CBSE Set Qa1 Physics Sample Test Papers For Class 12th for students online

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Physics Class- XII (CBSE)
You are on Set no 1 Answer 1 to 13

Q1) Horizontal component of Earth's magnetic field at a place is 3 times the vertical component. What is the value of angle of dip at this place?  (Marks 1)
Ans1)  tan = BV/BH = 1/3
or = 30o

Q2) Force between two point electric charges kept at a distance d apart in air is F. If these charges are kept at the same distance in water, how does the force between them change?  (Marks 1)
F = (1/4or) q1q2/r2
where r is the relative permittivity.
r = 1 for air & 81 for water.
Fa/Fw = 81/1
or Fw = 81 Fa

Q3) Give any two factors on which thermo-electric emf produced in a thermo-couple depends. (Marks 1)
Thermoemf depends on
(i) Temp difference between the junctions
(ii) Nature of metals forming thermocouple

Q4) The electric current in a wire in the direction from B to A is decreasing. What is the direction of induced current in the metallic loop kept above the wire as shown in the figure?  (Marks 1)


Induced current will be clockwise

Q5) Name the electromagnetic radiations used for viewing objects through haze and fog.  (Marks 1)
Infrared rays.

Q6) Give the ratio of the number of holes and the number of conduction electrons in an intrinsic semiconductor.  (Marks 1)

Q7) In the given diagram, is the diode D forward or reversed biased?  (Marks 1)

Ans7) The diode is reverse biased since n-type is at a potential higher than p-type.

Q8) Name the planet which has maximum value of albedo. 
(Marks 1)

Q9) Two point electric charges of unknown magnitude and sign are placed a distance 'd' apart. The electric field intensity is zero at a point, not between the charges but on the line joining them. Write two essential conditions for this to happen.  (Marks 2)
(i) They should be of opposite signs
(ii) They must be of unequal magnitude.

Q10) The variation of potential difference V with length l in case of two potentiometers X and Y is as shown in the given diagram. Which one of these two will you prefer for comparing emf's to two cells and why?  (Marks 2)

Ans10) Potentiometer Y has less potential gradient (V/l). It will require more length for measuring same potential difference & hence will be more sensitive and accurate. So Y will be prefered

Q11) ) Name any one material having a small value of temperature coefficient of resistance. Write one use of this material.  (Marks 2)
Constantan (or manganin) has low temp coefficient of resistance. It is used for making standard resistors.

Q12) Write two advantages and two disadvantages of a secondary cell over a primary cell.  (Marks 2)
 Advantages of secondary cell over primary cell:
(i) It can be recharged and hence its life is long.
(ii) It has low internal resistance & hence we can get a large current.
(i) It is bulky & heavy , so is not easily portable.
(ii) Its Initial cost is very high.

Q13) In the figure, the straight wire AB is fixed while the loop is free to move under the influence of the electric currents flowing in them. In which direction does the loop begin to move? Give reason for your answer.  (Marks 2)


Force on AB & CD are equal & opposite, so cancel out.
The side AD experiences a force of attraction towards EF as they carry current in same direction. The force is given by FAD = oI1I2/2x
The side BC experiences a force of repulsion towards EF which is given by  FBC = oI1I2/2y
Net force = FAD - FBC which is towards EF i.e. ,the loop begins to move towards left.

Q14) A radio frequency choke is air-cored coil whereas in audio frequency choke is iron-cored. Give reasons for this difference
(Marks 2)

Ans14) Inductance L of iron-cored chokes is much greater than that of air cored choke.
Now Irons = Vrms/XL = Vrms/2VL
For a required value of irons, VL should be constant. If V is high (radiofrequency), L should be small (air cored). If V is low (audio frequency), L should be high (iron - cored)

Q15) An astronomical telescope consists of two thin lenses set 36 cm apart and has a magnifying power 8. Calculate the focal lengths of the lenses.  (Marks 2)
Ans15) m = fo/fe or fo = mfe
Also fo + fe = L
or mfe + fe = L
(m + 1) fe = L
Given m = 8 , L = 36
fe (1 +  8) = 36 or fe = 4 cm
& fo = 8 fe = 8 x 4 = 32 cm.

