SAMPLE PAPER 14 209

14

Physics
Class 12th
1. (d) AC source voltage, 5. (b) In 10 s, the number of nuclei has been reduced
1
V = V0 sinω t (as, V = 0 at t = 0) from 25% to 6.25%, i.e. th of given nuclei.
4
 π
⇒ VR = V0 sinω t and VC = V0 sin ω t −  ∴ Half-life of nuclei, T1/ 2 = 5s
 2
T 5
V and VR are in same phase while VC lags V(or VR) by 90°. ∴ Mean life, T = 1/ 2 = = 7. 21 s
Now, VR is in same phase with initial potential difference 0.693 0.693
across the capacitor for the first time when q 2B 2r 2
6. (c) Kinetic energy is given as, KE =
π 3π 2m
ω t = − + 2π =
2 2 1
⇒ KE ∝
3π m
⇒ t =
2ω (KE)d mp
⇒ =
dV (KE)p md
2. (a) Since, electric field, E = −
dr md
dV ⇒ (KE) p = × (KE )d
∴ Ex = − = − (6 − 8 y 2 ) mp
dx
2 mp
dV ⇒ (KE)p = × 50 (Q md = 2 mp )
Ey = − = − (−16 xy − 8 + 6 z ) mp
dy
dV ∴ (KE)p = 100keV, having same radius and same B
Ez = − = − (6 y − 8 z )
dz inside.
At origin, x = y = z = 0 7. (a) Diodes D1 and D3 are forward biased, while diode
⇒ E x = − 6, Ey = 8 and Ez = 0 D2 is reverse biased so the circuit will become
R
∴Electric field, E = E 2x + Ey2 = (−6)2 + (8)2

= 10 N/C R
3. (d) Given, L = (0.4/ π ) H, V = 200 V, f = 50 Hz
0.4 i
∴ XL = ωL = 2 πfL = 2 π × 50 × = 40 Ω
π i R
and R = 30 Ω
∴ Impedance, Z = R 2 + XL2 E
E
∴ Current in circuit will be, i =
= (30 )2 + (40 )2 = 50 Ω R
Vrms 200 (Q wire with D3 is resistance less)
∴ Current, Irms = = = 4A
Z 50 8. (b) As we know, R = R0 A1/ 3
4. (a) As both points A and B lies on axial axis, so 1/ 3
R′  A′ 
1 ⇒ = 
B∝ 3 RHe  AHe 
r
1/ 3
3 3  A
B A  rB   48 ⇒ (14)1/ 3 =   ⇒ A = 56
⇒ =  =  =8  4
BB  r A   24
∴ Atomic number of nucelus, Z = 56 − 30 = 26
210 PHYSICS CLASS 12

