organisation.

2. Publication of a Code:
‘Code’ refers to a formal written document of the principles, values and standards that guide a firm’s actions. It may cover the areas of fundamental honesty and adherence to laws, product safety and quality, health and safety in the workplace, etc.

3. Establishment of Compliance Mechanism:
A suitable mechanism should be developed to comply with the ethical standards of the enterprise.

4. Employees Involvement:
To make ethical business a reality, employees at all levels must be involved.

5. Measuring Results:
Ethical results must be verified and audited that how far work is being carried according to ethical standards.

Plus One Social Responsibilities of Business and Business Ethics Eight Mark Questions and Answers

Question 1.
Build up arguments for and against social responsibilities. (8)
Answer:
Arguments in favour of Social Responsibility:

1. Justification for Existence and Growth:
The prosperity and growth of business is possible only through continuous service to society.

2. Long term Interest of the firm:
A firm can improve its image and builds goodwill in the long run when its highest goal is to serve the society.

3. Avoidance of government regulations:
Business can avoid the problem of government regulations by voluntarily assuming social responsibilities.

4. Maintenance of society:
Law alone can’t help out people with all the difficulties they face. A socially responsible business can contribute something for social peace and harmony.

5. Availability of resources with business:
Business has valuable financial and human resources which can be effectively used for solving problems of the society.

6. Better environment for doing business:
Social responsibility creates better environment for business operations as it improves quality of life and standard of living of the people.

7. Contribution to social problems:
Some of the social problems have been created by business firms themselves such as pollution, unsafe workplaces, discrimination, etc. Therefore, it is the moral obligation of business to solve such social problems.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 6 Social Responsibilities of Business and Business Ethics

Arguments Against Social Responsibility:

1. Violation of profit maximization objective:
According to this argument, business exists only for the maximum profit to its shareholders and do not have responsibility to the society as a whole.

2. Burden on consumers:
Involvement of business in social responsibilities involve a lot of expenditure which will ultimately be borne by the customers.

3. Lack of Social Skills:
The business firms and managers are not expert to tackle the social problems like poverty, overpopulation etc.

4. Lack of public support:
Business cannot fulfill social responsibility because of lack of public confidence & cooperation.

Question 2.
Mr. Biju, a readymade garment manufacturer, is of the view that “different interest groups always claim some sort of protection from the business as such. These interest groups, no doubt consist of persons connected with business in one way or other, such as shareholders, employees, society and so on”. Do you agree with this statement? Explain. (8) 

OR

“Business is a social institution having responsibilities towards different interest group.” Do you agree. Explain the responsibility towards each group.
Answer:
Social Responsibility towards different interest groups

1. Responsibility towards share holders or owners:

  1. Provide a fair and regular return on the investment of shareholders.
  2. Provide regular and accurate information on the financial position of the firm.
  3. To ensure the safety of their investment.

2. Responsibility Towards the workers:

  1. Providing fair wages
  2. Providing good working conditions and welfare amenities.
  3. Respect democratic rights of workers to form
    unions.

3. Responsibility toward consumers:

  1. Supply right quality and quantity of goods and services at reasonable prices.
  2. Avoiding unfair trade practices like adulteration, poor quality, misleading advertisement etc.
  3. Inform them about new products, its features, uses and other matters relating to the products.
  4. To handle the customers grievance promptly.

4. Responsibility Towards Government:

  1. Respect the laws of the country
  2. Pay taxes regularly and honestly.
  3. act according to the well accepted values of the society.

5. Responsibility towards community:

  1. Make employment opportunities
  2. Protect the environment from pollution.
  3. To uplift the weaker sections of society

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 6 Social Responsibilities of Business and Business Ethics

Question 3.
Prosperity and higher living standards are the hallmarks of industrial development for any nation. But during this course, it equally causes damages to the environment. (8)

  1. Explain the different types of pollution.
  2. Substantiate the above statement with regard to various types of pollution.

Answer:
Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 6 Social Responsibilities of Business and Business Ethics 3

Business & Environmental Protection Causes of Pollution: Many industrial organisations have been responsible for causing air, water, land and noise pollution.

1. Air Pollution:
Air pollution is mainly due to Carbon monoxide emitted by automobiles and smoke and other chemicals from manufacturing plants. It has created a hole in the ozone layer leading to global warming.

2. Water pollution:
Water becomes polluted primarily from chemical and waste dumping. It has led to the death of several animals and posed a serious problem to human life.

3. Land Pollution:
Dumping of toxic wastes reduces the quality of land and making it unfit for agriculture or plantation.

4. Noise Pollution:
Noise caused by the running of factories and vehicles create a serious health hazard such as loss of hearing, malfunctioning of the heart and mental disorders.

Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 6 Social Responsibilities of Business and Business Ethics

Question 4.
Explain the role of business in environmental protection. (8)
Answer:
Role of Business in Environmental Protection:

  1. A definite commitment by top management to create a work culture for environmental protection
  2. Ensuring that commitment of environmental protection is shared throughout the enterprise by all divisions and employees.
  3. Developing clear cut policies and programmes for purchasing good quality raw materials, introducing superior technology, using scientific techniques of disposal of waste and developing employee skills for pollution control
  4. Complying with the laws and regulations enacted by the Government for prevention of pollution.
  5. Participation in government programs relating to management of hazardous substances, cleaning up of polluted rivers, plantation of trees, and checking deforestation.
  6. Periodical assessment of pollution control programmes in terms of costs and benefits with a view to improve them.
  7. Arranging educational workshops and training materials to share technical information with everyone involved in pollution control.

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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) = \(\frac{P(E \cap F)}{P(E)}=\frac{\frac{1}{18}}{\frac{1}{2}}=\frac{1}{9}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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) = \(\frac{1}{2} \times \frac{25}{51}=\frac{25}{102}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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)
Plus Two Maths Probability 3 Mark Questions and Answers 1

2. P(A/B)
Plus Two Maths Probability 3 Mark Questions and Answers 2

3. P(B/A)
Plus Two Maths Probability 3 Mark Questions and Answers 3

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) = \(\frac{1}{4}\) ⇒ \(P(\bar{A} \cup \bar{B})=\frac{1}{4}\)
Plus Two Maths Probability 3 Mark Questions and Answers 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.

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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)
Plus Two Maths Probability 3 Mark Questions and Answers 5

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.

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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)
Plus Two Maths Probability 3 Mark Questions and Answers 6

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)
Plus Two Maths Probability 3 Mark Questions and Answers 7

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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
Plus Two Maths Probability 3 Mark Questions and Answers 8

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) = 10Cxq10-xpx = 10Cx\(\left[\frac{9}{10}\right]^{10-x}\left[\frac{1}{10}\right]^{x}\)
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.
Plus Two Maths Probability 3 Mark Questions and Answers 9

3.
Plus Two Maths Probability 3 Mark Questions and Answers 10

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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.

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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)
Plus Two Maths Probability 4 Mark Questions and Answers 11

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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)
Plus Two Maths Probability 4 Mark Questions and Answers 12

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)
Plus Two Maths Probability 4 Mark Questions and Answers 13

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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)
Plus Two Maths Probability 4 Mark Questions and Answers 14

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
Plus Two Maths Probability 4 Mark Questions and Answers 15
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)
Plus Two Maths Probability 4 Mark Questions and Answers 16

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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)
Plus Two Maths Probability 4 Mark Questions and Answers 17

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)
Plus Two Maths Probability 4 Mark Questions and Answers 18

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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
Plus Two Maths Probability 4 Mark Questions and Answers 19

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)
Plus Two Maths Probability 4 Mark Questions and Answers 20
P(X = 2) = P(2 doublet and 1 non-doublets)
Plus Two Maths Probability 4 Mark Questions and Answers 21
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
Plus Two Maths Probability 4 Mark Questions and Answers 22

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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)
Plus Two Maths Probability 4 Mark Questions and Answers 23
P(X=2) = P(2white, 1 red balls)
= P(WWR) + P(RWW) + P(WRW)
Plus Two Maths Probability 4 Mark Questions and Answers 24
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
Plus Two Maths Probability 4 Mark Questions and Answers 25

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))
Plus Two Maths Probability 4 Mark Questions and Answers 26
P(X = 2) = P(six, six) = \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\)
The required Probability Distribution is
Plus Two Maths Probability 4 Mark Questions and Answers 27
Variance = σ2 = E(X2) – [E(X)]2
Plus Two Maths Probability 4 Mark Questions and Answers 28

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

Question 16.
A random variable X has the following probability distribution
Plus Two Maths Probability 4 Mark Questions and Answers 29
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
Plus Two Maths Probability 4 Mark Questions and Answers 30

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)
Plus Two Maths Probability 4 Mark Questions and Answers 31
P(Problem is solved)
= P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Plus Two Maths Probability 4 Mark Questions and Answers 32

(iii) P(Exactly one of them solve)
Plus Two Maths Probability 4 Mark Questions and Answers 33

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)
\(=\frac{4}{52} \cdot \frac{3}{51} \cdot \frac{4}{50}=\frac{2}{5525}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

Question 2.
60 shirts of different colours are on sale. If one shirt is chosen at random.
Plus Two Maths Probability 4 Mark Questions and Answers 34

  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)
Plus Two Maths Probability 4 Mark Questions and Answers 35

4.
Plus Two Maths Probability 4 Mark Questions and Answers 36
∴ 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.

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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
Plus Two Maths Probability 4 Mark Questions and Answers 37

(iii) P (None of them will be selected)
Plus Two Maths Probability 4 Mark Questions and Answers 38

Question 5.
(i) Find P(A∩B) if A and B are independent events with P(A) = \(\frac{1}{5}\) and P(B) = \(\frac{5}{8}\)
Plus Two Maths Probability 4 Mark Questions and Answers 39
(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)
Plus Two Maths Probability 4 Mark Questions and Answers 40
∴ A and B are independent events.

