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) Draw an equipotential surface in a uniform electric field.  (Marks 1)

Equipotential surfaces are parallel planes perpendicular to the field lines.

Q2) If a wire is stretched to double its original length without less of mass, how will the  resistivity of the wire be influenced?  (Marks 1)
Ans2) The resistivity of the wire will remain same as it is the resistance of wire of unit length & unit cross-sectional area.

Q3) Why do magnetic lines of force prefer to pass through iron than through air?  (Marks 1)
Ans3) The magnetic lines of force prefer to pass through iron then through air as iron has very high  magnetic permeability.

Q4) What is the power factor of an LCR series circuit at resonance?  (Marks 1)
Ans4) Power factor cos = R/(R2 + (L - 1/c)2)
At resonance L = 1/c
... cos   = R/(R2 + 0) = 1

Q5) Why is the transmission of signals using ground waves restricted to frequencies upto 1500 kHz?  (Marks 1)
Ans5) The transmission of signals using ground wave is restricted to frequencies upto 1500 kHz because higher frequencies are damped by interaction with matter.

Q6) The polarising angle of a medium is 60o. What is the refractive index of the medium?  (Marks 1)
Ans6) = tan ip
ip = 60o
... = tan 60o = 3

Q7) How does the collector current change in a junction transistor, if the base region has larger width?  (Marks 1)
Ans7) If the base region has larger width collector current will decrease.

Q8) Two stars A and B have magnitudes -2 and +4 respectively. Which star appears brighter?  (Marks 1)
Ans8) mA - mB = - 2.5 log10 lA/lB
- 2 - 4 = - 2.5 log10 lA/lB
or log(lA/lB) = 6/2.5 = 2.4
or lA = (10)2.4 lB
lA lB
or A with magnitude - 2 is brighter than B.

Q9) An electric flux of -6 x 103 Nm2/C passes normally through a spherical Gaussian surface of radius 10 cm, due to a point charge placed at the center.
(i) What is the charge enclosed by the Gaussian surface?
(ii) If the radius of the Gaussian surface is doubled, how much flux would pass through the surface?  (Marks 2)

Ans9) (i) Flux E = - 6 x 103 Nm2/c
E = q/Eo
or q = E Eo = -6 x 103 x 8.85 x 10-12
= -53.10 x 10-9 C
(ii) E = .
If radius of Gaussian surface is doubled area becomes four times and flux also increase four times
i.e. , -6 x 103 x 4
= -24 x 103 Nm2/C

Q10) Three identical resistors, each of resistance R, when connected in series with a d.c. source, dissipate power X. If the resistors are connected in parallel to the same d.c. source, how much power will be dissipated?  (Marks 2)
Ans10) In series, total resistance
Rs = R + R + R = 3R
power dissipated X = V2/Rs where V is emf of d.c. source.
= V2/3R
When connected in parallel, equivalent resistance Rp is given by
1/Rp = 1/R + 1/R + 1/R = 3/R 
or Rp = R/3
power dissipated = V2/(R/3) = 3V2/R
9(V2/3R) = 9X
Hence in the second case, power dissipated will be nine times of that dissipated in the first case.

Q11) Define mutual inductance. State two factors on which the mutual inductance between a given pair of coils depends.  (Marks 2)
Ans11) Mutual induction is the property of a pair of two coils placed close to each other due to which each coil opposes any change in the strength of current flowing in the other by producing an e.m.f. in itself. It depends upon (give any two) 
(i) Distance between the two coils.
(ii) medium on which the coils are wound
(iii) Geometry of the two coils i.e. size, no of turns in each, area, shape of coil
(iii) Orientation of the two coils.