Q16) Use the mirror formula to show that for an object lying between the pole and focus of a concave mirror, the image formed is always virtual in nature. (Marks 2)
Ans16) According to mirror formula
1/f = 1/v + 1/u
using sign convention
-1/|f| = 1/v - 1/|u|
or 1/v = 1/|u| - 1/|f|
object lies between pole & focus. So |u| < |f|
= 1/|u| 1/|f|
or 1/v is (+ ve) or v is (+ ve). Hence a virtual image is formed.


Q17) An -particle and a proton are accelerated through the same potential difference. Calculate the ratio of velocities acquired by the two particles.  (Marks 2)
Ans17) When a charge particle of charge q is accelerated through a potential difference of V, it acquires a velocity v given by
qV = 1/2 mv2
For particle
(2e)V = 1/2 mv2  - (i)
For proton
(e)V = 1/2 mpvp2  - (ii)
Dividing (i) by (ii)
2 = mv2/mpvp2
or v2/vp2 = 2mp/m
mp/m = 1/4
= v2/vp2 = 2 x 1/4 =1/2
or v/vp = 1/2

Q18) Describe the method used for determination of distance of a planet by parallax method.  (Marks 2)

Parallax Method: A planet P is observed simultaneously from two different observation centres A & B on surface of earth. The angle between two directions of observations along which the planet is viewed, is measured. The distance AB ( = b) is called basis. Let D be the distance of the planet from the surface of the earth i.e, AP = D = BP.
(radians) = Arc(AB)/radians(PA) = b/D
or D = b/

Q19) Explain the principle of a tangent galvanometer. How does the reduction factor of the galvanometer change, when (i) number of turns of the coil is increased and (ii) radius of the coil is decreased? Give reason for your answer in each case.  (Marks 3)
Tangent galvanometer is based on Tangent law which states that when a magnet is suspended under the combined action of two uniform magnetic fields of intensities F & H perpendicular to each other, the magnet comes to rest making an angle with the direction of H such that F=H tan
Reduction factor of a tangent galvanometer is given by
K = 2rBH/no
where n : no of turns of coil
BH : Horizontal component of earth's field
r : radius of coil
(i) K 1/n, so the reduction factor decreases when the no. of turns of the coil increases
(ii) K r, so reduction decreases when the radius of the coil is decreased.

Q20) A wire of uniform cross-section and length l has a resistance of 16 . It is cut into four equal parts. Each part is stretched uniformly to length l and all the four stretched parts are connected in parallel. Calculate the total resistance of the combination so formed. Assume that stretching of wire does not cause any change in the density of its material.  (Marks 3)
Let l be the length of unstretched wire & A be its area. Its resistance is 16
When cut into 4 equal parts, resistance of each small part will be 16/4 = 4 , & length of each small part = l/4
Now R = sl/A
For each small part 4 = (s(l/4))/A or 4 = sl/4A  -  (i)
When each part of length l/4 is stretched to length l, area of each part becomes 1/4th i.e, A/4,
... resistance of each stretched part R' = sl/(A/4) = 4sl/A - (ii)
From (i) & (ii)
R' = 64
Now four wires each of resistance 64 are connected in parallel. Their resultant Rp is given.
1/Rp = 1/64 + 1/64 + 1/64 + 1/64 = 1/16
or Rp = 16

Q21) An electric heater and an electric bulb are rated 500 W, 220 V and 100 W, 220 V respectively. Both are connected in series to a 220 V a.c. mains. Calculate the power consumed by (i) the heater and (ii) electric bulb.  (Marks 1+2)
We know P = V2/R or R =V2/P
For 500 W heater - resistance R1 = (220)2/500 = 484/5
Resistance of 100 W bulb R2 = (220)2/100 = 484
When connected in series, then total resistance R is
R = R1 + R2 = 484/5 + 484 = (484 x 6)/5
As they are in series, same current flows through them
I = V/R = (220 x 5)/(6 x 484) = 25/66 A
Power consumed by 500 W, 220V heater = I2R1
= (25/66)2 x  484/5 = 13.89 W
Power consumed by 100 W, 220 bulb = I2R2
= (25/66)2 x 484 = 69.44 W 

