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) Name the physical
quantity whose SI unit is Coulomb/Volt. (Marks 1)
Ans1) Capacitance
Q2) Write the
frequency limit of visible range of electromagnetic spectrum in kHz.
(Marks 1)
Ans2) 8 x 10^{11} HHz to 4 x 10^{11} KHz
Q3) How does the
conductance of a semiconducting material change with rise in temperature?
(Marks 1)
Ans3) Increases.
Q4) The force
experienced by a particle of charge e moving with velocity
in a magnetic field is given
by. = e(
x ) Of these, name the pairs
of vectors which are always at right angles to each other. (Marks 1)
Ans4) = e(
x )

and 
Q5) Two wires A
and B are of same metal, have the same area of crosssection and have their
lengths in the ratio 2 : 1. What will be the ratio of currents flowing through
them respectively when the same potential difference is applied across length of
each of them?
(Marks 1)
Ans5) Resistance length
R_{A}/R_{B} = 1/2
Current is inversely proportional to resistance
.^{.}. I_{A}/I_{B} = 1/2
Q6) Calculate
the rms value of the alternating current shown in the figure. (Marks 1)
Ans6) Irms = i^{2}
= ((2^{2} + (2)^{2}
+2^{2})/3) = (12/3)
= 2A
Q7) The image of
an object formed by a lens on the screen is not in sharp focus. Suggest a method
to get clear focusing of the image on the screen without disturbing the position
of the object the lens or the screen. (Marks 1)
Ans7) The reason for such image is spherical aberration. Clearer focusing
can be got by using the other central position of the lens & blocking the
inter portion or using other proton & blocking central portion.
Q8) Two points A
and B are placed between two parallel plates having a potential difference V as
shown in the figure.
Will these protons experience equal or unequal force? (Marks 1)
Ans8) Both A and B will experience equal force as field between the
plates is uniform.
Q9) Define the
terms 'threshold frequency' and 'stopping potential' for photoelectric effect.
Show graphically how the stopping potential, for a given metal, varies with
frequency of the incident radiations. Mark threshold frequency on this graph.
(Marks 2)
Ans9) Threshold frequency: It is the maximum frequency of the
incident radiation below which no emission of photoelectrons takes place.
Stopping potential : It is the maximum negative potential required to
stop the fastest photoelectrons i e, photoelectric current becomes zero.
Q10) Draw
labelled diagram of a Leclanche cell. Write the function of charcoal powder and
manganese dioxide used in its porous pot. (Marks 2)
Ans10)
Charcoal powder makes manganese dioxide conducting and manganese dioxide acts as
depolarizer.
Q11) How does
the mutual inductance of a pair of coils change when :
(i) the distance between the coils is increased?
(ii) the number of turns in each coil is decreased?
Justify your answer in each case. (Marks 2)
Ans11) (i) Mutual inductance decrease with the increase of the distance
between the coils. This is due to lower coupling between the coils.
(ii) With the decreases in the no of turns in each coil, the mutual inductance
of a pair of coils is given by
_{n}N_{s}N_{p}A/l
Q12) Define the
terms magnetic inclination and horizontal component of Earth's magnetic field at
a place. Establish the relationship between the two with the help of the
diagram. (Marks 2)
Ans12) Magnetic inclination : at a place is defined as the
angle when the direction of total intensity of earth's magnetic field makes with
a horizontal line in the magnetic median.
Horizontal component of earth's magnetic field : if the component of
total intensity of earth's magnetic field in the horizontal direction in
magnetic meridian.
: magnetic inclination
H : Horizontal component of earth's magnetic field 
Q13) An
electron in an atoms revolves around the nucleus in an orbit of radius 0.53 A^{o}
. Calculate the equivalent magnetic moment if the frequency of revolution of
electron is 6.8 x 10^{9} MHz. (Marks 2)
Ans13) r = 0.53A^{o} = 0.53 x 10^{10} m
=6.8 x 10^{9}
MHz = 6.8 x 10^{15} Hz
e = 1.6 x 10^{19} c
Now I = e
& magnetic moment M = IA = er^{2}
= 6.8 x 10^{15} x 1.6 x 10^{19} x 3.14 (0.53 x 10^{10})^{2}
= 9.596 x 10^{24} Am
Q14) Write
the function of base region of a transistor. Why is the region made thin and
slightly doped? (Marks 2)
Ans14) The base provides an interaction between emitter &
collection of transistor. The base is thin & lightly doped to reduce
recombination of holes & electrons in this region & so there is
appreciable collector current.
