Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

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Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits NCERT Text Book Questions and Answers

Question 1.
In an n-type silicon, which of the following statement is true.
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c)  Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer:
(c) Holes are minority carriers and pentavalent atoms are the dopants.

Question 2.
Carbon, silicon, and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)c, (Eg)Si and (Eg)Ge. Which of the following statements is true?
(a) (Eg)Si < (Eg)Ge < (Eg)c
(b) (Eg)c < (Eg)Ge < (Eg)Si
(c) (Eg)c > (Eg)Si > (Eg)Ge
(d) (Eg)c = (Eg)Si = (Eg)Ge
Answer:
(c) (Eg)c > (Eg)Si > (Eg)Ge

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 3.
When a forward bias is applied to a p-n junction, it.
(a) raise the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above
Answer:
(c) lowers the potential barrier.

Question 4.
For transistor action, which of the following statements are correct:
(a) Base, emitter and collector regior should have similar size and dopin concentrations.
(b) The base region must be very thin are lightly doped.
(c) The emitter junction is forward biase and collector junction is reverse biased.
(d) Both the emitter junction as well as to collector junction are forward biased.
Answer:
(b) The base region must be very thin are lightly doped,
(c) The emitter junction is forward biase and collector junction is reverse biased.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 5.
For transistor amplifier, the volta gain.
(a) remains constant for all frequencies.
(b) is high at high and low frequencies.
(c) is low at high and low frequencies a constant in the middle frequency large.
(d) None of the above.
Answer:
(c) is low at high and low frequencies a constant in the middle frequency large.

Question 6.
In half-ware rectification, what is the output frequency if the input frequency is 50Hz. What is the output frequency of a fullwave rectifier for the same input frequency.
Answer:
Given Input frequency = 50Hz
Output frequency
For Halfwave rectifier = 50 Hz
For full wave rectifier = 50 × 2 = 100 Hz.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 7.
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8eV. Can it detect a wavelength of 6000 nm?
Answer:
Band gap Eg = 2.8 eV
Energy band gap corresponding to wavelength 6000 nm is given by
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 1
Since Eg, < Eg, so it can not detect a wave length of 6000 nm.

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits One Mark Questions and Answers

Question 1.
The zenerdiode works in______bias.
Answer:
reverse bias.

Question 2.
A transistor is operated in common emitter configuration at Vc = 2 V, such that a change in the base current from 100 µA to 200 µA produces a change in the collector current from 5 mA to 10 mA. The current gain is
(a) 100
(b) 150
(c) 75
(d) 50
Answer:
(d) 50
Explanation: Current gain,
 Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 2

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 3.
The electrical conductivity of an intrinsic semiconductor at 0 K is
(a) less than that an insulator
(b) equal to zero
(c) equal to infinity
(d) more than that of an insulator
Answer:
(b) equal to zero.

Question 4.
The voltage between the terminals A and B is 17 V and Zener breakdown voltage is 9 V. Find the potential across R is
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 3
Answer:
The potential across R = 17V – 9V = 8V.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 5.
Hole is
(a) an antiparticle of electron.
(b) a vacancy created when an electron leaves a covalent bond.
(c) absence of free electrons.
(d) an artificially created particle.
Answer:
(b) a vacancy created when an electron leaves a covalent bond.

Question 6.
A circuit is constructed by using certain gates is given below
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 4
1. Each gate is a…….. gate
2. Complete the truth table of above circuit
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 5
Answer:
1. NAND gate

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 6

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 7.
In both p and n-type semiconductor, actually electrons are flowing. What difference do you observe in the motion of electrons in these semiconductors?
Answer:
Electrons in valence band are flowing. Electrons in conduction band are flowing.

Question 8.
Unidirectional property of diode; Rectification. Then the break down action of Zener diode:…….
Answer:
Voltage regulation.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 9.
For an input frequency 50Hz, the output frequency……..of hall wave rectifier is and the output of full
wave rectifier for the same input frequency is………
Answer:
50 Hz, 100 Hz

Question 10.
The following questions consists of two statements. Assertion: Zener diode works as a voltage regulator Reason: Zener voltage is independent of the Zener current variations and change of load resistance. Write the correct response from the following.
(a) Both assertion and reason are true and the rea¬son is not a correct explanation of the assertion.
(b) Assertion is true, but reason is false.
(c) Both assertion and reason are true and reason is correct explanation for the assertion.
(d) Both assertion and reason are false.
Answer:
(c) Both assertion and reason are true and reason is correct explanation for the assertion.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 11.
Assertion: Semiconductors have -ve temperature co-efficient of resistance.
Reason: As temperature of a semiconductor increases, number density of charge carriers also increases.
(a) Both assertion and reason are correct, but reason is not proper explanation.
(b) Both assertion and reason are correct and reason is proper explanation.
(c) Assertion is correct but reason is wrong.
(d) Assertion is correct, and reason also is correct.
Answer:
(a) Assertion js correct, but reason is incorrect

Question 12.
Correct the following CE amplifier circuit.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 7
Answer:
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 8

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Two Mark Questions and Answers

Question 1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 9
Answer:
Semiconductor – P-type, n-type.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 2.
Fill in the blanks with appropriate word given below. (Base, collector, emitter, bias-collector junction, collector-emitter junction, emitter bias junction) Structurally, a bipolar junction transfer consists of emitter, base and………Out of these regions……….is the most heavily doped. For proper functioning of a transistor………..is forward biased and………….is reverse biased.
Answer:

  1. Collector
  2. Emitter
  3. EB junction
  4. CB junction

Question 3.
Classify the following into conductors, insulators, and semiconductors.
Ga, As, Ni, Calcite, Graphite
Answer:

  1. Conductor – Graphite, Ni
  2. Insulator – Calcite
  3. Semiconductor – Ga,As.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 4.
Construct truth table for following logic circuit.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 10
Answer:
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 11

Question 5.
State whether true or false and justify.

  1. Zener diode are used under forward bias.
  2. In n-p-n transistor current conduction is primarily due to electrons.
  3. Transistor amplifier do not strictly obey law of conservation of energy since output power is greater than input power.
  4. In a transistor amplifier all the frequency will have exactly equal gain.

Answer:

  1. False, Zener diodes are used under reverse bias
  2. True
  3. False
  4. False

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 6.
Match the following
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 12
Answer:
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 13

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Three Mark Questions and Answers

Question 1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 14
An electric circuit containing a battery, a bulb, and two switches is give above
1. Identify the gate an alogues to the above electric circuit
2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 15
If the above two input signal are applied to the gate what will be the shape of out put wave Draw the out put wave.
Answer:
1. OR gate

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 16

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 2.
A boy designs a circuit to study the input and output characteristics of an npn transistor
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 17

  1. Identify the transistor configuration, input current and output current
  2. By keeping the output voltage constant, the boy measures the input current by varying the input voltage. If a graph is drawn, what is the nature of the input characteristic? Justify your answer.

