33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Telegram @unacademyplusdiscounts
54 NEET-AIPMT Chapterwise Topicwise Solutions Physics
Final potential energy of dipole,
MBH
Uf
=− MBH
cos60°=−
2
MBH
MBH
W = Uf
− Ui
=− −− ( MBH) =
...(i)
2 2
τ= MBH sin60 °= 2 3
W × [Using eqn. (i)]
2
= 3W
3. (c) : The direction of magnetic dipole moment is from
south to north pole of the magnet.
In configuration (1),
mnet = 2
m + 2
m + 2mmcos90
°
2 2
= m + m = m
In configuration (2),
2
m net = m – m = 0
In configuration (3),
m net = 2
m + 2
m + 2mmcos
30 °
2 2 ⎛ ⎞
= 2m + 2m ⎝
⎜
3
⎠ ⎟ = m 2+
3
2
In configuration (4),
mnet = 2
m + 2
m + 2mmcos60
°
2 2 ⎛
= 2m + 2m 1
⎝
⎜ ⎞ ⎠ ⎟ = m 3
2
4. (d) : Let m be strength of each pole of bar magnet of
length l. Then
M = m × l
...(i)
When the bar magnet is bent in the form of an arc as
shown in figure. Then
π πr
l= × r =
3 3 or r l
= 3 π
New magnetic dipole moment
M′ = m × 2r sin30°
= m × 2 × 3 l ml M
× 1 3 3
2 = = (Using (i))
π π π
5. (a)
6. (b) : Work done in changing the orientation of a
magnetic needle of magnetic moment M in a magnetic
field B from position q 1 to q 2 is given by
W = MB(cosq 1 – cosq 2 )
Here, q 1 = 0°, q 2 = 60°
= MB ⎛ ⎞
−
⎝
⎜
⎠
⎟ = MB
1 1 ...(i)
2 2
The torque on the needle is
τ= M × B
In magnitude,
3
τ= MBsinθ= MBsin60°=
MB
...(ii)
2
Dividing (ii) by (i), we get
τ
=
W
3 or τ= 3W = 3 × 3J = 7. (b) : Here, Magnetic moment, M = 0.4 J T –1
Magnetic field, B = 0.16 T
When a bar magnet of magnetic moment is placed in a
uniform magnetic
field , its potential energy is
U =−M⋅ B=−MB
cosθ
For stable equilibrium, q = 0°
\ U = –MB = –(0.4 J T –1 )(0.16 T) = –0.064 J
8. (d) : The time period T of oscillation of a magnet is
given by
T = 2π
I
MB
where,
I = Moment of inertia of the magnet about the axis of
rotation
M = Magnetic moment of the magnet
B = Uniform magnetic field
As I and M remain the same
1 T2
B1
∴ T ∝ or =
B T1
B2
According to given problem,
B 1 = 24 mT, B 2 = 24 mT – 18 mT = 6 mT, T 1 = 2 s
( 24 µ T)
∴ T 2 = ( 2 s)
= 4 s
( 6 µ T)
9. (a) : Magnetic field due to bar magnets exerts force
on moving charges only. Since the charge is at rest, no
force acts on it.
10. (c) : Magnetic moment of the loop
M = NIA = 2000 × 2 × 1.5 × 10 –4 = 0.6 J/T
Torque t = MBsin30°
= × × − 2 1 −2
06 . 5 10 × = 15 . × 10 Nm
2
11. (b) : Here, M = 2 × 10 4 J T –1
B = 6 × 10 –4 T, q 1 = 0°, q 2 = 60°
W = MB(cosq 1 – cosq 2 ) = MB(1 – cos60°)
4 −4
⎛ ⎞
W = 2× 10 × 6× 10 −
⎝
⎜1 1 ⎠
⎟
2
= 6 J.
12. (b) : Initial mass of the magnet (m 1 ) = m and final
mass of the magnet (m 2 ) = 4m.
2
I mk
The time period, T = 2π = 2π
∝ m
MB MB
Therefore T 1 m1
m 1
= = =
T2
m2 4m
2
or T 2 = 2T 1 = 2T
13. (a) :