Q22) Why is diffraction of sound waves easier to observe than diffraction of light waves? What two main changes in diffraction pattern of a single slit will you observe when the monochromatic source of light is replaced by a source of white light?   (Marks 3)
For diffraction to occur, the size of obstacle should be comparable to wavelength. Wavelength of Sound Waves is very large compared to light waves, hence bending of light waves is very small as compared to sound waves. Changes in diffraction pattern of a single slit when monochromatic source of light is replaced by white light:
1. In case of monochromatic light, diffraction pattern consists of alternate bright and dark bands of unequal widths. In case of white light the diffraction pattern is coloured & central maximum is white.
2. As band width is proportional to wavelength, red band with larger wavelength is wider than violet band with smaller wavelength.

Q23) Explain surface wave and sky wave propagations of radio waves. Why is short wave communication over long distances not possible by surface wave propagation.  (Marks 3)
Surface wave propagation: The amplitude modulated radiowaves which are travelling directly following the surface of the earth are called surface waves. They can have frequency upto 1500KHz.
Sky wave propagation: The amplitude modulated radiowaves which are received after being reflected from the ionoshere are called sky waves. Their frequency is from 1500KHz to 40MHz. Short wave communication over long range is not possible by surface wave propagation as the surface wave band around the corners of the objects on the earth and hence their intensity falls with distance.

Q24) Give reasons for following observations on the surface of moon:
(i) Sun-rise and sun-set are abrupt
(ii) Sky appears dark
(iii) A rainbow is never observed.  (Marks 3)
(i) As there is no atmosphere on the moon, the phenomenon of refraction does not occur due to which sun-rise and sun-set are abrupt.
(ii) There is no scattering of light on moon due to absence of atmosphere & hence sky around moon appears black.
(iii) Rainbow is the result of refraction and total internal reflection of light by water molecules in air. As there is no water in the atmosphere of moon, rainbow is never observed.

Q25) The energy levels of an atom of element are shown in the following diagram. Which one of the level transitions will result in the emission of photons of wavelength 620 nm? Support your answer with mathematical calculations.  (Marks 3)

Ans25) E2 - E1 = h  = hc/
= 620 nm = 620 x 10-9 m
E2 - E1 = (6.6 x 10-34 x 3 x 108)/(620 x 10-9) J = 3.2 x 10-19 J = (3.2 x 10-19)/(1.6 x 10-19) eV = 2 eV
In transition D E2 - E1 = - 1 - (- 3) = 2 ev
Hence D emits radiation of wavelength 620 nm .


Q26) Give the logic symbol and truth table for AND gate. Explain, with the help of a circuit diagram, how this gate is realised in practices.  (Marks 3)
Symbol of AND Gate:

Truth Table of AND Gate: 

0 0 0
1 0 0
0 1 0
1 1 1

Realization of AND Gate:

(i) When both A and B are connected to earth (i.e. A = 0 and B = 0) both the diodes D1 and D2 get forward biased and hence conduct. The diodes being ideal, no voltage drop takes place across either diode. Therefore a voltage drop of 5V takes place across R, with C at zero potential w.r.t. earth. Thus the output y (which is the voltage at C) is 0 (in level).

(ii) When A is earthed and B is connected to positive terminal of battery 5V (i.e. A = 0 and B = 1), the diode D1 will conduct while D2 will not conduct. Since diode D1 is ideal, no voltage drop takes place across D1 . Therefore a voltage drop of 5V takes place across R, having D at + 5V and C at zero volt  w.r.t. earth. Now the output y is (0 in level).

(iii) When A is connected to positive terminal of battery 5 V and B is earthed (i.e. A = 1 and B = 0) the diode D2 will conduct while D1 will not conduct. Since D2 is ideal, no voltage drop takes place across D2. Therefore, a voltage drop of 5V takes place across R, having terminal D at + 5V and C at zero volt w.r.t. earth . Now the output Y is 0 (in level).

