1 MODEL TEST PAPER

(ANSWERS)

SECTION A
1. (b) zero
1
2. (a) ∈0 E 2 Ad
2
3. (b) 14 V
4. (a) 105 V m– 1
5. (d) mB (1 – cos θ)
6. (a) 250 J
7. (a) 3.54 Hz
1
8. (d)
µ 0 ∈0
2 d1d2
9. (d)
d1 – d2
10. (b) increase by 589 Å
11. (a) 1.1 eV
12. (b) Circuit (B)
13. (c) Assertion (A) is true, but Reason (R) is false.
14. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of
Assertion (A).
15. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of
Assertion (A).
16. (c) Assertion (A) is true, but Reason (R) is false.

SECTION B
17. (a) In reverse bias the junction width increases. The higher junction potential restricts
the flow of majority charge carriers. However, such a field favours flow of minority
carriers. Thus, reverse bias current is due to flow of minority carriers only. Since the
number of minority carriers is very small, the current is small and almost independent
of the applied potential upto a critical (before breakdown) voltage.

model test Paper – 1 1

 (b) At a critical (breakdown) voltage the reverse bias current shows a sudden increase.
Under high reverse bias, the high junction field may strip an electron from the valence
band, which can tunnel to the n-side through the thin depletion layer. This mechanism
of emission of electrons after a critical applied voltage leads to a high reverse
(breakdown) current.
18. (a) Long distance radio broadcasts use short wave bands because wavelengths of short
wave bands is less and so their frequency and consequently energy is higher. Due to
higher energy these waves can propagate through a long distance and still be detected
successfully.
(b) Experimental demonstration of electromagnetic waves is possible only in the low
frequency (radio wave) region because these waves can be easily produced by
oscillatory electronic circuits.
19.(A) VA – VB = VD – VC because no current can flow through 5 W and 10 W resistors.
Between D and C, three cells are in parallel, hence
ε ε ε ε 1 2 3 6
= 1 + 2 + 3 = + + = =6
r r1 r2 r3 1 1 1 1
1 1 1 1 1 1 1 3
and = + + = + + =
r r1 r2 r3 1 1 1 1
1
or r= W
3
1
⇒
e = 6r = 6 × =2V
3
Or
(B) As per Kirchhoff’s 1st law :
I = I1 + I2 ...(i)
Applying Kirchhoff’s 2nd law to mesh ACDBA, we have
– 12I – 10I2 – 30I – 3I + 16 = 0
⇒
45I + 10I2 = 16 ...(ii)
and for mesh CFHDC, we have
– 20I1 – 5I1 – 25I1 + 10I2 = 0
⇒
10I2 – 50I1 = 0 ...(iii)
Solving these three equations, we get
I = 0.3 A, I1 = 0.05 A and I2 = 0.25 A
20. Here magnetic moment m = 1.2 A m2 and distance r = 0.1 m
Strength of magnetic field at the axial point

µ0 2m 10 −7 × 2 × 1.2
B= × = = 2.4 × 10–4 T
4π r 3 (0.1) 3

2 U-LIKE Physics–XII
 13.6
21. Energy of electron in nth orbit En = −
eV
n2
\ Energy required to take an electron from ground state (n1 = 1) to the second excited state

(n2 = 3) is :
 13.6   13.6 
E = En2 − En1 =  − 2  −  − 2  = 12.1 eV
 (3)   (1) 

SECTION C
22. (a) (i) Nuclear forces are extremely short range forces and do not operate beyond a
distance of 3 fm.
(ii) The forces are strongly attractive in nature and about 100 times stronger than the
electrostatic forces.
(b) Constancy of binding energy per nucleon in the range of mass number ‘A’ ranging from
30 to 170 can be explained on the basis of short ranged nature of nuclear force.
Inside a sufficiently large nucleon a particular nucleon is under the influence of only
some of its neighbours which come within the range of the nuclear force. If a nucleon
can have a maximum of p neighbours within the range of nuclear force, its binding
energy would be proportional to p or the binding energy of the nucleon will be pk
where k is a constant having the dimensions of energy.
Thus, it is clear that the binding energy per nucleon is a constant and equal to pk.
→ →
23.(A) (a) Force F acting on a particle of mass m and charge q moving with velocity v in a
→
magnetic field B is given as :
→ → →
F = q( v × B )
→ →
(i) The charged particle will move along a circular path if B and v are in mutually
perpendicular directions.
→ →
(ii) The charged particle will move in a helical path if angle q between v and B is
finite but not 90°.
(b) Since force due to magnetic field acts perpendicular to the direction of motion of a

charged particle, hence work done by the force is zero. As a result, in accordance with
work-kinetic energy theorem the kinetic energy of charged particle remains constant.
Or
φ
i.e., deflection per unit current is known as the current sensitivity of a
(B) (a) The term
I
φ NAB
galvanometer. Current sensitivity =
I k
Thus, to enhance the current sensitivity we use a strong magnetic field B and
suspension spring of small value of torsional constant ‘k’.

model test Paper – 1 3

 Theoretically current sensitivity can also be increased by increasing N and by
increasing A. But these are not practical options beyond a limit.
(b) If a moving coil galvanometer of resistance RG gives full scale deflection for a current
Ig, it can be converted into an ammeter of maximum range I by connecting a shunt of
resistance S, where
RG . I g
S= ⇒ SI = (RG + S) Ig ...(i)
I − Ig
To convert the given galvanometer into an ammeter of maximum range 2I we shall
connect a shunt of resistance S′, where
RG . I g
S′ = ⇒ S’.2I = (RG + S’)Ig ...(ii)
2I − I g

2S′ R + S′ S ⋅ Rg
From (ii) and (i), we get = G ⇒ S’ =
S RG + S (S + 2 Rg )
24. (a) V-I characteristics of a p-n junction diode are drawn here.
IF(mA)

80
60
Forward
40 Bias
Characteristics
20
100 80 60 40 20
VR(V) VF(V)
Vbr 0.2 0.4 0.6 0.8 1.0
Reverse 10
Bias
Characteristics 20
30
IR(µA)
(b) Threshold voltage for a diode is the voltage in forward bias beyond which current
increases rapidly with voltage. For a silicon diode threshold voltage is about 0.7 V.
Breakdown voltage on the other end is that limiting voltage applied in reverse bias
arrangement, beyond which current suddenly increases manyfold with any increase
in voltage. Its value ranges from 50 to 100 V for a p-n diode.
(c) For rectification of a.c. voltages we make use of the property that a p-n junction offers
very small resistance and allows flow of current easily in forward bias but does not
allow easy flow of current in reverse bias arrangement.
3 2
25. (a) Q Net outward flux φE = 8.0 × 10 N m /C
q
From the mathematical expression of Gauss’ theorem φE = , we find that
∈0
3 −12
Net charge inside the box q = φE . ∈0 = 8.0 × 10 × 8.85 × 10 C
= 7.08 × 10 −8 C = 0.07 µC.

4 U-LIKE Physics–XII
 (b) If the net outward flux through the surface of the box were zero then in accordance
with Gauss theorem we conclude that the net charge inside the box were zero. It
means that either no charge was present in the box or charges were present inside
the box but their algebraic sum was zero.
26. (a) When an alternating current flows in the circuit is given by I = Im sin ωt.
∴ Average power dissipated per complete cycle of a.c. is given by
1 T 1 T V I T
Pav = ∫ V I dt = ∫ Vm sin ωt . I m sin ω t dt = m m ∫ sin 2 ωt dt
T 0 T 0 T 0

T
Vm I m T V I  sin 2 ω t 
= ∫ (1 − cos 2ω t) dt = m m t − 
2T 0 2T  2ω 0

V I   sin 2 ω T − sin 0  
= m m
(T − 0) −  
2T   2ω 

V I   sin 4 π − sin 0   Vm I m   0 − 0 
= m m
T −    = 2 T T −  2 ω  
2T   2ω     
V I 1 2
= m m = Im R
[Q Vm = RIm ]
2 2
2
2 Vrms
(b) Here P = 100 W, Vrms = 220 V. As P Irms R=
R
2
Vrms (220)2
⇒ Resistance R =
= = 484 Ω.
P 100
27. (i) Let the capacitance of given capacitor be C mF. Then as per data given in the question :
CV = 360 µC …(1) and C(V – 120) = 120 µC …(2)
Subtracting (2) from (1), we have
240 µC
C × 120 = 240 µC ⇒ C= = 2 µF
120 V
360 µC 360 µC
and from (1) V= = = 180 V
C 2 µF
(ii) When the potential applied is increased by 120 V, the potential becomes
V’ = (V + 120) volt = (180 + 120) volt = 300 volt
∴ Charge on capacitor Q’ = CV’ = 2 µF × 300 volt = 600 µC
28.(A) For refraction through the convex lens, u1 = – 40 cm, f1 = + 10 cm
1 1 1 1 1 3
\
= + = + =
v1 f1 u1 (+ 10) (− 40) 40
40
⇒
v1 = + cm
3

model test Paper – 1 5

 The image formed by convex lens acts as an object for concave lens.
40 25
Now, u2 = + −5 = + cm , f2 = – 10 cm
3 3
1 1 1 1 3 1
\
= + = + = +
v2 f 2 u2 (− 10) 25 50
⇒ v2 = + 50 cm
So the final image is a real image formed on right side of concave lens at a distance 50 cm
from it.
Or
(B) As per question, path S1P = 2d and path S2P = (2 d )2 + (d )2 = d 5
\ Path difference D = S2P – S1P = d 5 − 2 d = d( 5 − 2)
λ
As we get first minima at point P, hence D =
2
λ
⇒ d( 5 − 2) =
2
λ
⇒ d=
2( 5 − 2)

SECTION D
29. [VERY IMPORTANT NOTE : In the language of Q.29. there is a mistake. In Q.(iii) - option (c)
must be 0.5 A instead of 0.8 A.]
(i) (b) 3 W
(ii) (a) 7.2 W
(iii) (c) 0.5 A
(iv) (A) (a) zero
Or
(B) (c) Wheatstone’s bridge remains balanced as previously and there is no deflection
in galvanometer.
30. (i) (d) all the above type of waves.
(ii) (c) Both produce waves of same wavelength having a constant originating phase
difference.

( )
2
(iii) (c) I1 + I 2
(iv) (A) (c) 1.0 mm
Or

(B) (a) 0.2 mm
SECTION E
31.(A) (a) Photoelectric effect is the phenomenon of emission of electrons from the surface of
metals when light radiations of suitable high frequency are incident on them. The
emitted electrons are called photoelectrons and the current so obtained is called the
photoelectric current.
6 U-LIKE Physics–XII
 (b) (i) On increasing the intensity of incident light magnitude of photoelectric current
increases but stopping potential remains unchanged as shown in graph (i).
(ii) On increasing the frequency of incident light, the saturation current remains
unchanged but the value of stopping potential i.e., maximum kinetic energy of
photoelectrons increases as shown in graph (ii).

