Solution

MEGA TEST - KE20240711

Class 12 - Physics
Section A
1.
(b) −(0.8mN)^i
Explanation:
^
−(0.8mN) i

2.
(d) e (v B + v x y y Bx )
^
k

Explanation:
^
e (vx By + vy Bx ) k

The force is given as -e (v×B).

3.
2
a
(b) ( b
)

ad
Explanation:
2
a
( )
b
Ac
4.
(c) 4πμ V
Explanation:
ow

To find the induced emf (E), use Faraday's law:

dA
E = −B
dt

The area A = πr . The radius shrinks at 1 mm/s, and at r = 4 cm :

2

dA dr 2
Kn

= 2πr = 2π(0.04)(−0.001) = −0.08π m /s

dt dt

Now,
−3
E = 50 × 10 × 0.08π = 4πμ V

Thus, the induced emf is 4πμ V

5. (a) resistance (r)

Explanation:
current depends on resistance (r), but emf doesn't.
6. (a) 2200 V - 50 Hz
Explanation:
In a transformer with 500 turns in the primary coil and 5000 turns in the secondary coil, the voltage transformation can be
determined using the formula:
Ns
Vs = Vp ×
Np

Here, V is 220 V , N is 500 , and N is 5000.

p p s

Calculating the secondary voltage:

5000
Vs = 220 V × = 220 V × 10 = 2200 V
500

The frequency remains unchanged at 50 Hz. Therefore, the output across the secondary coil is 2200 V 50 Hz.
7.
(d) 3.6 W
Explanation:
3.6 W

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 8.
(d) Ultraviolet rays
Explanation:
Ultraviolet rays

9.
(d) Infrared waves
Explanation:
Infrared waves

Section B
10. a. No, The magnetic force is always normal to B
⃗
.
b. Gauss's law of magnetism states that the flux of B⃗ through any closed surface is always zero, ∑ B⃗ ⋅ ΔS ⃗ = 0
s

If monopoles existed, then Gauss' law ⃗ ⃗

∑ B ⋅ ΔS ≠ 0
s

11. Vertical component of magnetic field BV is

BV = B sin δ

Given B = 0.3, δ = 30o

ad
BV = 0.3 sin 30o
BV = 0.15 G
12. As the point O lies on the straight part AB, So
Ac
BAB = 0
μ I
BBCD =
4πR
0
⋅
3π

2
, acting normally outward
μ0 I μ0 I
BDE = (sin 90
∘ ∘
+ sin 0 ) = , acting normally outward
ow

4πR 4πR

Total magnetic field at the centre O

B = BAB + BBCD + BDE
μ0 I 3π μ0 I
= 0 + ⋅ +
Kn

4πR 2 4πR

μ0 I
or B = 4πR
(
3π

2
+ 1) , acting normally outward.

13. i. Magnetic moment M = 0.9 J/T

τ= 0.063 J, θ = 30o
We know τ = M × B
= MB sin θ
0.063 = 0.9 × B × sin 30o
B= = 0.14 T
2×0.063

0.9

ii. Stable equilibrium is position of minimum energy. Since U = −M⃗ ⋅ B

⃗

U = - MB cos θ
Where U is the energy stored or P.E. of the magnet inside magnetic field B.
So, when θ = 0, U = -MB is the minimum energy.
Thus, when M⃗ and B⃗ are parallel to each other bar magnet is in stable equilibrium.
14. For the generation of motional emf: B, l and v must be in mutually perpendicular directions. If any two of these quantities are
parallel, emf is not induced.

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 i. In Fig. (a), the velocity of any conductor is parallel to field B, so no emf is induced in any conductor.
ii. (if) In Fig. (b), the arms AE, BF, CG and DH are parallel to the velocity v, no emf is induced in these arms. Also, the arms AB,
DC, EF and HG are parallel to field B, so no emf is induced in these arms.
The arms AD, BC, EH and FG are perpendicular to both B and v. Hence emf is induced in each of these arms and is given by
−2 −2
ε = Blv = 0.05 × 10 × 10 × 5 = 2.5 × 10 V

15. Consider a solenoid of length 2l, radius r and carrying a current I and having n turns per unit length.
Consider a point P at a distance a from the center O of solenoid.
Consider an element of solenoid of length dx at a distance x from its center.
This element is a circular current loop having (ndx) turns. The magnetic field at axial point P due to this current loop is
2
μ0 (ndx)I r
dB = 3

2 2
2( r +(a−x ) ) 2

The total magnetic field due to entire solenoid is

ad
+1 2
μ0 (ndx) I
∴ B= ∫
3
−1 2 2
2( r +(a−x ) ) 2

For a > 1 and a >> r, we have

Ac
3

{r
2
+ (a − x) }
2
2 = a3
2 +1 2
μ0 nL r μ0 nI r (2l)
∴ B= 3
∫ dx = 3
2a 2a
ow

−1

The magnetic moment of solenoid

m(NIA) = (n2I) I ⋅ π r2
μ0 2m
∴ B= 3
4πa
Kn

This is also the far axial magnetic field of a bar magnet. Hence, the magnetic field, due to current carrying solenoid along axial
line is similar to that of a bar magnet for far off axial points.
16. A coil B kept near another coil A has magnetic flux passing through it when kept near coil A. The ratio of magnetic flux through
the coil B to the current in the coil A is called as mutual inductance of coils. or Mutual Inductance is the interaction of one coils
magnetic field on another coil as it induces a voltage in the adjacent coil
ϕ2
M12 = i

Let S1​carries a current i.

