CBSE Set Qa1 Physics Sample Test Papers For Class 12th for students online
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 90^{o} to field lines) 
Q5) If
the number of turns of a solenoid is doubled, keeping the other factors
constant, how does the selfinductance of the solenoid change ? (Marks 1)
Ans5) The self inductance becomes four times on doubling the no of turns
of solenoid since L = oN^{2}A/e
, L N^{2}
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)^{4
}S_{1}/S_{2} = (T_{1}/T_{2})^{4} =
(1/2)^{4} = !/16
or S_{1} : S_{2} = 1 : 16
Q7) The
wavelength of light coming from a distant galaxy is redshifted. 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 = B_{V}/B_{H}
= 1
= 45^{o}
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 r_{1} and r_{2} around the sun.
As earth and planet revolve around
the sun. The distance r_{1 }remains constant, but r_{2 }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 90^{o}. Let o
be the maximum value of planet's elongation. Then in right triangle SPE
r_{1}/r_{2} = cos o
r_{2} = 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
C_{1}, C_{2}, C_{3} are in parallel. There equivalent
capacitance = C' =C_{1} + C_{2} + C_{3} = 2 + 3 + 5 = 10
F
This equivalent capacitance is in series with C_{4}.
Net capacitance 'C' is given by
1/C = 1/C' + 1/C_{4} = 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 2input 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 = 60^{o}
e = 3/4 A = i = e = 3/4 x 60 = 45^{o}
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 = 30^{o}
= (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/R_{1}
 1/R_{2})
As increase, f decreases.
Q17) A
rectangular coil N turns and area of crosssection A, is held in a timevarying
magnetic field given by B = B_{0} 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 = B_{0} sin t
flux = N.
=NBA cos
Here = 0^{o}
= NBA = NAB_{0}
sin t
induced emf =
d/dt
=  d/dt (NAB_{0} sin t
) =  NB_{0}A cos
t
Q18)
Draw the graph showing variation of thermo e.m.f. of a thermocouple 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
V_{A} = Kq/(r + a)
Potential at P due to q charge at B
V_{B} = Kq/(r  a)
Total Potential at P = V_{A} + V_{B}
= Kq/(r + a)  Kq/(r  a) = 2kqa/(r^{2}  a^{2})
If the point P lies to the left of A then potential at P = Kq/(r  a)  Kq/(r +
a) = 2kqa/(r^{2}  a^{2})
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 electrochemical 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 = I^{2}Rt
=(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 (i_{c}) , the angle
of refraction is 90^{o} ^{d}_{r}
= sin i/sin r = sin i_{c}/sin 90^{o} = sin i_{c}
or sin i_{c} = 1/^{r}_{d}
= 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
(= l_{1}). Now e.m.f. of the cell o
= P.D across the length l_{1} of the wire.
or o = kl_{1}
 (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 J_{1}, Note the length of the wire AJ_{1}
(=l_{2})
Then P.D. between terminals of cell = P. D across length l_{2}
or V = kl_{2}  (ii)
From (i) & (ii)
/V = l_{1}/l_{2 }
 (iii)
Now internal resistance r is given by
r = ((o/V)  1)R
r = ((l_{1}/l_{2})  1)R
Q25) A short bar
magnet of magnetic moment 0.9 joule/tesla, is placed with its axis at 45^{o}
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
= 45^{o}
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
V_{d} = (eE/m)
E = Potential difference/length = V/l
V_{d} = ev/ml
If l is tripled, V_{d} becomes 1/3rd (as V_{d}
1/l )
(ii) The resistance of conductor is given by
R = ml/ne^{2}A
R 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 = 90^{o} ,
so F = qvB
This force provides the centripetal force
qvB = mv^{2}/r
or r = mv/qB
m_{p} : m_{}
= 1 : 4
q_{p} : q_{}
= 1 : 2
r_{p}/r_{}
= m_{p}/q_{p} 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 nonradiating 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 E_{1} is total energy of electron in inner orbit & E_{2}
in outer orbit then frequency
of radiation emitted in jumping from outer to inner orbit is given by
h = E_{2}  E_{1}
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/
= E_{1}  E_{2}  (1)
E_{1} = 1.5 eV
E_{2} = 13.6 eV
E_{1}  E_{2} = 12.1 eV = 12.1 x 1.6 x 10^{19} J =
19.36 x 10^{19} J
From (1) = hc/(E_{1}
 E_{2}) = 6.6 x 10^{34} x 3 x 10^{8}/ 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
= 2
= 100 rad/s
L = 50 mH = 0.05 H
For maximum brightness, impedance should be minimum.
c =1/(^{2}L) =
1/(2)^{2}L
c =1/(2 x 50)^{2}L
= 1/(10^{4} x ^{2}
x 0.05) = 2 x 10^{4} F = 200 F
Q30)
Drawing a labelled circuit diagram, explain the working principle of a
commonemitter 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 npn transistor as a common emitter amplifier is shown above.
The input (emitterbase) circuit is forward biased with battery voltage V_{EB},
and the output circuit (collectoremitter) is reserve biased with battery
voltage V_{CE}. The low input a.c voltage is applied across baseemitter
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 I_{e}, I_{b} & I_{c} be the
emitter current, base current and collector current resp.
Then according to kirchoff's law
I_{e} = I_{b} + I_{c}
I_{b} = 5% of I_{e}, I_{c} = 95% of I_{e}
Due to collector current I_{c}, voltage drop across load resistance R_{L}
= I_{c} R_{L}
If V_{c} is collector voltage ie, P.D. between collector & emitter ,
then
V_{CE} = V_{c} + I_{c} R_{L
}or V_{c} = V_{CE}  I_{c} R_{L } (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 commonemitter transistor amplifier curcuit, the input signal voltage &
the output collector voltage are in opposite phase i.e., 180^{o} out of
phase.
Or
Transistor as an oscillator
The circuit diag in shown below
L.C. circuit is inserted in emitterbase circuit of transistor which is forward
biased with battery voltage B_{1}. The collector emitter circuit is
reserves biased with battery B_{2.} A coil L_{1} is inserted in
collector emitter circuit. It is coupled with L in such a way that if magnetic
flux linked will L_{1} increases it induces e.m.f. in L which supports
the forward bias of the emitterbase circuit.
When key K is closed, the collector current begins to increase. As a result flux
linked with L_{1} 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 emitterbase circuit. This
increases the emitter current and hence an increase in the collector current.
Due to it, flux linked with L_{1} 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 emitterbase
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 emitterbase circuit is withdrawn, thereby emitter
current & hence collector current decreases. Due to it, a decreasing
magnetic flux is linked with L_{1} & 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 emitterbase 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
emitterbase 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))
Boarding Schools By State

Boarding Schools Top Cities

Boarding Schools By Board
