CBSE Set Qa1 Physics Sample Test Papers For Class 12th for students online
Physics
Class- XII
CBSE)
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Q 1. What is the work done in moving a
charge of 10 nC between two points on an equipotential surface ? (1 mark)
Ans 1. dw = qdv
on equipotential surface dv = 0
dw = 0
i.e,no work is done in moving a change on an equipotential surface.
Q 2. Name the device used for measuring the internal
resistance of a secondary cell ? (1 mark)
Ans 2. Internal resistance of a secondary cell can be measured
by a potentiometer.
Q 3. What is the nature of magnetic field in a
moving coil galvanometer ? (1 mark)
Ans 3. A radial magnetic fields exists in a
moving coil galvanometer.
Q 4.If a rate of change of current of 2 A/s induces an
e.m.f.of 10mV in a solenoid, what is the self -inductance of the solenoid ? (1
mark)
Ans 4. Given dI = 2 A/S ,
10mv
= 10-3V
dt
| | = L dI
dt
or 10 x 10
-3 = L x 2
or L = 10-2 =
5 x 10-3 H = 5mH
2
Q 5. What type of magnetic material is used in making
permanent magnets ? (1 mark)
Ans 5. A substance like steel which as a large retentivity and a
large coercivity is used in making permanent magnets.
Q 6. Two metals A, B have work-functions 2eV, 4eV
respectively. Which metal has lower threshold wavelength for photoelectric
effect ? (1 mark)
Ans 6. Work function W = h v
0 = hc
where 
is
threshold wavelength.
x 1
W
Metal B has higher value of W & hence lower threshold wavelength.
Q 7. Which have greater ionising power : alpha particles
or beta particles ? (1 mark)
Ans 7. Alpha particles have more ionising power than
beta particles.
Q 8. Which part of the electromagnetic spectrum is
used in operating a RADAR ? (1 mark)
Ans 8. Microwaves are used in operating a RADAR.
Q 9. Which type of biasing gives a semiconductor diode very
high resistance ? (1 mark)
Ans 9. Reverse biasing gives a semiconductor diode very high
reistance.
Q 10.How is the luminosity of a star related to its radius ?
(1 mark)
Ans 10. The luminosity 'L' is related to the radius 'r' by the
equation.
L = 4
o
T4
where T is the absolute temperature of the star and 0 is the Stefan's
constant.
Q 11. An electric dipole, when held at 300
with respect to a uniform electric field of 104N/C,
experiences a torque of 9 x 10-26Nm. Calculate the
dipole moment of the dipole. (2 marks)
Ans 11. Torque z = pE since
Given z = 9 x 10-26 Nm , E
= 104 N/C , O = 300,
9 x 10-26 = p x 104 x sin 30 = p x 104
x 1
2
or p = 9 x 10-26 x 2
= 18 x 10-30 c - m
104
Q 12.A set of n identical resistors, each of resistance
R ohm, when connected in series have an effective resistance X ohm and when the
resistors are connected in parellel, their effective resistance is Y ohm. Find
the relation between R, X and Y. (2 marks)
Ans 12. When connected in series
X = R + R + ............ n
times = nR ---(1)
When connected in parellel
1 = 1
+ 1 + ........ n times = n
Y
R R
R
or Y = R
---(2)
n
Multiplying eqns (1) & (2)
XY = nR x R =
R2
n
or R = 
XY
Q 13.State Kirchhoff's rules for eletrical networks. (2
marks)
Ans 13. Kirchoffs rules for elecrical networks :
1. First law or Junction law : It states that the algebraic sum of
the currents meeting at a juction in a closed circuit is zero.
2. Second law or loop rule : It states
that in any closed path of an electrical circuit, the algebraic sum of the
emfs is equal to the algebraic sum of the resistance & the respective
currents flowing through them.
Q 14.Show that Lenz's law is in accordance with the law of
conservation of energy. (2 marks)
Ans 14. Lenz's law is in accordance with the law of
conservation of energy. For this, consider a coil connected to a galvanometer as
shown.
When the N-pole of the magnet is
moved towards the coil then according to Lenz's law the end of coil facing the
magnet acquires north polarity.Therefore work has to be done against the force
of repulsion in bringing the magnet closer to the coil. Similarily, when the
N-pole of a magnet is moved away, south polarity develops on the end of the coil
facing the agnet.Therefore work has to be done against the force of attraction
in taking the magnet away. In both the cases, it is mechanical work done in
moving the magnet wrt the coil that changes into electrical energy,
producing induced current and the galvanometer shows deflection.Thus energy is
only being transformed.