(iii)
Plus Two Maths Probability 4 Mark Questions and Answers 41
Probability of exactly one of them solves the problem = P(A)P(B’) + P(B)P(A’)
Plus Two Maths Probability 4 Mark Questions and Answers 42

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

Question 6.
(i) A set of events E1 + E2,…….En are said to be a partition of the Sample Space, then which of the following conditions is always not true (1)
(a) E1∪ E2 ∪……..∪ En = S,
(b) E1 ∩ En = Φ,
(c) P(E1) > 0,
(d) P(E1) ≥ P(En)
(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(En).

(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
Plus Two Maths Probability 4 Mark Questions and Answers 43

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
Plus Two Maths Probability 4 Mark Questions and Answers 44

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
Plus Two Maths Probability 4 Mark Questions and Answers 45
When a diamond card is lost, there are 12 diamond cards in 52 cards. Then
Plus Two Maths Probability 4 Mark Questions and Answers 46
When a non diamond card is lost, there are 13 diamond cards in 51 cards. Then
Plus Two Maths Probability 4 Mark Questions and Answers 47

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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
Plus Two Maths Probability 4 Mark Questions and Answers 48

(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)
Plus Two Maths Probability 4 Mark Questions and Answers 49
P(X = 2) = P(TT) = P(T).P(T) = \(\frac{1}{16}\)
Then the Probability distribution is
Plus Two Maths Probability 4 Mark Questions and Answers 50

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) = nCxqn-x Px
Plus Two Maths Probability 4 Mark Questions and Answers 51
1. P(x = 6)
Plus Two Maths Probability 4 Mark Questions and Answers 52

2. P(at least 6 heads) = P(X ≥ 6)
= P(X = 6) + P(X = 1)+P(X = 8)
Plus Two Maths Probability 4 Mark Questions and Answers 53

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)
Plus Two Maths Probability 4 Mark Questions and Answers 54
Plus Two Maths Probability 4 Mark Questions and Answers 55

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

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
Plus Two Maths Probability 4 Mark Questions and Answers 56
1. P(all 5 are spades) = P(X = 5)
Plus Two Maths Probability 4 Mark Questions and Answers 57

2. P(3 are spades) = P(X = 3)
Plus Two Maths Probability 4 Mark Questions and Answers 58

3. P(non-spade) = P(X = 0) =
Plus Two Maths Probability 4 Mark Questions and Answers 59

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
Plus Two Maths Probability 4 Mark Questions and Answers 60

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
Plus Two Maths Probability 4 Mark Questions and Answers 61

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

Question 12.
(i) If A and B are two events such that A ⊂ B and P(A) ≠ 0 then P(A/B) is (1)
Plus Two Maths Probability 4 Mark Questions and Answers 62
(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)
Plus Two Maths Probability 4 Mark Questions and Answers 63
Plus Two Maths Probability 4 Mark Questions and Answers 64

Question 13.
Consider the following probability distribution of a random, variable X.
Plus Two Maths Probability 4 Mark Questions and Answers 65

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

Answer:
(i) We have; Σpi = 1
Plus Two Maths Probability 4 Mark Questions and Answers 66

(ii)
Plus Two Maths Probability 4 Mark Questions and Answers 67

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

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

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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.

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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

Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img1
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 qD = 60 – P for 0 = P = 60 qS = 30 + P for P > 10 Calculate equilibrium price.
Answer:
The demand and supply functions are given as
qD= 60 – P for 0 = P = 60
qS = 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.

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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.
qD = 200 – P for 0 = P = 200
qS = 120 + P for P > 10
Answer:
We find equilibrium price by equating market demand function and market supply functions as shown below.
qD = qS
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.
qD= 200 – P
qD =200 – 40 = 160
Therefore equilibrium price is ₹40 and equilibrium quantity is 160.

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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, qD (P*) = qS (P*) where P* denotes the equilibrium price and qD (P*) and qS (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
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img2
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)

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Question 9.
Calculate equilibrium price and quantity based on the following information.
qd= 400 – P (1)
qs= 240 + 3 (p – 4) (2)
Answer:
At equilibrium,
qd = qs
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,
qd =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.

Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img3
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.

Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img4
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
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img5

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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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img6

  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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img7
Answer:
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img8

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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?

Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img9
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.

HSSLive.Guru

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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img10
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.

Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img11

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

HSSLive.Guru

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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img12

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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img13

Question 11.
Suppose we have two equations, one for demand and other for supply.
Demand equation: Qxd = 100 – 10Px
Supply equation : Qxs = 60 + 10Px

  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
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img14

HSSLive.Guru

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 Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img16

Plus Two Economics Market Equilibrium Eight Mark Questions and Answers

Question 1.
Observe the following table.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img17

  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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img18
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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img19

HSSLive.Guru

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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img20

2. When demand curve shift to left (decrease in demand), both equilibrium quantity and equilibrium price falls. This is shown in the diagram.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img21

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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img22

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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img23

Question 3.
Suppose the demand and supply curves of salt are given by. qD = 1000 – P qS = 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 qS = 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
qD = 1000 – P
qS = 700 + 2P
For equilibrium
qD = qS
1000 – P = 700 + 2P
1000 – 700 = 3 P
3P = 300
P = 300/3 = 100
Put the value of P in supply equation
qS = 700 + 2P
qS = 700 + 2×100
qS =700 + 200 = 900
Therefore the equilibrium price = ? 100 and the equilibrium quantity is = 900 units

HSSLive.Guru

2. For equilibrium
qD = qS
1000 – P = 400 + 2P
1000 – 400 = 3P
600 = 3 P
P = 600 / 3 = 200
Put the value of P in demand equation
QD= 1000 – P
QD = 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. qD= 1000 – P
qS= 700 + 2P
When ₹3 as tax is imposed on sale of salt the new demand and supply function will change
qD= 1000 – (P + 3)
qS= 700 + (2P +3)
In part A equilibrium price was ₹100 which goes up to ₹103 with imposition of tax
qD = 1000 – (100 + 3) = 1000 – 103 =897
qS = 700 + (2P + 3)
= 700 + 2(100 + 3)
= 700 + 2×103
= 700 + 206 = 906
qD < qS
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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img25
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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img26
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.

Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img27

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.
Plus Two Economics Chapter Wise Questions and Answers Chapter 5 Market Equilibrium img28
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

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Students can Download Chapter 3 Components of the Computer System Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Plus One Components of the Computer System One Mark Questions and Answers

Question 1.
The Tangible parts of a computer is _________
Answer:
Hardware

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 2.
The instructions that tell the hardware to perform a task is __________
Answer:
Software

Question 3.
The brain of the computer is __________
Answer:
CPU

Question 4.
CPU means _________
Answer:
Central Processing Unit

Question 5.
ALU is
Answer:
Arithmetic Logic Unit

Question 6.
I am an input device. I can read text or picture on paper and translate into computer usable form. Who am I?
Answer:
Scanner

Question 7.
Odd man out.
(a) Trackball
(b) Joy Stick
(c) Scanner
(d) LCD
Answer:
(d) LCD. It is an output device. Others are input device

Question 8.
Odd man out.
(a) Inkjet
(b) Laser
(c) Dot Matrix Printer
(d) Thermal
Answer:
(c) Dot Matrix Printer. It is impact printer others are non impact printers.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 9.
The storage capacity of a CD is ________
(a) 1.44 MB
(b) 700 GB
(c) 700 MB
(d) 650 GB
Answer:
(c) 700 MB

Question 10.
__________ sheet is used to write answers in Kerala Entrance Exam.
Answer:
OMR Sheet

Question 11.
ABC textile uses reader to input the item and its price.
Answer:
Bar code reader

Question 12.
_________ device senses the presence or absence of a pencil mark.
Answer:
OMR

Question 13.
Name any two printing devices.
Answer:
Mouse and touchpad

Question 14.
__________ is a device that draws pictures on a paper.
Answer:
Plotter

Question 15.
Primary memory is classified into two. What are they?
Answer:
RAM and ROM