9. (b) Statement given in option (b) is incorrect and it can 14. (b) The magnetic field (B) at the centre of circular
be corrected as, current carrying coil of radius R and current I,
The lines of constant magnitude of magnetic field form µ I
B= 0 …(i)
concentric circles called magnetic field lines. They 2R
originate from positive charges and ends at negative Similarly, if current = 3I, then
charges.
µ 3I
Rest statements are correct. magnetic field = 0 = 3B
2R
h
10. (c) Since, λ = ...(i) So, resultant magnetic field
p
0.5 h = B 2 + (3B)2 = 10 B 2
⇒ λ− λ=
100 p+ ∆p
µ 0I 10
199λ h = 10 B = [from Eq. (i)]
⇒ = 2R
200 p + ∆p
15. (b) The system is in equilibrium, so, it can be shown as
199h h +
or = [using Eq. (i)]
200 p p + ∆p +
200 P +θ
⇒ p + ∆p = p + T cos θ
199 + T θ
+
⇒ p = 199 ∆p + qE
T sin θ
11. (c) Using mass action law, ni2 = ne nh
Here, ni = 10 16 /m 3
mg
⇒ T sinθ = qE ...(i)
and ne = 10 21 /m 3
and T cosθ = mg ...(ii)
n 2 (10 16 )2
∴ nh = i = = 10 11 /m 3 Dividing Eq. (i) by Eq. (ii), we get
ne 10 21  σ  
qE q σ 
tan θ = = ×  Q E = 2 ε 
12. (b) If charge on capacitor at time t is Q, then electric mg mg  2 ε 0   0
field between plates of capacitor will be
⇒ σ ∝ tan θ
Q
E= 16. (a) At resonance, XL = XC
ε0 A
The flux through the given area will be ∴ Z = R 2 + ( XL − XC )2 = R
Q A Q Vm Vm
φE = × = ∴Current, I = = (QVL = VC )
ε0 A 2 2ε0 Z R
dφ ε dQ Therefore, A and R are true and R is the correct
∴ Displacement current, id = ε 0 = 0
dt 2 ε 0 dt explanation of A.
i  dQ  17. (a) In a detector loop, when the light falls on the metal
= Q = i surface, i.e. the emitter plate, some electrons near the
2  dt 
surface absorb enough energy from incident radiations,
φ i.e. from UV light, to overcome the attraction of the
13. (c) As we know, intensity, I = 4I0 cos 2
2 positive ions in the material of the surface.
φ  After gaining sufficient energy from UV-light, the
cos 2  1
I1 2 electrons escape from the emitter plate into the
⇒ = surrounding space, thus enhancing the high voltage
I2 φ 
cos 2  2  spark across detector loop.
2
Therefore, A and R are true and R is the correct
cos 2(0 ° )
= explanation of A.
 90 °
cos 2   18. (a) As resistance of depletion region is large, potential
 2 
drop occurs mainly in depletion region.
1 1
= = So, the applied voltage mostly drop across the
cos 2 45°  1
2
depletion region and the voltage drop across the p-side
 
 2 and n-side of the junction is negligible.
=2:1 Therefore, A and R are true and R is the correct
explanation of A.
 SAMPLE PAPER 14 211

19. (i) Q Force on a charge in a magnetic field is F = qvB (ii) 10 3m - 10 1m → radio waves
F 1N It is used in radio communication.
⇒ B= ⇒ 1T =
qv (1 C ) (1 ms −1) 23. (i) According to Gauss’s theorem, φ = Σq
ε0
Thus, if a charge of 1C moving with a velocity of
1 ms −1 perpendicular to a uniform magnetic field, φ ∝ Σq
experiences a force of 1 N, then the magnitude of φS1 2Q 2Q 1
∴ = = =
the field is 1T. φS 2 2Q + 4Q 6Q 3
(ii) Yes, we can increase the range of voltmeter by
(ii) If a medium of dielectric constant ε r is introduced in
connecting additional resistance in series with the
the space inside S 1 in place of air, then
voltmeter. We can also decrease the range of
voltmeter by connecting a suitable resistance in Σq 2Q
φS1 = =
parallel with the voltmeter. ε 0ε r ε 0ε r
Or 24. The number of electrons thermally generated
Given, pole strength, m = 6.64 A -m, (ni ~ 10 6m − 3 ) are negligibly small as compared to
r = 3 mm = 3 × 10 −3m those produced by doping.
µ 0 m1m2 ∴ ne ~
− ND
∴ Force, F= × 2
4π r 1ppm = 1part per million
µ0 m 2 5 × 10 28
⇒ F= × (Qm 1 = m2 = m) ⇒ ND = = 5 × 10 22m − 3
4π r 2 10 6
µ0 (6.64)2 ∴ ne ~
= 5 × 10 22m − 3
= ×
4π (3 × 10 −3 )2 Since, we know that, ne nh = ni2
4π × 10 −7 44.0896 The number of holes, nh = ni2 / ne
= ×
4π 9 × 10 −6 . × 10 16 )2
(15
−1 =
= 10 × 4.8988 5 × 10 22
F = 0.49 N ~
= 4.5 × 10 9m − 3
20. By conservation of charge and mass, given equation
can be written as Or
1 235 88 136 As it is given that, both the diodes are identical and
0n + 92 U → 38Sr + 54 Xe + 12 10 n + Q (energy)
they are forward biased, so both will conduct.
For amount of energy released, we use following formula The diodes are ideal, so the equivalent resistance of
Q = ∆ m × 931MeV arm AB and DC is
where, ∆ m = mass defect. 3×2 6
R1 = = Ω
21. For glass prism, critical angle is 3 +2 5
 1  1  This is in series with 1Ω resistance, so total resistance
ic = sin −1   = sin −1   of circuit,
 µ  3 / 2
6 11
 2 R2 = + 1 = Ω
= sin −1   = 42 ° 5 5
 3
V 6 30
A ∴Current across 1Ω resistor is, I = = = A
R2 11 11
60º 5
25. V
90º
30
º