Answer:
1. Common Emitter, ib, and ic

2. Input Characteristics (CE configuration):
The graph connecting base current with base emitter voltage (at constant VCE) is the input characteristics of the transistor.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 18
To study the input characteristics, the collector to emitter voltage (VCE) is kept at constant. The base current IB against VBE is plotted in a graph. The ratio ∆ VBE /∆IB at constant VCE is called the input resistance.
i.e.,Input resistance = \(r_{i}=\frac{\Delta V_{B E}}{\Delta I_{B}}\).

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 3.
Truth table of a logic gate is given below:
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 19
1. Identify the logic gate.
2. Explain the working of this gate using diode and battery.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 20
3. If these input signals are applied across the gate, what will be the shape of the output wave form.
Answer:
1. OR gate

2. Out of syllabus
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 21

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 4.
A circuit using two switches (A and B), cell and bulb is shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 22
1. The above circuit is equivalent to

  • OR gate
  • AND gate
  • NOT gate
  • NOR gate

2. Give symbol and truth table of the above gate
Answer:
1. OR gate

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 23

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Four Mark Questions and Answers

Question 1.
Forward biased pn junction diodes are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 24

  1. Identify the figure, which shows the correct direction of flow of charges.
  2. What do you mean by barrier potential and depletion region of a pn junction?
  3. When forward bias is applied to a p-n junction, what happens to the potential barrier and the width of depletion region.

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 25

2. The potential developed across the junction, which tends to prevent the movement of electron from the n region into the p region, in semiconductor is called a barrier potential.

The space-charge region on either side of the junction at which there is no free charge carriers is known as depletion region.

3. The potential barrier decreases and depletion region gets reduced.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 2.
The symbol of a n-p-n transistor is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 26

  1. Redraw the symbol and mark emitter, collector and base of the transistor.
  2. Arrange the doping concentration and width of emitter, collector and base regions in ascending order.
  3. What happens when both the emitter and the collector of a transistor are forward biased?

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 27
2. Doping concentration of base < doping concentration of collector < doping concentration of emitter. Width of base < Width of emitter < Width of collector.

3. The transistor will work as two p-n junctions with common base terminals.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 3.
A green house has an electric system, which automatically switches ON a heater if the air temperature in the green house drops too low. A manual switch is included so that the automatic system can be switched off.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 28

  1. What is meant by 1 and 0 in digital circuit?
  2. Name logic gate X. Why is it used?
  3. Name the logic gate Y?
  4. Construct a truth table of this electronic system by taking A and B as inputs and D as output.

Answer:
1. 1 – means maximum, 0 – means minimum voltage

2. NOT gate

3. AND gate

4.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 29

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 4.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 30
The forward-biased diode is wrongly given above.

  1. Redraw the above circuit correctly.
  2. Draw the graph of current I with voltage v in forward bias.
  3. Classify the following circuit into forward bias, reverse-bias, unbias.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 31
Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 32
2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 33

3. circuit into forward bias, reversebias, unbias:

  • Reverse bias
  • Forward bias
  • Reverse bias
  • Unbias

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 5.
A car stereo working at stabilized voltage supply of 9 v DC and has a zener diode of 9V, 0.25W. But the voltage supply inside the car is 12V DC.

  1. Which mode of bias will you suggest to connect zener diode voltage regulator?
  2. Draw a circuit diagram of voltage regulation to help the boy.
  3. Which device is essential for circuit diagram? Find the value of that device.

Answer:
1. Reversebias

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 34

3. Resistance: R can be calculated using the equation
Vs = IR + Vz , 12 = \(\left(\frac{0.25}{9}\right)\)R + 9, R = 108Ω.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 6.
Electric current is the flow of charges along a definite direction and take place through metals as well as semiconductors.

  1. Mention the charge carriers in the above cases.
  2. Give the sketch of graph with V along X-axis and I along Y-axis for a metal at room temperature.
  3. Give the physical significance of the slope of the graph.
  4. If the above graph is drawn at 100°C, compared the nature of the graph with the graph at room temperature.

Answer:

  1. Metals – electrons
    Semiconductor – Electrons and holes
  2. Straight line graph
  3. Slope gives conductance
  4. When the temperature increases, resistance is also increased and hence slope decreases.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 7.
A transistor in the common-emitter mode can be used as an amplifier

  1. Design a circuit to amplify an ac signal given in the input region
    [Hint: Give forward biasing to input region, reverse biasing to output region and take output across a resistor]
  2. Derive expressions for voltage gain, current gain and power gain in the above transistor configuration.

Answer:
1.

2. When we apply an AC signal as input, we get an AC base current denoted by iB. Hence input AC voltage can be written as
Vi = iBr ……..(1)
where ‘r’ is the effective input resistance.
This AC input current produces an AC output current (ic) which can flow through a capacitor. Hence the output voltage can be written as
V0 = ic × output resistance
If we take output resistance as RL then v0 becomes
V0 = ic RL
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 36
Substituting eq(1) and eq(2),in the above equation we get
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 37

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits
Power gain:
The power gain Ap can be expressed as the product of the current gain and voltage gain.
ie. power gain A = βac × Av.

Question 8.
The circuit diagram of a full wave rectifier is shown
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 38

  1. Explain how its works? Also draw the output wave form
  2. If the frequency of a.c. at the input is 50Hz what will be the output frequency?

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 39
Full wave rectifier consists of transformer two diodes and a load resistance RL. Input a.c signal is applied across the primary of the transformer. Secondary of the transformer is connected to D1, and D2. The output is taken across RL.
Working:
During the +ve half cycle of the a.c signal at secondary, the diode D1 is forward biased and D2 is reverse biased. So that current flows through D1 and RL.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

During the negative half cycle of the a.c signal at secondary, the diode D1 is reverse biased and D2 is forward biased. So that current flows through D2 and RL.

Thus during both the half cycles, the current flows through RL in the same direction. Thus we get a +ve voltage across RL for +ve and -ve input. This process is called full wave rectification.

2. 100Hz

Question 9.
Forward biased pn junction diodes are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 40
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 41

  1. Identify the figure, which shows the correct direction of flow of charges. (1)
  2. What do you mean by barrier potential and depletion region of a pn junction? (2)
  3. When forward bias is applied to a p-n junction, what happens to the potential barrier and the width of depletion region. (1)

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 42

2. The potential developed across the junction .which tends to prevent the movement of electron from the n region into the p region, in semiconductor is called a barrier potential.

The space charge region on either side of the junction at which there is no free charge carriers is known as depletion region.

3. The potential barrier decreases and depletion region gets reduced.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 10.
LEDs that can emit red, yellow, orange, etc. commercially available.

  1. How these colours are obtained in a LED. (1)
  2. Write any two uses of LED. (1)
  3. What are its advantages over ordinary bulbs? (2)

Answer:
1. Different colours are obtained by changing the concentration of arsenic and phosphors in Gallium Arsenide Phosphide.