When we don't move the magnet, work done is zero.Hence no induced current is
produced.
Q 15.Draw a labelled diagram of Hertz's experiment for
producing electromagnetic waves. (2 marks)
Ans 15. Hertz's expt :
Ans 16. Differences between fringes formed in single slit diffraction & Young's double slit expt :
(1) In Young's double slit experiment all the bright fringes are of the same intensity, whereas in single ;slit diffraction pattern all the bright fringes are not of same intensity.
(2) In Young's double slit experiment, the intensity of minima is generally zero or very small & there isgood contrast between bright and dark fringes. In single slit diffraction pattern the intensity at minima is never zero & there is poor contrast between bright & dark fringes.
Q 17.Light of
wavelength 600nm is incident on an aperture of size 2mm. Calculate the distance
up to which the ray of light can travel such that its spread is less than the
size of the aperture. (2 marks)
Ans 17. The
distance upto which a ray of light can travel that its spread is less than the
size of the aperture is Fresnel distance which is given by Z
F
= a2
Given a = 2mm = 2 x 10-3m
= 600nm =600 x 10-9
m
Q 18.Explain the principle used in Bunsen's grease photometer
to compare the illuminating powers of two light sources. (2 marks)
Ans 18. Bunsen's
grease photometer is based on the principle of photometry which states that if
the two sources produce equal illuminations on a surface, then the ratio of the
luminious intensities of the two sources is equal to the ratio of the square of
the respective distances of the two sources from that surface.
...............The two sources whose luminious
intensities are to be compared are placed on the two sides of the screen of the
Bunsen's photometer & their distances from the screen are so adjusted
that the grease spot becomes equally bright on its two sides. If E1
& E2 are the illuminations on the screen
due to the respective sources, then:
E1
= E2  |
 |
Q 19.A ray of light
while travelling from a denser to a rarer medium undergoes total internal
reflection. Derive the expression for the critical angle in terms of the speed
of light in the respective media.
(2 marks)
Ans 19. When the the
angle of incidence is equal to the critical angle 'ic' then angle of
reflection = 900
= sin i = sin
ic = sin ic
sin
r sin90
= Speed of light is denser medium
Speed
of light is rarer medium
sin ic = Speed of light is denser medium
Speed
of light is rarer medium
Q 20.Draw a graph
showing the variation of binding energy per nucleon with mass number of
different nuclei. Mark the region where the nuclei are most stable. (2 marks)
Ans 20.
Q 21.Name
the logic gate shown in the diagram and give its truth table. (2 marks)
Ans 21. And Gate
is shown in the given diagram
...............Its truth table is :
|
A
|
B
|
C
|
|
0
1 0 1 |
0
0 1 1 |
0
0 0 1 |
Q 22. State and explain
Seebeck effect. (2 marks)
Ans 22. Seebeck is the phenomenon
of generation of electric current in a thermocouple by keeping its two junctions
at different temperatures.
............... When two dissimilar metals are
joined to form thermocouple, then due to diffusion of free electron from metal,
with greater free electron density to lower free electron density, a potential
difference is developed at the functions called contact potential. When there is
temperature difference at the two
junctions, contact potentials at the two fuctions become different & there
is an effective potential difference in the circuit called thermoems &
current flows in the circuit.
Q 23. Explain
with the help of a circuit diagram how the value of an unknown resistance can be determined
using a Wheatstone bridge. Give the formula used. (3 marks)
Ans 23
|
|
|
Slide wire bridge (Wheatstone bridge )
to field unknown resistence
|
|
...............In
figure X is unknown resistance and R.B. is resistance box. After
inserting the key K, jockey is moved on wire AC till
the galvanometer shows no deflection (Point B).
|
Q 24 Earth receives an amount of heat
radiaton 1.4 x 103 W/m2 from the sun.Assume that earth
re-emits all the radiation recieved from the sun.Calculate the surface
tempratueof the earth. (3 marks)
Ans 24 Amount of
radiation recieved by earth per unit area per second = 1.4 x 103
w/m2
..........................................................
= solar constant
Energy recieved per second by earth
= 4 S —
(1)
..............................where r is radius
of earth.