Q12) Light from a galaxy, having wavelength of 6000 Ao, is found to be shifted towards red by 50 Ao. Calculate the velocity of recession of the galaxy.  (Marks 2)
= 6000 Ao
d = 50 Ao
we know that d/ = V/C
or V = d x C/ = 50 x 3 x 108/6000
= 25 x 105 m/s

Q13) A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. If it is immersed in a liquid index 1.3, what will be its new focal length?  (Marks 2)
 Focal length in air f = 20 cm
ag = 1.6
1/f = (ag - 1)(1/R1 - 1/R2)
1/20 = (1.6 - 1)(1/R1 - 1/R2)
or 1/R1 - 1/R2 = 1/20 x 1/0.6 = 1/12
Now let the new focal length when immersed in liquid be f '
1/f ' = (lg - 1)(1/R1 - 1/R2)
= (1.6/1.3 - 1)(1/12) = 1/52
or f ' = 52 cm

Q14) Draw a labelled ray diagram to show the image formation in astronomical telescope for normal adjustment position. Write down the equation for its magnifying power.  (Marks 2)
Ans14) Ray diagram of image formation by Astronomical telescope in normal adjustment.

Magnifying power = -fo/fe
where fo = focal length of objective lens.
fe = focal length of eye piece.

Q15) The half-life of a radioactive sample is 30 seconds. Calculates (i) the decay constant, and (ii) time taken for the sample to decay to 3/4th of its initial value.  (Marks 2)
Ans15) (i) T1/2 = 30 s
Decay constant = 0.693/T1/2 = 0.693/30
= 0.0231 s-1
(ii) N = No e -t
N/No = 3/4
... 3/4 = e-t
or et = 4/3
t = 2.3026 log10(4/3)
t = 2.3026 log10 (4/3)/0.0231 = 2.3026 x 0.1249/0.0231
or t = 12.45 s

Q16) Draw a logic circuit diagram showing how a NAND gate can be converted into a NOT gate.  (Marks 2)


When both the inputs are joined i.e., A = B NAND becomes NOT

NAND becomes NOT


Q17) What is an ideal diode? Draw the output waveform across R, for the input waveform given below :  (Marks 2)

An ideal diode is one which conducts only in forward bias, i.e., it has zero resistance in forward bias and infinite resistance in reverse bias.

The diode conducts only when forward biased i.e. in position BC. Therefore, the output will be as follows.

Q18) Write, in brief, the method to determine the distance of an inferior planet from the sun.  (Marks 2)
The distance of an inferiors form the earth can be measured by radio echo method. A radiowave is directed towards the planet & the time taken by it to produce an echo is noted. If t is the time taken, then distance of the inferior planet from earth = ct/2, where c is speed of light.

Q19) Explain, with the help of a circuit diagram, the use of potentiometer for determination of internal resistance of a primary cell. Derive the necessary mathematical expression.  (Marks 3)
Ans19) A circuit is set up as shown in diagram below :

Close key K and maintain suitable constant current in the potentiometer wire with the help of rheostat Rh. Adjust the position of jockey at different points of wire and find a point J on the wire where if jockey is pressed, galvanometer shows no deflection. Note the length AJ (= l1)of the potentiometer wire. Now e.m.f. of the cell, E = potential differences across the length l1, of the potentiometer wire.
or  E = Kl1   ...(i)
Where K is the potential gradient across the wire. 
Close key K1, so that the resistance R is introduced in the cell circuit. Again find the position of the jockey on the potentiometer wire where galvanometer shows no deflection. Let it be at J1. Note the length of the wire AJ1 (= l2 say).
Then, potential difference between two poles of the cell, V = potential difference across the length l2 of the potentiometer wire
i.e. V = Kl2   ...(ii)
Dividing (i) by (ii), we have 
E/V = l1/l2   ...(iii)
We know that the internal resistance r1 of a cell of e.m.f. E, when a resistance R is connected in its circuit is given by 
r1 = (E - V)/ V x R = (E/V - 1)R   ...(iv)
Putting the value from (iii) and (iv), we get
r1 = (l1/l2 - 1)R = (l1 - l2)/l2 x R
Thus, knowing the value of l1, l2 and R, the internal resistance r1 of the cell can be determined.