2. LEDs find extensive use in remote controls, burglar alarm systems, optical communication, etc.

3. LEDs have the following advantages over conventional incandescent low power lamps:

  • Low operational voltage and less power.
  • Fast action and no warm-up time required.
  • The bandwidth of emitted light is 100A° to 500A° or in other words it is nearly (but not exactly) monochromatic.
  • Long life and ruggedness Fast on-off switching capability.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 11.
The symbol of a n-p-n transistor is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 43

  1. Redraw the symbol and mark emitter, collector and base of the transistor. (1)
  2. Arrange the doping concentration and width of emitter, collector and base regions in ascending order. (2)
  3. What happens when both the emitter and the collector of a transistor are forward biased? (1)

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 44

2. Doping concentration of base < doping concentration of collector < doping concentration of emitter. Width of base < Width of emitter < Width of collector.

3. The transistor will work as two p-n junctions with common base terminals.

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Five Mark Questions and Answers

Question 1.
The transfer characteristic of n-p-n transistor in CE configuration is shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 45

  1. Identify the cut-off region, active region, saturation region from the figure.
  2. In which of these regions, a transistor is said to be switched off.
  3. For a CE transistor amplifier, the audio signal voltage across collector resistance of 2.0 kΩ is 2.0V. Suppose the current amplification factor of the transistor is 100. What should be the value of RB in series with VBB supply of 2.0V, if DC base current has to be 10 times the signal current?
  4. In the working of a transistor, the emitter-base (EB) junction is forward biased while collector-base (CB) junction is reverse biased. Why?

Answer:
1.

  • I – cut off region
  • II – active region
  • III-saturation region

2. Region I

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 46
Idc =10-5 × 10 = 10-4A
Vbb = Vbe + IbRb
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 47

4. Only forward biased emitter-base junction can send the majority charge carriers from emitter to base and only reverse biased collector can collect these majority charge carriers form the base region.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 2.
The basic building blocks of digital electronic circuits are called Logic Gates. Some logic gates and their names are given in the table.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 48
1. Match the symbols of logic gates with their names.
2. Draw the output wave form, from the given input wave form of a NAND gate as shown in figure (input terminals are A and B)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 49
3. Write the truth table forthe given circuit.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 50
Answer:
1. Match the symbols of logic gates with their names:

  • A – 5
  • B – 1
  • C – 2
  • D – 3
  • E – 4

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 51

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 52

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 3.
Following figure is an incomplete circuit of common emitter transistor in CE configuration with the input forward biased.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 53

  1. Identify the transistor an NPN or PNP.
  2. Complete the above circuit diagram by giving proper bias in the output and connect load resistance of 4 KΩ.
  3. When the base current changes by 20µ Afor VBE = .02 V. What is the voltage gain of the amplifier, if Ic = 2mA
  4. npn transistors are preferred in devices with very high-frequency source. Why?

Answer:
1. npn transistor

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 54

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 55

4. The resistance of a semiconductor decreases with rise in temperature.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 4.
Diodes are one of the building elements of electronic circuits. Some type of diods are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 56

  1. Identify rectifier diode from the figure.
  2. Draw the circuit diagram of a forward biased rectifier diode by using a battery.
  3. Draw the forward and reverse bias characteristics of a rectifier diode and mark threshold voltage or cut in voltage.
  4. What happens to the resistance of a semiconductor on heating?

Answer:
a.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 57

b.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 58

c. The forward and reverse characteristics of a silicon diode is as shown in the fig.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 59

d. The resistance of a semiconductor decreases with rise in temperature.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 5.
A full wave rectifier circuit is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 60
1. Draw the output wave form of the rectifier.
2. If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be

  • 50 Hz
  • 70.7 Hz
  • 100 Hz
  • 25 Hz

3. In a zener regulated power supply, a Zener diode with Vz = 6.0V is used for regulation. The load current is to be 4.0 mA and the unregulated input 10.0V. What is the value of series resistor R?
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 61
Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 62

2. 100 Hz

3. Input Voltage = 10V, Zener voltage Vz = 6V The voltage drop on the resistor should be 4V
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 63

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 6.
A P N junction diode is connected to a cell as a shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 64

  1. Name the type of biasing used here
  2. Design a circuit diagram to draw the characteristics of the diode in above biasing.
  3. Trace the characteristics curve if the polarity of battery is reversed

Answer:
1. forward biasing

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 65

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 66
In forward bias, current first increases very slowly up to a certain value of bias voltage. After this voltage, diode current increases rapidly. The diode offers low resistance in forward bias.

In reverse bias, current is very small. It remains almost constant upto break down voltage (called reverse saturation current). After this voltage reverse current increases sharply.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 7.
The circuit diagram of a full wave rectifier is shown.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 67
1. Explain how its works? Also draw the output wave form.
2. If another diode is connected in series with D2, as shown below what will happen to the out put wave form?
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 68
3. If the frequency of a.c. at the input is 50Hz what will be the output frequency of full wave rectifier?
Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 69
Full wave rectifier consists of transformer two diodes and a load resistance RL. Input a.c signal is applied across the primary of the transformer. Secondary of the transformer is connected to D1, and D2. The output is taken across RL.
Working:
During the +ve half cycle of the a.c signal at secondary, the diode D1 is forward biased and D2 is reverse biased. So that current flows through D1 and RL.

During the negative half cycle of the a.c signal at secondary, the diode D1 is reverse biased and D2 is forward biased. So that current flows through D2 and RL.

Thus during both the half cycles, the current flows through RL in the same direction. Thus we get a +ve voltage across RL for +ve and -ve input. This process is called full wave rectification.

2. Out put will be halfwave.

3. 100 Hz.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 8.
Diodes are one of the building elements of electronic circuits. Some type of diods are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 70

  1. Identify rectifier diode from the figure. (1)
  2. Draw the circuit diagram of a forward biased rectifier diode by using a battery. (1)
  3. Draw the forward and reverse bias characteristics of a rectifier diode and mark threshold voltage or cut in voltage. (2)
  4. What happens to the resistance of a semiconductor on heating? (1)

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 71

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 72

3. The forward and reverse characteristics of a silicon diode is as shown in the fig.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 73
4. The resistance of a semiconductor decreases with rise in temperature.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 9.
A full wave rectifier circuit is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 74
1. Draw the output wave form of the rectifier. (1)
2. If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be (2)

  • 50 Hz
  • 70.7 Hz
  • 100 Hz
  • 25 Hz (2)

3. In a zener regulated power supply, a zenerdiode with Vz = 6.0V is used for regulation. The load current is to be 4.0 mA and the unregulated input 10.0V. What is the value of series resistor R?(2)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 75
Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 76

2. 100 Hz

3. Input Voltage = 10V, Zener voltage Vz = 6V The voltage drop on the resistor should be
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 77

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 10.
The following diagram shows energy bands in a semiconductor.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 78
(a) Which diagram shows energy band positions at OK? (1)
(b) What do you mean by energy gap? Match the elements /compounds with their respective energy gap values. (1)

Diamond 6 eV
Aluminium 0.03 eV
Germanium 1.1 eV
Silicon 0.71 eV

(c) Classify solids into conductors, semiconductors, and insulators by drawing energy diagram. (3)
Answer:
(a)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 79

(b) The gap between the top of the valence band and bottom of the conduction band is called the energy band gap.