I I
I1
I2
I3 ν1 > ν2 > ν3
I1 > I2 > I3
1
ν2
ν3
– V03
– ve – V0 0 + ve – ve – V – V01 0 + ve
02
V V
(i) (ii)
(c) As per Einstein’s photoelectric equation, energy of incident photon
E = f0 + Kmax = f0 + eV0
⇒ h(n – n0) = Kmax = eV0
Here f0 = work function, Kmax = maximum K.E. of photoelectrons, n = frequency of
incident photon and n0 = threshold frequency.
(d) Planck’s constant h = Slope of V0 – n graph × e = (4.15 × 10–15) × (1.6 × 10–19)
= 6.64 × 10–34 J-s
Or
(B) (a) As per de Broglie hypothesis matter particles exhibit wave character, the wavelength
λ associated with a particle of mass m moving with speed v is given by
h
             λ =
mv
(b) Consider an electron of mass m and charge e and initially at rest. If it is accelerated

through a potential V volts, then kinetic energy gained by electron = work done on the
electron
1 2 eV
∴ m v 2 = eV ⇒ v =
2 m
∴ de Broglie wavelength of electron

h h h h 1.227 nm
λ= = = = =
p mv 2e V 2m e V V volt
m⋅
m
(c) de Broglie argued that the electron in its circular orbit must be seen as an electron

wave. In terms of electron wave, he said that only those stationary Bohr orbits are
possible in which total distance (i.e., the circumference of the orbit) contains a definite
number of these electron waves.

model test Paper – 1 7

 Mathematically,
total path length of orbit = n (λelectron wave)
∴ 2π rn = n λ,
where n is an integer and may have values 1, 2, 3, 4, ……. etc.
From de Broglie concept of matter waves, we know that wavelength of electron waves
may be expressed as
h h
λ= = ,
pn m vn
where m = mass of electron and vn = velocity of electron in state corresponding to n.
Substituting the value of λ in de Broglie’s quantum condition, we get
h h
2π rn = n . ⇒ m vn rn = n
m vn 2π
which is Bohr’s quantum condition (on the second postulate for quantisation of
angular momentum.
32.(A) (a) On the basis of his experimental study, Faraday gave a law of electromagnetic
induction, which has following two parts :
(i) Whenever the amount of magnetic flux linked with a circuit (or a coil) changes
(either increases or decreases), an emf is induced in the circuit. The induced emf
lasts so long as the change in magnetic flux continues.
It indicates that the real cause of electromagnetic induction is the change in magnetic
flux linked with a circuit with time.
(ii) The magnitude of induced emf in a circuit is equal to the rate of change of magnetic
flux linked with the circuit. Mathematically,
dφ
Induced emf ε = − B .
dt
Here – ve sign has been incorporated because induced emf always opposes the
change in magnetic flux.
(b) Direction of induced current in a circuit is given by Lenz’s law which states that the
induced current always opposes the change in magnetic flux that produced it.
As an illustration consider a bar magnet NS
S N S N
pushed towards a conducting loop. When
N pole of magnet moves towards the loop
[Fig. (a)], the face of the loop facing the north
pole N develops N polarity as per Lenz’s law
so as to oppose the motion of magnet. As a
G G
result, an anticlockwise current is setup in
(a) (b)
the coil when viewed from that face.
Again, when N-pole of magnet moves away from the loop [Fig. (b)], the nearby face of
loop develops south pole S which opposes the motion of magnet away from the loop.
Consequently an induced current in clockwise direction, when viewed from that face
is setup.

8 U-LIKE Physics–XII
 (c) Here self inductance with iron core Liron = 2.4 H
and self inductance with air core Lair = 3 mH = 3 × 10–3 H
Liron 2.4
\ Relative permeability of iron mr =
= = 800
Lair 3 × 10 −3
Or
(B) (a) When a coil of N turns and each turn enclosing area A is placed in a uniform magnetic
field of strength B making an angle q from normal to the direction of magnetic field,
magnetic flux linked with the coil is φ = N B A cos θ.
As the coil is rotated about its own axis with an angular speed w then value of angle
θ = ω t and hence, magnetic flux changes and an induced emf is developed across the
ends of coil.
As φ = N B A cos ωt
dφ
∴ Induced emf ε = −
= N B A ω sin ωt = ε0 sin ωt
dt
where ε0 = N B A ω = maximum (peak) value of induced emf.
It is obvious that induced emf is sinusoidal in nature.
(b) (i) As magnetic flux of inductor fB = LI, hence fB versus I graph is a straight line
whose slope gives the value of self inductance L.

φB

O
I

dI dI
(ii) As induced emf e = – L , hence e versus graph
dt dt
dI
is a straight line parallel to axis as shown here.
dt
1 2
(iii) As magnetic potential energy UB =
LI , hence
2
potential energy UB versus current I graph is a
parabolic curve as shown below.

UB

O
I

model test Paper – 1 9

33.(A) (a) Consider a point object O situated on the principal axis of a biconvex lens, whose two
surfaces have radii of curvature R1 and R2, respectively.
As shown in figure due to refraction at 1st surface of lens an image I′ is formed for
the object O. If OC = u, CI′ = v′, then using the refraction formula at a single spherical
surface, we have
n2 n1 n − n1
− = 2 ...(i)
v′ u R1
A B

The image I′ behaves as a R1 R2

virtual object for refraction
C I I
at the second surface of O u v
v′
the lens and the final real n1 n1
image is formed at I. Thus, 1st 2nd
surface surface
for second surface applying
n1
refraction formula, we have
n1 n2 n − n2 n2 − n1
− = 1 = ...(ii)
v v′ R2 (− R2 )
Adding (i) and (ii), we have

n1 n1  1 1 
− = (n2 − n1 )  − 
v u R
 1 R2 

1 1 n  1 1   1 1 
− =  2 − 1 
or −  = (n21 − 1)  − 
v u n
 1 R
 1 R 2  R
 1 R2 
If u = ∞, then by definition v = f and, hence,

1 1  1 1  1  1 1 
− = (n21 − 1)  −  ⇒ = (n21 − 1)  − 
f ∞  R1 R2  f  R1 R2 
This relation is the requisite lens maker’s formula.
(b) As per lens maker’s formula the focal length of a given lens depends on its refractive
index, which in turn depends on the wavelength (colour) of incident light. Thus, it is
clear that focal length of a lens depends upon the colour of incident light. In fact, as
nv > nR, hence focal length for violet colour fV is least and that for red colour fR is
maximum.
(c) When a convex lens is immersed in water, value of refractive index of lens with respect
to surrounding (i.e., water) n21 changes and hence its focal length change. Actually
fwater is found to be much greater than fair.
However, when a concave mirror is immersed in water, its focal length remains
unchanged at same value as in air.
Or
(B) (a) Path of ray PQ inside the prism and on emergence from face AC is drawn here. Here
∠e is the angle of emergence and ∠d is the angle of deviation.
In figure in DPOR,

10 U-LIKE Physics–XII
 ∠r1 + ∠O + ∠r2 = 180° ...(i)
Again in quadrilateral AQRA,
∠A + 90° + ∠O + 90° = 360° or ∠A + ∠O = 180° ...(ii)
Comparing (i) and (ii), we get
∠A = ∠r1 + ∠r2
Now in DQRT, we have
∠d = ∠TQR + ∠TRQ = (∠i – ∠r1) + (∠e – r2)
= ∠i + ∠e – (∠r1 + ∠r2) = ∠i + ∠e – ∠A
⇒ ∠d + ∠A = ∠i + ∠e

(b) Graph showing variation of the angle of deviation ∠d as a
function of angle of incidence ∠i is drawn here.
From the graph, we find that in general there are two angles
of incidence ∠i1 and ∠i2 for same value of deviation angle
∠d. However position of minimum deviation ∠dm is unique
in the sense that it corresponds to only one particular value
of ∠i. It is possible only if ∠i = ∠e in this case and the light
rays passes through the prism symmetrically.
(c) In minimum deviation position ∠i = ∠e, hence we must have
∠r1 = ∠r2 = ∠r (say). Then

 A + δm 
∠i + ∠i = 2∠i = ∠A + ∠dm or ∠i =  
 2 
A
and ∠r1 + ∠r2 = 2∠r = ∠A or ∠r =  
2

 A + δm 
sin 
 2 
\ Refractive index of prism n = sin i =

A
sin  
sin r
 2
If dm = A, then
 A + A
sin 
 2  sin A A
n= = = 2 cos
 A A 2
sin sin
 2 2
and if angle of prism be A = 60°, then

3
n = 2 cos 30° = 2 × = 3
2

model test Paper – 1 11

 2 MODEL TEST PAPER
(ANSWERS)

SECTION A

r1
1. (a)
r2
2. (b) 0.1 J m–2
3. (b) zero at the centre of loop.
4. (d) R=0
5. (a) 1.0 H
6. (b) IR = IG
7. (d)
8. (d) 2π
9. (d) 11
10. (b) 3.2 × 10–19
11. (a) 1:2
12. (c)
13. (b)
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation
of Assertion (A).
14. (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation
of Assertion (A).
15. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of
Assertion (A).
16. (c) Assertion (A) is true, but Reason (R) is false.

SECTION B
17.(A) In the arrangement shown, capacitors C1 = 1 mF and C2 = 5 mF are in parallel and their
combined capacitance C12 = 1 + 5 = 6 mF.
Now, C12 is in series with capacitor C3 = 4 mF
6×4
Hence their combined capacitance C0 = = 2.4 mF
6+4
\ Charge on C3(= 4 mF) = Charge on C0 = 10 V × 2.4 mF = 24 mC

12 U-LIKE Physics–XII
 Or
1
CV 2
Energy 2 CV 2 (40 × 10 – 6 ) × (3000)2
(B) Power P = = = =
Time t 2t 2 × (2 × 10 − 3 )
= 9 × 104 W or 90 kW
18. (a) Magnetic susceptibility of a magnetic material is defined as the ratio of magnetisation
vector to the magnetic intensity.
M
\ Magnetic susceptibility c =

H
Since both M and H have the same unit A m–1, hence susceptibility is a unitless quantity.
(b) For diamagnetic materials susceptibility is small and negative.
19. Here u = – 20 cm and f = – 15 cm
1 1 = 1 , we have
From mirror formula +
v u f
1 1 1 1 1 1
= − = − = − or v = – 60 cm
v f u (− 15) (− 20) 60
h′ v
\ Linear magnification m =
= −
h u
h′ (− 60)
⇒
= − =–3 or h′ = – 3 cm
1 cm (− 20)
\ Area enclosed by the image of the wire = A′ = (h′)2 = 9 cm2.
20. In a Young’s double-slit experiment if intensity of light due to each slit be I then resultant
π
sum intensity IR = 4I cos 2  
 8
At centre of bright fringe, f = 0° and hence Imax = 4I

λ 2π λ π
If at a given point P path difference D = , then phase difference f = . =
8 λ 8 4
 π  π
\
Intensity IP = 4I cos 2   = Imax. cos 2   = Imax (cos 22.5°)
2
 8 8
= Imax.(0.9239)2 = 0.854 Imax
IP
⇒
= 0.854
I max
21. (a) Depletion region in a p-n junction is a small region on either side of junction which
consists of immobile ions only and is devoid of active charge carriers i.e., free electrons
and holes.
(b) (i) In forward bias width of depletion region decreases.
(ii) In reverse bias the width of depletion region increases.

model test Paper – 2 13

 SECTION C

22. (a) Field lines are drawn here :

Conducting surface
(b) Let charges Q and – 3Q be placed at points A and B separated by a distance d. The third
charge 2Q should be placed at point P on extension of line BA nearer the smaller charge
of Q at a distance x from it. As net force on 2Q charge is zero.
→ → → → →
∴ Fp = FA + FB = 0 or FA = – FB
1 Q(2Q) 1 ( − 3Q)(2Q)
⇒ ⋅ 2 = − ⋅
4 π ∈0 x 4 π ∈0 ( x + d )2
1 3 d
⇒ 2 = 2
⇒ x=
x ( x + d) ( 3 − 1)
23. Electromagnetic waves of wavelength l1, used in radar system, are microwaves. These can
be produced by special vacuum tubes e.g., klystrons, magnetrons and Gunn diode.
Electromagnetic waves of wavelength l2, used in water purifiers, are ultraviolet rays.
Mercury lamp and hydrogen tube are good sources of UV rays.
Electromagnetic waves of wavelength l3, used in remote switches of TV, are infrared
waves and are produced by any hot body.
24.(A) When an electron in a hydrogen atom is de-excited from a state n to the first excited state
(i.e., n′ = 2), the energy of emitted photon is
 13.6   13.6   13.6 
E =  − 2  −  − 2  =  3.4 − 2  eV
 n   (2)   n 