Q20) Calculate the resistance between A and B of the given network.  (Marks 3)


The equivalent circuit is 

This is a balanced wheatstone bridge ... P/Q = R/S 
The resistance of 10 is ineffective. The circuit reduce to 

P & Q are in series & their equvalent = 1 + 2 = 3
R & S are also in series & their equvalent = 2 + 4 = 6
3 & 6 are in parallel. Their equivalent Req is given by 
1/Req = 1/3 + 1/6 = 1/2
or Req = 2

Q21) State Faraday's laws of electrolysis. Write down the relation connecting chemical equivalent and electro-chemical equivalent.  (Marks 3)
Ans21) Faraday's laws of electrolysis
I law : The mass of the substance liberated or deposited at an electrode during electrolysis is directly proportional to the quantity of charge passed through the electrolyte.
m q
or m = Zq
where Z is called the electrochemical equivalent of the substance.
II Law : When the same amount of charge is made to pass through any number of electrolytes, the masses of the substances liberated or deposited at the electrode are proportional to their chemical equivalents i.e.,
m1/m2 = E1/E2
where m1, m2 = masses of the substance liberated or deposited at various electrodes when same charge flows through their electrolytes
E1,E2 = chemical equivalents of the substances deposited or liberated.
Relation between chemical equivalent & electro-chemical equivalent :
E/Z = F
where F is Faraday's constant = 96500C 

Q22) An electron is moving at 106m/s in a direction parallel to a current of 5 A, flowing through an infinitely long straight wire, separated by a perpendicular distance of 10 cm in air. Calculate the magnitude of the force experienced by the electrons.  (Marks 3)
Ans22) The field due to infinitely long wire carrying current is given by
B = oI/2r
I = 5 A
r = 10 cm = 10 x 10-2 m
B = 4 x 10-7 x 5/2 x 10-1 Tesla
Force experienced by electrons
= qVB sin
= 90o
F = qVB
= 1.6 x 10-19 x 106 x 4 x 10-7 x 5/2 x 10-1
= 1.6 x 10-18 N

Q23) A bar magnet, held horizontally, is set into angular oscillations in Earth's magnetic field. It has time periods T1 and T2 at two place, where the angles of dip are 1 and 2 respectively. Deduce an expression for the ratio of the resulting magnetic field at the two places.  (Marks 3)
Ans23) The time period of a bar magnet oscillating in horizontal field H is given by 
T = 2 (I/MH)
Where I = moment of inertia
M = magnetic moment 
H = R cos

is dip angle 
& R is resultant magnetic field
At two places
T1 = 2 (I/MH1) = 2 (I/MR1 cos
T2 = 2 (I/M R2 cos
T1/T2 = ((R2 cos
2)/(R1 cos 1))   = T12/T22 = (R2 cos 2)/( R1 cos 1)
or R1/R2 = (T22 cos
2)/(T12 cos 1)  

Q24) Verify Snell's law of refraction using Huygens' wave theory.  (Marks 3)
Ans24) Refraction on basis of Huygen's wave theory : 
Let XY be a plane surface separating two media 1 and 2, the velocity of light being c1 in the first medium and c2 in the second medium. Suppose a plane wave front AB is incident on surface XY. It first strikes at A and then the successive points towards c. Let i be the angle of incidence. According to Huyghen's principle, from each point on AC, the secondary wavelets start growing in the second medium with speed c2. Let the wave disturbance take time t to travel from B to C, then BC = c1t. During the time, the disturbance from B reaches the point C, the secondary wavelets from point A must have spread over a hemisphere of radius AD = c2t in the second medium. The tangent plane CD drawn from point C over this hemisphere of radius c2t will be the new refracted wave front.

Wave fronts and corresponding rays for refraction by a plane surface separating two media.

