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

Latest for students online. All these are just samples for prepration for exams only. These are not actual papers.

Physics Class- XII (CBSE)
You are on Set no 1 Answer 1 to 12

Q1) Give a reason to show that microwaves are better carriers of signals for long range transmission than radio waves.  (Marks 1)
Ans1)  Microwaves have smaller wavelength due to which they can be transmitted as beam signals in a particular direction, better than radiowaves because microwaves do not spread or bend around the corners of any obstacle coming in their way.

Q2) How does the energy gap in an intrinsic semiconductor vary, when doped with a pentavalent impurity ?  (Marks 1)
Ans2) The energy energy gap in semiconductor decreases when doped with a petavalent impurity.

Q3) State the condition in which terminal voltage across a secondary cell is equal to its e.m.f  (Marks 1)
Ans3) The terminal voltage across a secondary cell is equal to its e.m.f. when no current is drawn from it i.e., it is in an open circuit.

Q4) Draw an equipotential surface in a uniform electric field.
(Marks 1)

Ans4)

 (Equipotential surfaces are parallel planes at 90o to field lines)

Q5) If the number of turns of a solenoid is doubled, keeping the other factors constant, how does the self-inductance of the solenoid change ?  (Marks 1)
Ans5) The self inductance becomes four times on doubling the no of turns of solenoid since L = oN2A/e , L N2

Q6) What is the ration of solar constants on the surfaces of two planets, whose surface temperatures are in the ratio 1 :2 ?  (Marks 1)
Ans6) Solar constant (Temp)
S1/S2 = (T1/T2)4 = (1/2)4 = !/16
or S1 : S2 = 1 : 16

Q7) The wavelength of light coming from a distant galaxy is red-shifted. Is the galaxy receding or approaching the earth ?  (Marks 1)
Ans7) The red shift indicates that the galaxy is receding

Q8) What is the angle of dip at a place where the horizontal and vertical components of earth's magnetic field are equal ?  (Marks 1)
Ans8) tan = BV/BH  = 1
= 45o

Q9) Briefly explain how the distance of an inferior planet from earth can be determined ?  (Marks 2)
Ans9) The distance of an inferior planet from earth can be determined by Copernicus method.
Consider that an inferior planet P and the earth E revolve in circular orbits of radius r1 and r2 around the sun.

As earth and planet revolve around the sun. The distance r1 remains constant, but r2 goes on changing. Also the planet's elongation keeps on changing. The planet's elongation will be maximum when the any subtended by the earth and the sun on the planet is 90o. Let o be the maximum value of planet's elongation. Then in right triangle SPE
r1/r2  = cos o
r2 = r cos o where r = 1 A

Q10) Using Gauss' law, show that no electric field intensity exists inside a hollow charged conductor.  (Marks 2)
Ans10) Consider a hollow charged conductor. Charge reside only on the surface of conductor.

Consider a gaussian surface which lies inside the conductor as shown just touching the conductor. According to Gauss theorem

 . = q/o

where q is the charged enclosed inside the gaussian surface
Here q = 0
Hence E = 0

Q11) Four capacitors are connected as shown in the figure given below :

Calculate the equivalent capacitance between the points X and Y.  (Marks 2)
Ans11) The equivalent circuit is

C1, C2, C3 are in parallel. There equivalent capacitance = C' =C1 + C2 + C3 = 2 + 3 + 5 = 10 F
This equivalent capacitance is in series with C4.
Net capacitance 'C' is given by
1/C = 1/C' + 1/C4 = 1/10 + 1/10 = 1/5
or C = 5 F

Q12) Draw the graph showing the variation of binding energy per nucleon with the mass number of different nuclei. State two inferences from this graph.  (Marks 2)
Ans12)

Inferences from the graph
1. The decrease in binding energy/nucleon at high mass numbers indicates that the nucleons are more tightly bound when they are assembled into two middle size nuclei than in a single high mass nucleus. Therefore energy can be released in nuclear fission of a single massive nucleus into smaller fragments.

2. Similarly, smaller values of binding energy/nucleon for nuclei of low mass numbers indicate that energy can be released if two nuclei of small mass numbers combine to form a single middle class nucleus.