As work function of photo cell φ0 = 2 eV and stopping potential V0 = 0.55 V, the maximum
kinetic energy of photoelectron Kmax = eV0 = 0.55 eV
As per Einstein’s photoelectric equation = E = φ0 + Kmax
 13.6  13.6
⇒
 3.4 − 2  = 2 + 0.55 = 2.55 ⇒ = 3.4 – 2.55 = 0.85
 n  n2

13.6
⇒
n= =4
0.85

14 U-LIKE Physics–XII
 Or
(B) (i) Here energy of proton K = 4.1 MeV = 4.1 × (1.6 × 10– 13) J

1 2K 2 × 4.1 × (1.6 × 10 − 13 )

Q K= mv 2 hence v = = = 2.8 × 107 m s–1.
2 m 1.67 × 10 − 27
(ii) When proton beam (Z1 = 1) is bombarded on lead sheet (Z2 = 82), the distance of closest
approach r0 in given as :

1 Z Z e2 (9 × 10 9 ) × 1 × 82 × (1.6 × 10 − 19 )2
r0 = . 1 2 = − 13
= 2.88 × 10–14 m
4 π ∈0 K (4.1 × 1.6 × 10 )
25.(A) (a) Consider a battery connected to an inductor L. As the battery circuit is completed,
current starts growing in inductor and consequently an induced emf is set up in the
reverse direction given by
dI
|ε| = L
dt
If the battery supplies a current I through the inductor for a small time dt, the elementary
work done by emf source (battery) against the induced emf will be
dI
dW = ε I dt = L ⋅ I ⋅ dt = L I dI
dt
Hence, the total amount of work done by emf source till the current increases from
initial zero value to its final steady value I0 will be
I0 I
1 2 0 1 2
W= ∫ L I dI =  2 L I  0 = 2 L I0
0
This work done is stored in the inductor in the form of its magnetic energy. Thus,
1 2
Energy stored in an inductor U = W = L I0 .
2
dI
(b) Here self inductance L = 5 H and = – 2 A s–1
dt
dI
\
Emf induced in the coil e = − L = – 5 × (– 2) = 10 V
dt
Or
(B) (a) Mutual inductance of a pair of two coils is equal to magnetic flux linked with one coil
when a unit steady current flows through the other coil.
Alternately, mutual inductance is numerically equal to the induced emf in one coil
when current in the neighbouring coil is changing at a unit rate.
SI unit of mutual inductance is a henry (H).
(b) Here I = 0.5 A and φ2 = 0.5 × 10–3 Wb

model test Paper – 2 15

 φ2 0.5 × 10 −3 Wb
∴ M= = = 10–3 H or 1 mH
I1 0.5 A
If current through the smaller coil suddenly falls to zero, an instantaneous emf is
produced in the larger coil on account of mutual inductance.
26. (a) Ray diagram is drawn here.

(b) Here f0 = 140 cm and fe = 5.0 cm

f0 140
(i) For normal adjustment, magnifying power |m| = = = 28
fe 5.0
(ii) When final image is formed at a distance D = 25 cm, the magnifying power will be
f0  f e  140  5 
|m| =
fe  1 + D  = 5  1 + 25  = 33.6.
   
27. (a) Out of X, W and S, X is most stable, W is less stable and S is least stable.
A A2 A1
(b) ZS → Z2 X + Z1 W

(c) For nuclei of high mass number, the number of protons inside the nucleus is more.
These protons repel each other. Due to larger force of repulsion on account of greater
number of protons present in the nucleus, the value of binding energy per nucleon
decreases.
28. (a) A pure semiconductor material is called an intrinsic semiconductor e.g., silicon and
germanium.
(b) In an intrinsic semiconductor ne = nh.
(c) Electrical conductivity of a pure semiconductor depends on
(i) the energy band gap between the valence band and conduction band of that semi-
conductor, and
(ii) the temperature of the semiconductor.

16 U-LIKE Physics–XII
 SECTION D
29. (i) (c) particle velocity will increase in magnitude with time.
(ii) (d) a circle
(iii) (c) An electron moving in the direction of the electric field gains K.E.
E
(iv) (A) (c) E, B and v are mutually perpendicular and v =
B

Or
(B) (a) K
30. (i) (b) nucleus
(ii) (b) not stationary
(iii) (a) spiral
h
(iv) (A) (c)
π
Or
–11
(B) (c) 10 m

SECTION E
31.(A) (a) The conductivity of a material is given as :
J ne 2 τ
Conductivity s = = ,
E m
where n = number density of conduction electrons and t = mean relaxation period.
(i) In case of semiconductors the number density (n) of conduction electrons and holes
is usually very small but it increases rapidly with increase in temperature of semi-
conductor because with rise in temperature kinetic energy of electrons increases and
more electrons pass over from valence band to conduction band. So the conductivity
increases with increase in temperature.
(ii) In conductors the number density of conduction electrons is extremely large
(n  1028 m– 3) and is practically independent of temperature. However, with rise in
temperature, amplitude of atomic/molecular vibrations increases and consequently
frequency of collisions with drifting electrons increases. As a result, the relaxation
period t decreases and as a result the conductivity decreases.
 5 
(b) It is given that l’ = l + 5% of l = l  1 + .
 100 
As on stretching the wire its volume remains constant, hence A.l = A’.l’
Al Al A
⇒ A’ = = =
l′  5   5 
l 1 +   1+ 
 100   100 

model test Paper – 2 17

 2
 5 
ρl 1 +  2
ρ l′  100   5   10 
∴ New resistance R’ = = = R ⋅1 +  = R  1 +  (approx.)
A′ A  100  100 

 10 
R 1 +  −R
R′ − R  100 
∴ Percentage increase in resistance ∆R = × 100% = × 100%
R R
= 10%.
Or
(B) (a) To analyse complicated electrical networks Kirchhoff gave two rules :
(i) Junction rule or current rule – The algebraic sum of all the currents flowing into a
junction is zero. In other words, at any junction, the sum of currents entering the
junction must be equal to the sum of currents leaving it. Kirchhoff’s first rule is
based on the principle of conservation of electric charge. Mathematically, Σ I = 0.
While applying Kirchhoff’s current rule the current flowing towards the junction
is taken +ve and the current flowing away from the junction is taken –ve.
(ii) Loop rule or voltage rule – The algebraic sum of changes in potential around any
closed loop involving resistors and cells in the loop must be zero. Mathematically,
ΣV = 0 or Σ(ε + R I ) = 0
Kirchhoff’s second rule is based on the principle of conservation of energy.
While applying Kirchhoff’s voltage rule the change in potential in traversing a
resistance in the direction of current is –IR while in the opposite direction +IR.
Again the change in potential in traversing an emf source from negative to positive
terminal is +ε while in the opposite direction – ε.
(b) (i) As per Kirchhoff’s first law current flowing through unknown resistor R is
(I + 1) A.

As per Kirchhoff’s second law for mesh ACDBA, we have

– (I + 1)R – 1.2 + 3 = 0 or (I + 1)R = 1 ...(i)
∴ Voltage drop across R = V = (I + 1)R = 1 Volt
(ii) Again for mesh ECDFE, we have
– (I + 1)R – I⋅3 + 5 = 0 or (I + 1)R = 5 – 3I ...(ii)
Comparing (i) and (ii), we get
4
5 – 3I = 1 ⇒ I = A = 1.33 A
3

18 U-LIKE Physics–XII
32.(A) (a) Consider an alternating voltage given by
V = Vm sin ωt applied to a capacitor C and
resistor R joined in series. As C and R are in
series, same current I flows in the entire
→ → →
circuit. Let VC , VR and V be the magnitudes
of instantaneous values of voltage across the
capacitance, resistance and the source. Then, as
shown in phasor diagram.
→ → → → →
VC = I XC lagging behind in phase by π
as compared to I , and VR = I R in same
→ 2
phase as that of I .
These are being represented by phasors OB
and OA, respectively. Then, total instantaneous
voltage V is given by phasor OC such that

V = OA2 + OB2 = VR2 + VC2 = R 2 + XC2

(i) ∴ Impedance offered by the circuit,
2
V  1 
Z = = R 2 + XC2 = R 2 +   .
I C ω
→
(ii) Moreover, the phasor diagram indicates that source voltage V lags behind the
→
current I (or the circuit current leads the a.c. supply voltage) by a phase angle f,
where
OB I XC X 1/C ω 1
tan φ = = = C = = .
OA IR R R R ⋅C ω
(b) (i) As L = 5.0 H and C = 80 µF = 80 × 10–6 F
1 1
∴ Resonant angular frequency ωr = = = 50 s–1
LC 5 × 80 × 10 –6
(ii) As Vrms = 230 V and R = 40 Ω
∴ Impedance at resonance condition Z = R = 40 Ω
Vrms 230
and current Irms = = = 5.75 A
Z 40

∴ Current amplitude Im = 2 × Irms = 2 × 5.75 = 8.1 A.

(iii) RMS potential drops

across R, VR = Irms × R = 5.75 × 40 = 230 V

1 5.75
across C, VC = Irms × XC = Irms ×
= = 1437.5 V
C ωr 80 × 10 –6 × 50

model test Paper – 2 19

 and across L, VL = Irms × XL = Irms. L ω0 = 5.75 × 5 × 50 = 1437.5 V

and the potential drop across the LC combination VLC = VL – VC
= 1437.5 – 1437.5 = 0 V.
Or
(B) (a) A labelled schematic diagram of an a.c. generator is drawn here.
Armature coil
C B

Field magnet

N S

D A Brushes
Slip rings B1
R1
–ve

B2 R2
+ve

Load
→
Consider a coil of N turns, each of area A, held in a uniform magnetic field B . Let the
coil be rotated at a steady angular velocity ω about its own axis.
Let at any instant t, normal to the area (i.e., the area
→
vector A ) subtends an angle θ = ω t from direction of
→
magnetic field B . Then, at that moment, the magnetic
flux linked with the coil is
→ →
φB = N B ⋅ A = N B A cos θ = N B A cos ωt

d φB d
∴ Induced emf ε = − = − ( N B A cos ωt)
dt dt
d
= − N B A (cos ωt) = N B A ω sin ωt
dt
= ε0 sin ωt
where ε0 = NBAω = maximum value of induced emf.
(b) Here cross-sectional area of coil A = 200 cm2 = 200 × 10– 4 m2, no. of turns in coil N = 20.
Angular speed of rotation of coil w = 50 rad s– 1 and magnetic field B = 3.0 × 10­– 2 T.
\ Maximum value of induced emf, e0 = N A B w
= 20 × (200 × 10– 4) × (3.0 × 10– 2) × 50
= 0.6 V

20 U-LIKE Physics–XII
 If resistance of the coil be R, then maximum value of current in the coil
ε 0.6
I= = A
R R
33.(A) (a) When a plane wavefront WW’ is incident normally on a narrow single-slit AB, each
point on unblocked portion ADB of wavefront sends out secondary wavelets in all
the directions. For secondary wavelets meeting at symmetrical central point O on
the screen, the path difference between wave is zero and hence the secondary waves
reinforce each other giving rise to central maximum at point O.
Consider secondary waves travelling in a direction making an angle with DO
and reaching the screen at point P. Obviously, path difference between extreme
secondary waves reaching P from A and B
∆ = BC = AB sin q = a sin q
If this path difference a sin q = l, then point P will have minimum intensity. In this situation,
wavefront may be supposed to consist of two equal halves (known as half period elements)
AD and DB and for every point on AD there will be a corresponding point on DB having

a path difference λ . Consequently, they nullify effect of each other and point P behaves
2
as first secondary minimum.
In general, if path difference a sin qn = nl, where n = 1, 2, 3, .........., then we have nth
secondary minima corresponding to that angle of diffraction qn.