60º Input
waveform Input frequency
Time = 60 Hz
B C

Now, consider the diagram, angle of incidence on the (i) Output waveform Output frequency
of half-wave = 60 Hz
face AC, i = 60 > ic . rectifier
So, the ray will be totally reflected inside the prism. (ii) Output waveform Output frequency
of full wave rectifier
22. (i) 10 −12m - 10 −8m → X-ray = 120 Hz

It is used in crystallography.
212 PHYSICS CLASS 12

26. Given, T = 127 °C = 127 + 273 = 400 K 2

q1  x 
= 
de Broglie wavelength, λ =
h q2  r + x 
p
A x q1 r q2
where, p = momentum.
3  3  where, x is the distance of zero field point from q 1.
Q p = mv = 2 mE = 2 m × kT Q E = kT
2  2  Or
= 3 mkT On introducing the dielectric slab to fill the gap between
h plates of capacitor completely, when capacitor is
∴ λ=
3 mkT connected with battery. Then
6.63 × 10 −34 Aε 0
= (i) Capacitance of capacitor, C = K
d
. × 10 −27 × 138
3 × 166 . × 10 −23 × 400
Here, K = 10 and d ′ = 3d
6.63 × 10 − 10 Aε
= New capacitance, C′ = 10 × 0
5243
. 3d
= 1264
. Å 10
⇒ C′ =
C
27. (i) Velocity of light in vacuum, c = 1 3
µ 0ε 0 (ii) The potential difference V between capacitor is same
1 due to the connectivity with battery and charge is
and velocity of light in medium, v = q′ = C′ V ′
µε
 10  10
c  µε 
1/ 2 =  C (V ) = CV
So, index, n = =  3  3

v  µ 0ε 0 
(iii) The electric field E is same due to the connectivity
(ii) Given, velocity of electromagnetic wave in vacuum with battery. The energy density of the capacitor is
= 3 × 10 8 m/s given by
1
Relative electric permittivity, ε r = 2 U = ε 0E 2
2
and magnetic permeability, µ r = 1
Now, the energy density will become
Since, velocity of electromagnetic wave in a medium
1 1
can be calculated by U ′ = K ε 0E 2 = × 10 × ε 0E 2
1 2 2
v= 1
ε 0ε rµ 0µ r U ′ = 10 × ε 0E 2
2
1
= U ′ = 10 U
ε 0µ 0 × µ r ε r
mv 2 mK
1 29. We know that, R = = [Q mv = 2 mK ]
where, =c qB 2qB
ε 0µ 0
2mqV
c 3 × 10 8 = [Q K = qV ]
So, v= = qB
µ r εr 2 ×1
∴ R∝ m
3
⇒ v= × 10 8 m/s The radius of heaviest particle, R1 = a + b
2
and the radius of lightest particle, R2 = a, then
28. (i) The electric force on the particle always acts in the
direction of a line of force. So, if the particle is initially R1
=
m1
at rest, then it will move along the direction of line of R2 m2
force. But if the particle enters the field at an angle 2
a+ b m1 m1  a + b
with its direction, then it will not move along the line = ⇒ = 
of force. a m2 m2  a 
(ii) Two essential conditions are Or
(a) The charges (q 1 and q 2) should be of opposite The magnetic field due to arc of current carrying coil
nature. which subtends an angle θ at centre is given by
(b) The magnitude of the charge nearer to the point µ Iθ
B= 0
should be smaller than that of the other. Thus, if r 4π R
be the distance between them, then
 SAMPLE PAPER 14 213