Q15) Derive
an expression for the energy stored in a charged parallel plate capacitor with
air as the medium between its plates. (Marks 2)
Ans15) Suppose the capacitor is charged gradually at any stage the
charge on the capacitor is q
Potential = q/c
small amount of work done in giving an additional charge dq is
dW = q/c dq
Total work done in giving a charge Q
= Energy stored
For a parallel plate capacitor
Energy stored
Q16) In
the diagram given below for the stationary orbit of the hydrogen atom, mark the
transitions representing the Balmer and Lyman series. (Marks 2)
Ans16)
Q17) The
given figure shows an inductor L and resistor R connected in parallel to a
battery B
through a switch S. The resistance of R is the same as that of the coil that
makes L. Two identical bulbs, P and Q are put in each arm of the circuit as
shown. When S is closed, which of the two bulbs will light up earlier? Justify
your answer. (Marks 2)
Ans17) P will light up earlier in the induced I, will be
induced current which will oppose the growth of current through it & hence
reduce the current
Q18) Two
points electric charges of value q and 2q are kept at a distance d apart from
each other in air. A third charge Q is to be kept along the same line in such a
way that the net force acting on q and 2q is zero. Calculate the position of
charge Q in terms of q and d. (Marks 2)
Ans18)
For equilibrium of charge q Force bet Q and q should be equal & apposite to
force between q & 2q
i.e. (1/4o)(Qq/x^{2})
= (1/4o)(2qq/d^{2})
 (i)
For equilibrium of charge 2q, for a between 2q & Q should be equal &
apposite to force between 2q and 2q.
i.e. (1/4o)(2qQ/(d
 x)^{2}) = (1/4o)(2qq/d^{2})
 (ii)
Equating (i) & (ii)
2x^{2} = (d  x)^{2}
or 2x = (d  x) or x = d /
(2 +1)
i.e. , at a distance d/(2
+ 1) from charge q
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) To find internal resistance of a cell E, a current is set up
as shown 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 difference across the length 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) in (iv) we get
r_{1} = (l_{1}/l_{2}  1) R = (l_{1}  l_{2})/l_{2}
x R
Thus, knowing the values of l_{1}, l_{2} and R the internal
resistance r_{1} of the cell can be determined.
Q20) Sketch
the wavefronts corresponding to converging rays. Verify Snell's law of
refraction using Huygens' wave theory. (Marks 3)
Ans20)
Snell's law of refraction using 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.
Consider a ray POQ normal to both the incident and the refracted wave fronts.
Let the angle 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 traveling from the
incidence wave front AB and 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.
Q21) An
electric dipole is held in a uniform electric field.
(i) Show that no translatory force acts on it.
(ii) Derive an expression for the torque acting on it. (Marks 3)
Ans21)
Consider an electric dipole
consisting of two equal and opposite point charges q and +q separated
by a small distance 2a, having dipole moment p = q x 2a Let this dipole be held in a uniform external electric field at an angle with the direction of . Force on charge +q = q, along the direction of Force on charge q = q, in a direction opposite to . 
As the force on the two charged of
the dipole are equal and opposite, therefore, net force on the electric dipole
is zero.
These forces being equal, unlike and parallel form a couple, which rotates the
dipole is clockwise direction, lending to align its axis along the direction of
the field.
Draw AC 
and BC 
.^{.}. perpendicular distance between the forces = arm of couple = AC
As Torque = moment of the couple
= force x arm of couple
.^{.}. = F x AC
= F x AB sin
= F x 2a sin
= (q x 2a) E sin
= p E sin
.^{.}. in the vector form, we can rewrite this equation. As
=
x
Q22) Derive
an expression for the width of the central maxima for diffraction of light at a
single slit. How does this width change with increase in width of the slit?
(Marks 3)
Ans22)
The set of parallel rays falling on the slit form a plane wave from WW'.
According to Huyghen's principle, each point on the unblocked portion of
plane wave front AB sends out secondary wavelets in all the direction.
The secondary waves, from points equidistant from the center C of the slit lying
in the portion CA and CB of wave front travel the same distance in reaching O,
and hence the path difference between them is zero. These secondary waves
reinforce each other, resulting in the maximum intensity at point O.