Question 16.
The storage capacity of a DVD is _________.
Answer:
4.7 GB

Question 17.
Winzip is a ___________ utility.
Answer:
Compression utility

Question 18.
Win Rar is a ________ utility.
Answer:
Compression utility

Question 19.
Most commonly used input device is _________________
Answer:
Keyboard or mouse

Question 20.
Govt, decided to conduct a test that contains all objective type questions. Which device is most suitable for evaluation.
Answer:
OMR

Question 21.
__________ is used mostly for computer games.
Answer:
Joy stick

Question 22.
I am an input device. I have a stick with two buttons called triggers on the top. Who am I?
Answer:
Joy stick

Question 23.
I am a pointing device. I am a stationary device. Who am I?
Answer:
Touchpad

Question 24.
In portable computers which pointing device is suitable?
Answer:
Touchpad

Question 25.
State true or false.
Hard copy devices are very slow compared to soft copy devices.
Answer:
True

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 26.
________ is also called firrfi ware.
Answer:
ROM

Question 27.
__________ is acts as an interface between user and computer.
Answer:
Operating system

Question 28.
_________ is used to create and modify any type of document.
Answer:
Word Processor

Question 29.
_________ is a set of programs that manage the database.
Answer:
DBMS

Question 30.
__________ package contains rows & columns.
Answer:
Spread sheet

Question 31.
_________ is a presentation package.
Answer:
Power point

Question 32.
Your computer teacher asked you to explain the project work done by your group. Which package will help you to do so?
Answer:
Power Point

Question 33.
___________ is a DTP Package.
(a) Excel
(b) Powerpoint
(c) PageMaker
(d) None of these
Answer:
(c) PageMaker

Question 34.
DTP is ________
Answer:
Desk Top Publishing

Question 35.
_______ is designed to help computer for its smooth functioning.
Answer:
Utilities

Question 36.
Customised software is also called _________
Answer:
Tailor made software

Question 37.
__________ S/W is used to remove virus from a computer.
Answer:
Anti Virus S/W

Question 38.
Name any Antivirus S/W.
Answer:
Norton Antivirus, McAfee, Avira, AVG

Question 39.
Name the two classifications of output devices.
Answer:
Hard copy and soft copy

Question 40.
Name the two classifications of printers.
Answer:
Impact printer and non-impact printer

Question 41.
_______ gas is used in plasma panels
Answer:
(a) Oxygen
(b) Neon
(c) Mercury
(d) helium Neon
Answer:
(b) Neon

Question 42.
__________ printers are used in fax machine.
Answer:
Thermal Printers

Question 43.
_________ is read/write memory.
Answer:
RAM

Question 44.
________ is of volatile memory
(a) ROM
(b) RAM
(c) CD
(d) DVD
Answer:
(b) RAM

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 45.
From the following which is expensive?
(a) CD
(b) DVD
(c) HDD
(d) RAM
Answer:
(d) RAM

Question 46.
You want to input your photograph into computer. Which device is used for this?
(a) Scanner
(b) Mouse
(c) OMR
(d) OCR
Answer:
(a) Scanner

Question 47.
The primary memory which is commonly used in electronic billing machines to store price of products is
(a) P R O M
(b) E P R O M
(c) E E P R O M
(d) None of these
Answer:
(b) E P R O M

Question 48.
Name any three pointing devices,
Answer:
Mouse, Touchpad and light pen.

Question 49.
You want to print your brother’s resume which printer will you choose. Why?
Answer:
I will choose either Ink-jet or Laser printer. Because these produce less noise and produce high quality printing output. They are used to print characters as well as graphics (Photos) with very high quality.

Question 50.
The fastest memory in a computer is
Answer:
Registers

Question 51.
The storage capacity of a single layer DVD is _______
Answer:
4.7 GB

Question 52.
Give two examples for OS.
Answer:
Windows 7, Windows Vista

Question 53.
A program in execution is called _______
Answer:
Process

Question 54.
Name the software that translates assembly language program into machine language program.
Answer:
Assembler

Question 55.
DBMS stands for _________.
Answer:
Data Base Management System .

Question 56.
Duplicating disc information is called __________.
Answer:
Backup

Question 57.
An example of free and open source software is _______?
Answer:
Linux

Question 58.
The software that gives users a chance to try it before buying is __________
Answer:
Shareware

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 59.
An example of proprietary software is _________.
Answer:
Tally

Question 60.
Which software is used for calculation?
(a) Word processor
(b) Spreadsheet
(c) Presentation
(d) Multimedia
Answer:
(b) Spreadsheet

Question 61.
Accumulator stores ________
(a) address of data
(b) instruction to be executed
(c) address of next instruction to be executed
(d) intermediate result
Answer:
(d) intermediate result

Question 62.
If Tracks and Sectors : Hard disk, then
___________ : Compact disk
Answer:
Pits and Lands (OR) 0 and 1

Question 63.
Which one of the following file extensions is different from others?
(a) WAV
(b) MP3
(c) PNG
(d) MIDI
Answer:
(c) PNG, the others are audio files.

Question 64.
Which register holds the memory address of next instruction to the executed?
(a) Accumulator
(b) PC
(c) MBR
(d) MAR
Answer:
(b) PC

Question 65.
1. Write the following memory devices in the order of their speed. (fastest to slowest order)

  • Cache
  • RAM
  • Hard Disk
  • Registers

2. What do you mean by Freeware and Shareware?
Answer:
1. Memory devices in the order of their speed.

  • Registers
  • Cache
  • RAM
  • Hard Disk

2. Freeware and Shareware:

  • Freeware: A s/w with Copy right is available free of cost for unlimited use.
  • Shareware: It is an introductory pack distributed on a trial basis with limited functionality and period.

Plus One Components of the Computer System Two Mark Questions and Answers

Question 1.
The Higher Secondary Department wishes to conduct an examination for +1 students with multiple choice questions and publish results as soon as possible. Suggest a method to evaluate the answer scripts and publish the results quickly & correctly with the help of computers.
Answer:
OMR has to be used, it senses the presence or absence of a mark (bubbles) using a high density beam then converted into electric signals for computer. It needs good quality expensive paper and accurate alignment of printing on forms.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 2.
Remesh is a graphic designer who prepares his drawing using a computer. He desires for an alternative device by which he can draw directly on the screen. Suggest a device for this and explain it working.
Answer:
Light pens are used for this. It consists fo a photocell placed in a small tube, it is able to detect the light coming from the screen. Hence locate the exact position on the screen. It is used by graphic designers, illustrators and drafting engineers, with the help of CAD to draw directly on the screen.

Question 3.
You might have noticed that in some shops billing is done using computers without typing the item name, price, quantity, etc. Mention the device used for entering data and explain its working.
Answer:
A device called Bar Code Reader is used for this. It contains photoelectric scanner that read the bar code and input the information to the computer attached to it. It helps to reduce the errors and process the bills quickly.

Question 4.
Your scholl has arranged an excursion. You are having an ordinary camera whereas your friend has a digital camera. List the benefits your friend enjoys by using digital camera.
Answer:

  1. Digital camera does not need film.
  2. More number of shots can take
  3. Operational cost is less
  4. Very easy to manipulate images in digital form using computers.

Question 5.
A medical shop in your locality wishes to purchase a printer for their billing purpose. Which type of printer will you recommend if carbon copies are to be taken. Justify.
Answer:
Dot Matrix Printer. To take carbon copies impact printer is a must, operational cost is less and it can print bills in a moderate speed.

Question 6.
Find the extact match.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 1
Answer:

  1. D
  2. C
  3. B
  4. A

Question 7.
Your friend wishes to start a DTP centre with facilities to design posters and notices, to scan pictures and modify them and to print them. What would be your suggestions regarding the computer and peripherals?
Answer:
The requriements are computer, scanner, printer and software.

Question 8.
Find the most appropriate match.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 2
Answer:
1 – D
2 – C
3 – A
4 – B

Question 9.
Suggest a suitable device for the following.

  1. High quality printing
  2. High quality drawing
  3. Printing with carbon copies
  4. Economical printing of small quantities of data

Answer:

  1. Non impact – Laser printers, Inkjet
  2. Plotter
  3. Impact (DMP(Dot Matrix Printer))
  4. Dot Matrix Printers

Question 10.
“Not all primary memory is volatile”. Justify this statement.
Answer:
Primary Memory (Main memory) is classified into two RAM and ROM. Out of this RAM is volatile but ROM is non-volatile.

Question 11.
Categorise the softwares in the list according to the appropriate classifications given below.

  1. Classification: OS, Compiler, DTP Software, Compression software, Word processor
  2. List: Open Office Writer, Photoshop, 7 Zip, MS Word, Unix, C++, PageMaker, Winzip, C, Windows 98.