(iv) When A and B both are connected to positive terminal of battery 5V (i.e. A = 1 and B = 1), none of the diodes will conduct. There will be no current through R. Now potential at C is equal to potential at D, which is equal to + 5V w.r.t. earth. Hence the output is 1 (in level).

Q27) Drawing a labelled circuit diagram, explain how a NPN transistor can be used as an amplifier in common base configuration.  (Marks 3)

The input (emitter base) circuit is forward biased by using a low battery voltage VEB. As a result, the resistance of input circuit is small . The output (collector base) circuit is reverse biased by using battery voltage VCB. Due to which, the resistance of output circuit is large. RL is a load resistance connected in collector circuit. The low input a.c. voltage signal is applied across emitter base circuit and the amplified a.c. voltage signal (i.e. output) is obtained as the change in collector voltage.
When no a.c. signal voltage is applied to input circuit but emitter base circuit is closed then let us consider that Ie, Ib and Ic be the emitter current, base current and collector current respectively. Then, according to Kirchhoff's first law.
Ie = Ib + Ic   - (i)
  Let us consider 5% of the emitter current appears as base current due to electron hole combination in base and 95% of the emitter current flows as a collector current. i.e., Ib = 5% of Ie = 0.05 Ie and Ic = 95% of Ie = 0.95 Ie . Due to collector current Ic voltage drop across RL = ICRL. If VC is collector voltage (i.e.,potential difference between collector and base ) then
or VC = VCB - ICRL   - (ii)
  When the input signal voltage is fed to the emitter base circuit, it will change the emitter voltage and hence to the emitter current; which in turn will change the collector current. Due to which the collector voltage VC will vary in accordance with relation (ii). This variation in collector voltage appears as an amplified output.
Phase relationship between input and output voltages.
  When the positive half cycle of input a.c. signal voltage comes, it opposes the forward biasing of the emitter base circuit. Due to which the emitter current decreases and consequently the collector current decreases. As a result of which, the collector voltage VC increases [from relation (ii)]. Since the collector is connected to the positive terminal of VCB battery, therefore the increase in collector voltage means the collector will become more positive. This indicates that during positive half cycle of input a.c. signal voltage, the output signal voltage at the collector also varies through a positive half cycle.
  When negative half cycle of input a.c. signal voltage comes, it supports the forwards biasing of emitter base circuit. Due to which the emitter current increases and consequently the collector current increases. As a result of which , the collector voltage VC decreases [from relation (ii)] i.e., the collector becomes less positive. This indicates that during negative half cycle of input a.c. signal voltage, the output signal voltage at the collector also varies through the negative half cycle.
  Thus in common base amplifier circuit, the input signal voltage and the output collector voltage are in the same phase, as shown in fig.

Q28) Explain the effect of introducing a dielectric slab between the plates of a parallel plate capacitor on its capacitance. Derive an expression for its capacitance with dielectric as the medium between the plates.
Give the principle and explain the working of a Van de Graaff generator with the help of a labelled diagram  (Marks 5)
Van de Graff Generator:
: This generator is based on
(i) the action of sharp points i.e. the phenomenon of corona discharge.
(ii) the property that charge given to a hollow conductor is transferred to outer surface and is distributed uniformly over it.

Working: The spray comb is given a positive potential (~ 104 Volt.) w.r.t. the earth by high tension source H.T. Due to discharging action of sharp points, a positively charged electric wind is set up, which sprays positive charge on the belt (corona discharge). As the belt moves, and reaches the sphere, a negative charge is induced on the sharp ends of collecting comb B2 and an equal positive charge is induced on the farther end of B2 . This positive charge shifts immediately to the outer surface of S. Due to discharging action of sharp points of B2, a negatively charged electric wind is set up. This neutralises the positive charge on the belt. The uncharged belt returns down, collects the positive charge from B1, which in turn is collected by B2. This is repeated. Thus the positive charge on S goes on accumulating.  