Consider the secondary waves traveling in a direction making and angle
with CO. All the secondary waves traveling in this direction reach a point P on
the screen. The intensity at P will depend on the path difference between the
secondary waves emitted from the corresponding points of the wave front. Draw AN
perpendicular to BK. Path difference between the secondary waves reaching P from
A and B = BN = AB sin = a
sin
If this path difference is ,(the
wavelength of light used), then P will be point of maximum intensity. This is
because the whole wavefront can be considered to be divided into two equal
halves CA and CB and if the path difference between the secondary waves from A
and B is , then the path
difference between the secondary waves from A and C reaching P will be
/2 and path difference between the secondary waves from B and C reaching P will
again be /2. Also for
every point in the upper half AC, there is a corresponding point in the lower
half CB for which the path difference between the secondary waves, reaching P is
/2 Thus, destructive
interference takes place at P and therefore, P is a point of first secondary
minimum. Thus for I minima
a sin =
= sin = /a
 (i)
The width of central maximum is the distance between first secondary minimum on
either side of O
Let I minima be at a distance x
If Q is small, sin ~
=x/D
From (i)
x/D = /a or x =D/a
Width of central maximum = 2x =2D/a
Q23) A
capacitor of capacitance 100F
and a coil of resistance 50
and inductance 0.5 H are connected in series with a 110 V, 50 Hz source.
Calculate the rms. value of the current in the circuit. (Marks 3)
Ans23) Given C =100 F
= 100 x 10^{6} F = 10^{4} F
R = 50 , L = 0.5 H, V_{rms}
= 110V
= 50 Hz
X_{L} = L = 2L
= 2 x 50 x 0.5 = 50
X_{C} = 1/c = 1/2c
= 1/2 x 50 x 10^{4}
= 100/
Z = (R^{2} + (X_{L}
 X_{C})^{2}) = (50^{2}
+ (50  100/)^{2})
= 134.7
I_{rms} =V_{rms}/Z = 110/134.7 = 0.816 A
Q24) Draw
a labelled ray diagram to show the image formation in a reflecting type
telescope. Write its two advantages over a refracting type telescope. On what
factors does its resolving power depend? (Marks 3)
Ans24)
Advantages:
1. There is no chromatic aberration as the objective is a mirror.
2. Image is brighter compared to that in a refracting telescope.
Its resolving power depend on
(i) Diameter of the objective mirror.
(ii) Wavelength of light with which the object is viewed.
Q25) Define
the terms 'solar constant' and 'solar luminosity'. Explain how their knowledge
helps us to calculate the surface temperature of the sun. Derive the necessary
mathematical expression. (Marks 3)
Ans25) Solar constant : is defined as the amount of radiant
energy received per second by a unit area of a perfectly black body surface
field at right angle to the direction of the sun rays at the mean distance of
the earth from the sun.
Solar luminosity : is defined as the amount of energy emitted per second
by the sun in all direction.
Expression for surface temperature of sun : Consider the sun to be a
black body at temperature T and of radius R at the center of a bottom sphere of
radius r, where r = 1A.U & r R
According to Stefan's law the energy emitted per second per unit area by the sun
is
E = T^{4} Where is Stefan's constant. Total energy emitted per second by sun = 4R^{2}E = 4R^{2}T^{4} = L_{s} where L_{s} is solar luminosity. Since solar luminosity is also equal to the total energy radiated per second by the sun 4R^{2}T^{4} = 4R^{2}S or T = (r^{2}S/R^{2})^{1/4} 
Knowing the value of S, , r & R, the value of T can be determined.
Q26) An
object is kept in front of a concave mirror of focal length 15 cm. The image
formed is three times the size of the object. Calculate two possible distance of
the object from the mirror. (Marks 3)
Ans26) Given f = 15 cm , m = + 3u
Now m = f/(f  u)
For a real image m = 3
So 3 = 15/(15  u) or 3 = 15/(15 + u)
45  3u = 15
3u = 60 or u = 20 cm
For virtual image m = 3
3 = 15/(15  u)
or 45 + 3u = 15 or u = 10 cm
.^{.}. The two possible distances of object from the mirror are 20 cm,
10 cm
Q27) A
voltmeter V of resistance 400
is used to measure the potential difference across a 100
resistor in the circuit shown here.
(a) What will be the reading on the voltmeter?
(b) Calculate the potential difference across 100
resistor before the voltmeter is connected. (Marks 3)
Ans27)
(i) Effective resistance of 400
& 100 'R' is given by
1/R = 1/100 + 1/400 or R = 80
Total distance of the circuit = R' = 80 + 200 = 280
Current drawn from cell = V/R' = 84/280 = 0.3 A
P.D across 80 resistance =
0.3 x 80 = 24V
P.D across 100 or 400
= 24V
(ii) When voltmeter is not connected, then the total resistance of the circuit
is
100 + 200 = 300
Current drawn from cell = 84/300 = 0.28 A
P.D across 100 = 100 x
0.28 = 28V
Q28) Derive
a mathematical expression for the force per unit length acting on each of the
two straight parallel metallic conductors carrying current in the same direction
and kept near each other. Why do such current carrying conductors attract each
other?