Answer:

  • OS – Unix, Windows 98
  • Compiler – C, C++
  • DTP Software – Photoshop, PageMaker
  • Compression – 7 Zip, Winzip
  • Word Processor – Open Office Writer, MS word

Question 12.
Your friend has just assembled a computer. Now he is provided with installation CD’s of MS Word and Microsoft Windows XP. In what order will he install them? Justify your answer.
Answer:
First he has to install the Microsoft Windows XP because it is the OS, it makes the computer to work other programmes. After that only he can install MS Word It is a package.

Question 13.
A group of 20 students is given a test in 6 subjects. The examiner wishes to prepare a neatly formatted mark list with total and Rank. Suggest a suitable software to serve this purpose. Give reasons.
Answer:
The spreadsheet Excel is a suitable software to serve this purpose. It consists of inbuilt functions, that facilitates to find total and rank easily.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 14.
A program is written in BASIC, C and assembly language. Mention the difference in converting these programs to machine language.
Answer:

  1. C – Compiler
  2. BASIC – Interpreter
  3. Assembly Language – Assembler

Question 15.
Your friend told you that he has a system. What is a system? Explain.
Answer:
A computer is also called a system. A computer is not a single unit. It consists of more than one unit such as input unit, output unit, memory unit, ALU and control unit. Therefore a computer is called a system.

Question 16.
What is cache memory?
Answer:
A cache (pronounced cash) memory is a high-speed memory placed in between the processor and primary memory to reduce the speed mismatch between these two.

Question 17.
What is the use of program counter register?
Answer:
This register stores the memory address of the next instruction to be executed by the CPU.

Question 18.
What is HDMI?
Answer:
Its full form is High Definition Multimedia Interface. Through this port we can connect high definition quality video and multi channel audio over a single cable.

Question 19.
Give two examples for customized software.
Answer:

  1. Payroll System: It keeps track of details of employee and their salary details in an organisation
  2. Inventory Management System: It keeps tack of all about inventory in a company

Question 20.
What do you mean by free and open source s/w?
Answer;
Here “free” means there is no copy right or licensing. That is we can take copies of the s/w or modify the source code without legal permission of its vendor (creator) we can use and distribute its copy to our friends without permission. That is Freedom to use, to modify and redistribute

Question 21.

  1. What do you mean by cache memory? (1)
  2. Write the names of the figures given below.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 3

Answer:
1. It is a high speed memory placed in between the CPU and RAM. CPU is a high speed memory compared to RAM. There is a speed mismatch between the CPU and RAM to resolve this problem a high speed memory called cache memory is placed in between the CPU and RAM

2. QR code and Bar Code

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 22.
What is the role of students in e-Waste disposal?
Answer:
Student’s role in e-Waste disposal

  • Stop buying unnecessary electronic equipments
  • Repair Faulty electronic equipments instead of buying a new one.
  • Give electronic equipments to recycle
  • Buy durable, efficient, quality, toxic free, good warranty products
  • check the website or call the dealer if there is any exchange scheme
  • Buy rechargeable battery products

Plus One Components of the Computer System Three Mark Questions and Answers

Question 1.
Why is the paper coming from the laser printer hot? Explain.
Answer:
Laser printer uses photocopying technology. It uses a positively charged drum and negatively charged toner (dry powder). A laser beam is used to scan the page to be printed on the drum with positive charge and then rolled through a reservoir of negatively charged toner.

It uses a combination of heat and pressure to adhere the dry powder to the paper. That is why, the paper coming from the laser printer is hot.

Question 2.
Explain the process how data from the hard disk is taken to the processor for processing.
Answer:
A processor is a high speed device. It can access data only from the Primary Memory (RAM). So we have to transfer data from hard disk to RAM. We know that a hard disk is a slow device also. So data is first transferred to RAM.

A RAM is comparatively slower than processor. To reduce the speed mismatch between the RAM and processor, the data has to transfer to CPU registers. Then the processor takes the data from the CPU register because CPU register has almost equal speed as processor.

Question 3.
Why computer is called as a system? (3)
Answer:
A computer is also called a system. A computer is not a single unit. It consists of more than one unit such as input unit, output unit, memory unit, ALU and control unit. Therefore a computer is called a system.

Question 4.
Differentiate hardware and software. (3)
Answer:
The tangible parts of a computer is called hard¬ware. We can see, touch and feel the hardware in a computer. The set of instructions that tell the hardware how to perform a task is called software. Without software computer cannot do anything.

Question 5.
We all have a brain. Just like this, is the computer has a brain? Explain it?
Answer:
Yes. CPU is the brain of a computer. The CPU comprises three parts ALU, Control Unit and Memory. The control unit control the overall functioning of a system. ALU performs all the arithmetic calculations and takes logical decisions. Memory is used for storage of data for future reference.

Question 6.
Explain the various functions of a control unit.
Answer:
The control unit performs the following functions.

  1. It controls data flow between input device ALU, memory and output devices.
  2. Normal execution of a program is line by line. The control unit controls this sequence with the help of ALU and memory.
  3. It controls the decoding and interpretation of in-structions.

Question 7.
Match the following

1. Input device a. Linux
2. Output device b. Java
3. Secondary Memory c. Joystick
4. System S/W d. Plotter
5. Application S/W e. PROM
6. Primary Memory f. HDD

Answer:
1 – c
2 – d
3 – f
4 – a
5 – b
6 – e

Question 8.
Write down the full form of the following.

  1. VDU
  2. OMR

Answer:

  1. VDU – Visual Display Unit
  2. OMR – Optical Mark Reader

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 9.
We know that a scanner is a hardware. What do you think of a virus scanner? Explain.
Answer:
We know that a scanner is a hardware but a virus scanner is not a hardware. It is a program. That is a virus scanner is an antivirus software. That scans your disk (HDD, CD, DVD, Pen Drive) for viruses and removes them (if it can), if any virus is found.

Question 10.
Your friend told you that a compiler is a hardware. Is it true? Justify your answer.
Answer:
It is not true. A compiler is not a hardware but it is a software. A compiler is a collection of programs that translates program written in HLL into machine language.

Question 11.
Anil purchased a product from a supermarket and he found that its wrapper contains light and dark bars. What is the purpose of this ?
Answer:
This light and dark bars are called bar code. It is used to record some details about the product such as item code, name, price etc. A device called Bar Code Reader contains photo electric scanner that read the bar code and input the information to the computer attached to it. It helps to reduce error and process the bills quickly.

Question 12.
What are the disadvantages of OMR? (2)
Answer:
The disadvantages are:

  1. It needs accurate alignment of printing on forms.
  2. It needs good quality expensive paper.

Question 13.
Differentiate CRT and LCD (OR) Your friend going to purchase a computer. He asked you which is better, CRT or LCD? What is your opinion?
Answer:
The difference between CRT and LCD is given below:

CRT LCD
It is heavy and bulky It is neither heavy nor bulky
It consumes more power and emits heat It consumes less power and does not emit heat
It is used in desk top computer It is used with laptop and desktop
It is cheaper It is expensive.

So LCD is more better than CRT

Question 14.
While you pressing “A” on the keyboard what is actually stored in the memory?
Answer:
The keyboard is an electro mechanical device that is designed to create electronic codes when a key is pressed and this code is transmitted to the memory through the cable. Here while you pressing “A” on the keyboard, the electronic codes corresponding to the ASCII value 01000001 is transmitting to the memory.

Question 15.
Your family friend started a supermarket. He asked you, which printer is suitable to print bills. Give your suggestion.
Answer:
According to my opinion, dot matrix printer is most suitable. Because they are capable of faster printing as well as it is cheap also. It’s printing quality is not good but cost per copy is very cheap. Dot matrix printer consists of a ribbon cartridge that is cheap and can be changed easily. Here we have to print more copies at a time. So dot matrix is suitable.

Question 16.
Suppose your brother is an engineer. He wants to draw some drawings. Which output device is suitable? Explain.
Answer:
Plotter is suitable for him. A plotter is a device that draws pictures ordiagrams on paper based on commands from a computer. Plotters draw lines using a pen. Pen plotters generally use drum or flat bed paper holders.

In a drum plotter the paper is mounted on the surface of a drum. Here the paper is rotated. But in a flatbed plotter the paper does not move and the pen holding mechanism provides the motion that draws pictures. Plotters are used in engineering applications where precision is needed.

Question 17.
Match the following.

1. Operating System a. Compiler
2. Language Processor b. Windows Vista
3. Package Administratin S/W c. Santhi Hospital
4. Utility d. Spreadsheet
5. Customised S/W e. Disk defragmenter

Answer:
1 – b
2 – a
3 – d
4 – e
5 – c

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 18.
There are special purpose storage locations within the CPU. What are they explain ?
Answer:
Registers are special purpose storage locations within the CPU. They are temporary storage locations. The processing power of a CPU depends on register. Registers appears with storage capacity of 8 bits, 16 bits, 32 bits an 64 bits. They accept, store and transfer data from the CPU at a very high speed.