According to Stephans law,energy
radiated per second per unit area = 
T4, where is
Stephans constant.
Energy radiated perb second by earth = 4
T4 — (2)
Equating (1) & (2)
4
T4 = 4S
or
T4 = S
...................................
T
= (S)1/4 = ( 1.4 x 103)1/4 = 396.4 k
..........................
5.67 x 10-8
Q 25. A straight wire carries a current of 3A.Calulate the
magnitude of the magnetic field at a point 10cm away from the wire.Draw
diagram to show the direction of the magnetic field. (3 marks)
Ans 25 I = 3A, r = 10cm = 0.1 m B = ?
| B = |  |
10-7
x 2 x 3 0.1 |
|
||
Q 26. Explain how a vibration
magnetometer can be used to determine the magnetic moment of a bar magnet. (3
marks)
Ans 26. A vibration magnetometer consists of wooden box with glass
slides.A plane mirror is placed at base & index line is marked along its
length.A long glass tube is fixed vertically at the center of the box.It has a
torsion head T from which hangs an unspun silk fibre.A stirrup of non-magnetic
material like brass is attached to the free of the fibre,so that it is
just above the base of the box.
Setting : Magnetometer is set in magnet meridian.
This is done by placing a compass needle on the index line & rotating
the box till the compass needle is parallel to the index line.Magnet is
placed on the stirrup so that the north-pole of the magnet can be set into
vibrations of small amplitude using another magnet from outside & its
time period can be determined.
| Knowing T, BH & I( I = mass |  |
) ,where l and b
are the length & breadth of the |
magnet,the magnetic moment M can be determined.
Q 27. Deduce lens maker's
formula for a thin biconvex lens.
(3 marks)
Ans 27. Consider a point object O lying on the principal axis of
the lens.A ray of light starting from O and incient normally on
the surface XP1Y along OP1 passes
straight.Another ray incident on XP1Y along OA is
refracted along AB.If the lens material were continuous and there were no boundry/second
surface XP2Y of the lens,the refracted ray AB would go
straight meeting the first refracted ray at I1.Therefore,I1
would have been a real image of O formed after refraction at XP1Y.
....................(1)
Actually,the lens material is not
continious.Therefore,the refracted ray AB sufers furthes refraction at B
and emerges along BI,meeting atually the principal axis at I.Therefore
I is the final real image of O,formed after refraction through the
convex lens.
For
refraction at the second surface XP2Y,we can regard I1
as a virtual object,whose real image is formed at I.
Therefore,
Let R2
be radius of curvature of second surface of the lens.
As
refraction is now taking place from denser to rarer medium,therefore,
using eqn.(21),we get
|
..............(3)
| Put |  |
reflective index of material of lens w.r.t. surrounding medium. |
image
is formed at the principal focus of the lens.
when u = ,v = f = focal length of the
lens.putting this in eqn. 3
1 (M-1) ( 1 - 1 )
f R1 R2
Q 28. Explain the
origin of spectral lines of hydrogen using Bohr's theory. (3 marks)
Ans 28.Bohr's postulates of hydrogen
atom :
1. There is a positively charged nucleus at the center of the atom
around which electrons resolve in circular orbits.
...............The
centripetal force is provided by coulomb's force of attraction exerted by
nucleus.
Kze2 = mv2
r2 r
Kze2 =
mv2
r
2. Electrons revolve in certain discrete,non-radiating orbits
called statimary orbits,for which the angular momentum is an integral
multiple of h
........................................................ 
3. When electron jumps from one stationary orbit of higher energy E2
to another lower energy E1 it emits energy of frequency
to where
...............E2-E1
= hv.
...............From
eqns of (1) & (2)
............... r
= x2h2
............... mke2
............... Total
energy of electron = K.E + P.E
1
mv2 + (-ke2)
2
r
............... using
equation (1)
Total energy = 1
(kze2) - kze2 = kze2
...............................
2 r
r 2r
...............substituting
r from eqn (3)
...............Total
energy =
When electron jumps from outer level (energy E2)
to inner level (energy E1), then from Bohr's theory:
h v = E2-E1
=
hv =
|
|
hv = hc = |
 |
 |
Depending m values of n1 & n2,radiations
of different wave lengths are emitted.