μ N1 i
Magnetic field inside S1 will be B = o

Flux through each turn of S2​is ϕ = B × πr 2

2

2
μo N1 π r iN2

ϕ = l
2

Flux through each turn of S2​is ϕ = ϕN B 2

2
μo N1 π r iN2

ϕB = l
2

2
ϕB μo N1 N2 π r
M= i
= l
2

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17. i. Magnetic field of coil 'X'
N1 μ0 I1
B1 = 2r1

As the inner coil placed co-axially has very small radius, therefore B1 may be taken as constant over its cross-sectional area.
Hence flux associated with inner coil is
N1 μ I1
Φ2 = N2 πr B1 = N2 πr
2
2
2
2
(
2r1
0
) ...(i)
But Φ = M21I1 ...(2)
2

From (1) & (2)

2
μ π N1 N2 r

M21 =
o 2

2r1
2
μo π N1 N2 r

M12 = M21 = 2r1

2

18. i. Here, L = 2.0 H, C = 32μF = 32 × 10

−6
F , R = 10 Ohm
3
10
ωr =
1
=
1
=
8
= 125 rad/s
√LC √2.0×32×10−6

ii.

d
−
− −−−−−−
Q =
1

R
√
L

C
=
10
1
√
2

−6
= 25

19. a. λ =
λα
h

mp vp
32×10 ca
λp
= mα vα
h
×
h
= 1

4
A
−−−−−−−−
b. p = √2m(K . E)
√2mp (K.E⋅ )p −−
m
−
λα h p 1
= =√ =
ow

×
λp h mα 2
√2mα (K⋅E. )α

−−−
2qV
c. v = √ m
−−−−
λα mp vp mp 2q V −−−
mα
−
= =
h p
× √ × √
Kn

λp mα vα h mα 2qα V
mp

−−
− −
q
−
mp m p
= mα
×√
mp
α
× √
qα
= 1

2√2

20. a. Gamma rays, use: it is used in cancer treatment.

b. Ultraviolet rays use: it is used to sterilize surgical instruments.
c. Infrared radiation uses: it is used in thermal imaging cameras.
21. i. Magnetic field induction at a point P on conductor C2 D2 due to current I1 passing through C1 D1 is
μo I1
B1 = 2πr
where r is the separation between two conductors.
According to right hand thumb rule, the direction of magnetic field B1 is perpendicular to the plane of paper, directed inwards.
As the current carrying conductor C 2 D2 lies in the magnetic field B1 (produced by the current through C1D1 therefore , the
length of C2 D2 will experience a force given by
F2 = B1 I2 × 1 = B1 I2 1
F2 μ I1 I2
Putting the value of B1, we have l
= o

2Π
⋅
r

The ampere is the value of that steady current which, when maintained in each of the two very long straight, parallel
conductors of negligible cross section and placed in one meter apart in vacuum would produce of each of these conductors a
force equal to 2 × 10-7 Newton per meter of length.
μo I1 α μo I1 l1
ii. B1 = 2r
= 2
vertically outward
2r
μ I2 β μ I2 l2
B2 = o

2r
= o

2
vertically inward
2r

From calculation, using potential difference is same in parallel,I 1 l1 = I2 l2

⇒ B1 = B2

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 ⇒ Net magnetic field B = B 1 + B2 =0
BACB − BADB =0

22. i.

The working principle of transformer is mutual induction.

When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links

d
the secondary and induces an emf in it.
Causes of energy losses ca
a. Copper Losses (I2R Losses):
Explanation: These losses occur due to the resistance of the wire in the coils. When current flows through the wire, some
A
energy is dissipated as heat due to the resistance.
Mitigation: Using thicker wires with lower resistance and cooling mechanisms can help reduce these losses.
ow

b. Core Losses (Iron Losses):

Explanation: Core losses include hysteresis loss and eddy current loss. Hysteresis loss occurs due to the lagging of
magnetic domains behind the changing magnetic field. Eddy current loss is caused by circulating currents induced within
the core material due to the changing magnetic field.
Kn

Mitigation: Using laminated silicon steel cores reduces eddy currents, and using materials with low hysteresis loss
minimizes core losses.
c. Leakage Flux:
Explanation: Not all magnetic flux produced by the primary coil links with the secondary coil. Some flux "leaks" and
does not contribute to energy transfer between coils, leading to inefficiency.
Mitigation: Improving core design and using better core materials can help reduce leakage flux.
ii. A step-up transformer does not violate the conservation of energy. It increases voltage by decreasing current proportionally.
The power input (low voltage, high current) equals the power output (high voltage, low current), ensuring energy
conservation. The transformer merely transfers energy, with losses typically due to inefficiencies.
Vs Ns
iii. 1. VP
=
NP

Ns 3000
Vs = × VP = × 90
NP 200

Vs = 1350 V
IP Ns
2. Is
=
N
P
3000
IP = × 2 = 30 A
200

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23. i. Diagram

Principle - It is based on the principle of electromagnetic induction. Whenever there is a change in magnetic flux linked with
a coil , an emf is induced in the coil.
Working - When a rectangular coil is rotated in a magnetic field, the magnetic flux changes continuously which induces an
emf and the direction of current changes periodically.
−N dϕ
ε=
dt
d
= −N BA (cos ωt)
dt

ε = N BAω sin ωt

ii. ε = N BAω
= 100 × 0.8 × 0.5 × 60
= 2400 V
hence, the maximum emf generated will be 2400 V.
2
εrms
Power dissipated, P =

d
R
2
2400
( )
√2

=
100

= 28.8 kW
A ca
ow
Kn

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