  • Program Counter: This register stores the memory address of the next instruction to be executed by the CPU.
  • Instruction Register: The instruction to be executed is stored in this register.
  • Memory Address Register: The address on the memory location from which data has to be read is stored here.
  • Memory Buffer Register: The data read from the location specified by the MAR is stored in this register.
  • General Purpose Registers: These are used to store the result and intermediate results during a processing.

Question 19.
What is an operating system?
Answer:
An operating system acts as an interface between user and computer without an operating system computer is a bare machine. That is without an OS computer cannot do anything. The OS not only makes the system convenient to use but also use hardware in an efficient manner,
eg: Windows XP, Vista, Linux, MS Dos, Windows 7.

Question 20.
We know that a computer only knows low level language and human beings use high level language. So how is it possible to communicate? Explain.
Answer:
The language processors translate the programs written in HLL into machine language which is understood by the computer. The different language processors are given below:
1. Assembler:
This language processor translates programs written in assembly language into machine language.

2. Interpreter:
This language processor translates programs written in HLL into machine language by converting and executing it line by line. If there is any error, the execution is stopped we, can continue after the correction of the program.

3. Compiler:
This language processor is same as interpreter. But it translates HLL into machine language by converting whole lines at a time. If there is any error, correcting all the errors then only it will execute.

Question 21.
Normally a CD contains 700 MB. Is it possible to store a file with size 1 GB? Explain.

OR

Normally a Car has a seating capacity of 5 persons including the driver. But some adjustments more persons can be accommodated in Car. This is connected with a utility. Which is the utility? Explain.
Answer:
Compression utility is used for this. By using compression utility programs we can reduce the file size upto the one third of the file size. So by using this we can reduce 1GB file and store in a CD. It is provided by the OS.

The other compression utility programs are Winzip, WinRar etc. It is possible to compress the files and when needed, these compressed files can be uncompressed and it is restored to their original form.

Question 22.
What is a Virus?
Answer:
A virus is a bad program or harmful program to damage routine working of a computer system. It reduces the speed of a computer. It may be delete the useful system files and make the computer useless.

Question 23.
What do you mean by Utilities?
Answer:
Utilities are useful programs which are designed to help computer for its smooth functioning. Some utilities are back up utility, Disk defragmentation. Virus scanner, etc. It is provided by the O.S.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 24.
Differentiate RAM and ROM.
Answer:
The difference between RAM and ROM is given below.

RAM ROM
1. It is Random Access Memory 1. It is Read only Memory
2. It is Read/Write memory 2. We can’t write but we can only read memory.
3. It is temporary 3. It is permanently stored.
4. It is volatile 4. It is non volatile
5. RAM is faster 5. It is slower
6. It is used to store data and instructions needed by CPU for processing 6. It contains instructions to check the hardware components, BIOS operations etc.
7. It is also called firmware.

Question 25.
Mention any two functions of OS.
Answer:
Major functions of an operating System

  1. Process management: It includes allocation and de allocation of processes(program in execution) as well as scheduling system resources in efficient manner
  2. Memory management: It takes care of allocation and de allocation of memory in efficient manner
  3. File management: This includes organizing, naming, storing, retrieving, sharing, protecting and recovery of files.
  4. Device management: Many devices are connected to a computer so it must be handled efficiently.

Question 26.
Give two examples of human ware.
Answer:
(Write any two from the following). The term refers the persons who use computer System Administrator: It is a person who has central control over the computer systems.

  • System Managers: He is responsible for all business transactions with all vendors and contractors.
  • System Analysts: He is responsible to improve the productivity and efficiency.
  • Database Administrator: It is a person who has a central control over the DBMS.
  • Computer Engineer: The person design either h/w or s/w of a computer system.
  • Application Programmer: These are computer professionals who interact with the DBMS through programs.
  • Computer operators: He is an end user and does not know computer in detail.

Question 27.
Explain how e-waste creates environmental and health problems. What are the different methods for e-waste disposal? Which one is the most effective in your point of view? Why? (5)
Answer:
e-Waste(electronic waste): It refers to the malfunctioning electronic products such as faulty computers, mobile phones, tv sets, toys, CFL, batteries etc. It contains poisonous substances such as lead, mercury, cadmium etc and may cause diseases if not properly managed.

A small amount is recycled. Due to this our natural resources are contaminated(poisoned). Some of them can .recycle properly. But it is a very big problem in front of the Government to collect segregate, recycle and disposal of e-Waste.
e-Waste disposal methods:

  1. Reuse: Reusability has an important role of e-Waste management and can reduce the volume of e-Waste
  2. Incineration : It is the process of burning e-Waste at high temperature in a chimney
  3. Recycling of e-Waste: It is the process of making new products from this e-Waste.
  4. Landfilling: It is used to level pits and cover by thick layer of soil.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 28.
What do you mean by e-waste? Explain the role of students in e-waste disposal.
Answer:
e-Waste(electronic waste): It refers to the malfunctioning electronic products such as faulty computers, mobile phones, tv sets, toys, CFLetc. It contains poisonous substances such as lead, mercury, cadmium etc and may cause diseases if not properly managed.
Student’s role in e-Waste disposal

  1. Stop buying unnecessary electronic equipments
  2. Repair Faulty electronic equipments instead of buying a new one.
  3. Give electronic equipments to recycle
  4. Buy durable, efficient, quality, toxic free, good warranty products
  5. check the website or call the dealer if there is any exchange scheme
  6. Buy rechargeable battery products

Plus One Components of the Computer System Five Mark Questions and Answers

Question 1.
Write short notes about input devices.
Answer:
An input device is used to supply data to the computer. They are given below:
1. Key board:
It is the most widely used device to input information in the form of words, numbers etc. There are 101 keys on a standard key board. The keys on the key board are often classified into alpha numeric keys (A to Z, Oto 9), function keys (F1 to F12), special purpose keys (Special characters), cursor movement keys (arrow keys). While pressing a key, the corresponding code’s signal is transmitted to the computer.

2. Mouse:
It is a pointing device, that controls the movement of the cursor, or pointer as a display screen. A mouse has two or three buttons, it is often used in GUI oriented computers. Under the mouse there is a ball, when the mouse moves on a flat surface this ball also moves. This mechanical motion is converted into digi¬tal values that represents x and y values of the mouse movement.

3. Optical Mark Reader (OMR):
This device identifies the presence or absence of a pen or pen¬cil mark. It is used to evaluate objective type exams. In this method special preprinted forms r.e designed with circles can be marked with dark pencil or ink.

A high intensity beam in the OMR converts this into computer usable form and detects the number and location of the pencil marks. By using this we can evaluate easily and reduce the errors.

4. Bar code / Quick Response (QR) code reader:
Light and dark bars are used to record item name, code and price is called Bar Code. This information can be read and input into a computer quickly without errors using Bar Code Readers.

It consists of a photo electric scanner and it is used in super market, jewellery, textiles etc. QR codes are similar to barcodes but it uses two dimensional instead of single dimensional used in Barcode.

5. Joy Stick:
It is a device that lets the user move an object quickly on the screen. It has a liver that moves in all directions and controls the pointer or object. It is used for computer games and CAD / CAM systems.

6. Light Pen:
It is an input device that use a light sensitive detector to select objects directly on a display screen using a pen. Light pen has a photocell placed in a small tube. By using light pen, we can locate the exact position on the screen.

7. Scanner:
It is used to read text or pictures printed on paper and translate the information into computer usable form. It is just like a photostat machine but it gives information to the computer.

8. Digital Camera:
By using digital camera, we can take photographs and store in a computer. Therefore we can reduce the use of film. Hence it is economical.

9. Touchpad:
It is a pointing device found on the portable computers(laptop). Just like a mouse it consists of two buttons below the touch surface to do the operations like left click and right click. By using our fingers we can easily operate.

10. Microphone:
By using this device we can convert voice signals into digital form.

11. Biometric sensor:
It is used to read unique human physical features like fingerprints, retina, iris pattern, facial expressions etc. Most of you give these data to the Government for Aadhaar.

12. Smart card reader:
A plastic card(maybe like your ATM card) stores and transmit data with the help of a reader.

13. Digital Camera :
By using digital camera, we can take photographs and store in a computer. Therefore we can reduce the use of film. Hence it is economical.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 2.
Briefly explain the various visual display units.
Answer:
The visual display units are given below:
1. Cathode Ray Tube (CRT):
There are two types of CRT’s, monochrome (Black and white) and colour. Monochrome CRT consists of one elec-tron gun but colour CRT consists of 3 electron guns (Red, Green and Blue) at one end and the other end coated with phosphor. It is a vacuum tube. The phosphor coated screen can glow when electron beams produced by electron guns hit.

It is possible to create all the colours using Red, Green and Blue. The images produced by this is refreshed at the rate of 50 or 60 times each second.Its disadvantage is it is heavy and bulky. It consumes more power and emits heat. But it is cheap. Nowadays its production is stopped by the company.