For the current carrying loop of quarter circle of radius Case II If v, E and B are mutually perpendicular, i.e.
R, lying in the positive quadrant of the X-Y plane, θ = 90 °, then Lorentz force is zero which means
µ I(π / 2 ) $ µ 0 I $ particle will pass through the field without any
B1 = 0 k= k
4π R 4 2R deviation.
(ii) To calculate the induced emf, the rate of change of
For the current carrying loop of quarter circle of radius
flux needs to be calculated first. Here, magnetic field
R, lying in the positive quadrant of the Y-Z plane,
is uniform, so change in flux takes place due to
µ I $
B2 = 0 i change in area swept by loop. By calculating the
4 2R rate of change of area, we can calculate the rate of
For the current carrying loop of quarter circle of radius change of flux, i.e. induced emf.
R, lying in the positive quadrant of the Z-X plane, Let the lengths of horizontal arms of circuit be x1 and
µ I $ x2 at instants t 1 and t 2, respectively.
B3 = 0 j
4 2R × × Q × ×
Current carrying loop consists of three identical quarter
circles of radius R, lying in the positive quadrants of the × × ×
X-Y, Y-Z and Z-X planes with their centres at the origin,
× l l v
joined together is equal to the vector sum of magnetic
field due to each quarter and given by × × ×
1 $ $ $ µ 0I
B= (i + j + k)
4π 2R × × P × ×
30. (i) As, 1u = 931 MeV/c 2 x1
x2
Energy released = ∆ m × 931MeV
Area of loop inside the magnetic field,
∆ m = 4m(11H ) − m(42He ) A1 = lx1, A2 = lx2
∴Energy released ∴ ∆ A = A2 − A1
= [4 ⋅ m(11H ) − m(42He )] × 931 MeV = l ( x2 − x1) = l ∆ x
= [4 × 1007825
. − 4.002603] × 931MeV = 2672
. MeV ∴ ∆φ = B∆ A = Bl ∆ x
hc ∆φ ∆x
(ii) Energy of incident photon, E = hν = = 13.6 eV ⇒ = Bl = Blv
λ ∆t ∆t
hc 6.6 × 10 −34 × 3 × 10 8 ∆φ
⇒ λ= = By Faraday’s law, induced emf, e = = vBl
E . × 10 −19
13.6 × 16 ∆t

⇒ λ = 0.910 × 10 −10 m Or
This radiation lies in ultraviolet region. (i) As, Pav = Vrms Irms cos φ
In ideal inductor, current Irms lags behind applied
31. (i) During motion, free electrons are shifted at one end
π
due to magnetic force. So, due to motion of rod, voltage Vrms by .
2
electric field is produced which applies electric force
π
on free electrons in an opposite direction. ∴ φ=
2
+ π
Fe Thus, Pav = Vrms × Irms cos
2
e = Vrms × Irms × 0 = 0
l v
(ii) Let initially Ir be current flowing in all the three
Fm circuits. If frequency of applied AC source is
increased, then the change in current will occur in
following manner.
At equilibrium, Lorentz force, Fe + Fm = 0 (a) AC circuit containing resistance only
where, Fe = force due to electric field = qE
and Fm = force due to magnetic field = q(v × B ). I
∴ qE + q (v × B ) = 0
Ir
⇒ E = −v ×B=B × v
⇒ |E|= B v sin θ
Case I If B, E and v are collinear, then charged
ν
particle is moving parallel or anti-parallel.
214 PHYSICS CLASS 12