Column I Column II
A. Diamond 1. 1.1 eV
B.  Aluminium 2. 0.71 eV
C. Germanium 3. 0.03 eV
D. Silicon 4. 6 eV

(c) For conductors
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 80
For Insulators
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 81
For Semiconductors
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 82

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 11.
The given figure shows an npn transistor.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 83

  1. Redraw the figure and show the biasing voltage, direction of current and direction of flow of electrons and holes. (2)
  2. Draw the input and output characteristics of transistor connected in common emitter configuration. (2)
  3. In a transistor, a change of 7.9mA is observed in the collector current for a change of 7.99mA in the emitter current. Determine the change in base current. (1)

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 84

2. Input characteristic
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 85
Output characteristic
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 86

3. ∆IE = ∆Ic + ∆IB, ∆Ic = 7.9mA, ∆IE = 7.99mA
∆IB = ∆IE – ∆Ic
= 7.99 – 7.9 = 0.09mA.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 12.
The transfer characteristic of n-p-n transistor in CE configuration is shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 87

  1. Identify the cut-off region, active region, saturation region from the figure. (1)
  2. In which of these regions, a transistor is said to be switched off. (1)
  3. For a CE transistor amplifier, the audio signal voltage across collector resistance of 2.0 kΩ is 2.0V. Suppose the current amplification factor of the transistor is 100. What should be the value of RB in series with VBB supply of 2.0V, if DC base current has to be 10 times the signal current? (2)
  4. In the working of a transistor, the emitter-base (EB) junction is forward biased while collector-base (CB) junction is reverse biased. Why? (1)

Answer:
1. cut-off region, active region, saturation region from the figure:

  • I – cut off region
  • II – active region
  • III – saturation

2. Region I

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 88

4. Only forward biased emitter-base junction can send the majority charge carriers from emitter to base and only reverse biased collector can collect these majority charge carriers form the base region.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 13.
The basic building blocks of digital electronic circuits are called Logic Gates. Some logic gates and their names are given in the table.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 89
1. Match the symbols of logic gates with their names. (2)
2. Draw the output wave form, from the given input wave form of a NAND gate as shown in figure (input terminals are A and B) (1)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 90
3. Write the truth table for the given circuit. (2)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 92
Answer:
1. Match the symbols of logic gates with their names:

  • A – 5
  • B – 1
  • C – 2
  • D – 3
  • E – 4

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 93

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 94

Question 14.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 95

  1. According to energy gap, Classify them as metal, Insulator, and semiconductor.
  2. From which of the above material we can eject electrons with minimum effort Explain
  3. In Photo electric effect, while we are measuring photo current by varying retarding potential the variations is as shown in graph.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 96
Answer:

  1. Insulator, semiconductor, and conductor
  2. Metals
  3. This graph shows that photocurrent increases and reaches saturation with anode potential. This increase in current shows that electrons are emitted from the metal surface with different energies.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Question 15.
A diode can be properly doped at the time of its manufacture, so that it have a shape break down voltage
1. The above diode is called

  • Zener diode
  • Photo diode
  • Light emitting diode
  • Solar cell

2. Compare V-l Characteristics of above diode with that of an ordinary diode

3. Explain how the above diode can be used as an voltage regulator.

Answer:
1. Zener diode

2. Zener diode has sharp breakdown voltage than ordinary diode

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 97
The zener diode is connected to a fluctuating voltage supply through a resistor Rz. The out put is taken across RL.
Working:
When ever the supply voltage increases beyond the breakdown voltage, the current through zener increases (and also through Rz).

Thus the voltage across Rz increases, by keeping the voltage drop across zener diode as a constant value. (This voltage drop across Rz is proportional to the input voltage)

Similarly, when supply voltage decreases beyond a certain value, the current through the zener diode decreases. Thus the voltage across Rz decreases, by keeping the voltage drop across zener diode as constant (Zener diode as a voltage regulator).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Students can Download Chapter 11 Three Dimensional Geometry 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 11 Three Dimensional Geometry

Plus Two Maths Three Dimensional Geometry Three Mark Questions and Answers

Question 1.
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 1

  1. Write the Cartesian equation. (1)
  2. Find the angle between the line. (2)

Answer:
1.
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 2

2. cosθ
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 3

Question 2.
Find the vector equation of the plane passing through the intersection of the planes \(\bar{r}\).(i + j + k) = 6 and \(\bar{r}\).(2i + 3 j + 4k) = -5 at the point (1,1,1).
Answer:
The Cartesian equation of the planes are x + y + z = 6 and 2x + 3y + 4z = – 5. Therefore the equation of the plane passing through the intersection of these planes is
x + y + z – 6 + k(2x + 3y + 4z + 5) = 0
Since it pass through (1, 1, 1) we get,
1 + 1 + 1 – 6 + k(2 + 3 + 4 + 5) = 0 ⇒ -3 + k14 ⇒ k = \(\frac{3}{14}\)
∴ the equation is
x + y + z + -6 + \(\frac{3}{14}\) (2x + 3 y + 4z + 5) = 0
14x + 14y + 14z – 84 + 6x + 9y + 12z + 15 = 0
20x + 23y + 26z = 69
Vector equation is \(\bar{r}\). (20i + 23j + 26k) = 69.

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 3.
Find the equation of the plane passing through the intersection of the planes x + y + 4z + 5 = 0 and 2x – y + 3z + 6 = 0 and contains the point (1, 0, 0).
Answer:
The equation of the planes passing through the intersection of the planes
x + y + 4z + 5 = 0 and 2x – y + 3z + 6 = 0 is
x + y + 4z + 5 + k(2x – y + 3z + 6) = 0 ____(1)
Since (1) pass through (1, 0, 0)
⇒ 1 + 0 + 0 + 5 + k(2 – 0 + 0 + 6) = 0
⇒ 6 + 8k = 0 ⇒ k = –\(\frac{3}{4}\); Then (1)
⇒ x + y + z + 5 + \(\frac{3}{4}\)(2x – y + 3z + 6) = 0
⇒ 4x + 4y + 16z + 20 – 6x + 3y – 9z – 18 = 0
⇒ 2x – 7y – 7z = 2.

Plus Two Maths Three Dimensional Geometry Four Mark Questions and Answers

Question 1.
Consider the point (-1, -2, -3).

  1. In which octant, the above point lies.(1)
  2. Find the direction cosines of the line joining (-1, -2, -3) and (3, 4, 5). (1)
  3. If P is any point such that OP = \(\sqrt{50}\) and direction cosines of OP are \(\frac{3}{\sqrt{50}}\), \(\frac{4}{\sqrt{50}}\) and \(\frac{5}{\sqrt{50}}\), then find the co-ordinate of P. (2)

Answer:
1. The point lies in the octant X’OY’Z’.

2. Direction ratios of the line joining (-1, -2, -3) and (3, 4, 5) are (3 + 1), (4 + 2), (5 + 3) ⇒ 4, 6, 8 ⇒ 2, 3, 4.
Therefore direction cosines are
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 4

3. Given, OP = \(\sqrt{50}\)
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 5
Therefore the point is (3, 4, 5).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 2.
Consider a cube of side ‘a’ unit has one vertex at the origin O.
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 6

  1. Write down the co-ordinate of 0, 0′, A and A’ (1)
  2. Find the direction ratios of OO’ and AA’. (2)
  3. Show that the angle between the main diagonals of the above cube is cos-1\(\left(\frac{1}{3}\right)\) (1)

Answer:
1. O(0, 0, 0), O'(a, a, a), A(a, 0, 0) and A'(0, a, a).

