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 13
Q1) Draw
an equipotential surface in a uniform electric field. (Marks 1)
Ans1)
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 crosssectional 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/(R^{2} + (L
 1/c)^{2})
At resonance L = 1/c
.^{.}. cos =
R/(R^{2} + 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 60^{o}. What is the refractive index of
the medium? (Marks 1)
Ans6) = tan i_{p}
i_{p} = 60^{o}
.^{.}. = tan 60^{o}
= 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) m_{A}  m_{B} =  2.5 log_{10} l_{A}/l_{B}
 2  4 =  2.5 log_{10} l_{A}/l_{B}
or log(l_{A}/l_{B}) = 6/2.5 = 2.4
or l_{A} = (10)^{2.4} l_{B}
l_{A} l_{B}
or A with magnitude  2 is brighter than B.
Q9) An
electric flux of 6 x 10^{3} Nm^{2}/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 10^{3} Nm^{2}/c
_{E} = q/E_{o}
or q = _{E} E_{o}
= 6 x 10^{3} 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 10^{3} x 4
= 24 x 10^{3} Nm^{2}/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
R_{s} = R + R + R = 3R
power dissipated X = V^{2}/R_{s} where V is emf of d.c. source.
= V^{2}/3R
When connected in parallel, equivalent resistance R_{p} is given by
1/R_{p} = 1/R + 1/R + 1/R = 3/R
or R_{p} = R/3
power dissipated = V^{2}/(R/3) = 3V^{2}/R
9(V^{2}/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 A^{o},
is found to be shifted towards red by 50 A^{o}.
Calculate the velocity of recession of the galaxy. (Marks 2)
Ans12) = 6000
A^{o}
d = 50 A^{o}
we know that d/
= V/C
or V = d x C/
= 50 x 3 x 10^{8}/6000
= 25 x 10^{5} 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)
Ans13) Focal length in air f = 20 cm
^{a}_{g} =
1.6
1/f = (^{a}_{g}
 1)(1/R_{1}  1/R_{2})
1/20 = (1.6  1)(1/R_{1}  1/R_{2})
or 1/R_{1}  1/R_{2} = 1/20 x 1/0.6 = 1/12
Now let the new focal length when immersed in liquid be f '
1/f ' = (^{l}_{g}
 1)(1/R_{1}  1/R_{2})
= (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 = f_{o}/f_{e}
where f_{o} = focal length of objective lens.
f_{e} = focal length of eye piece.
Q15)
The halflife 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) T_{1/2} = 30 s
Decay constant = 0.693/T_{1/2}
= 0.693/30
= 0.0231 s^{1}
(ii) N = N_{o} e ^{t}
N/N_{o} = 3/4
.^{.}. 3/4 = e^{t}
or e^{t} = 4/3
t = 2.3026 log_{10}(4/3)
t = 2.3026 log_{10} (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)
Ans16)
NAND Gate
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)
Ans17) 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)
Ans18) 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 (= l_{1})of the potentiometer
wire. Now e.m.f. of the cell, E = potential differences across the length l_{1},
of the potentiometer wire.
or E = Kl_{1} ...(i)
Where K is the potential gradient across the wire.
Close key K_{1}, 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 J_{1}. Note the length of
the wire AJ_{1} (= l_{2} say).
Then, potential difference between two poles of the cell, V = potential
difference across the length l_{2} of the potentiometer wire
i.e. V = Kl_{2} ...(ii)
Dividing (i) by (ii), we have
E/V = l_{1}/l_{2} ...(iii)
We know that the internal resistance r_{1} of a cell of e.m.f. E, when a
resistance R is connected in its circuit is given by
r_{1} = (E  V)/ V x R = (E/V  1)R ...(iv)
Putting the value from (iii) and (iv), we get
r_{1} = (l_{1}/l_{2}  1)R = (l_{1}  l_{2})/l_{2}
x R
Thus, knowing the value of l_{1}, l_{2} and R, the internal
resistance r_{1} of the cell can be determined.
Q20) Calculate
the resistance between A and B of the given network. (Marks 3)
Ans20)
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 R_{eq} is given by
1/R_{eq} = 1/3 + 1/6 = 1/2
or R_{eq} = 2
Q21) State
Faraday's laws of electrolysis. Write down the relation connecting chemical
equivalent and electrochemical 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.,
m_{1}/m_{2} = E_{1}/E_{2}
where m_{1}, m_{2} = masses of the substance liberated or
deposited at various electrodes when same charge flows through their
electrolytes
E_{1},E_{2} = chemical equivalents of the substances deposited
or liberated.