However, if for some point P1 on the screen secondary waves BP1 and AP1 differ in path
3λ
by , then point P1 will be the position of first secondary maxima. In this situation,
2
wavefront AB may be divided into three equal parts such that path difference between

corresponding points on first and second parts will be λ . and they will nullify. But
2
secondary waves from third part remain as such and give rise to first secondary maxima,
whose intensity will be much less than that of central maxima.

λ
In general, if path difference a sin θn = (2n + 1) , where n = 1, 2, 3, .........., then we have
2
nth secondary maxima corresponding to these angles.

model test Paper – 2 21

 (b) Here λ1 = 590 nm = 5.90 × 10–7 m, λ2 = 596 nm = 5.96 × 10–7 m, aperture of slit a = 2 × 10–4 m
and distance of screen from the slit D = 1.5 m
3 λD
As first maxima is obtained at a distance x =
2a
3D 3 × 1.5
∴ ∆x = x2 − x1 = (λ 2 − λ 1 ) = −4
× (5.96 – 5.90) × 10–7 = 6.75 × 10–5 m
2a 2 × (2 × 10 )
Or
(B) (a) Refraction of light rays through prism has been shown here.
Here ∠i is the incidence angle at first surface of prism and ∠e is the angle of
emergence from the second surface. ∠r1 and ∠r2 are the respective refraction angles
at the two faces and ∠δ is the angle of deviation.
Now in ∆QMR, ∠r1 + ∠r2 + ∠M = 180° …(i)
and in quadrilateral AQMR, ∠A + 90° + ∠M + 90° = 360°
or ∠A + ∠M = 180° …(ii)

Comparing (i) and (ii), we get

∠r1 + ∠r2 = ∠A …(iii)
Now in ∆JQR ∠δ = ∠JQR + ∠JRQ
= (∠i – ∠r1) + (∠e – ∠r2) = ∠i + ∠e – (∠r1 + ∠r2) = ∠i + ∠e – ∠A
⇒ ∠i + ∠e = ∠A + ∠δ ...(iv)
(b) A graph showing variation in angle of deviation (δ) with variation in angle of
incidence (i) is drawn. As minimum deviation position corresponds to only one
angle of incidence, it means that for minimum deviation position ∠i = ∠e and also
∠r1 = ∠r2 = r (say).
In minimum deviation position
 A
∴ ∠A = ∠r1 + ∠r2 = ∠r + ∠r = 2∠r or ∠r =  
2
 A + δm 
and a ∠A + ∠δm = ∠i + ∠e = ∠i + ∠i = 2∠i or ∠i =  
 2 
 A + δm 
sin  
sin i  2 
∴ Refractive index of prism n =
=
sin r  A
sin  
2
22 U-LIKE Physics–XII
 3 MODEL TEST PAPER
(ANSWERS)

SECTION A
1. (c) A3
2. (b) 0.15 mV
3. (b) 2 µF
4. (d) zero
5. (a) zero
3I
6. (c) f and
4
7. (a) 0.30 mm
8. (d) Scattering of IR radiation is less than visible light.
9. (b) 2
10. (b) 3.12 eV
11. (b) kinetic energy is high enough to overcome the coulombian repulsion between the
nuclei.
h
12. (c) integer multiple of
2π
13. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of
Assertion (A).
14. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of
Assertion (A).
15. (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation
of Assertion (A).
16. (d) Both Assertion (A) and Reason (R) are false.

SECTION B
17. Here nX = 2nY and AX = AY = A (say)
In a series circuit the electric current flowing through the entire circuit is exactly same,
hence

model test Paper – 3 23

 I = nXAe(vd)X = nYAe(vd)Y
(vd )X n n 1
⇒
= Y = Y =
(vd )Y nX 2nY 2
18.(A) No. of small drops n = 27, potential of each drop V = 10 V and let radius of each drop = r,
then capacitance of each drop C = 4p∈0r when these drops are combined to form a bigger
drop of radius r′, then
4 3 4
πr′ = 27 × πr 3 ⇒ r′ = 3r
3 3
\ Capacitance of bigger drop, C′ = 4p∈0r′ = 3 × 4p∈0r = 3C
and total charge on bigger drop Q′ = Total charge on 27 smaller drops
= 27CV = 27 C × 10 = 270C
Q′ 270C
\ Potential of bigger drop V′ = = = 90 V
C′ 3C
Or
(B) Electric potential at point P, V = VA + VB + VC + VD
1  +q q q q 
=
4 π ∈0  AP − BP − CP + DP 
 

1 q q q q
=  − − + 
4 π ∈0  a 5a 5a a 

1 2q  1 
= . 1–
4 π ∈0 a  5 
19. Here distance from current carrying wire, r = 10 cm = 0.1 m and magnetic field strength
B = 10–5 T.
µ0 I
As B= , hence
2 πr

2 πrB 2 π × 0.1 × 10 − 5
I= = = 5.0 A
µ0 (4 π × 10 − 7 )
20. (a) Displacement current is the current that comes into play whenever the electric field
and consequently electric flux is changing with time.
Mathematically,
dφ
Displacement current Id = ε0 = E
dt
d φE
where e0 = permittivity of free space and = rate of change of electric flux with
dt
time.

24 U-LIKE Physics–XII
 (b) If an open surface is bounded by a closed loop then Ampere-Maxwell’s law states that
 → →
the line integral  ∫ B . dl  of the magnetic field along the loop is equal to m0 times
 
the sum of conduction current and displacement current. Mathematically,
→ →
∫ B . dl = m0(Ic + Id)
21. (a) The symbol ‘b’ is known as the “impact parameter” and symbol ‘θ’ is named as “angle
of scattering”.
(b) (i) If θ 0° then ‘b’ has a large value.
(ii) If θ = π radian then ‘b’ has almost a zero value.

SECTION C
22. The de Broglie wavelength of a charged particle is given as :
h h h h
l= = = =
mv p 2mK 2mqV
h
(a) If an a-particle and a proton have same speed v, then as per relation l = , we have
mv
λα mp 1
= =
λp mα 4
h
(b) If a-particle and proton have same kinetic energy K then as per relation l = ,
2mK
we have
λα mp 1 1
= = =
λp mα 4 2
(c) If the two particles are accelerated through same potential difference V, then we have
λα mp . qp 1 1 1
= ×= =
λp mα . qα 4 2 2 2
23.(A) (a) Two essential conditions for total internal reflection are :
(i) The light ray should travel in a denser medium towards a rarer medium.
(ii) Angle of incidence in the denser medium should be greater than the critical angle
for the pair of media in contact.
(b) (i) A graph showing the variation of apparent depth (h′) with
the real depth (h) of the coin inside the fluid is drawn here.
It is a linear graph.
(ii) The slope of the graph is related to the refractive index of the
transparent liquid w.r.t. the surroundings. In fact,
∆h′ 1
slope of graph = ,
∆h n
where n is the refractive index of given liquid with respect to
the surrounding medium.

model test Paper – 3 25

 Or
(B) Here R1 = 10 cm, R2 = 20 cm and refractive index of lens n = 1.5
(a) As per sign convention R1 is +ve and R2 is –ve. Hence, focal length of lens in air is given
as :
1  1 1   1 1  3
= (n − 1)  −  = (1.5 − 1)  −  = 0.5 ×
fa  R1 R2   (+10) (− 20)  20
40
⇒
fa = cm = 13.3 cm
3
(b) When the lens is immersed in a transparent liquid of refractive index n′ = 1.25,

1 n  1
=  − 1   −
1   1.5 − 1   1 − 1 
 =   
fe  n′  1R R2   1.25   (+ 10) (− 20) 
100
⇒
fe = cm = 33.3 cm
3
24.(A) In the circuit R2 = 15 W, R3 = 15 W and R4 = 30 W are joined in parallel between points A and
B and their combined resistance R0 is given as :
1 1 1 1 1 1 1 1
= + + = + + =
R0 R2 R3 R4 15 15 30 6
⇒
R0 = 6 W
Now the total equivalent resistance of the circuit R = R1 + R0 = 4 + 6 = 10 W
ε 10 V
\ Current through R1, I1 == = 1.0 A
R 10 Ω
Let currents through R2, R3 and R4 be I2, I3 and I4 respectively. Then
I2 + I3 + I4 = I1 = 1.0 A
and I2 R2 = I3 R3 = I4 R4 or 15 I2 = 15 I3 = 30 I4
I2
⇒
I2 = I3 and I4 =
2
I2 5
\
I2 + I2 + = 1.0 or I 2 = 1.0 or I2 = 0.4 A
2 2
and then I3 = 0.4 A and I4 = 0.2 A
Or
(B) The circuit diagram of network of resistors can be redrawn as shown in the figure.
Obviously, here resistance of branch ACB is :
1 1 r r
r′= + = + =r
 1 1  1 1 2 2
 +   + 
r r r r
Thus, effectively between A and B, we have three resistors each
of value r. Hence effective resistance R between A and B will be
1 1 1 1 1 1 1 3 r
= + + = + + = ⇒ R=
R r r′ r r r r r 3

26 U-LIKE Physics–XII
 As internal resistance of cell is also r, hence
E E 3E
(a) the current drawn from the cell I = = =
R + r r + r 4r
3
2
 3E  r 3E 2
and (b) power consumed in the network P = I 2 R =   × = .
 4r  3 16r

25. (a) Circuit diagrams are given below :

(i) Forward bias (ii) Reverse bias

(b) Characteristics of a p-n junction are drawn below :

VF (V)
100 50 0
120
IF (mA)

10

80
20

IR (µA)
40
30
0 0.4 0.8 1.2
VF (V)

(i) Forward bias characteristics (ii) Reverse bias characteristics

(c) Breakdown voltage is that value of reverse bias for a p-n junction at which the reverse
bias current suddenly increases without increasing the reverse voltage.
26. (a) Mass defect of a nucleus is the difference between the sum of the masses of constituent
nucleons of a nucleus and the mass of that nucleus. Mass defect of a nucleus of element
M
ZX is :
DM = Z mp + (A – Z)mn – M­x, where Mx is the mass of given nucleus.
Binding energy of a nucleus is defined as the amount of energy needed to separate a
nucleus into its constituent nucleons.
Binding energy of a nucleus is the energy equivalent of its mass defect.
\ Binding energy EB = DM.c2, where c = speed of light.

model test Paper – 3 27

 (b) The graph is plotted below :

27. (a) The root mean square (rms) value of an a.c. is that steady current which, when passed
through a resistance, produces exactly the same amount of heat in given time as is
produced by actual a.c. when flowing through the same resistance for same time. It can
be shown that
Im
Irms = Ieff = = 0.707 Im
2
(b) Here L = 200 mH = 0.2 H, C = 5mF = 5 × 10–6 F and V = 70.7 sin(1000 t) means that
ω = 1000 s–1.
Vm V 70.7
For inductive circuit, Im = = m = = 0.354 A
XL Lω 0.2 × 1000
 π  π
So expression for current will be, IL = Im sin  ωt −  = 0.354 sin  1000t −  A
 2  2
Vm
For capacitive circuit, Im = = Vm.C.ω = 70.7 × 5 × 10–6 × 1000 = 0.354 A
Xc
π π
∴ Expression for current will be, Ic = Im sin  ωt +  = 0.354 sin  1000t +  A
 2  2
28. (a) A wavefront is defined as a surface of constant phase.
(b) Consider a plane surface XY separating a denser medium of refractive index n1 from a
rarer medium of refractive index n2. Let c1 and c2 be the values of speed of light in the
two media, where c2 > c1. AB is a plane wavefront incident on XY at an angle i. Let at a
given instant the end A of the wavefront just strikes the surface XY but the other end
B has still to cover a path BC. If it takes time t, then BC = c1t.

