33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Telegram @unacademyplusdiscounts
36 NEET-AIPMT Chapterwise Topicwise Solutions Physics
kx mg
aA = − 3
3m
mg mg
aA = 4 − 3
3m
mg g
= =
3m
3
and a B = g.
64. (a) : Let F be the upthrust of the air. As the balloon
is descending down with an acceleration a,
\ mg – F = ma ... (i)
Let mass m 0 be removed from the balloon so that it starts
moving up with an acceleration a. Then,
F – (m – m 0 )g = (m – m 0 )a
F – mg + m 0 g = ma – m 0 a
... (ii)
Adding eqn. (i) and eqn. (ii), we get
m 0 g = 2ma – m 0 a ; m 0 g + m 0 a = 2ma
m 0 (g + a) = 2ma
2ma
m0
=
a + g
65. (c) : As all blocks are moving with constant speed,
therefore, acceleration is zero. So net force on each block
is zero.
66. (c) : Here, Mass of a person, m = 60 kg
Mass of lift, M = 940 kg,
a = 1 m/s 2 , g = 10 m/s 2
Let T be the tension in the supporting cable.
\ T – (M + m)g = (M + m)a
T = (M + m)(a + g)= (940 + 60)(1 + 10)
= 11000 N
67. (a) : F – Mg = Ma
a
8000 = 2000a
\ Acceleration is 4 m s –2 upwards.
68. (d) : The wedge is given an acceleration to the left.
\ The block has a pseudo acceleration to the right,
pressing against the wedge because of which the block is
not moving.
\ mg sinq = ma cosq
g
or a = sin θ
cosθ
Total reaction of the wedge
on the block is
N = mg cosq + ma sinq
mg sinθ⋅sinθ
or N = mg cosθ+
cosθ
2 2
mg(cos θ+
sin θ)
mg
or N =
= ⋅
cosθ
cosθ
69. (d) : When the lift is accelerating upwards with
acceleration a, then reading on the scale
R = m (g + a) = 80 (10 + 5) N = 1200 N
70. (d) : Let T be the tension in the rope when monkey
climbs up with an acceleration a. Then,
T – mg = ma
5
25 g – 20g = 20 a ⇒ a = × 10
= 25 . m/s 2 .
20
71. (b) : For a lift which is moving in upward direction
with an acceleration a, the tension T developed in the
string connected to the lift is given by T = m (g + a).
Here m = 1000 kg, a = 1 m/s 2 , g = 9.8 m/s 2
\ T = 1000(9.8 + 1) = 10,800 N
72. (b) : The force equations are
T – 5g = 5a,
10g – T = 10a
Adding, 10g – 5g = 15a
5g
g
or a = =
15 3 .
73. (c) : Upward acceleration, ma = T 1 – mg
T 1 = m(g + a)
Downward acceleration, ma = mg – T 2
or, T 2 = m(g – a)
T1
g + a 98 . + 49 . 3
= = =
T2
g − a 98 . − 49 . 1
74. (a) : The pseudo acceleration for the body a′ = a
If the pseudo force Macosq =
Mgsinq, then the body will be at
rest,
a = gtanq
This horizontal acceleration should be applied to the
wedge to the left.
75. (c) : Let T be the tension in the branch of a tree
when monkey is decending with acceleration a
Thus, mg – T = ma
also, T = 75% of weight of monkey
T = ⎛ mg mg
⎝ ⎜ 75 ⎞ 3
⎠
⎟ =
100 4
∴ = − ⎛ ⎝ ⎜ 3 ⎞ 1
g
ma mg
⎠
⎟ mg = mg or a =
4 4 4
vvv