2. Direction ratios along OO’ is a – 0, a – 0, a – 0
⇒ a, a ,a ⇒ 1, 1, 1

3. cosθ
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 7

Question 3.
Consider two points A and B and a line L as shown in the figure.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 8

  1. Find \(\overline{A B}\) (1)
  2. Find the Cartesian equation of the line L.  (1)
  3. Find the foot of the perpendicular drawn from ( 2, 3, 4 ) to the line L. (2)

Answer:
1. \(\overline{A B}\) = (3 – 1)i + (- 3 – 3)j + (3 – 0)k = 2i – 6j + 3k.

2. The Cartesian equation of a line passing through the point (4, 0, -1 ) and parallel to the vector \(\overline{A B}\) is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 9

3. We take, \(\frac{x-4}{2}=\frac{y}{-6}=\frac{z+1}{3}\) = r then any point of the line can be taken as (2r + 4, -6r, 3r – 1). Assume that this point be the foot of the perpendicular drawn from (2, 3, 4 ). The dr’s of the line is 2 : – 6 : 3 and dr’s of the perpendicular line L is
2r + 4 – 2 : -6r – 3 : 3r – 1 – 4 ⇒ 2r + 2: -6r – 3 : 3r – 5
Since perpendicular,
2(2 r + 2) – 6(-6 r -3) + 3(3r – 5) = 0
49r = -7 ⇒ r = \(\frac{-7}{49}\) = –\(\frac{1}{7}\). Therefore the foot of the
perpendicular is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 10

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 4.
Cartesian equation of two lines are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 11
(i) Write the vector equation of the lines. (2)
(ii) Shortest distance between the lines. (2)
Answer:
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 12
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 13
Question 5.
Consider the points (1, 3, 4) & (-3, 5, 2)

  1. Find the equation of the line through P and Q. (1)
  2. At which point that the above line cuts the plane 2x + y + z + 3 = 0. (3)

Answer:
1. Equation of a line passing through( 1, 3, 4) and (-3, 5, 2) is given by,
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 14
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 15

2. Let \(\frac{x-1}{-2}=\frac{y-3}{1}=\frac{z-4}{-1}\) = λ
Then any point on the line is (-2λ + 1, λ + 3, -λ + 4)
Since the plane 2x + y + z + 3 = 0 cuts the aboveline. We have,
⇒ 2(-λ + 1) + λ + 3 – λ + 4 + 3 = 0
⇒ -2λ + 2 + λ + 3 – λ + 4 + 3 = 0
⇒ -2λ = -12 ⇒ λ = 6
∴ point of intersection is (-2 × 6 + 1, 6 + 3, -6 + 4)
⇒ (-11, 9, -2).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 6.
Let the equation of a plane be \(\bar{r}\). (2i – 3j + 5k) = 7, then

  1. Find the Cartesian equation of the plane. (1)
  2. Find the equation of a plane passing through the point (3, 4, -1) and parallel to the given plane. (2)
  3. Find the distance between the parallel planes. (1)

Answer:
1. Given, \(\bar{r}\).(2i – 3j + 5k) = 7 and if we substitute \(\bar{r}\) = xi + yj + zk Then we get the Cartesian equation as 2x – 3y + 5z – 7 = 0.

2. The equation of a plane parallel to the above plane differ only by a constant, therefore let the equation be 2x – 3y + 5z + k = 0.
⇒ 6 – 12 – 5 + k = 0 ⇒ k = 11
Therefore the equation is 2x – 3y + 5z + 11 = 0

3. The distance between the parallel planes
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 16

Question 7.

  1. State the condition for the line \(\bar{r}\) = \(\bar{a}\) + λ \(\bar{b}\) is parallel to the plane \(\bar{r}\).\(\bar{n}\) = d. (2)
  2. Show that the line \(\bar{r}\) = i + j + λ(2i + j + 4k) is parallel to the plane \(\bar{r}\). (-2i + k) = 5. (1)
  3. Find the distance between the line and The Plane in (ii). (1)

Answer:
1. The line \(\bar{r}\) = \(\bar{a}\) + λ \(\bar{b}\) is parallel to the plane \(\bar{r}\).\(\bar{n}\) = d, if the normal of the plane is perpendicularto the line.
∴ \(\bar{b}\).\(\bar{n}\) = 0.

2. Given,
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 17
The line \(\bar{r}\) = i + j + λ(2i + j + 4k) is parallel to the plane \(\bar{r}\). (-2i + 4k) = 5 ⇒ -2x + 4y = 5.

3. Distance = Distance between – 2x + 4y = 5 and point (1, 1, 0) on the line
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 19

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 8.
Choose the correct answer from the bracket,
(i) If a line in the space makes angle α, β and γ with the coordinates axes, then cos2α + cos2β + cos2γ is equal to (1)
(a) 1
(b) 2
(c) 0
(d) 3
(ii) The direction ratios of the line are \(\frac{x-6}{1}=\frac{2-y}{2}=\frac{z-2}{2}\) (1)
(a) 6, -2, -2
(b) 1, 2, 2
(c) 6, 1, -2
(d) 0, 0, 0
(iii) If the vector equation of a line is \(\bar{r}\) = i + j + k + µ(2i – 3j – 4k), then the Cartesian equation of the line
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 20
(iv) If the Cartesian equation of a plane is x + y + z =12, then the vector equation of the line is (1)
(a) \(\bar{r}\).(2i + j + k) = 12
(b) \(\bar{r}\).(i + j + k) = 12
(C) \(\bar{r}\).(i + y + 2k) = 12
(d) \(\bar{r}\).(i + 3j + k) = 12
Answer:
(i) (a) 1

(ii) (b) 1, 2, 2

(iii) (b) \(\frac{x-I}{2}=\frac{y-1}{-3}=\frac{z-1}{-4}\)

(iv) (b) \(\bar{r}\).(i + j + k) = 12.

Question 9.
Consider the lines \(\bar{r}\) = (i + 2j – 2k) + λ(i + 2 j) and \(\bar{r}\) = (i + 2j – 2k) + µ(2j – k)

  1. Find the angle between the lines.
  2. Find a vector perpendicular to both the lines.
  3. Find the equation of the line passing through the point of intersection of lines and perpendicular to both the lines.

Answer:
1. \(\bar{b}_{1}\) = i + 2j; \(\bar{b}_{2}\) = 2j – k
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 21

2. Perpendicular vector = \(\bar{b}_{1}\) × \(\bar{b}_{2}\)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 22
= i(-2 – 0) -j(-1 – 0) + k(2 – 0)
= -2i + j + k.