Consider a ray POQ normal to both the incident and the refracted wave fronts. Let the angles of incidence and refraction be i and r, which can be taken as the angles made respectively by the incident wave front AB and refracted wave front CD with the surface of separation XY. For CD to be the true refracted wave front, all the wave disturbances must take the same time in travelling from the incident wave front AB to wave front CD. 
Now the total time taken by the ray to travel from P and Q
= PO/c1 + OQ/c2
= (AO sin i)/c1 + (OC sin r)/c2
= (AO sin i)/c1 + ((AC - AO) sin r)/c2
= (AC sin r)/c2 + AO[(sin i)/c1- (sin r)/c2]
Different rays from the incident wave front have different values of AO. Since the time taken by these rays to reach the wavefront CD is same, the coefficient of AO must vanish i.e., sin i/c1 = sin r/c2
or sin i/sin r = c1/c2 = Constant, n21
This proves Snell's law of refraction. The constant n21 is called the refractive index of second medium with respect to the first medium. 
Further, since the incident ray SA, the normal AN and refracted ray AD are respectively perpendicular to the incident wave front AB, the dividing surface XY and the refracted wavefront CD (all perpendicular to the plane of the paper), therefore, they all lie in the plane of the paper, i.e., in the same plane. This proves another law of refraction.

Q25) Find the position of an object which when placed in front of a concave mirror or focal length 20 cm, produces a virtual image, which is twice the size of the object.  (Marks 3)
Here f = -20 cm 
m = +2 (+ve for virtual image)
2 = -20/(-20 - u)
or 2 = 20/(20 + u)
or 20 + u = 10 or u = -10 cm


Q26) If the frequency of the incident radiation on the cathode of a photo cell is doubled, how will the following change :
(i) Kinetic energy of the electrons,
(ii) Photoelectric current,
(iii) Stopping potential.
Justify you answer.  (Marks 3)

Ans26) (i) Let E1 & E2 be K.E. corresponding to frequency & 2 and W be the work function. According to Einsteins photoelectric equation
h = E1 + W   -(i)
& 2h = E2 + W   -(ii)
From (i) and (ii)
E2 - E1 = h = E1 + W
or E2 = 2E1 + W
Hence K.E. will become more then double if the frequency of the incident radiation is doubled.
(ii) Photoelectric current remains the same as it depends on intensity & not on frequency of incident radiation.
(iii) As K.E. increases, stopping potential also increases i.e., it becomes more negative.

Q27) Explain, with the help of a circuit diagram, why the output voltage is out of phase with the input voltage in a common emitter transistor amplifier.  (Marks 3)

When no a.c signal voltage is applied to the input circuit but emitter base circuit is closed let us consider that le, lb and lc be the emitter current, base current and collector current respectively. Then according to Kirchhoff's first law.
Ie = Ib + Ic   ...(i)
Due to current lc, voltage drop across RL = IcRL
If Vc is collector voltage then
VCE = Vc + IcRL
or Vc = VCE - IcRL   ...(ii)
When the positive half cycle of input a.c, signal voltage comes, it supports the forward biasing of the 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)]. Since the collector is connected to the positive terminal of VCE battery, therefore decrease in collector voltage means the collector will become less positive, which means negative w.r to initial value. This indicates that during positive half cycle of input a.c signal voltage, the output signal voltage at the collector varies through a negative half cycle.
When negative half cycle of input a.c signal voltage comes, it opposes the forward biasing of emitter-base circuit, due to which the emitter current decreases and hence collector current decreases; consequently the collector voltage Vc increases [from relation (ii)] i.e., the collector becomes more positive. This indicates that during the negative half cycle of input a.c signal voltage, the output signal voltage varies through positive half cycle.
Thus in a common emitter amplifier circuit, the input signal voltage and the output collector voltage are in opposite phase i.e., 180o out of phase

Q28) With the help of a labelled diagram, describe Millikan's oil-drop experiment for determining the charge of an electron.  (Marks 5)
Ans28) Principle : The working of Millikan oil drop method is based on the measurement of the terminal velocity of the oil droplet under the action of gravity alone and under the combined action of gravity and an electric field opposed of gravity.