28 U-LIKE Physics–XII
 Meanwhile, point A begins to emit secondary wavelets which cover a distance c2t in
second medium in time t. Draw a circular arc with A as centre and c2t as radius and
draw a tangent CD from point C on this arc. Then CD is the refracted wavefront in the
rarer medium which advances in the direction of rays 1′2′. The refracted wavefront
subtends an angle r from surface XY.

BC ct
Now in ∆ABC, sin i = = 1
AC AC
AD c2 t
and in ∆ADC, sin r = =
AC AC
sin i c t/ AC c1 n
∴
= 1 = = a constant = 2 = n21
sin r c2 t/ AC c2 n1

which is Snell’s
law of refraction.
SECTION D
29. (i) (c) H
(ii) (d) 5 × 10– 3 H
(iii) (d) increasing the number of turns in it.
(iv) (A) (c) M = L1L2
Or
(B) (a) M12 = M21
30. (i) (a) transforms alternating current into unidirectional current.
(ii) (c) 0.1 A
(iii) (b)
(iv) (A) (a) In a full-wave rectifier two diodes work in a complimentary mode.
Or
(B) (c) 100 Hz

model test Paper – 3 29

 SECTION E
31.(A) (a) Equipotential surfaces are drawn below :

(i) For an electric dipole (ii) For two identical charges placed near each other
(b) Here area of each plate A = 6 × 10– 3 m2 and separation between the capacitor plates
d = 3 mm = 3 × 10– 3 m
∈0 A (8.85 × 10 − 12 ) × (6 × 10 − 3 )
(i) Capacitance C = =
d (2 × 10 − 3 )
= 17.7 × 10– 12 F or 17.7 pF
(ii) If supply voltage V = 100 V, then
Charge on each plate of capacitor q = CV = (17.7 × 10– 12) × 100
= 17.7 × 10– 10 C or 1.77 nC
(iii) On introducing a mica sheet of thickness t = d = 3 mm = 3 × 10– 3 m and dielectric
constant K = 6, new capacitance
K ∈0 A
C′ = = KC = 6 × 17.7 pF = 106.2 pF
d
\ New charge on each plate q′ = C′V = (106.7 pF) × 100 V
= 106.7 × 10– 10 C = 10.67 nC
Or
(B) (a) Let three charges be placed as shown here.
Then total potential energy of system :

1  − qQ qQ q 2 
U= . − +  =0
4 π ∈0  r r 2r 

q
⇒
.[− 4Q + q] = 0 ⇒ q – 4Q = 0 or q = 4Q
2r

Q
⇒
=1:4
q

(b) (i) Consider a uniformly charged thin spherical shell of radius R and having a charge
Q. To find electric field at a point P outside the shell situated at a distance r (r > R)
from the centre of shell, consider a sphere of radius r as the Gaussian surface.
All points on this surface are equivalent relative to given charged shell and, thus,

30 U-LIKE Physics–XII
 electric field at all points of Gaussian surface has same magnitude and are
parallel to each other.
Total electric flux over the Gaussian surface
→
φE = ∫ E ⋅ nˆ ds = E.4 π r 2 ...(i)

According to Gauss’s theorem,

1 Q
φE = (charge enclosed) = ...(ii)
∈0 ∈0
Comparing (i) and (ii), we get
Q Q
E.4pr2 = or E = .
∈0 4 π ∈0 r 2
(ii) The E(r) – r graph is plotted below :
Q
E
4  0 R2

E (r)
E  1h2
E0
O r<R A r>R
32.(A) (a) Magnetic moment of original coil of N turns and radius R carrying a current I
m1 = N1A1I = N⋅π⋅R2⋅I
R1 R
When the coil is unwound and is rewound to make another coil of radius R2 =
= ,
2 2
the total number of turns in the coil will change to N2 = 2N1 = 2N. As same current I
flows through the coil even now, hence new magnetic moment is given by
→
2
R 1 |m1|
m2 = N2A2I = 2N⋅π⋅   ⋅ I = NπR2I =
 2 2 2
(b) We know that when a particle of mass ‘m’ and charge ‘q’
→
moving with a velocity v enters a region of magnetic
→
field B acting normal to the direction of velocity, it
mv
describes a circular path of radius r = .
qB
As per question vproton = vdeuteron and charges are also same.
But mdeuteron = 2mproton. As a result, rdeuteron = 2rproton.
The trajectories have been traced here.

model test Paper – 3 31

 Or
→
(B) (a) When a charged particle having charge q and moving with a velocity v enters a
→ → → →
uniform magnetic field B , it experiences a Lorentz force F = q ( v × B ) . The force is
maximum when charged particle is moving perpendicular to the magnetic field. In
that case
F = q v B sin 90° = q v B
→ →
As the force is acting perpendicular to both v and B , it
does not change the magnitude of velocity but changes
its direction from a straight line path to a circular path of
radius r such that the magnetic force = centripetal force
needed.
m v2 mv
∴
Bqv= r =
r Bq

The time period of the charged particle in its circular path will be
 mv
2π 
2π r  B q  2π m 2π qB
T= = = , and angular frequency ω = = .
v v Bq T m
(b) Here current I = 5 A, radius of semi circular arc R = 2 cm = 0.02 m.
The magnetic field due to long straight wire is zero because the centre point of the arc
lies on the extended line of wire itself.
1  µ0 I  µ0 I (4 π × 10 −7 ) × 5
And field due to semi circular arc B = = = = 7.85 × 10–5 T
2  2 R  4 R 4 × 0.02
→
As per right hand thumb rule, the magnetic field B points perpendicular to plane of
paper directed into the paper.
33.(A) (a) The ray diagram is drawn here. The im-
age A′B′ is real, inverted and magnified
one.
As ∆A′B′C and ∆ABC are similar, hence
A′B′ CB′
= ...(i)
AB CB
Again as ∆A′B′F and ∆LCF are similar, hence
A′B′ B′F A′B′ B′F
= or = ...(ii) [∵ LC = AB]
LC CF AB CF
From (i) and (ii), we have
CB′ B′F CB′ − CF
= =
CB CF CF

32 U-LIKE Physics–XII
 As per sign conventions used, CB = – u, CB′ = + v and CF = + f
v v− f
∴ =   or  vf = – uv + uf
−u f
Dividing both sides by uvf, we get
1 1 1
= − +   or   1 − 1 = 1 .
u f v v u f
(b) Here ∠A = 60° and ∠i = 45°
(i) As the light ray is passing symmetrically through the prism it undergoes minimum
deviation.
\
∠e = ∠i = 45°
and ∠Dmin = ∠i + ∠e – ∠A = 45° + 45° – 60° = 30°

A + Dmin  60° + 30° 

sin sin  
2  2 
(ii) Refractive index of prism material n = =
A  60 ° 
sin sin  
2  2 
sin 45°
= = 2
sin 30°
Or
(B) (a) The speed as well as wavelength of a monochromatic light depend on optical density
of medium and change from one medium to another. However, frequency of a
monochromatic light is a characteristic of the source emitting the light and does not
change with change in medium.
(b) In a single-slit diffraction experiment if the width of the slit (i.e., ‘a’) is doubled, then
2λ
width of central diffraction band x = .D is reduced to one half of its previous value
a
and accordingly the intensity of the central diffraction band will become 4 times of its
original intensity.
(c) The diffraction from each slit is on account of superposition of a continuous family of
waves originating from each point on the single-slit. But interference pattern is due to
superposition of two waves originating from the two narrow slits. However, both the
patterns are governed by the same ideas of superposition and Huygen’s secondary
wavelets. Actually in a Young’s double-slit experiment, the pattern observed is the
combination of diffraction patterns due to individual slits and the interference pattern.

model test Paper – 3 33

 4 MODEL TEST PAPER
(ANSWERS)

SECTION A
1. (a) no net charge is enclosed by the surface.
2. (d) 4 µF
3. (c) 2R W
4. (b) A m2
5. (a) 0.80
6. (b) 2
7. (d) real, inverted, height = 4 cm
8. (a) 180° – 2A
9. (b) a wavefront.
10. (c) 23.8 MeV
11. (b) Only statement 1 is true.
5R
12. (c)
36
13. (d) Both Assertion (A) and Reason (R) are false.
14. (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation
of Assertion (A).
15. (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation
of Assertion (A).
16. (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation
of Assertion (A).
SECTION B
17. Magnetic moment of circular coil of N turns and radius R carrying a current I
m1 = NA1I = NπR2I ...(i)
When the coil is unwound and is rewound to make another square coil of side ‘a’ and
number of turns N, as total length of wire remains constant, we have
πR
N.2πR = N4a ⇒ a =
2

34 U-LIKE Physics–XII
 ∴ Magnetic moment of square coil for same current I
2 2 2
m2 = NA2I = Na2I = N  πR  I = N π R I ...(ii)
 2  4
m2 π
⇒ =
m1 4
18. (a) Magnetic susceptibility c of a material is defined as the ratio of magnetisation vector m
to the magnetic intensity H. Thus,
m
c=
H
Magnetic susceptibility c = mr – 1, where mr is the relative magnetic permeability of that
substance.
(b) Material A is diamagnetic and material B is ferromagnetic in nature.
19. Here refracting angle of prism A = 30° and refractive index n = 3.
As light ray incident at the silvered surface, retraces its path after reflection, hence ∠r2 = 0°.
\ Angle of refraction at 1st surface ∠r1 = A – r1 = 30° – 0° = 30°

1 3
\ sin i = n sin r1 = 3 .sin 30° = 3× =
2 2
 3
⇒ i = sin −1   = 60°
 2 

20.(A) When particle of mass m, initially at rest, disintegrates into two parts of masses m1  = 
m
 3
and m2  =  2m 
 respectively then as per conservation law of momentum :
 3 
m1v1 + m2v2 = m × 0 = 0
⇒ |m1v1| = |m2v2|
h h
\ l1 = and l2 =
m1v1 m2 v2
λ1
⇒ l1 = l2 or =1
λ2
Or
(B) Here wavelength l = 500 nm = 5 × 10–7m and power of laser P = 2 mW = 2 × 10–3 W
P Pλ
\ Number of photons emitted per second N = =
hc hc
λ
(2 × 10 −3 ) × (5 × 10 −7 )
= = 5 × 1015
(6.63 × 10 −34 ) × (3 × 108 )

model test Paper – 4 35

 21.
Half-wave rectifier Full-wave rectifier
1. The output voltage is unidirectional, 1. The output voltage is unidirectional,
intermittent and varying in magnitude. continuous and varying in magnitude.
2. The frequency of output signal is same 2. The frequency of output signal is
as that of input signal wave. double as compared to that of input
signal wave.

