33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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122 NEET-AIPMT Chapterwise Topicwise Solutions Physics
The sign is negative inside the bracket. Therefore this
wave travels in the positive x-direction.
7. (a) : y 1 = 10 –6 sin[100t + (x/50) + 0.5]
y 2 = 10 –6 cos[100t + (x/50)]
= 10 –6 sin[100t + (x/50) + p/2]
= 10 –6 sin[100t + (x/50) + 1.57]
The phase difference = 1.57 – 0.5 = 1.07 radians
8. (c) : The equation of progressive wave travelling in
positive x-direction is given by
2π
y= asin ( vt −x)
λ
Here a = 0.2 m, v = 360 m/s, l = 60 m
∴
2π
⎡ ⎛ x ⎞ ⎤
y= 02 . sin ( t− x = ⎢ −
⎝
⎜ t
⎠
⎟
60 360 ) 02 . sin
⎥
⎣
π 6 60 ⎦
9. (c) : Displacement, y max = a, y min = 0
Time taken = T/4
\ T/4 = 0.170 \ T = 0.68
The frequency of wave = 1/T = 1.47 Hz
10. (c) : Sound wave equation is
y = 0.0015 sin (62.4 x + 316 t)
Comparing it with the general equation of motion
⎡ x t ⎤ 2π
y= Asin 2π ⎢ +
⎣ T
⎥ , we get = 62.
4
λ ⎦ λ
2π
or λ = = 01 . unit
62.
4
π
11. (a) : Phasedifferenceθ
= 60°=
rad
Phasedifference
() θ
3
2π
= × Pathdifference
λ
π λ λ
ThereforePathdifference = × =
3 2π
6 .
12. (a, b) : Option (a) represents a harmonic progressive
wave in the standard form whereas (b) also represents a
harmonic progressive wave, both travelling in the positive
x- direction. In (b), a is the angular velocity, (w) and b is
k; c is the initial phase. (d) represents only S.H.M.
13. (d) : Compare with the equation,
y = acos(2put + f)
This give 2pu = 2000; υ = 1000 Hz
π
14. (b) : The standard equation of a progressive wave is
⎡ ⎛ t x ⎞
y= a ⎢ −
⎝
⎜
T ⎠
⎟
⎣
+ ⎤
sin 2π φ
λ
⎥
⎦
The given equation can be written as
⎡ ⎛ t x ⎞
y = ⎢ −
⎝
⎜
⎠
⎟
⎣
+ π⎤
4sin
2π
10 18 6
⎥
⎦
\ a = 4 cm, T = 10 s, l = 18 cm and f = p/6.
15. (d) : Wavelength of pulse at the lower end,
λ1 ∝ velocity( v1)
=
Similarly, λ2 ∝ v2
=
λ2
T2
∴ = =
λ1
T1
T1
µ
T2
µ
( m1+
m2)
g
=
mg 2
m1+
m2
m2
16. (c) : Since 4.0 g of a gas occupies 22.4 litres at NTP,
so the molecular mass of the gas is M = 4.0 g mol –1
RT
As the speed of the sound in the gas is v = γ
M
where g is the ratio of two specific heats, R is the universal
gas constant and T is the temperature of the gas.
2
Mv
∴ γ =
RT
Here, M = 4.0 g mol –1 = 4.0 × 10 –3 kg mol –1 , v = 952 m s –1 ,
R = 8.3 J K –1 mol –1 and
T = 273 K (at NTP)
−3 −1 −12
(. 40×
10 kg mol )( 952 ms )
∴ γ =
= 16 .
−1 −1
(. 83JK mol )( 273 K)
Cp
By definition, γ= or Cp
= γCv
Cv
But g = 1.6 and C v = 5.0 J K –1 mol –1
\ C p = (1.6)(5.0 J K –1 mol –1 ) = 8.0 J K –1 mol –1
17. (b) : The given wave equation is
y= 3 sin π ( t−x)
2 50
⎛ π ⎞
y= 3sin
t−
x
⎝
⎜25π
2 ⎠
⎟
...(i)
The standard wave equation is
y = Asin(wt – kx)
...(ii)
Comparing (i) and (ii), we get
π
ω= 25π, k =
2
ω 25π
−
Wave velocity, v = = = 50 ms 1
k ( π / 2)
dy d
Particle velocity, vp = sin t x
dt
= ⎛ ⎛
dt ⎝
⎜ − π ⎞⎞
⎠
⎟
⎝
⎜3 25π
2 ⎠
⎟
⎛ π ⎞
= 75πcos −
⎝
⎜25πt x
2 ⎠
⎟
Maximum particle velocity, (v p ) max = 75p m s –1
( vp)
max 75π
3
∴ = = π
v 50 2
18. (c) : Here, v air = 350 m/s, v brass = 3500 m/s
When a sound wave travels from one medium to another
medium its frequency remains the same