3. Equation of line is \(\bar{r}\) = (i + 2j – k) + µ(-2i + j + 2k).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 10.
Consider the line \(\bar{r}\) = (2i – j + k) + λ(i + 2j + 3k)

  1. Find the Cartesian equation of the line.
  2. Find the vector equation of the line passing through A (1, 0, 2) and parallel to the above line.
  3. Write two points on the line obtained in (ii) which are equidistant from A.

Answer:
1. \(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-1}{3}\).

2. \(\bar{r}\) = (i + 2k) + λ(i + 2j + 3k).

3. Put λ = a and λ = -a for any real value ‘a’.
Let us put λ = 1 and λ = -1
\(\bar{r}\) = (i + 2k) + 1(i + 2j + 3k) = 2i + 2j + 5k
⇒ (2, 2, 5)
\(\bar{r}\) = (i + 2k) – 1(i + 2j + 3k) = 0i – 2j – k
⇒ (0, -2, -1)
The equidistant points are(2, 2, 5) and (0, -2, -1).

Question 11.

  1. Find the equation of the plane through the point(1, 2, 3) and perpendicular to the plane x – y + z = 2 and 2x + y – 3z = 5 (2)
  2. Find the distance between the planes x – 2y + 2z – 8 = 0 and 6y – 3x – 6z = 57 (2)

Answer:
1. Required equation is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 23
(x – 1)(3 – 1) – (y – 2)(-3 – 2) + (z – 3)(1 + 2) = 0
2(x – 1) + 5(y – 2) + 3(z – 3) = 0
2x + 5y + 3z – 2 – 10 – 9 = 0
2x + 5y + 3z – 21 = 0

2. The planes are
x – 2y + 2z – 8 = 0 and 3x – 6y + 6z + 57 = 0
ie, 3x – 6y + 6z – 24 = 0 and 3x – 6y + 6z + 57 = 0
Distance
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 24

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 12.
Consider the Cartesian equation of a line \(\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-5}{-2}\)

  1. Find the vector equation of the line. (1)
  2. Find its intersecting point with the plane 5x + 2y – 6z – 7 = 0 (2)
  3. Find the angle made by the line with the plane 5x + 2y – 6z – 7 = 0 (1)

Answer:
1. The vector equation is \(\bar{r}\) = (3i – j + 5k) + λ(2i + 3 j – 2k).

2. Any point on the line is
\(\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-5}{-2}\) = λ
x = 2λ + 3, y = 3, λ – 1, z = -2λ + 5
Since this lies on the plane ,it satisfies the plane
5(2λ + 3) + 2(3λ – 1) -6(-2λ + 5) – 7 = 0
10λ + 6λ + 12λ + 15 – 2 – 30 – 7 = 0
28λ = 24
λ = 6/7
The point of intersection is \(\left[\frac{33}{7}, \frac{11}{7}, \frac{23}{7}\right]\).

3. Let θ be the angle between the line and the plane. The direction of the line and the plane
\(\bar{b}\) = 2i + 3j + k; \(\bar{m}\) = 5i + 2j – 6k
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 25

Question 13.
From the following figure
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 26

  1. Find \(\overline{A B}\). (1)
  2. Find the vector equation of line L. (1)
  3. Find a point on line L other than C. (2)

Answer:
1. P.v of A = i – j + 4k,
P.v. of B = 2i + j + 2k
\(\overline{A B}\) = p. v. of B – p. v. of A
= 2i + j + 2k -(i – j + 4k) = i + 2j – 2k.

2. The line L passes through (1, -2, -3) and parallel to \(\overline{A B}\)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 28
∴ Vector equation of line L is \(\bar{r}=\bar{a}+\lambda \bar{m}\)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 29

3. From (1) of part (ii), we have
xi + yj + zk = (l + λ)i + (-2 + 2λ)j + (-3 – 2λ)k
Put λ = 1
⇒ xi +yj + zk = (1 +1)i + (-2 + 2)j + (-3 – 2 )k
⇒ xi + yj + zk = 2i + 0j – 5k
Therefore a point on line L is (2, 0, -5).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 14.
Find the vector equation of the plane which is at a distance of \(\frac{6}{\sqrt{29}}\) from the origin with perpendicular vector 2i – 3j + 4k. Convert into Cartesian form. Also, find the foot of the perpendicular drawn from the origin to the Plane.
Answer:
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 30
Perpendicular distance from origin = d = \(\frac{6}{\sqrt{29}}\)
The equation of the Plane is \(\bar{r}\).\(\hat{n}\) = d
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 31
Cartesian equation is 2x – 3y + 4z = 6
The direction cosines perpendicular to the Plane is \(\frac{2}{\sqrt{29}},-\frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}\).
Perpendicular distance to the Plane is as \(\frac{6}{\sqrt{29}}\)
Hence the foot of the perpendicular is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 32

Question 15.
Consider the Plane \(\bar{r}\).(-6i -3j – 2k) + 1 = 0, find the direction cosines perpendicular to the Plane and perpendicular distance from the origin.
Answer:
Convert the equation of the plane into normal form
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 33
Direction cosines perpendicular to the Plane is \(\frac{6}{7}, \frac{3}{7}, \frac{2}{7}\)
Perpendicular distance from the origin is \(\frac{1}{7}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 16.
Consider three points (6, -1, 1), (5, 1, 2) and (1, – 5, -4) on space.

  1. Find the Cartesian equation of the plane passing through these points. (2)
  2. Find direction ratios normal to the Plane.(1)
  3. Find a unit vector normal to the Plane. (1)

Answer:
1. Equation of a plane passing through the points (6, -1, 1),(5, 1, 2) and (1, -5, 4)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 34
⇒ (x – 6)(-10 + 4) – (y + 1)(5 + 5) + (z – 1)(4 + 10) = 0
⇒ (x – 6)(-6) – (y + 1)(10) + (z – 1)(14) = 0
⇒ -6x + 36 – 10y – 10 + 14z – 14 = 0
⇒ 6x +10y – 14z -12 = 0.

2. Dr’s normal to the plane are 6 : 10 : -14 ⇒ 3 : 5 : -7.

3. Since the dr’s normal to the plane are 3 : 5 : -7, a unit vector in this direction is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 35

Question 17.
Consider a straight line through a fixed point with position vector 2i – 2j + 3k and parallel to i – j + 4k.

  1. Write down the vector equation of the straight line. (1)
  2. Show that the straight line is parallel to the plane \(\bar{r}\).(i + 5y + k) = 5 (1)
  3. Find the distance between the line and plane. (2)

Answer:
1. Vector equation of a straight line is \(\bar{r}=\bar{a}+\lambda \bar{b}\) where a is \(\bar{a}\) fixed point and \(\bar{b}\) is a vector parallel to the line. Here \(\bar{a}\) = 2i – 2y + 3 it and \(\bar{b}\) = i – j + 4k. Therefore vector equation of the line \(\bar{r}\) = 2i – 2j + 3k + λ(i – j + 4k).

2. The vector parallel to the line is i – j + 4k and vector normal to the plane is i + 5j + k.
Then, (i – j + 4k). (i + 5j + k) = 1 – 5 + 4 = 0
implies that straight line and plane are parallel.