SECTION C
→
22. (a) The magnetic moment m of a current loop is given by
→ →
m = IA
→ → →
and for a coil of N turns m = N I A, where A = area vector of the loop and
NI = number of ampere turns in the coil.
The direction of magnetic moment is given by right hand rule. SI unit of magnetic
moment is A–m2.
(b) Here current = I and area of current loop A = area of a square of side ‘a’ + area of 4
semicircles each of radius r = a .
2
2
π  a πa 2 π
\
A = a2 + 4 × × = a2 + = a2  1 + 
2  2 2  2
π
\ Magnetic moment of current loop m = IA = Ia 2  1 + 
 2
23.(A) (a) An equipotential surface is that, electric potential at every point of which is the same.
By definition potential difference between two points = work done in carrying unit
+ ve test charge from one point to another. As for an equipotential surface potential
difference is zero hence no work is to be done in moving a test charge from one point
to another of an equipotential surface. Thus,
→ →
dW = E ⋅ dr = E dr cos θ = 0 and it leads to cos q = 0 or q = 90°
→
Thus, electric field intensity E on the surface of a conductor is always perpendicular
to the surface.
(b) Two equipotential surfaces cannot intersect each other. Because if they do then at the
point of intersection there are two possible values of electric potential, which is not
possible.
Or
(B) (a) Electric field at a point may be defined as the negative of the rate of change of electric
potential with position (i.e., the negative of the potential gradient) at that point i.e.,

36 U-LIKE Physics–XII
 → → dV
 dV  ˆ
E = −
 r or E = | E | =
 dr  dr
Moreover, direction of electric field is the direction in which electric potential is
decreasing at a maximum rate i.e., where decrease of potential is steepest.
(b) In the circuit capacitors of capacitance C1 = 4 mF and C2 = 2 mF are joined in parallel and
their combined capacitance C12 = 4 + 2 = 6 mF
Now, C12 = 6 mF is in series with the capacitor C3 = 6 mF. Hence the net capacitance of
C12 × C3  6 × 6 
the circuit C = =  µF = 3 mF
C12 + C3  6 + 6 
\ Total charge drawn from battery Q = CV = 3 mF × 12 V = 36 mC
Since in series arrangement same charge exists in each capacitor, hence charge on 6 mF
capacitor = Q = 36 mC
24.(A) (a) According to Lenz’s law, the direction of the induced current in a closed circuit is such
so as to oppose the change in magnetic flux, that produces it.
Lenz’s law is a consequence of the law of conservation of energy.
(b) The induced emf is expected to be constant in case of rectangular loop only.
Induced emf depends upon rate of change of magnetic flux, which in present case is
proportional to change in area of the loop. For a rectangular loop the rate of change in
area is constant, hence induced emf is expected to be constant. However, for a circular
loop, the rate of change of area of the loop during its passage out of the field region is
not constant. Hence, induced emf will also vary accordingly.
Or
(B) (a) Due to flow of high frequency a.c. the magnetic flux linked with metallic piece changes
rapidly and consequently a strong current is induced in it. This strong current heats
up the metallic piece on account of transformation of electrical energy into thermal
energy.
(b) Here length of solenoid l = 0.5 m, number of turns per unit length n = 10 turns/cm
= 10 × 100 m–1 = 1000 m–1, area of cross-section A = 1 cm2 = 10–4 m2, change in current
∆I = I2 – I1 = (2 – 1) A = 1 A and time ∆t = 0.1 s.
∴ Self-inductance of solenoid L = µ0n2lA
= (4π × 10–7) × (1000)2 × 0.5 × (10–4) = 6.28 × 10–5 H
∆I 1
∴ Magnitude of induced voltage, |ε| = L = (6.28 × 10–5) × = 6.28 × 10–4 V
∆t 0.1
or 0.628 mV
25. In general, applied voltage and current flowing in a LCR series a.c. circuit be given by
V = Vm sin ωt and I = Im sin (ωt – φ)
∴ Instantaneous power P = V I = Vm sin ωt . Im sin (ωt – φ)
= Vm Im sin ωt [sin ωt cos φ – cos ωt sin φ]
= Vm Im [sin2ωt cos φ – sin ωt cos ωt sin φ]

model test Paper – 4 37

 1
= Vm Im [sin2ωt cos φ – sin 2 ωt sin φ]
2
∴ Average power for one complete cycle of a.c.
 2 1 
P = Vm Im (sin ωt) cos φ − (sin 2 ωt) sin φ 
 2 
1
But we know that average value of sin2 ωt for a complete cycle is , whereas average
2
value of sin 2ωt for one complete cycle is zero. Hence,

1
P or Pav = Vm Im  cos φ  = Vm Im cos φ = Vrms Irms cos φ
1

2  2
π
(i) If phase difference φ between V and I be (as for a pure inductor or a pure capacitor
2
π
or a series LC circuit) then cos φ = cos = 0 and hence average power dissipated is
2
zero even though the current flows through the circuit.
(ii) If phase difference φ = 0° (as for a pure resistor or for a resonant circuit) then
cos φ = cos 0° = 1 and hence average power Pav = Vrms Irms, which is maximum power
dissipated in the circuit.
26. When a plane wavefront of monochromatic light is incident on a narrow slit, each point of
slit behaves as a secondary source of light in accordance with Huygen’s principle. These
secondary sources emit wavelets which travel in different directions. Naturally wavelets
emitted by different parts of the single narrow slit superpose. As a result of superposition,
we obtain a broad pattern with central bright region and on both sides there are alternate
dark and bright regions. Thus, intensity distribution on screen is redistributed. This
phenomenon is called diffraction of light due to a narrow slit.
nDλ
(i) It can be easily shown that positions of minimas are given on the screen by x = ± ,
a
where n is any integer, a = slit width and D = distance of screen from the slit.

(n + 1)Dλ nDλ Dλ
∴ Width of nth maximum β = xn+1 – xn =
− =
a a a
2Dλ
However, width of central maximum = x1 – (– x1) = 2x1 = = 2β
a
Thus, it is clear that central maximum is twice as wide as the other maxima.
(ii) In terms of diffraction angle ‘θ’, nth diffraction maxima is obtained when
λ
a sin θ = (2n + 1)
2
In this situation the diffracted wavefront can be divided into (2n + 1) ‘half period
λ
elements’ such that any two consecutive elements differ in path by (or differ in phase
2

38 U-LIKE Physics–XII
 by π rad). Being in mutually opposite phase, amplitudes (and hence intensity) due
to consecutive half period elements nullify one another. In this way effect of 2n half
period elements is nullified and we get diffraction maxima due to the effect of last one
1
half period element. As n increases, we get intensity only due to th part of the
(2n + 1)
wavefront and consequently intensity successively falls rapidly.
27. Work function of a metal is the minimum amount of energy of an incident photon for
which an electron can be pulled out from the surface of that metal.
According to Einstein’s photoelectric equation, we know that
1
h(ν – ν0) = m v 2 max
2
In present problem ν0 = f0 and when ν = 2 f0 , vmax = v1
1 1
∴
h(2f0 – f0) = m v12 or h f0 = m v12 ...(i)
2 2
Again, when ν = 5 f0 , vmax = v2. Hence, we have
1 1
h(5f0 – f0) = m v22 or 4 h f0 = m v22 ...(ii)
2 2
Dividing (i) by (ii), we get

1 v2 v1 1 1.
= 12 or = =
4 v2 v2 4 2
28. (a) Two processes that take place in the formation of a p-n junction are (i) diffusion, and
(ii) drift of charge carriers i.e., electrons and holes across the junction.
(b) Depletion region is a thin layer on either – +
side of the p-n junction which is devoid of – +
–
free electrons and holes (i.e., free charge p – + n
–
carriers) and contains only immobile ions. –
+
– +
In the neighbouring region, region PQ is the
depletion region. P Q
Depletion Region
(c) The width of the depletion region in a p-n junction (i) decreases in forward bias
arrangement, and (ii) increases in reverse bias arrangement.
SECTION D
29. (i) (b) 8.0 × 10–28 C m
→
→ p
1
(ii) (a) E = . 3
2 π ∈0 r
1
(iii) (c) V ∝
r2

model test Paper – 4 39

 E
(iv) (A) (c)
2
Or
(B) (a) pE sin q, – pE cos q
30. (i) (a) 3 × 108 m s–1
(ii) (c) 2.39 × 1010 Hz
(iii) (d) 1.26 cm
(iv) (A) (c) along Z-axis.
Or
–7 –1
(B) (a) 2 × 10 V m

SECTION E
31.(A) (a) On applying a potential difference V across the ends of a conductor of length l, the
electric field →
E
V l
E= .
l vd e
–

eV
Force experienced by an electron F = e E = A B
l + –
in a direction opposite to that of E.
V
∴ Acceleration of electron in a direction opposite to that of E will be
F e V
a=
= ⋅
m m l
If the average time between two successive collisions suffered by an electron be t, the
drift velocity of electron will be
e V
vd = a τ =
⋅ ⋅τ ...(i)
m l
But in terms of drift velocity magnitude of electric current is given by
e V n A e2 V
I = n A e vd = n A e ⋅ ⋅ ⋅τ = τ⋅
m l m l
V ml
∴ Resistance of conductor R = = ...(ii)
I n A e2 τ
RA m
and resistivity of the material of conductor ρ = = . ...(iii)
l n e2 τ
The resistivity of a material thus depends on (i) the number density of electrons, and
(ii) its relaxation time.
(b) Here number density of conduction electrons n = 8.5 × 1028 m–3, and mean free period
t = 25 fs = 25 × 10–15 s
m 9.1 × 10 −31
\ Resistivity r =
= = 1.67 × 10–8 Wm
ne 2 τ (8.5 × 10 28 ) × (1.6 × 10 −19 )2 × (25 × 10 −15 )

40 U-LIKE Physics–XII
 Or
(B) (a) To analyse electrical networks Kirchhoff gave following two rules :
(1) Current rule – The algebraic sum of all the currents flowing into a junction is zero.
In other words, at any junction, the sum of currents entering the junction must be
equal to the sum of currents leaving it. Mathematically, ∑ I = 0
(2) Voltage rule – The algebraic sum of changes in potential around any closed loop
involving resistors and cells in the loop must be zero. Mathematically,
ΣV = 0 or Σ(ε + R I ) = 0
(b) Kirchhoff’s first rule is based on the law of conservation of electric charge and the
second rule is based on the law of conservation of energy.
(c) [VERY IMPORTANT NOTE : In the language of Q.31Or part (c) there is a mistake in
the question. The last line must be values of currents I2, I3 and I6 respectively.]
Applying Kirchhoff’s first rule we can say that
I6 = I5 = 0.4 A
Again I5 + I3 = I4 ⇒ I3 = I4 – I5 = 0.8 – 0.4 = 0.4 A
and finally I4 = I1 + I2 ⇒ I2 = I4 – I1 = 0.8 – (– 0.3) = 1.1 A
32.(A) (a) Consider an object AB placed on the principal N
axis of a concave mirror NMP between its
principal focus F and centre of curvature C.
A M
As shown in diagran, its real, inverted and
enlarged image A′B′ is formed beyond the B C
B F P
centre of curvature C. f
We consider aperture of mirror to be small u
and hence MP may be considered as a
A
straight line. Then consider right angled v
triangles A’B’F and MPF, which are similar
triangles. Therefore,
B'A' B'F
=
PM FP
B'A' B'F
But as PM = AB, hence, we have = ...(i)
BA FP
Again triangles APB and A’PB’ are also similar triangles and we have
B'A' B'P
= ...(ii)
BA BP
B'P B'F B'P − FP
Comparing (i) and (ii), we get = =
BP FP FP
As per sign convention followed : BP = – u, B’P = – v, and FP = – f. Therefore, we
get
( − v) ( − v) − ( − f ) v v− f
= ⇒ =
( − u) (− f ) u f
⇒ vf = uv – uf or uf + vf = uv

model test Paper – 4 41

 Dividing throughout by uvf, we get
1 1 1
+ = , which is the requisite mirror formula.
v u f
(b) When a parallel beam of light falls on convex lens L1 of focal length f1 = 30 cm, the lens

converges the light beam at its focus point i.e., at a distance 30 cm from convex lens
(or v1 = + 30 cm).
As the final emergent beam leaving the concave lens L2 is a parallel beam, the light
beam must originate from its focus point F2 i.e., the image formed by convex lens must
be the focus point of concave lens (i.e., u2 = – 10 cm).
\ Distance between two lenses, x = v1 + u2 = 30 – 10 = 20 cm

Or
(B) (a) A neat labelled ray diagram of compound microscope is drawn here.