The current through the resistor is As both faces of convex lens have same radius of
V curvature, so R1 = R, R2 = − R.
I=
R Focal length of lens, f = + 30 cm
As, R is independent of frequency of AC. Using lens Maker’s formula,
So, there is no effect on current, of the increase in 1 a  1 1
= ( µ g − 1)  − 
frequency. f  R1 R2 
(b) AC circuit containing inductance only With the 1  1 1
increase of frequency of AC source, inductive ⇒ = (155
. − 1)  + 
30  R R
reactance increases as
1 2
XL = 2πνL ⇒ = 0.55 ×
30 R
V V
∴ I = rms = rms ⇒ R = 0.55 × 2 × 30 = 33 cm
XL 2πνL
This is the required radius of curvature.
1
So, for given circuit, I ∝ (ii) Given, radius of curvature of objective mirror,
ν
R1 = 220 mm
Hence, current decreases with the increase in
R
frequency. and f1 = 1 = 110 mm
2
I Radius of curvature of secondary mirror,
R2 = 140 mm
Ir
and f2 = R2 / 2 = 70 mm
Distance between the two mirrors,
νi
d = 20 mm from objective mirror
ν For secondary mirror,
where, ν i = initial frequency of AC source. u = f1 − d = 110 − 20 = 90 mm
(c) AC circuit containing capacitor only 1 1 1
For mirror formula, + =
With the increase in frequency of AC source, the v u f2
capacitive reactance decreases as 1 1 1 1 1
⇒ = − = −
1 1 v f 2 u 70 90
XC = =
ωC 2 πν C ⇒ v = 315 mm = 315 . cm
V Vrms ∴The final image is at 31.5 cm to the right of
∴ Current, I = rms =
XC  1  secondary mirror.
 
 2πνC  Or
⇒ I = 2πν CVrms (i) Given, wavelength of red light,
So, for given circuit, I ∝ ν λ = 650 nm = 650 × 10 −9m
Hence, current increases with the increase in (a) For first minimum of the diffraction pattern,
frequency.
d sin θ = λ ; θ = 60 ° (given)
λ 650 × 10 −9
I ∴ d = =
sin θ sin 60 °
Ir
650 × 2 × 10 −9  3
= Q sin 60 ° = 
3  2 
νi
ν = 750.55 × 10 −9m = 7.5 × 10 −7 m
32. (i) Given, the refractive index of glass with respect to air, (b) For first maximum of the diffraction pattern,
a
µ g = 155
. 3λ
d sin θ =
2
3λ
⇒ d =
R1 R2 2 sin θ
Here, θ = 60 ° (given)
 SAMPLE PAPER 14 215

3 × 650 × 10 −9 3 × 650 × 10 −9 v A nB n 1
⇒ d = = Given, n A = 2 nB ⇒ = = B =
2 × sin 60 ° 3 vB n A 2 nB 2
2×
2 (iii) Current drawn from cell, I =
E
= 3 × 650 × 10 −9 = 1125.8 × 10 −9 R+r

= 1125
. × 10 −6 m For maximum current, R = 0
E 19
.
(ii) Distance between the two sources, ∴ Imax = = = 0.005 A
r 380
. × 10 −4 m
d = 0.15 mm = 15
For driving the starting motor of a car, a large current
Wavelength, λ = 450 nm = 4.5 × 10 −7 m
of the order of 100 A is required, so the cell cannot
Distance of screen from source, D = 1m
drive the starting motor of the car.
(a) I. The distance of nth order bright fringe from
central fringe is given by Or
nDλ (i) Consider the circuit. Let the emf of the battery is Eeq .
yn =
d It is clear from the figure that the cells E1, E2 and E3
For second bright fringe, are in series.
2 Dλ 2 × 1 × 4.5 × 10 −7 1V 2V 3V
y2 = =
d 1.5 × 10 −4 E3 E2 E1
−3
⇒ y2 = 6 × 10 m 6V

