33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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60 NEET-AIPMT Chapterwise Topicwise Solutions Physics
dφ
d 2
Inducedemf , ε =− =− ( 50t
+ 4)
=− 100t
V
dt dt
At t = 2 s, e = – 200 V; |e| = 200 V
Induced current in the coil at t = 2 s is
| ε | 200 V 1
I = = = A=
05 . A
R 400 Ω 2
5. (b) : Here, Magnetic field, B = 0.025 T
Radius of the loop, r = 2 cm = 2 × 10 –2 m
Constant rate at which radius of the loop shrinks,
dr
dt = 1 × 10 −3 ms
−1
Magnetic flux linked with the loop is
f = BAcosq = B(pr 2 )cos0° = Bpr 2
The magnitude of the induced emf is
dφ
d
| ε | = = ( π ) = π
dt dt B r 2
B 2 r dr
dt
= 0.025 × p × 2 × 2 × 10 –2 × 1 × 10 –3
= p × 10 –6 V = p mV
6. (a) : Once a rectangular loop or a square loop is
being drawn out of the field, the rate of cutting the lines
of field will be a constant for a square and rectangle, but
not for circular or elliptical areas.
7. (d) : Rate of decrease in the radius of the loop is
2 mm/s.
Final radius = 2 cm = 0.02 m
Initial radius = 2.2 cm = 0.022 m, B = 0.04 T
dφ
ε =− =−B dA
dt dt
e = –p (0.022 2 – 0.02 2 ) × 0.04 = –p × 3.36 × 10 –6 V
|e| = p × 3.36 × 10 –6 V = 3.4p mV
8. (a) : Work done due to a charge W = QV
NAdB
9. (c) :
ε
i = =
dt
R R
−4
20 × ( 25 × 10 ) × 1000
=
= 05 . A
100
−( φ2 −φ1) −( 0 −NBA)
NBA
10. (b) : ε = = =
t
t t
−2 −2
NBA 50 × 2× 10 × 10
t = = = 01 . s
ε
01 .
11. (a) :
When the electron moves from X to Y, the flux linked
with the coil abcd (which is into the page) will first
increase and then decrease as the electron passes by. So
the induced current in the coil will be first anticlockwise
and will reverse its direction (i.e., will become clockwise)
as the electron goes past the coil.
12. (c) : When the magnet is dropped through the
ring, an induced current is developed into the ring in the
direction opposing the motion of magnet (Lenz’s law).
Therefore this induced current decreases the acceleration
of bar magnet.
13. (a) : According to Faraday’s law, it is the
conservation of energy.
14. (b) : Here, B = 0.1 T, r = 0.5 m, w = 10 rad/s
So, the emf generated between its centre and rim is,
e = 1 2 1
2
Bω r = × 01 . × 10 × ( 05 . ) = 0. 125 V
2 2
15. (b) : Here, PQ = RS = PR = QS = a
Emf induced in the frame, e = B 1 (PQ)V – B 2 (RS)V
µ 0I
µ 0I
=
aV −
aV
2π( x−
a/ 2) 2π( x+
a/
2)
µ 0I
⎡ 2 2 ⎤
= ⎢ −
2π
⎣( 2x−
a) ( 2x+
a)
⎥ aV
⎦
µ 0I
⎡ 2a
⎤
= × 2
2π
⎢
⎣( 2x− a)( 2x+
a) ⎥ aV
⎦ ∴ ∝ 1
ε
( 2x− a)( 2x+
a)
16. (d) : Motional emf induced in the semicircular ring
PQR is equivalent to the motional emf induced in the
imaginary conductor PR.
i.e., e PQR = e PR = Bvl = Bv(2r)(l = PR = 2r)
Therefore, potential difference developed across the ring
is 2rBv with R at higher potential.
17. (d) : Length of conductor (l) = 0.4 m,
Speed (v) = 7 m/s and magnetic field (B) = 0.9 Wb/m 2
Induced e.m.f. (e) = Blv = 0.9 × 0.4 × 7 = 2.52 V.
18. (b) : Given n = 2 × 10 4 , I = 4 A
Initially I = 0 A
\ B i = 0 or f i = 0
Finally, the magnetic field at the centre of the solenoid
is given as
B f = m 0 nI = 4p × 10 –7 × 2 × 10 4 × 4 = 32p × 10 –3 T
Final magnetic flux through the coil is given as
f f = NBA = 100 × 32p × 10 –3 × p × (0.01) 2
f f = 32p 2 × 10 –5 T m 2
| ∆φ | | φf
− φi
| 2
× − 5
32π
10
Induced charge, q = = =
R R
2
10π
= 32 × 10 –6 C = 32 mC
19. (b) : q = Area under i-t graph
1
= × 4× 0. 1=
0. 2 C
2
As q = ∆φ
R