Or
Derive a mathematical expression for the force acting on a current carrying
straight conductor kept in a magnetic field. State the rule used to determined
the direction of this force. (Marks 5)
Ans28) Consider AB and CD, two infinite long straight conductor
carrying I_{1}, I_{2} in the same direction. They are separated
by distance d
field B_{1} on CD due to current I_{1} in AB is given by
B_{1} = _{o}I_{1}/2d Force on CD due to field B_{1} = B_{1}I_{2}l F_{2} = _{o}II_{2}l/2d According to Fleming's left hand rule F_{2} is perpendicular to both direction of the current I_{2} and B_{1} and is acting as shown in the figure. Again field B_{2} on AB due to current I_{2} in CD is B_{2} = _{o}I_{2}/2d Force F_{1} on AB due to B_{2} is F_{2} = B_{2}I_{1}l or F_{2} = _{o}I_{2}I_{1}l/2d 
It is directed inwards in accordance
with Fleming's left hand rule.
Hence wires AB and CD attract each other with a force F = F_{1} = F_{2}
= _{o}I_{1}I_{2}l/2d
Force per unit length = _{o}I_{1}I_{2}/2d
Whenever a current carrying conductor is placed with a magnetic field, moving
electrons experience a force. Since the electrons are not free to move outside
the body of the conductor, they transmit, their force to the conductor through
collisions with the atoms of the conductor. This results in a force on
conductor.
Let us consider a wire of length L and area crosssection A placed in magnetic
field B acting perpendicular to the plane of the paper and directed inwards. If
n is the number of free electrons per unit volume and V_{d} is the drift
velocity, then
Force on each electron m
= e(_{d} x )
In magnitude F_{m} = e V_{d} B sin 90^{o }= e V_{d}
B
Volume of wire = Al
Number of electrons in the wire N = n Al
Total force on the wire F = N F_{m}
Or F = (n Al) e V_{d} B = (neV_{d}A)l B
But current I = neV_{d}A
.^{.}.F = I/B
In vector form = I (
x )
Direction of the force can be determined by using right hand rule.
Q29) Draw a
labelled diagram of Thomson's experimental setup to determine e/m of electrons.
Explain by deriving the necessary mathematical expression how of electron can be
determined by this method. (Marks 5)
Ans29) Construction : it consist of a discharge tube. A P.D. of
a few thousand volts is maintained between C and A. Small hole in the anode
produces a fine beam of cathode rays which strike the screen coated with ZnS.
Two plates are given negative & positive potential resp. They produce
electric field that is perpendicular to the path of electrons. A uniform
magnetic field is applied perpendicular to both electric field and the flow of
electrons is directed into the paper.
Theory : If no electric or magnetic field is applied, the electron beam meets the screen at point M. Apply both the electric and magnetic fields and adjust the strength of electric field and magnetic field B so that the electron beam meets the screen S at its undeflected position M. In this situations the forces on the electron due to electric field magnetic field balance each other.  Thomson's method for the determination of e/m of an electron 
Let m and e be mass and charge of
electron and v be the velocity of the electron when it comes out of the hole of
the anode.
Force on the electron due to electric field = Ee
Force on the electron due to magnetic field = Bev
For the undeflected position of the spot on the screen S, the two forces must be
equal and opposite, therefore
Ee = Bev or v = E/B  (i)
As the electron beam is accelerated from cathode to anode, its potential energy
at the cathode appears as gain in its kinetic energy at the anode. If V is the
potential difference between cathode and anode, then potential energy of
electron at cathode = charge x potential difference = eV
Gain in Kinetic Energy of electron at anode
= 1/2 mv^{2}
So, we have eV = 1/2 mv^{2}
or e/m = 1v^{2}/2V = (1/2V)(E/B)^{2}
= E^{2}/2VB^{2}  (ii)
Q30) Define the
terms 'potential harrier' and 'depletion region' for a pn junction. Explain
with the help of a circuit diagram, the use of pn diode as a full wave
rectifier. Draw the input and output waveforms. (Marks 5)
Ans30) Potential Barrier : The potential difference created
across the junction due to diffusion of free electrons & holes across the
junction is called potential barrier.
Depletion region : It is the region around the junction which is devoid
of free charge carriers Diode as full wave reflector.
During the positive half of the
input A.C. the upper pn junction diode is forward biased as shown in Fig (a)
and the lower pn junction diode is reverse biased. The forward current flows on
account of majority carriers of upper pn junction diode in the direction shown.
During the negative half cycle of input A.C. the upper pn junction diode is
reverse biased, and the lower pn junction diode is forward biased, Fig (b) The
forward current flows on account of majority carriers of lower pn junction
diode. We observe that during both the halves, current through R flows in the
same direction.
Boarding Schools By State

Boarding Schools Top Cities

Boarding Schools By Board