∴The distance of the second bright fringe,

y2 = 6 mm So, equivalent emf, E4 = E1 + E2 + E3
II. The distance of nth order dark fringe from central = 3V + 2 V + 1V = 6V
fringe is given by As the two cells are in parallel, so equivalent emf
Dλ 6V
y′n = (2 n − 1)
2d
For second dark fringe, n = 2
Dλ 3Dλ 6V
y′n = (2 × 2 − 1) =
2d 2d Eeq = emf of single cell = 6 V
−7
3 1 × 4.5 × 10 (ii) Currents and emfs are shown in the given circuit.
⇒ y′n = ×
2 1. 5 × 10 − 4 Now, applying Kirchhoff’s second law, in closed loop
ABEFA,
= 4.5 × 10 −3m
4 = (I1 + I2 ) 4 + I1 × 2
∴The distance of the second dark fringe,
⇒ 4 = 4I1 + 4I2 + 2 I1
y′n = 4.5 mm
⇒ 4 = 6I1 + 4I2 …(i)
(b) With increase of D, fringe width also increases as Similarly, in closed loop BCDEB,
Dλ 6 = 4 (I1 + I2 ) + 12 I2
β= or β ∝ D
d
6 = 4I1 + 16I2
33. (i) Current flowing through both the bars is equal. ⇒ I1 + 4I2 = 3 / 2 …(ii)
Now, the heat produced is given by On solving Eqs. (i) and (ii), we get
H = I 2Rt 5
5I1 = ⇒ I1 = 0 .5 A
2
⇒ H∝R
Substituting I1 = 0.5 A in Eq. (i), we get
H AB R AB (1 / 2 r )2  1 1
∴ = = 2 
QR ∝ ∝ 2 ∴ I2 = 0.25 A
HBC RBC (1 / r )  A r 
34. (i) The statement given is false and it can be corrected
1 as,
=
4 For microscope, the magnification in case of image
(ii) In series, current is same.  D
formed at infinity  m =  is one less than the
∴ I A = IB = I = neAvd  f
For same diameter, cross-sectional area is same. magnification when the image is at the near point
∴ AA = AB = A   D 
m = 1 + f  .
∴ n A eAv A = nB eAvB
216 PHYSICS CLASS 12

(ii) A simple microscope has a limited maximum exchanged between both electrodes and
magnification (≤ 9) for realistic focal length. electrolytes.
(iii) (a) For least distance of distinct vision, the angular (ii) If V is the potential difference across R , then from
magnification of simple microscope is Ohm’s law,
D  1 V = IR
M = 1+ ⇒ M = 1 + DP Q Power (P ) = 
f  f Combining equations V = IR and V = ε − Ir , we get
D IR = ε − Ir
and for normal adjustment, M = ⇒ M = DP.
f ε
or I=
(b) If the angular magnification of simple microscope R+r
can be increased, by increasing the power of
(iii) (a) The maximum current that can be drawn from a
lens.
cell is when internal resistance equal to external
Or resistance.
(iii) (a) Objective of a compound microscope is a (b) When R is infinite, so that I = V / R = 0, where V is
convex lens. Convex lens forms real and enlarged the potential difference between P and N, then
image when an object is placed between focus and the circuit is equivalent to open circuit, hence
radius of curvature. PD = emf.
(b) The magnification due to objective is Or
m0 = 20 E
(iii) (a) Current, i =
and due to eyepiece is me = 12.5 R+r
The magnification of compound microscope is Putting numerical values, we have
|m|= m0me = 20 × 12.5 = 250 E = 2 V, r = 0.2 Ω and R = 3.8 Ω
35. (i) When two electrodes of the cell are immersed in an 2 2
⇒ i= = = 0.5 A
electrolytic solution as shown in the figure given, 3.8 + 0.2 4
then the circuit represented is closed. Since, (b) Also, V = E − ir = 2 − 0.5 × 0. 2
electrolytic solution also allow the flow of current
⇒ V = 1.9 V
through ions and electron. Thus, the charges are