Q13) In a single slit diffraction experiment, if the width of the slit is doubled, how does the (I) intensity of light and (ii) width of the central maximum change. Give reason for your answer.  (Marks 2)
Ans13) If the width of the slit is doubled, then
(i) the intensity of central maximum will becomes four times, because the area of central diffraction band would become 1/4th.
(ii) the width of central maximum will become half since width = 2/d, where d is width of slit

Q14) Draw the logic symbol of a 2-input NAND gate. Write down its truth table.  (Marks 2)
Ans14) Logic symbol of 2 - input NAND gates :

Truth Table

 A B Y 0 0 1 1 0 1 0 1 1 1 1 0

Q15) A ray of light passes through an equilateral glass prism, such that the angle of incidence is equal to the angle of emergence. If the angle of emergence is 3/4 times the angle of the prism, calculate the refractive index of the glass prism.  (Marks 2)
Ans15) Given i = e
A = 60o
e = 3/4 A = i = e = 3/4 x 60 = 45o
when i = e, = min
= (Sin ((A + min)/2))/Sin (A/2)
i + e = A +
2i = A + min ( = min as i = e)
min = 2i - A = 90 - 60 = 30o
= (Sin ((60 + 30)/2))/Sin (60/2) = (1/2)/2 =2

Q16) A concave mirror and a convex lens are held separately in water. What changes, if any, do you expect in the focal length of either ? (Marks 2)
Ans16) The focal length of concave mirror would remain same as focal length is equal to half the radius of curvature which does not change in changing the surrounding medium.
For convex lens, the focal length will decrease when it is held in water since from len maker's formula
1/f = (  - 1)(1/R1 - 1/R2)
As increase, f decreases.

Q17) A rectangular coil N turns and area of cross-section A, is held in a time-varying magnetic field given by B = B0 sin t, with the plane of the coil normal to the magnetic field. Deduce an expression for the e.m.f. induced in the coil.  (Marks 2)
Ans17) B = B0 sin t
flux = N. =NBA cos
Here = 0o
= NBA  =  NAB0 sin
induced emf  = -d/dt
= - d/dt (NAB0 sin t ) = - NB0A cos

Q18) Draw the graph showing variation of thermo e.m.f. of a thermo-couple with the temperature difference of its junction. How does its neutral temperature vary with the temperature of the cold junction ?  (Marks 2)
Ans18)

 Thermo e.m.f. Tn : Neutral temp Ti : Temp of inversion

Neutral temp is independent of temp of cold junction

Q19) Derive an expression for the electric potential at a point along the axial line of an electric dipole.  (Marks 3)
Ans19) Consider charges +q &  -q constituting an electric dipole Let the separation between then be 2a

Let P be a point lying on axis of dipole at a distance r from the centre of dipole.
Potential at P due to q charge at A
VA = Kq/(r + a)
Potential at P due to -q charge at B
VB = -Kq/(r - a)
Total Potential at P = VA + VB
= Kq/(r + a) - Kq/(r - a) = -2kqa/(r2 - a2)
If the point P lies to the left of A then potential at P = Kq/(r - a) - Kq/(r + a) = 2kqa/(r2 - a2)

Q20) A copper voltameter is in series with a heater coil resistance 0.1 ohm. A steady current flows in the circuit for 20 minutes, and a mass of 0.99 gm of copper is deposited at the cathode. If the electro-chemical equivalent of copper is 0.00033 gm/coulomb, calculate the heat generated in the coil.  (Marks 3)
Ans20)
Given R = 0.1
t = 20 min = 20 x 60 sec = 1200 sec
m = 0.99 g
z = 0.00033 g/c
According to Faraday's I law of electrolysis
m = ZIt
substituting values of m, z, t in the above eqn.
0.99 = 0.00033 x I x 1200
or I = 2.5 A
Heat generated in the coil = I2Rt
=(2.5)2 x 0.1 x 1200 = 750 J

Q21) State Huygen's postulates of wave theory. Sketch the wavefront emerging from a
(i) point source of light and (ii) linear source of light like a slit.  (Marks 1+2)
Ans21)
Huygen's postulates of wave theory
(1) Every point on the given wavefront (called primary wavefront) acts as a fresh source of new disturbance called secondary wavelets, which travel in all directions with the velocity of light in the medium.
(2) A surface touching these secondary wavelets tangentially in the forward direction at any given instant gives the new wavefront at that instant.
(i) For point source of light : spherical wavefront

(ii) For linear source of light : Cylindrical wavefront

Q22) State the conditions for total internal reflection of light to take place at an interface separating two transparent media. Hence derive the expression for the critical angle in terms of the speeds of light in the two media.  (Marks 3)
Ans22) Conditions for total internal reflection of light
1. Light should travel from a denser medium to a rarer medium.
2. Angle of incidence in denser medium should be greater than the critical angle for the pair of media in contact.
When angle of incidence is equal to critical angle (ic) , the angle of refraction is 90o dr = sin i/sin r = sin ic/sin 90o = sin ic
or sin ic = 1/rd = speed of light in denser medium/speed of light in rarer medium

Q23) State the dependence of work function on the kinetic energy of electrons emitted in a photocell. If the intensity of incident radiation is doubled, what changes occur in the stopping potential and the photoelectric current.  (Marks 1+2)
Ans23) The work function does not depend on kinetic energy of emitted electrons If the intensity of incident radiation is doubled then
(i) the stopping potential will remain uncharged as it is independent of intensity of incident radiation but depends on the frequency of incident light.
(ii) the photoelectric current will double as on doubling the intensity no of photons incedent will be double and hence no of photoelectrons emitted per second will double .