Q 29. A beam of electrons passes undeflected through
mutually perpendicular electric and magnetic fields E and B respectively.If the
electric field is cut-off ,the electron beam moves in a circular path
of radius 'r'.Derive the expression for e/m of electrons in terms of r,E
and B. (3 marks)
Ans 29. In the first eam when the electron beam goes
undeflected
...............evb = eE
...............or V = E ——
(1)
.......................B
.When electric field is cut off ,only magnetic
force acts on it which provides thecentripetal force.
e
B
= m2
r
or e
= m 2
m rB
substituting from eqn (1)
e = E2
m rB3
Q 30. Draw a labelled diagram of Van de Graff generator.Give
its principle and explain its working.
Ans 30 Van the Graft generator Principle.This generator is based on : (i) The
actions of sharp points i.e. the phenomenon of corona discharge.
(ii) The property that charge given to hollow conductor is transferred
to outer surface and is distrbuted uniformily over it. Working :The spray
comb is given a positive potential (~104volt) w.r.t. the
earth by high tension source H.T.Due discharging action of sharp points,
a positively charged electric wind is set up, which sprays positive charge
on belt (corona discharge). As the belt moves,and reaches the sphere,a
negative charge is induced on the sharp ends of the 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
thepositive charge on S goes on accumalating.
Now the capacity of spherical shell = 
where R is radius of the shell. As V = Q
Q
............ C..........
Hence the potential V of the spherical
shell goes on increasing with increase in Q.
...............OR
The Capacitance of a capacitor is defined as
the ratio of charge on the plates.Its S.I. unit is Farad.
Electrostatic energy stored in a capacitor :
Let the capacitor be charged gradually.At any stage,change on it is q.
Therefore P.D
between the plates of capacitor = q
C
small amount of
work done in giving an additional charge dq is
dw
= q dq
C
Total work done in
giving a charge Q to the capacitor
But Q = CV
Energy stored = W = Q2 = 1 CV2
2C
2
Energy
density = Total energy = 1/2 CV2
............... ...............
........volume ............Ad
where d :
distance between plates
A
: Coin - sectioned area of plate
C
: 
Energy
density (u) =
V
=Ed
u =
Q 31.
Prove that the power dissipated in an ideal
resistor connected to an a.c. source is V2eff/R.Acapacitor,a
resistor and a 40 mH inductor are connected in series to an a.c. source
of frequency 60 Hz.Calculate
the capacitance of the capacitor,if the current is in phase with the
voltage. (2+3
marks)
Ans 31. let V = V0 sin wt
In an ideal resistor connected to an
a.c source, current & voltage are in phase.
I = I0 sin wt
small work done oin time dt
dw =IV dt = I0 V0
sin2 wt dt
Total work done = 
dw
= 
I0V0
sin2 wt dt
=
I0V0 T
2
Numerical : L = 40 MH =
40 x 10 -3H
v
= 60 Hz.
If
current is in phase with the voltage
WL = 1
WC
or
C = 1
= 1
W2L
2 v2L
1
= 176
x 10-6F
4
x (3.14)2 x (60) x 40 x 10 -3
= 176
MF
Q 32. Explain with the help of a circuit diagram the working principle of a transistor as an amplifier in the common emmitter configuration .Derive the expression for the voltage gain of the amplifier.(5 marks)
...The input (emitter base) circuit is forward
biased with battery voltage VEB, and the output(collector-emitter)
circuit is reversed biased with battery voltage Vce.Due
to which the resistance of input circuit is low and that of output circuit is
high.RL is a load resistance conected in collector
circuit.The low input a.c voltage sgnal is applied across base-emitter circuit
and the amplified a.c. voltage signal (i.e. output) is obtained as the change
in collector volta.
....When no a.c. signal voltage is applied to
input circuit but emitter base circuit is closed 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)
If VC is
collector voltage,then
VCE = VC + IcRL
or
Vc =VCE - IcRL ----------(ii)
When the input signal is fed to the emitter base circuit,it
will change the emitter voltage and hence the emitter current,which in
turn will change the collector current.Due to this the collector voltage Vc
will vary in accordance with relation (ii).These variations in the
collector volage appear as an amplified output.
...............To find voltage gain :
...............change in output = 
Vc
= Vce -
IcRL
............... ...............
........ = -
IcRL (
Vce = 0).
...............Voltage gain = Av = V
0 =
V c
............... ...............
............ VL
VL
............... ......................
.. 
............... ......................Av
= -13ac R L
............... ......................
...............RL
Now RL RL & 13 ac
has very large value, there is
large voltage gain