3. A point on the line is 2i – 2j + 3k. Then the distance of 2i – 2j + 3k to the given plane
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 36

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 18.
Consider the vector equation of two planes \(\bar{r}\).(2i + j + k) = 3, \(\bar{r}\).(i – j – k) = 4

  1. Find the vector equation of any plane through the intersection of the above two planes. (2)
  2. Find the vector equation of the plane through the intersection of the above planes and the point (1, 2, -1 ) (2)

Answer:
1. The cartesian equation are 2x + y + z – 3 = 0 and x – y – z – 4 = 0 Required equation of the plane is
(2x + y + z – 3) + λ(x – y – z – 4) = 0
(2+ λ)x + (1 – λ)y + (1 – λ)z + (-3 – 4λ) = 0.

2. The above plane passes through (1, 2, -1)
(2+ λ)1 + (1 – λ)2 + (1 – λ)(-1) + (-3 – 4λ) = 0
3 – 3 + 4λ = 0
λ = 0
Equation of the plane is 2x + y + z – 3 = 0
\(\bar{r}\).(2i + j + k) = 3.

Question 19.
(i) Distance of the point(0, 0, 1) from the plane x + y + z = 3
(a) \(\frac{1}{\sqrt{3}}\) units
(b) \(\frac{2}{\sqrt{3}}\) units
(c) \(\sqrt{3}\) units
(d) \(\frac{\sqrt{3}}{2}\) units
(ii) Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to x – y + z = 0 (3)
Answer:
(i) (b) \(\frac{2}{\sqrt{3}}\) units.

(ii) Equation of the plane passing through the intersection is of the form
x + y + z – 1 + λ(2x + 3y + 4z – 5) = 0 _____(1)
(1 + 2λ)x + (1 + 3λ)j + (1 + 4λ)z – 1 – 5λ = 0
Thr Dr’s of the required plane is
(1 + 2λ), (1 + 3λ), (1 + 4λ)
Thr Dr’s of the Perpendicular plane is 1, -1, 1
⇒ (1 + 2λ)(1) + (1 + 3λ)(-1) + (1 + 4λ)(1) = 0
⇒ 1 + 2λ – 1 – 3λ + 1 + 4λ = 0
⇒ 3λ + 1 = 0 ⇒ λ = \(\frac{-1}{3}\)
(1) ⇒ x + y + z – \(\frac{1}{3}\)(2x + 3y + 4z – 5) = 0
⇒ 3x + 3y + 3z – 2x – 3y – 4z + 5 = 0
⇒ x – z + 2 = 0.

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 20.
Consider a plane \(\bar{r}\).(6i – 3j – 2k) + 1 = 0

  1. Find dc’s perpendicular to the plane. (2)
  2. Find a vector of magnitude 14 units perpendicular to given plane. (1)
  3. Find the equation of a line parallel to the above vector and passing through the point (1, 2, 1 ). (1)

Answer:
1. Given, \(\bar{r}\).(6i – 3j – 2k) + 1 = 0 ____(1)
Now, |6i – 3j – 2k| = \(\sqrt{36+9+4}\) = 7
∴ \(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\) is a unit perpendicular to the plane (1)
⇒ the dc’s perpendicular to the plane (1) are \(\frac{6}{7},-\frac{3}{7},-\frac{2}{7}\).

2. We have, \(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\) is a unit perpendicular to the Plane (1). Therefore, a vector of magnitude 14 units perpendicular to the Plane (1) is 14(\(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\))
⇒ 12i – 6j – 4k.

3. Equation of a line parallel to the vector 12i – 6j – 4k and passing through the point (1, 2, 1 )is given by
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 37

Plus Two Maths Three Dimensional Geometry Six Mark Questions and Answers

Question 1.
Consider the pair of lines whose equations are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 38

  1. Write the direction ratios of the lines. (1)
  2. Find the shortest distance between the above skew lines. (4)
  3. Find the angle between these two lines. (1)

Answer:
1. The direction ratios are 2, 5, – 3 and – 1, 8, 4.

2. The given lines are \(\bar{r}\) = (2i + j – 3k) + λ(2i + 5j – 3k)
i.e. \(\bar{r}\) = \(\overline{a_{1}}+\lambda \overline{b_{1}}\),
where \(\overline{a_{1}}\) = 2i + j – 3k) + λ(2i + 5j – 3k)
and \(\bar{r}\) =(-i + 4j + 5k) + µ(-i + 8j + 4k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 39

3. cosθ
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 40

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 2.
Consider the pair of lines \(\bar{r}\) = 3i + 4j – 2k + λ(-i + 2j + k) ——L1, \(\bar{r}\) = i – 7j – 2k + µ(i + 3j + 2k) ——L2

  1. Find one point each on lines L1 and L2. (1)
  2. Find the distance between those points. (2)
  3. Find the shortest distance between L1 and L2. (3)

Answer:
1. By putting λ = 0 in line L1 and µ = 0 in L2 we get the required points. L1 ⇒ \(\bar{r}\) = 3i + 4j – 2k
∴ Co-ordinate is (3, 4, -2)
L2 ⇒ \(\bar{r}\) = i – 7j – 2k
∴ Co-ordinate is (1, -7, -2).

2. Distance between (3, 4, -2) and (1, -7, -2)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 41

3. Let L1 ⇒ \(\bar{r}\) = 3i + 4j – 2k + λ(-i + 2j + k) is of the form \(\bar{r}=\overline{a_{1}}+\lambda \overline{b_{1}}\) where
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 42

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 3.
Consider the points A (2, 2, -1), B (3, 4, 2) and C (7, 0, 6).

  1. Are A, B, and C collinear? Explain.
  2. Find the vector and Cartesian equation of the plane passing these three points. (2)
  3. Find the angle between the above plane and the line \(\bar{r}\) = (i + 2j – k) + λ(i – j + k) (2)

Answer:
1. Direction ratios along A and B is 3 -2, 4 -2, 2 + 1 ⇒ 1, 2, 3
Direction ratios along B and C is
7 -3, 0 -4, 6 -2 ⇒ 4, -4, 4
Since the direction ratios are not proportional they are not collinear.

2. Cartesian equation of the Plane is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 43
⇒ (x – 2)(14 + 6) -(y – 2)(7 – 15) + (z + 1)(-2 -10) = 0
⇒ 20(x – 2) + 8(y – 2) – 12(z + 1) = 0
⇒ 20x – 40 + 8y – 16 – 12z – 12 = 0
⇒ 20x + 8y – 12z = 68
⇒ 5x + 2y – 3z = 17
Vector Equation is \(\bar{r}\).(5i + 2j – 3k) = 17.

3. Angle between the Plane and the Line
\(\bar{r}\) = (i + 2j – k) + λ(i – j + k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 44

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 4.
Consider three points on space ( 2, 1, 0 ), (3, -2, -2)and(3, 1, 7)

  1. Find the Cartesian equation of the plane passing through the above points. (2)
  2. Convert the above equation into vector form.
  3. Hence, find a unit vector perpendicular to the above plane and also find the perpendicular distance of the plane from the origin. (2)

Answer:
1. Equation of the plane is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 45
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 46
⇒ (x – 2)(-21) – (y – 1)(7 + 2) + z(0 + 3) = 0
⇒ 21x + 42 – 9y + 9 + 3z = 0 ⇒ -21x – 9y + 3z + 51 = 0
⇒ 7x + 3y – z = 17.