Q24) With the help of a labelled circuit diagram, explain how will you determine the internal resistance of a primary cell using a potentiometer. State the formula used.  (Marks 3)
Ans24)

A circuit is set as shown above. Close key k and maintain suitable constant current in the potentiometer wire AB with the help of rheostat Rh. Adjust the position of jockey at different position of wire and find a point j on the wire where if jockey is pressed, galvanometer shows no diflection. Note the length AJ (= l1). Now e.m.f. of the cell o = P.D across the length l1 of the wire.
or o =  kl1  - (i)
Where k is the potential gradient across the wire.
Now close key k, so that the resistance 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
Then P.D. between terminals of cell = P. D across length l2
or V = kl2  - (ii)
From (i) & (ii)
/V =  l1/l  - (iii)
Now internal resistance r is given by
r = ((o/V) - 1)R
r = ((l1/l2) - 1)R

Q25) A short bar magnet of magnetic moment 0.9 joule/tesla, is placed with its axis at 45o to a uniform magnetic field. If it experiences a torque of 0.063 joule, (i) calculate the magnitude of the magnetic field and (ii) what orientation of the bar magnet corresponds to the stable equilibrium in the magnetic field ?  (Marks 3)
Ans25) Given
(i) magnetic moment ' M' = 0.9 J/T
= 45o
Torque = 0.063 J
= MB Sin
or B = /(M Sin ) = 0.063/(0.9(1/2)) = 0.099 T
(ii) The magnet is in stable equilibrium when it is aligned parallel to the field direction.

Q26) A conductor of length 'I' is connected to a d.c source of potential 'V' . If the length of the conductor is tripled, by stretching it, keeping 'V' constant, explain how do the following factors vary in the conductor :
(i) Drift speed of electrons, (ii) Resistance and (iii) Resistivity.  (Marks 3)

Ans26) (i) The drift speed of electrons is given by
Vd = (eE/m)
E = Potential difference/length = V/l
Vd = ev/ml
If l is tripled, Vd becomes 1/3rd  (as Vd 1/l )

(ii) The resistance of conductor is given by
R = ml/ne2A
l, so resistance will also be tripled.

(iii) Resistively remains unchanged, as it depends on nature of material.

Q27) A proton and an alpha particle of the same velocity enter in turn a region of uniform magnetic field acting in a plane perpendicular to the magnetic field. Deduce the ratio of the radii of the circular paths described by the particles. Explain why the kinetic energy of the particle after emerging from the magnetic field remains unaltered.  (Marks 3)
Ans27) The magnetic force acting on a charged particle entering a uniform magnetic field is given by
F= q( x ) = qvB Sin
Here = 90o , so F = qvB

This force provides the centripetal force
qvB = mv2/r
or r = mv/qB
mp : m  = 1 : 4
qp  : q   = 1 : 2
rp/r = mp/qp x q/m = 1/4 x 2 = 1 : 2

After emerging from the magnetic field, no force act on the charged particles. So they continue to move with constant velocity & hence kinetic energy of particles remain uncharged.

Q28) State the postulates of Bohr's model of hydrogen atom. The electron, in a given Bohr orbit has a total energy of -1.5 eV. Calculate its (i) kinetic energy. (ii) potential energy and (iii) wavelength of light emitted, when the electron makes a transition to the ground state. (Ground state energy = -13.6 eV)  (Marks 5)
Ans28) Bohr's postulates :
1. The electrons revolve around the nucleus in circular orbits and the centripetal force required for revolution is provided by electrostatic force of attraction between electrons & protons.
2. Electron can revolve in those orbits only where its angular momentum is an integral multiple of h/2, Where h is plank's constant.
3. While revolving in circular orbits, electron does not radiate energy. These orbits are called Stationary orbits or non-radiating orbits.
4. The radiation of energy occurs only when an electron jumps from one permitted orbit to another. The difference in the total energy of electron in the two permitted orbits is absorbed when the electron jumps from inner to outer orbits & emitted when electron jumps from outer to inner orbits.
If E1 is total energy of electron in inner orbit & E2 in outer orbit then frequency of radiation emitted in jumping from outer to inner orbit is given by
h = E2 - E1
Numerical
Given total energy = -1.5 eV
(i) K.E. = -(Total Energy) = 1.5 eV
(ii) P.E. = -2(K.E.) = -3eV
(iii) h = hc/ = E1 - E2  - (1)
E1 = -1.5 eV
E2 = -13.6 eV
E1 - E2 = 12.1 eV = 12.1 x 1.6 x 10-19 J = 19.36 x 10-19 J
From (1)  = hc/(E1 - E2) = 6.6 x 10-34 x 3 x 108/ 19.36 x 10-19
= 1.022 x 10-7 m