2. Liquid Crystal Display (LCD):
It consists of two, electrically conducting plates filled with liquid crystal. The front plate has transparent electrodes and the back plate is a mirror. By applying proper electrical signals across the plates, the liquid crystals either transmit or block the light and then reflecting it back from the mirror to the viewer and hence produce images. It is used in where small sized displays are required.

3. Light Emitting Diocte(LED):
It uses LED behind the liquid crystals in order to light up the screen. It gives a better quality and clear image with wider viewing angle. Its power consumption is less.

4. Plasma Panels:
It consists of two glass plates filled with neon gas. Each plate has several parallel electrodes, right angles to each other. When low voltage is applied between two electrodes, one on each plate, a small portion of gas is glow and hence produce images. Plasma displays provide high resolution but are expensive. It is used in, where quality and size is a matter of concern.

5. Organic Light Emitting Diode(OLED) Monitors:
It is made up of millions of tiny LEDs. OLED monitors are thinner and lighter than LCDs and LEDs. It consumes less power and produce better quality images but it is very expensive.

Question 3.
Your friend wants to buy a printer. He wants to know more about printers. Explain different types of printers.
Answer:
Printer: There are two types of printers impact and non-impact printers. Printers are used to produce hard copy.
Impact Printers: There is a mechanical contact between print head and the paper.

1. Dot Matrix Printer:
Here characters are formed by using dots. The printing head contains a vertical array of pins. The letters are formed by using 5 dot rows and 7 dot columns. Such a pattern is called 5 × 7 matrix.

This head moves across the paper, the selected pins fire against an inked ribbon to form characters by dot. They are capable of faster printing, but their quality is not good.

2. Non-impact Printers:
There is no mechanical contact between print head and paper so carbon copies cannot be possible to take. They are inkjet, laser, thermal printers etc.

(a) Inkjet Printer:
It works in the same fashion as dot matrix printers, but the dots are formed with tiny droplets of ink to be fired from a bottle through a nozzle. These droplets are deflected by an electric field using horizontal and vertical deflection plates to form characters and images.

It is possible to generate colour output. They produce less noise and produce high quality printing output. The printing cost is higher. Here liquid ink is used.

(b) Laser Printer:
It uses photo copying technology. Here instead of liquid ink dry ink powder called toner is used. A drum coated with positively charged photo conductive material is scanned by a laser beam. The positive charges that are illuminated by the beam are dissipated. The drum is then rolled through a reservoir of negatively charged toner which is picked up by the charged portions of the drum.

It adheres to the positive charges and hence creating a page image on the drum. Monochrome laser printer uses a single toner whereas the colour, laser printer uses four toners. Its print quality is good less noise and printing cost is higher.

(c) Thermal Printers:
It is same as dot matrix printer but it needs heat sensitive paper. It produces images by pushing electrically heated pins to the special paper. It does not make an impact on the paper so we cannot produce carbon copies. It produce less noise, low quality print and inexpensive. It is used in fax machine.

3. Plotter:
A plotter is a device that draws pictures ordiagrams on paper based on commands from a computer. Plotters draw lines using a pen. Pen plotters generally use drum or flat bed paper holders. In a drum plotter the paper is mounted on the surface of a drum.

Here the paper is rotated. But in a flatbed plotter the paper does not move and the pen holding mechanism provides the motion that draws pictures. Plotters are used in engineering applications where precision is needed.

4. Three Dimensional (3D) printer: This device is used to print 3D objects.

Question 4.
Your school got two printers. One dot matrix and one Laser printer through ICT scheme of Central Govt. What is the difference between these two printers? Explain.
Answer:

Laser Printer Dot Matrix Printer
It is non-impact printer It is impact printer
Speed is high Speed is less
Good quality text and Low quality text and very
picture poor quality picture
Less noise More noise
Printing cost is high Printing cost is low
Not possible to take Possible to take carbon
carbon copy copy
Toner is used Ribbon is used

Question 5.
Explain Primary Memory in detail.
Answer:
Primary memory is classified into two, Random Access Memory (RAM) and Read Only Memory (ROM). The primary memory holds the data which is to be processed by the CPU and the set of instructions to be executed next. The CPU can access the instructions in the primary memory only.

1. Random Access Memory (RAM):
RAM is used to store data and instructions needed by the CPU for processing. RAM can be used for both reading and writing of data so it is called Read and Write memory. It is a volatile memory, that is contents of the RAM will be lost when the power is turned off.

The invention of integrated circuits (chips) increased the memory capacity and reduced the size and cost. Static RAM, Dynamic RAM, Synchronous Dynamic RAM (SDRAM) are the various types of RAM.

2. Read Only Memory (ROM):
We can read this memory but we cannot write into this memory. It is a nonvolatile memory, that is its contents will not lost when the power is turned off. This memory is stored in the ROM chip at the time of manufacture ing itself hence it is called firmware.

ROM contains instructions to check the hardware components connected to the system, perform some basic input/output operations (BIOS), initiates loading of essential software. The different categories are PROM, EPROM and EE PROM.

(a) PROM (Programmable Read Only Memory):
It is just like WORM. That means the instructions are write once but read many. But we cannot change the instructions.

(b) EPROM (Erasable Programmable Read Only Memory):
It functions just like PROM. But by using ultraviolet light we can erase the old data and can write new data.

(c) EEPROM (Electrically Erasable Programmable Read Only Memory):
Here instead of ultraviolet light electric signals are used to erase old data and write new data under software control. It is highly expensive than regular ROM chips.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 6.
Explain secondary memory in detail.

OR

To store large volume of data permanently. Which memory is used? Explain.
Answer:
Primary memory has a limited storage capacity and it is not permanent that is why to store large volume of data permanently secondary memory or storage devices are used. They are of many types and they vary in the capacity of storage, speed of data access and media of storage. Nowadays magnetic disks and optical disks are commonly used.

1. Magnetic Disk:
Magnetic disk allows the storage and retrieval for contents of the disk from anywhere at a moderate speed. Magnetic Disks available in various size and storage capacity but the storage media and data access mechanism are similar.

The storage media is circular platters or disks coated with magnetic material. It consists of a spindle capable of rotating with the help of an electrical motor and a read/write head.

(a) Floppy disk:
Floppy means flexible or soft, it uses flexible disk. Its storage capacity is 1,44MB and slower in data transfer rate.

(b) Hard Disk:
Instead of flexible or soft disk it uses rigid material hence the name hard disk. Its storage capacity and data transfer rate are high and low access time. These are more lasting and less error prone. The accessing mechanism and storage media are combined together in a single unit and connect to the mother board via cable.

Therefore we call it as hard disk drive (HDD). It contains one or more rigid platters coated both sides with a special magnetic material and a spindle. Datas are recorded on either surface of the disk except the outer side of last and first disk. Each surface will have one or more read/write heads (fixed head or movable head).

The spindle is attached to a motor that rotates at high speed typically 7200 rotation per minute (rpm). A floppy disk rotates only at 300 rpm.

2. Optical Disk:
The high power laser uses a concentrated, narrow beam of light, which is focuses and directed with lenses, prisms and mirrors for recording data. The optical disks are given below:

(a) Compact Disk Read Only Memory (CDROM):
The data in CDROM is imprinted.by the manufacturers. The user cannot erase or write on the disk but user can only read its contents. CDROM is written in a single continuous spiral unlike magnetic disks that uses concentric circles. Its storage capacity is 700MB.

(b) Erasable Optical Disk:
The disadvantage of CDROM is that we cannot change or erase the contents. But erasable disks can be changed and erased.

(c) Digital Versatile Disk: It is capable of storing upto 4.7GB and more faster.

(d) Blu-ray Disc:
It is used to read and write High Definition video data as well as to store very huge amount of data. While Cd and DVD uses red laser to read and write but it uses Blue-Violet laser, hence the name Blu ray disc. The blue-violet laser has shorter wavelength than a red laser so it can pack more data tightly.

3. Semiconductor storage (Flash memory): It uses EEPROM chips. It is faster and long lasting.

  1. USB flash drive: It is also called thumb drive or pen drive. Its capacity varies from 2 GB to 32 GB.
  2. Flash memory cards: It is used in Camera, Mobile phones, tablets etc to store all types of data.

Question 7.
What do you mean by a computer software? Mention its different classification.

OR

Your friend wants to know more about software. Explain more about different classification.

OR

Your friend told you that COBOL and MS Word are same softwares. Do you agree with him. Explain. What is a software and its classification?
Answer:
A Software is a collection of programs to perform a task. The softwares can be classified into two major groups.

  1. System software
  2. Application software

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 4
1. System Software:
It is a collection of programs used to manage system resources and control its operations. It is further classified into two.
(a) Operating System
(b) Language Processor
(a) Operating System:
It is collection of programs which acts as an interface between user and computer. Without an operating system computer cannot do anything. Its main function is make the computer usable and use hardware in an efficient manner,
eg: WindowsXP, Windows Vista, Linux, Windows 7, etc.

