33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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14 NEET-AIPMT Chapterwise Topicwise Solutions Physics
Retardation = 12 m s –2
18. (a) : Distance, x = (t + 5) –1 ...(i)
Velocity, v = dx d
( )
dt
= dt t + 5 −1
= – (t + 5) –2 ...(ii)
dv d
Acceleration, a = = [ − ( t + 5) −2
]
dt dt
= 2(t + 5) –3 ...(iii)
From equation (ii), we get
v 3/2 = – (t + 5) –3
...(iv)
Substituting this in equation (iii), we get
Acceleration, a = – 2v 3/2 or a ∝ (velocity) 3/2
From equation (i), we get
x 3 = (t + 5) –3
Substituting this in equation (iii), we get
Acceleration, a = 2x 3 or a ∝ (distance) 3
Hence option (a) is correct.
19. (c) : Given : At time t = 0, velocity, v = 0.
⎛ t ⎞
Acceleration f = f0 −
⎝
⎜ 1 T ⎠
⎟
⎛ t ⎞
At f = 0, 0= f0
1−
⎝
⎜
T ⎠
⎟
Since f 0 is a constant,
t
∴ 1− = 0 or t = T.
T
dv
Also, acceleration f =
dt
vx
t=
T T
⎛ t ⎞
∴ ∫ dv = ∫ fdt = ∫ f0
−
⎝
⎜1
⎠
⎟
T
dt
0 t=
0 0
T
⎡ ft ⎤
0 2 fT 0 2 1
∴ vx
= ⎢ f0t
− ⎥ = fT 0 − = fT 0 .
⎣⎢
2T
⎦⎥
2T
2
0
20. (b) : ds 2
= 9t
+ 14t+
14
dt
2
d s
⇒ = 18 t+ 14 = a
2
dt
a t = 1s = 18 × 1 + 14 = 32 m/s 2
21. (a) : Distance (x) = at 2 – bt 3
Therefore velocity (v) = dx d
dt
= dt ( at 2 −bt
3 ) = 2at – 3bt 2
Acceleration = dv d
= ( 2at
−3bt
2 ) = 2a – 6bt = 0
dt dt
2a
a
or t = =
6b
3b
22. (c) : Acceleration ∝ bt. i.e., d 2
x a bt
2
dt
= ∝
Integrating, dx 2
bt
= +C
dt 2
Initially, t = 0, dx/dt = v 0
Therefore, dx 2
bt
= +v0
dt 2
3
bt
Integrating again, x = + v0
t+
C
6
When t = 0, x = 0 ⇒ C = 0.
3
bt
i.e., distance travelled by the particle in time t = v0
t+ .
6
23. (c) : Displacement (s) = t 3 – 6t 2 + 3t + 4 m.
Velocity (v) = ds = 3t 2 – 12t + 3
dt
Acceleration (a) = dv = 6t – 12.
dt
When a = 0, we get t = 2 seconds.
Therefore velocity when the acceleration is zero is
v = 3 × (2) 2 – (12 × 2) + 3 = – 9 m/s
24. (d) : Here, u = 20 m/s, v = 80 m/s, g = 10 m/s 2 , h = ?
v 2 = u 2 + 2gh
⇒ 80 2 = 20 2 + 2 × 10 × h
Hence, h = 300 m
25. (d) : Distance covered by the stone in first 5 seconds
(i.e. t = 5 s) is
1 2 25
h1
= g() 5 = g ...(i)
2 2
Distance travelled by the
stone in next 5 seconds
(i.e. t = 10 s) is
1 2 100
h1+ h2
= g( 10) = g
...(ii)
2 2
Distance travelled by the stone in next 5 seconds
(i.e. t = 15 s) is
1 2 225
h1+ h2 + h3
= g( 15)
= g
...(iii)
2 2
Subtract (i) from (ii), we get
100 25 75
( h1+ h2)
− h1
= g − g = g
2 2 2
75
h2 = g = 3h1...(iv)
2
Subtract (ii) from (iii), we get
225 100
( h1 + h2 + h3) − ( h2 + h1
) = g − g
2 2
125
h3
= g = 5h 1 ...(v)
2
From (i), (iv) and (v), we get
h2 h3
h1
= =
3 5
26. (b) : Here, u = 0, g = 10 m s –2 , h = 20 m
Let v be the velocity with which the stone hits the ground.
\ v 2 = u 2 + 2gh
or v = 2gh
= 2× 10 × 20 = 20 m/s [ u = 0]
27. (a) : Let the two balls meet after t s at distance x
from the platform.
For the first ball, u = 0, t = 18 s, g = 10 m/s 2
Using h= ut + 1 2
gt
2
1
∴ x = × ×
2 10 182
For the second ball, u = v, t = 12 s, g = 10 m/s 2
...(i)