Q29) For a given a.c. circuit, distinguish between resistance, reactance and impedance. An a.c. source of frequency 50 hertz is connected to a 50 mH inductor and a bulb. The bulb glows with some brightness . Calculate the capacitance of the capacitor to be connected in series with the circuit , so that the bulb glows with maximum brightness.  (Marks 3+2)
Ans29)

 Resistance Reactance Impedance 1. The resistance of an a.c circuit is the ohmic resistances offered by a conductor in the circuit. 2. It does not depend on the frequency of the source. The reactance of an a.c circuit is the resistance offered by an inductor or capacitor connected in the circuit. It varies with the frequency of the source as inductive reactance =L  & Capacitve reactance =1/c The impedance of an a.c circuit is the effective resistance offered by L, R & C connected in the circuit. It varies with the frequency of source.

(ii) Given
= 50 Hz
L = 50 mH = 0.05 H
For maximum brightness, impedance should be minimum.
c =1/(2L) = 1/(2)2L
c =1/(2 x 50)2L = 1/(104 x 2 x 0.05) = 2 x 10-4 F = 200 F

Q30) Drawing a labelled circuit diagram, explain the working principle of a common-emitter transistor amplifier. State the phase relation between input and output signals.
Or
With the help of a labelled circuit diagram, explain how a transistor oscillator works.  (Marks 5)
Ans30)

The circuit using n-pn transistor as a common emitter amplifier is shown above. The input (emitter-base) circuit is forward biased with battery voltage VEB, and the output circuit (collector-emitter) is reserve biased with battery voltage VCE. The low input a.c voltage is applied across base-emitter circuit and the amplified a.c. voltage i.e. output is obtained as the change in the collective voltage.
When no a.c. signal voltage is applied to the input circuit but emitter base circuit is closed, let Ie, Ib & Ic be the emitter current, base current and collector current resp.
Then according to kirchoff's law
Ie = Ib + Ic
Ib = 5% of Ie, Ic = 95% of Ie
Due to collector current Ic, voltage drop across load resistance RL = Ic RL
If Vc is collector voltage ie, P.D. between collector & emitter , then
VCE = Vc + Ic RL
or Vc = VCE - Ic R -(i)
When the input signal voltage is feed to the emitter base circuit, it will change the emitter voltage and hence the emitter current which in turn will change collector current. Due to this the collector voltage will vary according to eqn(i) These variations in collector voltage appear as amplified output.
In common-emitter transistor amplifier curcuit, the input signal voltage & the output collector voltage are in opposite phase i.e., 180o out of phase.

Or

Transistor as an oscillator

The circuit diag in shown below

L.C. circuit is inserted in emitter-base circuit of transistor which is forward biased with battery voltage B1. The collector emitter circuit is reserves biased with battery B2. A coil L1 is inserted in collector emitter circuit. It is coupled with L in such a way that if magnetic flux linked will L1 increases it induces e.m.f. in L which supports the forward bias of the emitter-base circuit.
When key K is closed, the collector current begins to increase. As a result flux linked with L1 increases and due to mutual induction as induced L which will charge the upper plate of the capacitor positive and consequently there will be support to the forward biasing of emitter-base circuit. This increases the emitter current and hence an increase in the collector current. Due to it, flux linked with L1 further increase and hence more e.m.f. is induced in L, charging the upper plate of capacitor with more positive charge & hence providing more support to the forward biasing of emitter-base circuit. This further increases the emitter current due to which collector current also increases. The process continues till the collector current becomes saturated.
When collector current becomes maximum the mutual induction stops playing its part. The capacitor gets discharged through L. As a result of it, the support to the forward biasing of emitter-base circuit is withdrawn, thereby emitter current & hence collector current decreases. Due to it, a decreasing magnetic flux is linked with L1 & hence with L. An e.m.f. is induced in L which will charge the lower plate of capacitor with positive charge & this will oppose the forward biasing of emitter-base circuit. This results in further decrease in emitter current & hence in collector current. This process continues till the collector current becomes zero. Now again mutual induction stops playing its part. The capacitor gets discharged through inductance L. As a result of it the opposite to the forward biasing of emitter-base circuit is withdrawn, thereby the emitter current & collector current increases & the process is repeated. Thus the collector current oscillates between the maximum & zero value. The frequency of oscillations is given by.
= 1/(2(LC))