(b) Language Processes:
We know that a program is a set of instructions. The instructions to the computer are written in different languages. They are high level language (HLL) and low level language. In HLL english like statements are used to write programs. They are C, COBOL, PASCAL, VB, Java etc. HLL is very easy and can be easily understood by the human being.

Low level language are classifed into Assembly Language and Machine Language. In assembly language mnemonics (codes) are used to write programs
Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 5
In Machine Language 0’s and 1’s are used to write program. It is very difficult but this is the only language which is understood by the computer. Usually programmers prefer HLL to write programs because of its simplicity. But computer understands only machine language. So there is a translation needed. The program which perform this job are language processors.

The different language processors are given below:
1. Assembler:
This converts programs written in assembly language into machine language.

2. Interpreter:
This converts a HLL program into machine language by converting and executing it line by line. The first line is converted if there is no error it will be executed otherwise you have to correct it and the second line and so on.

3. Compiler:
It is same as interpreter but there is a difference, it translate HLL program into machine language by converting all the lines at a time. If there is no error then only it will executed.

2. Application Software:
Programs developed to serve a particular application is known as application software, eg:- MS Office, Compression Utility, Tally etc.
Application software can further be sub divided into three categories.
(a) Packages
(b) Utilities
(c) Customized Software

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

(a) Packages:
Application software that makes the computer useful for people to do every task. Packages are used to do general purpose application.
They are given below:
1. Word Processes:
This is used for creation and modification of text document. That means a word processor helps the people to create, edit and format a textual data with less effort and maximum efficiency.

By using word processor we can change font and font size of character, change alignment (left, right, center and justify), check spelling and grammar of the whole document etc. eg: MS Word.

2. Spread Sheets:
It contains data or information in rows and columns and can perform calculation (Arithmetic, Relational and logi¬cal operation). It helps to calculate results of a particular formula and the formula can apply different cells (A cell is the intersection of a row and column. Each column carries an alphabet for its name and row is numbered).

It is used to prepare budgets, balance sheets, P & L account, Payroll etc. We can easily prepare graphs and charts using data entered in a worksheet. A file is a workbook that contains one or more worksheets, eg: MS Excel is a spreadsheet software.

3. Presentation and Graphics:
You can present your idea with sound and visual effects with the help of presentation software by preparing slides. The application software that manipulate visual images is known as graphics software. eg: MS Power Point is a presentation package.

4. Data base package:
Database is a collection of large volume of data. DBMS is a set of programs that manages the datas are for the centralised control of data such that creating new records to the database, deleting, records whenever not wanted from the database and modification of the existing database. Example for a DBMS is MS Access.

(b) Utilities: Utilities are programs which are designed to assist computer for its smooth functioning.
The utilities are given below:

1. Text editor:
It is used for creating and editing text files.

2. Backup utility:
Creating a copy of files in another location to protect them against loss, if your hard disk fails or you accidently overwrite or delete data.

3. Compression Utility:
It is used to reduce the size of a file by using a program and can be restored to its original form when needed.

4. Disk Defragmenter:
It is used to speeds up disk access by rearranging the files that are stored in different locations as fragments to contiguous memory and free space is consolidated in one contiguous block.

5. Vims Scanner:
It is a program called antivirus software scans the disk for viruses and removes them if any virus is found.

(c) Customised software:
It is collection of programs which are developed to meet user needs to serve a particular application. It is also called tailor made software.

Question 8.
To use a computer not only the hardware but also software are required. Explain the classification of software.
Answer:
Software: The set of instructions that tell the hardware how to perform a task is called software. Without software computer cannot do anything. Two types System s/w and Application s/w System software.
It is a collection of programs used to manage system resources and control its operations. It is further classified into two.

  1. Operating System
  2. Language Processor

1. Operating System: It is collection of programs which acts as an interface between user and computer. Without an operating system computer cannot do anything. Its main function is make the computer usable and use hardware in an efficient manner, eg: Windows XP, Windows Vista, Linux, Windows 7, etc.
Major functions of an operating System:

  • Process management: It includes allocation and de allocation of processes(program in execution) as well as scheduling system resources in efficient manner.
  • Memory management: It takes care of allocation and de allocation of memory in efficient manner
  • File management: This includes organizing, naming , storing, retrieving, sharing , protecting . and recovery of files.
  • Device management: Many devices are connected to a computer so it must be handled efficiently.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

2. Language Processes:
We know that a program is a set of instructions. The instructions to the computer are written in different languages. They are high level language (HLL) and low level language. In HLL English like statements are used to write programs. They, are C, C++, COBOL, PASCAL, VB, Java etc. HLL is very easy and can be easily understood by the human being.

Low level language are classified into Assembly Language and Machine Language.
In assembly language mnemonics (codes) are used to write programs
Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 6
In Machine Language 0’s and 1’s are used to write program. It is very difficult but this is the only language which is understood by the computer. Usually programmers prefer HLL to write programs because of its simplicity. But computer understands only machine language. So there is a translation needed. The program which perform this job are language processors.

The different language processors are given below:
1. Assembler:
This converts programs written in assembly language into machine language.

2. Interpreter:
This converts a HLL program into machine language by converting and executing it line by line. The first line is converted if there is no error it will be executed otherwise you have to correct it and the second line and so on.

3. Compiler:
It is same as interpreter but there is a difference it translate HLL program into machine language by converting all the lines at a time. If there is no error then only it will executed.

Application Software:
Programs developed to serve a particular application is known as application software, eg:- MS Office, Compression Utility, Tally etc. Application software can further be sub-divided into three categories.
(a) Packages
(b) Utilities
(c) Customized Software

(a) Packages:
Application software that makes the computer useful for people to do every task. Packages are used to do general purpose application.
They are given below:

1. Word Processes:
This is used for creation and modification of text document. That means a word processor helps the people to create, edit and format a textual data with less effort and maximum efficiency. By using word processor we can change font and font size of character, change alignment (left, right, center and justify), check spelling and grammar of the whole document etc.
eg: MS Word.

2. Spread Sheets:
It contains data or information in rows and columns and can perform calculation (Arithmetic, Relational and logical operation). It helps to calculate results of a particular formula and the formula can apply different cells (A cell is the intersection of a row and column.

Each column carries an alphabet for its name and row is numbered). It is used to prepare budgets, balance sheets, P & L account, Pay roll etc. We can easily prepare graphs and charts using data entered in a worksheet. A file is a work book that contains one or more work sheets,
eg : MS Excel is a spread sheet software.

3. Presentation and Graphics:
You can present your idea with sound and visual effects with the help of presentation software by preparing slides. The application software that manipulate visual images is known as graphics software.
eg: MS Power Point is a presentation package.

4. Data base package:
Data base is a collection of large volume of data. DBMS is a set of programs that manages the datas are for the centralized control of data such that creating new records to the database, deleting, records whenever not wanted from the database and modification of the existing database. Example for a DBMS is MS Access.

DTP Packages: DTP means Desk Top Publishing. By using this we can create books, periodicals, magazines etc. easily and fastly. Now DTP packages are used to create in Malayalam also,
eg: PageMaker.

5. Utilities:
Utilities are programs which are designed to assist computer for its smooth functioning.
The utilities are given below:

  1. Text editor: It is used for creating and editing text files.
  2. Backup utility: Creating a copy of files in another location to protect them against loss, if your hard disk fails or you accidentally overwrite or delete data.
  3. Compression Utility: It is used to reduce the size of a file by using a program and can be restored to its original form when needed.
  4. Disk Defragmenter: It is used to speeds up disk access by rearranging the files that are stored in different locations as fragments to contiguous memory and free space is consolidated in one contiguous block.
  5. Virus Scanner: It is a program called antivirus software scans the disk for viruses and removes them if any virus is found.

(c) Specific purpose software (Customized software):
It is collection of programs which are developed to meet user needs to serve a particular application. It is also called tailor made software.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 9.
Describe the different types of memories and memory devices in computer with features and examples.
Answer:
Memory:
Storage Unit(Memory Unit): A computer has huge storage capacity. It is used to store data and instructions before starts the processing. Secondly it stores the intermediate results and thirdly it stores information(processed data), that is the final results before send to the output unit(Visual Display Unit, Printer, etc)
Memory measuring units are given below.

  • 1 bit = 1 or 0(Binary Digit)
  • 4 bits = 1 Nibble
  • 8 bits = 1 Byte
  • 1024 Bytes = 1 KB(Kilo Byte)
  • 1024 KB = 1 MB(MegaByte)
  • 1024 MB = 1 GB(Giga Byte)
  • 1024 GB = 1 TB(Tera Byte)
  • 1024 TB = 1 PB(Peta Byte)

Two Types of storage unit:
1. Primary Storage alias Main Memory:
It is further be classified into Two Random Access Memory (RAM) and Read Only Memory(ROM). The one and only memory that the CPU can directly access is the main memory at a very high speed.