2. Vector form is \(\bar{r}\).(7i + 3j – k) = 17 _____(1)

3. Now, |7i + 3j – k| = \(\sqrt{49+9+1}=\sqrt{59}\)
Dividing equation (1) by \(\sqrt{59}\), we get
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 47
Therefore the above equation is the normal form of the plane. Then \(\frac{7 i+3 j-k}{\sqrt{59}}\) is the unit vector perpendicular to the plane and \(\frac{17}{\sqrt{59}}\) is the perpendicular distance from the origin.

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 5.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 48
\(\overline{O A}\) = i + 2j + 3k
\(\overline{O B}\) = i – 2j + 4k
\(\overline{O C}\) = 2i + 3j + k
are adjacent sides of the parallelopiped.

  1. Find the base area of the parallelopiped. (2)
    (Base determined by \(\overline{O A}\) and \(\overline{O B}\))
  2. Find the volume of the parallelopiped. (2)
  3. Find the height of the parallelopiped. (2)

Answer:
1. \(\overline{O A}\) × \(\overline{O B}\) =
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 49
= 14i – j – 4k
Base area = |l4i – j – 4k|
\(=\sqrt{196+1+16}=\sqrt{213}\)

2. Volume of the parallelopiped is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 50
= (14i – j – 4k).(2i + 3 j + k)
= 28 – 3 – 4 = 21.

3. Height = \(\frac{\text {volume}}{\text {base area}}=\frac{21}{\sqrt{213}}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 6.

  1. Find the equation of the line passing through the point (2, 1, 0) and (3, 2, -1) (3)
  2. Find the shortest distance of the above line from the line \(\bar{r}\) = (i – j + 2k) + λ(2i + j – 3k) (3)

Answer:
1.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 51

2.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 52
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 53
= i(-3 + 1) – j(-3 + 2) + k(1 – 2)
= -2i + j – k
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 54

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 7.
The equation of two lines are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 55

  1. Find the dr’s of the given lines. (2)
  2. Find the angle between the given lines. (2)
  3. Find the equation of the line passing through (2, 1, 3) and perpendicular to the given lines. (2)

Answer:
1. The given lines are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 56
The dr’s of (1) are 2, 2, 3 and dr’s of (2) are -3, 2, 5.

2. The angle between (1) and (2) is given by
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 57

3. Let a, b, c be the dr’s of the line perpendicular to lines (1) and (2).
∴ 2a + 2b + 3c = 0, -3a + 2b + 5c = 0
Solving by the rule of cross-multiplication, we get
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 58
∴ dr’s of the required line are 4, -19, 10 and the line passes through (2, 1, 3).
∴ Equation of the required line is
\(\frac{x-2}{4}=\frac{y-1}{-19}=\frac{z-3}{10}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 8.

  1. Find the direction cosines of the vector 2i + 2j – k. (1)
  2. Find the distance of the point (2, 3, 4) from the plane \(\bar{r}\).(3i – 6j + 2 k) = -11. (2)
  3. Find the shortest distance between the lines \(\bar{r}\) = (2i – j – k)+ λ(3i – 5 j + 2k) an \(\bar{r}\) = (i+ 2 j + k)+ µ(i – j + k) (3)

Answer:
1. Direction ratios of the vector 2i + 2j – k is 2, 2, -1
Direction cosines of the vector 2i + 2j – k is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 59

2. The equation of the plane in the Cartesian form is 3x – 6y + 2z + 11 = 0 . Then distance from the point (2, 3, 4) is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 60

3. The given lines are \(\bar{r}\) = (2i – j – k) + λ(3i – 5j + 2k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 61

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 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

HSSLive.Guru

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.

HSSLive.Guru

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.

HSSLive.Guru

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.

HSSLive.Guru

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.

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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

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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

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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

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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

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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:

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

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

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

Plus Two Economics National Income Accounting One Mark Questions and Answers

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

Question 2.
The GDP deflator is equal to
i) Real GDP-Nominal GDP
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img1
Answer:
iii) \(\frac{{ No minal GDP }}{\text { Real GDP }} \times 100\)

Question 3.
NFIA is included in:
(a) NNPFC
(b) NDPFC
(c) GDPFC
(d) All the above
Answer:
(a) NNPFC

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

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

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

Plus Two Economics National Income Accounting Two Mark Questions and Answers

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

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

Final Goods Intermediary goods
Bench Wheat
Bread Wood
Tyre Rubber

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

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

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

Answer:
a. Stock

  • Wealth
  • Inventories
  • Capital
  • Money supply

b. Flow

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

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

Plus Two Economics National Income Accounting Three Mark Questions and Answers

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

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

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

Answer:

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

Question 3.
Provide appropriate term.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img2
Answer:

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

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

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

Question 5.
Classify the following under proper heads.
Flow of teacher services, Flow of subsidies and taxes, Flow of factor rewards, flow of finished goods, Flow of consumption expenditure, Flow of import goods.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img3
Answer:

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

Question 6.

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

Answer:
a. Consumer price index

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

b. GDP deflator

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

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Question 7.
Assume that there are three goods produced in an economy and they are sold at different prices in dif-ferent years. Calculate GDP Deflator.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img4
Answer:
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img5

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

2. Net Indirect tax = NDPMP – NDPFC
= 10300 – 10000 = 300

3. NNPFC = NDPFC + NFIA
= 10000 + 1500
= 11500

Plus Two Economics National Income Accounting Five Mark Questions and Answers

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

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

Answer:

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

Question 2.
Match the following.

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

Answer:

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

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

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

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Question 4.
The phase of circular flow of income in a two sector economy is given below.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img6

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

Answer:

1.

Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img7
2. Circular flow of income:
The concept that the aggregate value of goods and services produced in an economy is going around in a circular way. Either as factor payments, or as expenditures on goods and services, or as the value of aggregate production.

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

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

Answer:

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

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

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Question 6.
1. Estimate the NI of India and Pakistan from the data given below.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img8
2. Which method is used here?
3. What are the other methods of measuring national income?
Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 10.
Relate and complete the identities/equations in column A with column B.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img9
Answer:
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img10

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

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

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

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

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

Answer:

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

Question 13.
Fill in the blanks

  1. GNPMP – ……….. = NNPMP
  2. NNPMP – ………… = NNPFC
  3. GDPFC+ – ………… = GDPMP
  4. GDP + -………….. = GNP

Answer:

  1. GNPMP – depreciation = NNPMP
  2. NNPMP – net indirect tax = NNPFC
  3. GDPFC + net indirect tax = GDPMP
  4. GDP + net factor income from aborad = GNP

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

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

Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img11

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

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

Answer:

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

Plus Two Economics National Income Accounting Eight Mark Questions and Answers

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

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

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

2. NNP = GNP – Depreciation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img12

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

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