(b) The angular magnification of compound microscope is defined as the ratio of the angle
subtended at the eye by the final image to the angle subtended at the eye by the object
as viewed directly by placing it at near point of eye.
In figure C2 B ′′ = D = least distance of distinct vision. Let us imagine the object AB to
be shifted to A1B′′ so that it is also at a distance D from the eye.
If ∠A ′′ C2 B ′′ = β and ∠A1C2 B ′′ = α , then by definition
β tan β
m= = [Because α and β are extremely small]
α tan α
A′′B′′ A1B′′ AB
In ∆A ′′ B ′′ C2 , tan β = and in ∆A1B ′′ C2 , tan α = =
C2 B′′ C2 B′′ C2 B′′

A′′B′′/C2 B′′ A′′B′′ A′′B′′ A′B′

∴ m= = = × = me × m0 ,
AB / C2 B′′ AB A′B′ AB
where me = magnification produced by eyepiece and m0 = magnification by objective
lens.

42 U-LIKE Physics–XII
 A′B′ C1B′ v A′′B′′ C2 B′′ − D D
Now, m0 = = =− 0 and me = = = =
AB C1B u0 A′B′ C2 B′ − ue ue
Applying lens formula for eyelens, we have
1 1 1 1 1 1 D + fe
− = ⇒ = + =
(− D) (− ue ) fe ue fe D D ⋅ fe

 D + fe   D
me = D   = 1 + 
 D f e   fe 

v0  D
m = m0 × me = − 1 + 
u0  fe 
As a first approximation v0  L, distance between the objective and eye lens or
the length of the microscope tube and u0  f0, the focal length of objective lens.

L  D
Hence, m =− 1 + 
f0  fe 

(c) As magnifying power of objective lens m0 = 10

and magnifying power of eyepiece me = 15
\ Magnifying power of microscope m = m0 × me = 10 × 15 = 150

33.(A) (a) A nucleus A Z X of atomic number Z and mass number M consists of Z protons and
(M – Z) neutrons. Hence binding energy of nucleus
Eb = [Zmp + (M – Z)mn – M(M
Z X)]c
2

where mp = mass of a proton, mn = mass of a neutron and M(M Z X) is the mass of given
nucleus. If the masses are expressed in atomic mass units, then we know that energy
equivalent of 1 u mass is 931 MeV. Hence,
Eb = [Zmp + (M – Z)mn – M(M
Z X)u] × 931 MeV
(b) (i) In a fission process Uranium-235 nucleus breaks into two nearly equal fragments
releasing great amount of energy when bombarded by a slow moving (thermal)
neutron. The representative reaction is
1
0n + 235
92 U → ( 23692 U ) → 144
56 Ba + 89
36 Kr + 310 n + Q

The energy released (the Q-value) in the fission reaction is of the order of 200 MeV
per fissioning nucleus. This is estimated as follows :
Let us take a nucleus with mass number A = 240 breaking into two fragments each
of A = 120. Then binding energy per nucleon for A = 240 is about 7.6 MeV but that
for the two A = 120 fragments is about 8.5 MeV.
∴ Gain in binding energy per nucleon is about 0.9 MeV and hence the total gain in
binding energy is 240 × 0.9 or 216 MeV.
(ii) In fusion reaction four hydrogen nuclei fuse together to form a helium nucleus
alongwith release of about 26.7 MeV of energy. The actual reactions is quite complex

model test Paper – 4 43

 and takes place in four stages but the combined effect of all these stages may be
written in the form of following reaction :
4 11H + 2e – = 4
2 He + 2ν + 6γ + 26.7 MeV
The binding energy curve again helps in explaining the cause of energy generation
in fusion reaction. Binding energy per nucleon of 24He is about 6.7 MeV but bind-
ing energy of hydrogen is zero. As a result, gain in binding energy on fusion is
4 × 6.7 = 26.8 MeV.
Or
(B) (a) (i) Speed of electron : For revolution of an electron in nth Bohr orbit in a hydrogen
atom the centripetal force is provided by coulombian attractive force. Hence
m vn2 1 e2 e2
= ⋅ 2 ⇒ m v2n rn = ...(i)
rn 4 π ∈0 rn 4 π ∈0
Again from Bohr quantum condition, we have
nh
m vn r n = ...(ii)
2π
Dividing (i) by (ii), we get the expression for speed of electron in nth orbit as :

e2
vn = .
2 ∈0 n h
(ii) Radius of electron orbit :
From equation (i), we have

e2
rn =
4 π .∈0 m vn2
Substituting value of vn, we have
2
e2  2 ∈ nh  ∈ n2 h 2
vn = .  02  = 0 2
4 π .∈0 m  e  πme

(b) Third excited state in hydrogen atom corresponds to n = 4. Hence maximum number

n(n − 1) 4(4 − 1)
of spectral lines, which can be emitted, is = = = 6.
2 2

44 U-LIKE Physics–XII
 5 MODEL TEST PAPER
(ANSWERS)

SECTION A

C
1. (c) , 3V
3
2. (a) 0
3. (c) CD
4. (c) − 3ql ˆj
5. (a) a force and a torque.
6. (a) 100 V
7. (a) binding energy per nucleon increases.
8. (b) 24°
β
9. (b)
n
10. (a) 10.2 eV
h
11. (d) λ =
2mK
12. (c) 50°
13. (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation
of Assertion (A).
14. (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation
of Assertion (A).
15. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of
Assertion (A).
16. (c) Assertion (A) is true, but Reason (R) is false.

SECTION B
17. Here critical angle ic = 60°
1 1 2
\ Refractive index of medium n =
= =
sin ic sin 60° 3

model test Paper – 5 45

 c 3 × 108
\ Speed of light in given medium v =
= = 2.6 × 108 m s–1
n  2 
 3 

hc (6.63 × 10 − 34 ) × (3 × 108 )
18. Energy of light photon in eV = = = 3.01 eV
λe (412.5 × 10 − 9 ) × (1.6 × 10 − 19 )
As work function of Na and K are less than the photon energy, hence given light photon can
cause photoelectric emission in Na as well as K but cannot cause photoelectric emmision in
Ca and Mo.
19.(A) When the spring is stretched and released, it oscillates
to and fro simple harmonically. As a result, the wire
AB also oscillates to and fro about its equilibrium
position AB. Due to oscillation of wire AB, magnetic
flux linked with the rectangular wire frame changes
and hence an induced emf/current is set up in it. The
induced current also oscillates simple harmonically as
shown in the figure.
Or
5 5000
(B) Here XC = 100 Ω, n = kHz = Hz
π π
1 1
∴C=
= = 10–6 F = 1 µF.
XC .2 π ν 5000
100 × 2 π ×
π
20. Here n = 4, d = 0.28 mm = 2.8 × 10–4 m, D = 1.4 m and x = 1.2 cm = 1.2 × 10–2 m.
nDλ xn ⋅ d 1.2 × 10 −2 × 2.8 × 10 −4
For bright fringe xn = ⇒ λ= = = 6 × 10 −7 m or 600 nm.
d nD 4 × 1.4
21. (i) The semiconductor material X is behaving as p-type, because here the dopant indium is
trivalent. Material Y behaves as n-type because there the dopant arsenic is pentavalent.
As p-side of junction has been connected to positive terminal of battery, the junction is
forward biased.
(ii) V-I graph is shown in figure.

46 U-LIKE Physics–XII
 SECTION C
→ →
22.(A) (a) Electric field strength E at a point is defined as the ratio of force F experienced by a
small positive test charge q0 to the magnitude of test charge provided that test charge
is infinitesimally small one.
→
→
F
\ E = lim
q0 → 0 q0
SI unit of electric field strength is N C–1.
(b) The electric field at a point P on axial line of dipole due to +q charge at B
1 q
EB = ⋅ along BP
4 π ∈0 (r − a)2
and electric field due to –q charge at A is
1 q
EA = . along PB
4 π ∈0 (r + a)2
∴ Net electric field at point P is given by
q  1 1  q.4ar
E = EB − EA =  2
− 2
=
4 π ∈0  ( r − a ) ( r + a )  4 π ∈0 (r 2 − a 2 )2
2p r
⇒ E = , where p = q.2a = dipole moment of given dipole.
4 π ∈0 (r 2 − a 2 )2
For a short dipole or if the point P is situated far away, then a << r and hence
1 2p
E = . 3 along ABP
4 π ∈0 r
→
→ 1 2p
Vectorially E = . 3
4 π ∈0 r
Or
(B) (a) Electric potential energy of a system of point charges is equal to the amount of work
done in assembling the given system of charges by bringing them to their respective
positions from infinity.
(b) Consider two point charges q1 and q2 situated r distance
apart. If initially charge q1 is brought to its position at A,
work done W1, for it is zero because there is no electric field.
Then work done to bring charge q2 from ∞ to position B,
against the electric field of charge q1, will be
 1 q  qq
W2 = q2V1(r) = q2  . 1 = 1 2
 4 π ∈0 r  4 π ∈0 r
By definition, potential energy of the system is equal to total work done in bringing the
charges to their present positions.
qq qq
\ Potential energy U = W1 + W2 = 0 + 1 2 = 1 2
4 π ∈0 r 4 π ∈0 r

model test Paper – 5 47

 23. Let the net capacitance of the given infinite ladder of
capacitors between points A and B be C. As the ladder is
infinite, addition of one more element of two capacitors
(1 µF and 2 µF) across the points A and B does not change the
total capacitance C i.e., value of C should remain unchanged.
Then capacitor of 2 µF is in series with C, hence their
combined capacitance will be
C× 2 2C
C′ = = µF
C+2 C +2
This combination is in parallel with capacitor of 1 µF, hence, we have
2C
C = C′ + 1 = +1
C+2
or C(C + 2) = 2C + (C + 2) or C2 + 2C = 3C + 2 or C2 – C – 2 = 0
On solving the quadratic equation we find that C = + 2 or – 1 µF.
However, capacitance cannot be negative. Hence, net capacitance of infinite ladder of
capacitors
C = + 2 µF
dI dI
24. (a) (i) Here L1 = 12 mH, L2 = 30 mH and 1 = 2 = ‘k’ (say)
dt dt
dI1 dI
Then ε1 = − L1 = − kL1 and ε 2 = − L2 2 = − kL2
dt dt
Graphs showing variation of induced emf with
rate of change of current in each coil are shown
in figure.
1
(ii) Energy stored U = LI2
2
1 1
Hence U1 = L1I12 and U2 = L2I22. Graphs showing
2 2
variation of energy stored in each inductor with
the current flowing through them are shown in
figure.
(b) As power is defined as the rate of consumption of
energy and power dissipated in the two coils is the
same, hence energy of the two coils is also the same.
25.(A) (a) Photoelectric current in a photocell increases with increase in number of
photoelectrons ejected from the cathode plate which in turn depends on the number
of incident radiation photons. On increasing the intensity of incident radiation the
number of incident photons increases and hence photoelectric current increases.