It is expensive hence storage capacity is less. RAM is volatile(when the power is switched off the content will be erased) in nature but ROM is non volatile(lt is permanent). In ROM a “boot up” program called BIOS(Basic Input Output System) is stored to “boots up” the computer when it switched on. Some ROMs are given below.

  1. PROM(Programmable ROM): It is programmed at the time of manufacturing and cannot be erased.
  2. EPROM (Erasable PROM): It can be erased and can be reprogrammed using special electronic circuit.
  3. EEPROM (Electrically EPROM): It can be erased and rewritten electrically

Cache Memory:
The processor is a very high speed memory but comparatively RAM is slower than Processor. So there is a speed mismatch between the RAM and Processor, to resolve this a high speed memory is placed in between these two this memory is called cache memory. Commonly used cache memories are Level(L1) Cache(128 KB), L2(1 MB),L3(8 MB), L4(128MB).

2. Secondary Storage alias Auxiliary Memory :
Because of limited storage capacity of primary memory its need arises. When a user saves a file, it will be stored in this memory hence it is permanent in nature and its capacity is huge. Eg: Hard Disc Drive(HDD), Compact Disc(CD), DVD, Pen Drive, Blu Ray Disc etc.

(a) Magnetic storage device:
It uses plastic tape or metal/plastic discs coated with magnetic material.
Hard Disk: Instead of flexible or soft disk it uses rigid material hence the name hard disk. Its storage capacity and data transfer rate are high and low access time.

These are more lasting and less error prone. The accessing mechanism and storage media are combined together in a single unit and connect to the mother board via cable.

(b) Optical storage device:
Optical Disk: The high power laser uses a concentrated, narrow beam of light, which is focuses and directed with lenses, prisms and mirrors for recording data. This beams burns very very small spots in master disk, which is used for making molds and these molds are used for making copies on plastic disks.

A thin layer of aluminium followed by a transparent plastic layer is deposited on it. The holes made by the laser beam are called pits, interpreted as bit 0 and unburned areas are called lands interpreted as bit 1. Lower power laser beam is used to retrieve the data.

1. DVD(Digital Versatile Disc):
It is similar to CD but its storage capacity is much higher. The capacity of a DVD starts from 4.7 GB

2. Blu-ray Disc:
It is used to read and write High Definition video data as well as to store very huge amount of data. While Cd and DVD uses red laserto read and write but it uses Blue-Violet laser, hence the name Blu ray disc. The blue violet laser has shorter wave length than a red laser so it can pack more data tightly.

3. Semiconductor storage (Flash memory):
It uses EEPROM chips. It is faster and long lasting.

  • USB flash drive: It is also called thumb drive or pen drive. Its capacity varfes from 2 GB to 32 GB.
  • Flash memory cards: It is used in Camera, Mobile phones, tablets etc to store all types of data.

Question 10.
Explain how e-Waste creates environmental issues. Usually there are four methods for e-Waste dispoal. Which one is the most effective? Why? Write a slogan to aware the public about e-Waste hazards.
Answer:
e-Waste disposal methods:

  1. Reuse: Reusability has an important role of e-Waste management and can reduce the volume of e-Waste
  2. Incineration: It is the process of burning e-Waste at high temperature in a chimney
  3. Recycling of e-Waste: It is the process of making new products from this e-Waste.
  4. Land filling: It is used to level pits and cover by thick layer of soil.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 11.
With the help of a block diagram, explain the functional units of a computer.
Answer:
Functional units of computer:
A computer is not a single unit but it consists of many functional units(intended to perform jobs) such as Input unit, Central Processing Unit(ALU and Control Unit), Storage (Memory) Unit and Output Unit.
1. Input Unit:
Its aim is to supply data (Alphanumeric, image , audio, video, etc.) to the computer for processing. The Input devices are keyboard, mouse, scanner,mic, camera,etc

2. Central Processing Unit (CPU):
It is the brain of the computer and consists of three components

  • Arithmetic Logic Unit(ALU): As the name implies it performs all calculations and comparison operations.
  • Control Unit(CU): It controls overall functions of a computer
  • Registers: It stores the intermediate results temporarily.

3. Storage Unit(Memory Unit):
A computer has huge storage capacity. It is used to store data and instructions before starts the processing. Secondly it stores the intermediate results and thirdly it stores information(processed data), that is the final results before send to the output unit(Visual Display Unit, Printer, etc)

Two Types of storage unit
(a) Primary Storage alias Main Memory:
It is further be classified into Two- Random Access Memory(RAM) and Read Only Memory(ROM). The one and only memory that the CPU can directly access is the main memory at a very high speed.

It is expensive hence storage capacity is less. RAM is volatile (when the power is switched off the content will be erased) in nature but ROM is non volatile(lt is permanent)

(b) Secondary Storage alias Auxiliary Memory:
Because of limited storage capacity of primary memory its need arises. When a user saves a file, it will be stored in this memory hence it is permanent in nature and its capacity is huge. eg: Hard Disc Drive(HDD), Compact Disc(CD), DVD, Pen Drive, Blu Ray Disc etc.

4. Output Unit:
After processing the data we will get information as result, that will be given to the end user through the output unit in a human readable form. Normally monitor and printer are used.

Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

Students can Download Chapter 12 Introduction to Three Dimensional Geometry Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers

Question 1.
Prove by using distance formula that the A(1, 2, 3), B(-1, -1, -1) and C(3, 5, 7) are collinear.
Answer:
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 1
Now BC = AB + AC
Thus A, B, C are collinear.

Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

Question 2.
Verify the following: (3 score each)

  1. (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
  2. (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
  3. (-1, 2, 1), (1, -2, 5), (4, -7, 8) and (2, -3, 4) are the vertices of a parllelogram.

Answer:
1. Let A(0, 7, -10), B(1, 6, -6) and C(4, 9, -6) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 2
Now AB = BC, thus ABC is an isosceles triangle.

2. Let A(0, 7, 10), B(-1, 6, 6) and C(-4, 9, 6) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 3
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 4
Now AC2 = AB2 + BC2, thus ABC is a right triangle.

3. Let A(-1, 2, 1), B(1, -2, 5), C(4, -7, 8) and D(2, -3, 4) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 5
Now AB = CD, BC = AD, AC ≠ BD, thus A, B, C, D are vertices of a parallelogram.

Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

Question 3.
Find the equation of set points which are equidistant from the points (1, 2, 3) and (3, 2, -1).
Answer:
Let P(x,y,z) be any point which is equidistant from the point A(1, 2, 3) and B (3, 2, -1).
Given; PA = PB
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 6
(x – 1)2 + (y – 2)2 + (z – 3)2
= (x – 3)2 + (y – 2)2 + (z + 1)2
= x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9
= x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z + 1 – 2x + 14 – 6z = -6x + 14 + 2z
⇒ 4x – 8z = 0
⇒ x – 2z = 0.

Question 4.
Find the coordinate of the point which divides the line segment joining the points (3, -2, 5) and (3, 4, 2) in the ratio 2:1 (3 score each)

  1. 2:1 internally
  2. 2:1 externally

Answer:
1. Let P(x, y, z) be any point which divides the line segment joining points A(3, -2, 5) and B (3, 4, 2) in the ratio 2:1 internally.
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 7
Therefore coordinates of P are (3, 2, 3).

2. Let P(x, y, z) be any point which divides the line segment joining points A(3, -2, 5) and B (3, 4, 2) in the ratio 2:1 externally.
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 8
Therefore coordinates of P are (3, 10, -1).

Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

Question 5.
Find the ratio in which the line joining the points (1, 2, 3) and (-3, 4, -5) is divided by the xy-plane.
Answer:
Let the line joining the points A(1, 2, 3) and B(-3, 4, -5) is divided by the xy-plane in the ratio k:1.
Then the coordinate
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 9
Since the point lies on xy-axis, we have;
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 10
Thus the required ratio is \(\frac{3}{5}\); ie, 3:5.

Question 6.
Find the coordinates of the points which trisect the line segment joining the points P(4, 2, -6) and Q (10, -16, 6).
Answer:
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 11
Let R and S be two points which trisect the line join of PQ. Therefore PR = RS = SQ Then coordinate of R is
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 12
= (6, -4, -2)
Then coordinate of S is
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 13
= (8, -10, 2).

Plus One Maths Introduction to Three Dimensional Geometry Practice Problems Questions and Answers

Question 1.
Find the distance between the following pair of points: (1 score each)

  1. (2, 3, 5) and (4, 3, 1)
  2. (-3, 7, 2) and (2, 4, -1)
  3. (-1, 3, -4) and (1, -3, 4)

Answer:
1. Let A(2, 3, 5) and B(4, 3, 1) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Practice Problems Questions and Answers 14

2. Let A(-3, 7, 2) and B(2, 4, -1) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Practice Problems Questions and Answers 15

Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

3. Let A(-1, 3, -4) and B(1, -3, 4) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Practice Problems Questions and Answers 16