: It consists of two optical plane plates A and B 22 cm in diameter. They are parallel to each other at a distance of about 1.5 cm with the help of optically plane insulting strips of glass or ebonite. The plate A has a fine pin hole H in the center. A potential differences of about 10,000 volt is applied between the plates A and B by H.T and polarity of the plates can be reversed with the help of reversing key. The plates are arranged inside a double walled chamber which has three windows at equal angular separations of 120o.
Fine drops of low vapour density and non-volatile liquid (cloak oil) are sprayed by atomiser in the chamber above the hole H of plate A. These drops gets charged by friction during spraying. Some of them may enter into the space between the two plates A and B through the hole H.
The falling oil drops are illuminated by the light from the arc lamp through windows W1. X-rays, are allowed to enter the chamber, if the need arises, through another window W2. X-rays ionise the gas inside the chamber, resulting in more charge on oil drops.
The charged droplets are observed by a microscope having eye piece fitted with a graduated scale through third window (not shown in the figure.)
Working and theory. By using the electric field between the two plates A and B we select the negatively charged oil droplet moving along the scale of the eye piece. Its motion is studied as explained below :
(a) Motion under gravity alone. As the droplet falls under gravity, its velocity goes on increasing. According to Stoke' s law, the opposing viscous force on the droplet also goes on increasing. A stage comes when the viscous force on the oil droplet becomes just equal to its resultant weight. Then, the droplet moves with a constant velocity v1 is called terminal velocity. This is measured by measuring the time taken (t) by the droplet to travel a known distance (D) on the scale of microscope :
i.e. v1 = D/t
Let r = radius of droplet, s = density of oil, = density of air, 
= coefficient of viscosity of air.
The various forces acting in the droplet are :
(i ) wt of droplet (w) acting downwards = 4/3 r3sg
(ii) Bouyant force (B) acting upwards = 4/3 r3g
(iii) Viscous force (F) acting upwards = 6 rv1
When terminal velocity is attained
W = B + F
or 4/3 r3sg = 4/3 r3g + 6 rv1
or 4/3 r3(s - )g = 6 rv1
or r = ((9 v1)/(2(s - )g))   -(i) 
(b) Motion of the oil droplet when electric field is present :
Electric field is switched on and so adjusted that the force on negatively charged oil droplet acts upwards. Let v2 be the terminal velocity acquired by it. The various force acting on it are : 
(i) upward force due to electric field Fe = qE 
(ii) viscous force (F') acting downwards = 6 rv2
(iii) Effective weight acting downwards (W) = 4/3 r3(s - )g
When terminal velocity is attained.
Fe = W + F'
qE = 4/3 r3(s - )g + 6 rv2
or qE = 6 rv1 + 6 rv2 = 6 r(v1 + v2)
substituting the value of r from eqn (i)
q = 6 (v1 + v2)/E ((9 v1)/(2(s - )g))1/2
q = 6 (v1 + v2)d/V ((9 v1)/(2(s - )g))1/2
Millikan found that charge in each droplet was an integral multiple of some fixed basic charge which is charge in electron equal to 1.6 x 10-19 C

Q29) Draw the curves showing the variation of inductive reactance and capacitive reactance with applied frequency of an a.c source.
A capacitor, a resistor of 5 , and an inductor of 50 mH are in series with an a.c. source marked 100 V, 50 Hz. It is found that voltage is in phase with the current. Calculate the capacitance of the capacitor and the impedance of the circuit.  (Marks 2+3)


XL = L = 2L
XC = 1/C = 1/2c
XC 1/ 
R = 5 
L = 50 mH = 50 x 10-3 H
 = 50 Hz
If the voltage & current are in phase in LCR circuit then XL = XC
2L = 1/2c
or c = 1/422L
= 1/(4 x (22/7)2 (50)2 x 50 x 10-3)
= 202.4 x 10-6 F
= 202.4 F
Impedance Z = (R2 + (XL - XC)2)
... Z = R = 5 

Q30) Define capacitance of a capacitor. Give its unit. Derive an expression for the capacitance of a parallel plate capacitor in which a dielectric medium of dielectric constant K fills the space between the plates.
Explain the principle, construction and working of a Van de Graaff generator.  (Marks 5)

Ans30)  Capacitance of a capacitor is defined as the ratio of charge on the plate of the capacitor to the potential difference across its plate.
Its S.I. unit of Farad
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)]