48 U-LIKE Physics–XII
 (b) As per Einstein’s photoelectric equation, we have
h
h(ν – ν0) = eV0 ⇒ V0 = (ν − ν 0 )
e
Thus, V0 varies linearly with frequency ν of the incident radiation for a given
h
photosensitive surface. The slope of V0 – ν curve is which is a constant and
e
independent of the nature of photosensitive surface. Thus, the slope remains the same
for different surfaces.
(c) As per Einstein’s photoelectric equation, the maximum kinetic energy of ejected
photoelectron is given as
h(ν – ν0) = Kmax
Thus, Kmax depends only on the frequency of incident radiation but is independent
of its intensity.
Or
(B) de Broglie waves associated with a moving football are not visible because in accordance
h
with formula λ = , the value of wavelength is extremely small (of the order of 10–34 m).
mv
hc
Energy of a photon of wavelength λ, Ep = h ν =
λ
If de-Broglie wavelength of electron be λ, then
h h
λ = or v=
mv mλ
2
1 2 1  h  h2
∴ K.E. of an electron is Ke = m v = m ⋅ 
 =
2 2 mλ 2 m λ2
Ep h c/λ h c 2 m λ2 2 m λ c
⇒ = = × =
Ke h 2 /2 m λ 2 λ h2 h
2m λ c
or Ep = ⋅ Ke .
h
26. A pure semiconductor is called an intrinsic semiconductor. Si and Ge are two intrinsic
semiconductors.
(i) p-type semiconductor : When we
dope intrinsic silicon with a controlled Ec
Electron energy

quantity of trivalent (say aluminium)

impurity atoms, there will be one Eg
incomplete bond wherever impurity
EA
atom is present because it has only {
Ev
three valence electrons. Thus, there is a ~
~ 0.01 – 0.05 eV
deficiency of an electron. This deficiency
may be completed by taking an electron
from one of the Si-Si bonds. It will make

model test Paper – 5 49

 it ionised one (negatively charged) and creates an electron deficiency in that silicon
atom from where an electron was taken. Thus, electron deficiency, referred to as hole,
will continue to move. As a result, the semiconductor becomes p-type semiconductor
having a number of holes which form an energy level EA just above the valence
band as shown in energy band diagram.
(ii) n-type semiconductor : When we dope an
intrinsic semiconductor (say silicon), which Ec
has four valence electrons, with a controlled { ED

Electron energy
quantity of pentavalent atoms, say phosphorous, ~
~ 0.01 eV
Eg
which has five valence electrons, these impurity
atoms are adjusted in the crystal structure of
silicon. Four of the five valence electrons of Ev
phosphorous are shared in covalent bonding
with four neighbouring silicon atoms, while the
fifth electron is comparatively free to move.
Impurity atoms are therefore called the donor atoms and n-type semiconductor is
formed. As shown in the energy band diagram, these donor electrons form an energy
level slightly below the conduction band and on applying an external electric field
pass on to the conduction band and can freely conduct.
27. (i) g-rays are produced in radioactive decay of nucleus. These rays are used for treatment
of cancer.
(ii) Ultraviolet rays are produced during operation of welding arcs. These rays are used in
water purifiers and as disinfectant in hospitals.
(iii) Infra red rays are produced by hot bodies. These rays are widely used in remote switches
of household electronic devices.
28. (a) The net path difference between two waves, reaching on the screen through two slits,
is given as :
SS2P – SS1P = (SS2 – SS1) + (S2P – S1P)
λ xd
It is given that SS2 – SS1 = and we know that (S2P – S1P) = ∆ = , where ‘d’ is the
4 D
separation between slits S1 and S2 and ‘D’ the distance of screen from the double-slit.
λ xd
∴ Net path difference = +
4 D
λ xd
(i) For constructive interference + = nλ , where n = 0, 1, 2, 3, ....
4 D
xd λ λ (4n − 1) λ D
⇒ = nλ − = (4n − 1) ⇒ (xn)bright =
D 4 4 4d
λ xd λ
(ii) For destructive interference + = (2n − 1) , where n = 1, 2, 3, ....
4 D 2

50 U-LIKE Physics–XII
 xd λ (4n − 3) λ D
⇒ = (4n − 3) ⇒ (xn)dark =
D 4 4d
λD
(b) For central bright fringe n = 0 and hence (x0)bright = − . Thus, the observed central
4d
bright fringe shifts towards the line of slit S2.

SECTION D
29. (i) (b) 50 Hz
(ii) (c) rms value of supply voltage.
2 Im
(iii) (d)
π
(iv) (A) (d) Hot wire ammeter
Or
π

(B) (b)
2
30. (i) (a) Hydrogen
(ii) (c) neutron is unstable.
(iii) (b) a proton.
(iv) (A) (a) M – Z

Or

(B) (b) slow neutron.

SECTION E
31.(A) (a) Drift velocity of electrons is the steady velocity acquired by them in a conductor on
applying an external electric field on it.
Consider a conductor of uniform cross-section
area A, carrying a current I. Consider a small
section KL of the conductor having a length Dx
or having a volume A.Dx, then number of free
electrons present in this section = nA.Dx, where n = Number density of free electrons.
Total charge carried by these electrons while crossing the given section ∆Q = n A ⋅ ∆ x ⋅ e
∆x
and total time taken by the electrons to cross this section is ∆ t = where vd =
vd
drift velocity of electrons
∆Q n A ∆ x⋅e I
⇒ Current I = = = n A e vd , and the current density J = = ne vd
∆t  ∆ x A
 v 
d
→ →
Vectorially J = − ne vd

model test Paper – 5 51

 (b) Here I = 1.5 A, A = 5.0 × 10–6 m2 and electron density n = 9 × 1028 m–3
I 1.5
\ Drift speed vd =
= = 2.1 × 10–5 m s–1
nAe (9 × 10 28 ) × (5.0 × 10 − 6 ) × (1.6 × 10 − 19 )
Or
(B) (a) Resistance offered by the electrolyte and electrodes of a cell is known as the internal
resistance of the given cell.
(b) Internal resistance of a cell ‘r’ depends upon
(i) nature and concentration of electrolyte,
(ii) nature of electrodes,
(iii) surface area of electrodes immersed into the electrolyte, and
(iv) separation between the electrodes.
(c) No, internal resistance of an old cell is greater than that of a newly prepared cell. As the
cell is being used, electrolyte used in it is being continuously consumed and as a result
its internal resistance increases.
(d) The network is a Wheatstone’s bridge in balanced condition. Hence, resistance of 4 W
does not play any role and may be omitted.
1 1 1 1 1 15
Now, resistance of bridge = + = + =
R1 (4 + 2) (6 + 3) 6 9 54
54
⇒ = 3.6 W
R1 =
15
\ Net external resistance R = R1 + 2 = 3.6 + 2 = 5.6 W

ε 3
\ Circuit current I =
= = 0.5 A
R+r 5.6 + 0.4
→ →
32.(A) (a) Force F acting on an electric charge q moving in a uniform magnetic field B with a
→
velocity v is given by the expression
→ → →
FB = q ( v × B )
→
(i) If the charged particle enters a magnetic field such that its velocity v is in a
→
direction perpendicular to B , the particle moves in a circular path.
→ →
(ii) If angle q between the directions of B and v is other than 90°, the charged
particle move in a helical path.
→ → →
(b) The magnetic force F = q( v × B ) always acts normally to the direction of motion of
charge and does not do any work on it. So, the kinetic energy of charged particle does
not change at all and remains constant.
(c) Here B = 1.2 × 10–3 T, v = 6.0 × 106 m s–1, m = 9.1 × 10–31 kg and e = 1.6 × 10–19 C
mv (9.1 × 10 − 31 ) × (6.0 × 106 )
\ Radius of circular orbit r = =
eB (1.6 × 10 − 19 ) × (1.2 × 10 − 3 )
= 0.0284 m or 28.4 mm

52 U-LIKE Physics–XII
 Or
(B) (a) Current sensitivity of a galvanometer is defined as the ratio of deflection (f) given by
it and the current (I) passed through it.
φ NAB
\ Current sensitivity = =
I k
where N = no. of turns of galvanometer coil
A = area of coil
B = magnetic field, and
k = torsional constant of the galvanometer spring.
(b) Current sensitivity of a galvanometer can be increased by (i) using a strong magnetic
field B, and (ii) by using a fine spring having small value of torsional constant k.
Of course current sensitivity can also be increased by increasing the number of turns
N in the coil and by having a bigger sized coil. But these factors are not practicable and
hence not considered.
(c) Here RG = 30 W, Ig = 2 mA = 2 × 10–3 A and I = 3.0 A
I g .RG (2 × 10 − 3 ) × 30
\ Shunt resistance rs =
= = 0.02 W
I − Ig (3.0 − 2 × 10 − 3 )
33.(A) (a) An optical fibre consists of a core of high
refractive index and a cladding of lower
refractive index.
Optical fibre works on the principle of
total internal reflection. When a light signal
is incident on one end of the fibre at a small
angle, the light passes inside, undergoes repeated total internal reflections along the
fibre and finally comes out from the other end of fibre. The angle of incidence is always
greater than the critical angle of the core material with respect to the cladding. Even if
the fibre is bent, the light signal can easily travel through along the fibre.
1  1 1 
= (n − 1) 
(b) As per lens maker’s formula −  . As refractive index ‘n’ of a lens
f R
 1 R2 

depends upon the colour of incident light, it is obvious that the focal length ‘f’ of the
lens will depend on the colour of incident light.
(c) As per question we get image of same size for two different values of object distance
(namely 20 cm and 10 cm). It is possible only if in one case the image is real and in other
case virtual.
v f
\ Magnification of a lens m =
=
u u+ f

f f
Hence, – m = and + m =
(− 20 + f ) (− 10 + f )

model test Paper – 5 53

 −f f
⇒
=
(− 20 + f ) (− 10 + f )
⇒
f = 15 cm
Or
(B) (a) Consider two thin lenses A and B of focal lengths f1 and f2 placed in contact. Let a point
object be placed at O, beyond the focus of first lens A. Lens A forms a real image at I1.
This image serves as a virtual object for second lens B and the final real image is formed
at I.
For image I1 formed by first lens, we have
1 1 1
− = ...(i)
v1 u f1
and for the image I formed by the second lens, we have
1 1 1
− = ...(ii)
v v1 f2
Adding (i) and (ii), we have
1 1 1 1
− = + ...(iii)
v u f1 f 2
If the two lens system is considered as equivalent to a single lens of focal length f, then
1 1 1
− = ...(iv)
v u f

Comparing (iii) and (iv), we find that 1 = 1 + 1

f f1 f 2
and in terms of power, we have P = P1 + P2.
(b) If a biconvex lens is placed in a trough of fluid, whose refractive index is same as that
of lens, we shall feel that lens has become invisible. It is on account of the fact that light
beam does not bend while passing through the lens.
(c) Here f = – 0.4 m and m = + 5
v f
As m = − =
u f −u
(− 0.4)
Hence, we have 5=
(− 0.4) − u

⇒
u = – 0.32 m
So the object should be placed 0.32 m in front of given concave mirror.

54 U-LIKE Physics–XII