33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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114 NEET-AIPMT Chapterwise Topicwise Solutions Physics
∴ t0 = 2π
m/ keff
= 2π
m/( k1+
k2
) ...(iii)
1 1 k1
From (i),
t 2 =
1 4
2 ×
...(iv)
π m
From (ii),
1 1 k2
t 2 =
2 4
2 ×
π m
1 1 k1+
k2
From (iii),
t 2 =
0 4
2 ×
π m
From eqns (iv), (v) and (vi)
1 1 1
= + =
t1 2 t2 2 t0 2 ; − − −
∴ t0 2 = t1 2 + t2 2
50. (a) : f A = 2f B
1 g 1 g 1 1
⇒ = 2 × or, = 4 ×
2π
l 2π
l l l
A B A B
lB
or, l = , which does not depend on mass.
4
A
51. (a) : In DOAC, cosq = OA/l
or, OA = l cosq
\ AB = l (1 – cosq) = h
At point, C, the velocity of bob = 0.
The vertical acceleration = g
\ v 2 = 2gh
or, v= 2gl( 1−cos θ)
...(v)
...(vi)
52. (d) : Time period of a simple pendulum is given by
l
T = 2π ⇒ T ∝
g
T1
l1
1 1
∴ = = = or, T2 = 2T1
= 4sec.
T2
l2
4 2
53. (a) : Frequency of the pendulum υ
υl= 20 =
1 g
2π
20
l
l
= 5 =
1 g
2π
5 ;
υl=
5 20
∴ = = 2⇒ υl= 5 = 2υl=
20
υl=
20 5
As shorter length pendulum has frequency double the
larger length pendulum. Therefore shorter pendulum
should complete 2 oscillations before they will be again
in phase.
k
54. (a) : n 1
m n′=
1
;
2π
2π
\ n′ = n/2
4
km
55. (a) : l 2 = 1.02l 1 .
Time period (T ) = 2π ×
l
g
∝ l
Therefore T 2 l2
102 . l1
= = = 101 .
T1
l1
l1
Thus time period increased by 1%.
56. (b) : Potential energy at A or C = Kinetic energy at B.
1 2
Thus, mvB
= mgH or vB
= 2gH
2
57. (d) : Mass (m) = 5 kg and
time period (T) = 2p sec.
m
Thereforetimeperiod T = × k
⇒ 5
2π
k
= 1
or k = 5 N/m.
According to Hooke’s Law, F = –kl.
Therefore decrease in length (l)
F 5g
=− =− =−g
metres
k 5
l
58. (d) : Period of oscillation T = 2π . Therefore T
g
will decrease when acceleration (g) increases. And g
will increase when the rocket moves up with a uniform
acceleration.
59. (d) : The effective value of acceleration due to gravity
is 2 2
( a + g ) .
60. (d) : The effective spring constant of two springs in
kk
series is
1 2
k =
k1 + k2
m mk ( 1 + k2)
Time period, T = 2π
= 2π
k kk 1 2
61. (d) : This is a case of damped oscillation as the
amplitude of oscillation is decreasing with time.
Amplitude of oscillations at any instant t is given by
a = a 0 e –bt , where a 0 is the initial amplitude of oscillations
and b is the damping constant.
Now, when t = 100T, a = a 0 /3 [T is time period]
Let the amplitude be a′ at t = 200T i.e. after completing
200 oscillations.
\ a = a 0 /3 = a 0 e –100Tb ...(i)
and a′ = a 0 e –200Tb
...(ii)
1 −100Tb
−200Tb
1
From (i), = e ∴ e =
3
9
1 a0
From (ii), a′ = a0
× =
9 9
\ The amplitude will be reduced to 1/9 of initial value.
62. (a) : Smaller damping gives a taller and narrower
resonance peak.
63. (b) : The amplitude and velocity resonance occurs at
the same frequency.
At resonance, i.e., w 1 = w 0 and w 2 = w 0 , the amplitude and
energy of the particle would be maximum.
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