Now, the capacity of spherical shell = 4oR, where R is radius of he shell.
As V = Q/C ... V = Q/(4oR)
Hence the potential V of the spherical shell goes on increasing with increase in Q.


Capacitor of parallel plate capacitor with dielectric medium between the plates:
The capacitance of a parallel plate capacitor of plate area A and plate separation d with vacuum/air in between is
Co =oA/d
Suppose Q are the charges on the capacitor plates which produce a uniform electric field o in the space between the plates.

When a dielectric slab of thickness t < d is introduced between the plates, the molecules in the slab get polarized in the direction of o. The electric polarization vector in the direction of o induces an electric field p opposite to o. Therefore , the effective field inside the dielectric is
= o - p 
Outside the dielectric, field remains o only .

Therefore, potential difference between the two plates is
V = Eo(d - t) + Et
But Eo/E = r or K ... E = Eo/K
... V = Eo (d - t) + Eo/K t
V = Eo[d - t + t/K]
As Eo = /o = Q/Ao
... V = Q/Ao [d - t + t/K]
... Capacitance of the capacitor with dielectric in between is
C = Q/V = Ao/[d - t + t/K] =  oA/[d - t (1 - 1/K)]
i.e. C = oA/[d - t (1 - 1/K)]
Clearly C Co i.e. on introduction of a dielectric slab in between the plates of a capacitor, its capacitance increases.
If d = t
C = oA/[d - d + d/K)]
or C = KoA/d
or C = KCo

Q29) Distinguish between reactance and impedance. When a series combination of a coil of inductance L and a resistor of resistance R is connected across a 12 V, 50 Hz supply, a current of 0.5A flow through the circuit. The current differs in phase from applied voltage by /3 radian. Calculate the value of L and R.  (Marks 5)

Reactance is the resistance offered by an inductor or a capacitor to the flow of alternating current through them :
Inductive reactance  = XL = L
Capacitive reactance= XC = 1/
Impedance :
The combined oppositions offered by a resistor, inductor & capacitor to the flow of alternating current through them is called reactance
Numerically it is equal to Z = (R2 + (XL - XC)2)
(ii) Vrms = 12V ,  = 50Hz , Irms = 0.5 A
 = /3       
L = ? ,  R = ?
Z = Vrms/Irms = 12/0.5 =  24   - (i) 
Also tan   = XL/R
... XL/R = tan(/3) or XL = 3R  - (ii)
Z = (R2 + XL2)
Z2 = R2 + XL2  - (iii)
242 = R2 + 3R2
or R2 =144
R = 12 
XL = L = 2L = 3R
... L = 3R/(2) = (3 x 12)/(2 x 3.14 x 50) = 0.066 H

Q30) Explain the process of release of energy in a nuclear reactor. Draw a labelled diagram of a nuclear reactor and write the function of each part.  (Marks 5)
 Nuclear Reactor:
Principle: A nuclear reactor is based upon controlled nuclear chain reaction.

Construction: The main components of nuclear reactor are;
(1) Nuclear Fuel : It is a fissinable material like 92U235
(2) Moderator : Its function is to slow down the fast moving secondary neutrons produced during the fission. Suitable material used as moderator are heavy water, graphite.
(3) Control Rods : They have the ability to absorb the slow neutrons. To control the chain reaction, these rods cue inserted in the holes of reactor core upto a desirable length. As a result of it, the desired number of neutrons are absorbed and only limited number of neutrons are left to produce fission. These rods are made of boron or Cadmium.
(4) Coolant : It is used to remove the heat produced and transfer it from the core of the nuclear reactor to the surroundings. Generally coolant used is heavy water or liquid Sodium.
The coolant takes up the heat energy produced nuclear fission and passes on this energy to water in heat exchanger. As a result of it superheated steam is produced which drives a turbine coupled with an electric generator.
(5) Shielding : The whole reactor is protected with concrete walls, 2 to 2.5 metres thick, so that the radiations emitted during nuclear radiations may not produce harmful effects.