Relation between chemical equivalent & electrochemical equivalent :
E/Z = F
where F is Faraday's constant = 96500C
Q22) An
electron is moving at 10^{6}m/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
= 90^{o}
F = qVB
= 1.6 x 10^{19} x 10^{6} 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 T_{1} and T_{2} 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
where
is dip angle
& R is resultant magnetic field
At two places
T_{1} = 2 (I/MH_{1})
= 2 (I/MR_{1}
cos _{1})
T_{2} = 2 (I/M
R_{2} cos _{2})
T_{1}/T_{2} = ((R_{2}
cos _{2})/(R_{1}
cos _{1}))
= T_{1}^{2}/T_{2}^{2} = (R_{2} cos _{2})/(
R_{1} cos _{1})
or R_{1}/R_{2} = (T_{2}^{2} cos _{2})/(T_{1}^{2}
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 c_{1} in the first medium and c_{2} 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 c_{2}. Let the wave
disturbance take time t to travel from B to C, then BC = c_{1}t. 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 = c_{2}t in
the second medium. The tangent plane CD drawn from point C over this hemisphere
of radius c_{2}t 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/c_{1} + OQ/c_{2}
= (AO sin i)/c_{1} + (OC sin r)/c_{2}
= (AO sin i)/c_{1} + ((AC  AO) sin r)/c_{2}
= (AC sin r)/c_{2} + AO[(sin i)/c_{1} (sin r)/c_{2}]
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/c_{1} = sin r/c_{2}
or sin i/sin r = c_{1}/c_{2} = Constant, n_{21}
This proves Snell's law of refraction. The constant n_{21} 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)
Ans25) 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 E_{1} & E_{2} be K.E. corresponding to
frequency & 2
and W be the work function. According to Einsteins photoelectric equation
h = E_{1} + W
(i)
& 2h = E_{2}
+ W (ii)
From (i) and (ii)
E_{2}  E_{1 }= h
= E_{1} + W
or E_{2} = 2E_{1} + 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)
Ans27)
When no a.c signal voltage is applied to the input circuit but emitter base
circuit is closed let us consider that l_{e}, l_{b} and l_{c}
be the emitter current, base current and collector current respectively. Then
according to Kirchhoff's first law.
I_{e} = I_{b} + I_{c} ...(i)
Due to current l_{c}, voltage drop across R_{L} = I_{c}R_{L}
If V_{c} is collector voltage then
V_{CE} = V_{c} + I_{c}R_{L}
or V_{c} = V_{CE}  I_{c}R_{L}
...(ii)
When the positive half cycle of input a.c, signal voltage comes, it supports the
forward biasing of the emitterbase circuit. Due to which the emitter current
increases and consequently the collector current increases. As a result of
which, the collector voltage V_{c} decreases [from relation (ii)]. Since
the collector is connected to the positive terminal of V_{CE} 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 emitterbase circuit, due to which the emitter current
decreases and hence collector current decreases; consequently the collector
voltage V_{c} 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., 180^{o} out of
phase
Q28)
With the help of a labelled diagram, describe Millikan's oildrop 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.
Apparatus : 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 120^{o}.
Fine drops of low vapour density and nonvolatile 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 W_{1}. Xrays, are allowed to enter the chamber, if the need
arises, through another window W_{2}. Xrays 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 v_{1} 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. v_{1} = 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 r^{3}sg
(ii) Bouyant force (B) acting upwards = 4/3 r^{3}g
(iii) Viscous force (F) acting upwards = 6
rv_{1}
When terminal velocity is attained
W = B + F
or 4/3 r^{3}sg =
4/3 r^{3}g
+ 6
rv_{1}
or 4/3 r^{3}(s  )g
= 6
rv_{1}
or r = ((9
v_{1})/(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 v_{2} be the terminal velocity
acquired by it. The various force acting on it are :
(i) upward force due to electric field F_{e} = qE
(ii) viscous force (F') acting downwards = 6
rv_{2}
(iii) Effective weight acting downwards (W) = 4/3 r^{3}(s
 )g
When terminal velocity is attained.
F_{e} = W + F'
qE = 4/3 r^{3}(s  )g
+ 6
rv_{2}
or qE = 6
rv_{1} + 6
rv_{2} = 6
r(v_{1} + v_{2})
substituting the value of r from eqn (i)
q = 6
(v_{1} + v_{2})/E ((9
v_{1})/(2(s  )g))^{1/2}
q = 6
(v_{1} + v_{2})d/V ((9
v_{1})/(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)
Ans29)
X_{L} =
L = 2L X_{L} L 

X_{C} =
1/C = 1/2c X_{C} 1/ 
R = 5 L = 50 mH = 50 x 10^{3} H = 50 Hz If the voltage & current are in phase in LCR circuit then X_{L} = X_{C} 2L = 1/2c or c = 1/4^{2}^{2}L = 1/(4 x (22/7)^{2} (50)^{2} x 50 x 10^{3}) = 202.4 x 10^{6} F = 202.4 F Impedance Z = (R^{2} + (X_{L}  X_{C})^{2}) X_{L} = X_{C} .^{.}. 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.
Or
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
CAPACITANCE OF A PARALLEL PLATE CAPACITOR WITH A DIELECTRIC SLAB
The capacitance of a parallel plate capacitor of plate area A and plate
separation d with vacuum/air in between is
C_{o} = _{o}A/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 = E_{o}(d  t) + Et
But E_{o}/E = _{r}
or K .^{.}. E = E_{o}/K
.^{.}. V = E_{o} (d  t) + E_{o}/K t
V = E_{o}[d  t + t/K]
As E_{o} = /_{o}
= Q/A_{o}
.^{.}. V = Q/A_{o}
[d  t + t/K]
.^{.}. Capacitance of the capacitor with dielectric in between is
C = Q/V = A_{o}/[d
 t + t/K] = _{o}A/[d
 t (1  1/K)]
i.e. C = _{o}A/[d
 t (1  1/K)]
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