Practice Start Test

Waves

NEET Physics · 100 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is 5.0 J K −-1 mol −-1 . If the speed of sound in this gas at NTP is 952 m s −-1 , then the heat capacity at constant pressure is (Take gas constant R == 8.3 J K −-1 mol −-1 )
A 7.0 J K −-1 mol −-1
B 8.5 J K −-1 mol −-1
C 8.0 J K −-1 mol −-1
D 7.5 J K −-1 mol −-1
Correct Answer
Option C
Solution

Molar mass of the gas = 4g/mol Speed of sound

V=γRTm⇒952=γ×3.3×2734×10−3V = \sqrt {{{\gamma RT} \over m}} \Rightarrow 952 = \sqrt {{{\gamma \times 3.3 \times 273} \over {4 \times {{10}^{ - 3}}}}}
⇒γ=1.6=1610=85\Rightarrow \gamma = 1.6 = {{16} \over {10}} = {8 \over 5}

Also,

γ=CPCV=85\gamma = {{{C_P}} \over {{C_V}}} = {8 \over 5}
⇒CP=8×55=8 JK−1mol−1\Rightarrow {C_P} = {{8 \times 5} \over 5} = 8\,J{K^{ - 1}}mo{l^{ - 1}}

[C V = 5.0 JK –1 given]

Q22
The number of possible natural oscillations of air column in a pipe closed at one end length 85 cm whose frequencies lie below 1250 Hz are (Velocity of sound = 340 m s −-1 )
A 4
B 5
C 7
D 6
Correct Answer
Option D
Solution

Fundamental frequency of the closed organ pipe is

υ=v4L\upsilon = {v \over {4L}}

Here, v = 340 m s –1 , L = 85 cm = 0.85 m ∴\therefore

υ=340 ms−14×0.85 m=100 Hz\upsilon = {{340\,m{s^{ - 1}}} \over {4 \times 0.85\,m}} = 100\,Hz

The natural frequencies of the closed organ pipe will be

υn=(2n−1)υ =υ,3υ,5υ,7υ,9υ,11υ....{\upsilon _n} = \left( {2n - 1} \right)\upsilon \, = \upsilon ,3\upsilon ,5\upsilon ,7\upsilon ,9\upsilon ,11\upsilon ....

= 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz, 1300 Hz,... and so on Thus, the natural frequencies lies below the 1250 Hz is 6.

Q23
A speeding motorcyclist sees traffic jam ahead him. He slows down to 36 km hour −-1 . He finds that traffic has eased and a car moving ahead of him at 18 km hour −-1 is honking at a frequency of 1392 Hz. If the speed of sound is 343 m s −-1 , the frequency of the honk as heard by him will be
A 1332 Hz
B 1372 Hz
C 1412 Hz
D 1454 Hz
Correct Answer
Option C
Solution

Here, speed of motorcyclist, v m = 36 km hour –1

=36×518=10ms−1= 36 \times {5 \over {18}} = 10m{s^{ - 1}}

Speed of car,

vc=18 km hour−1=18×518ms−1=5ms−1{v_c} = 18\,km\,hou{r^{ - 1}} = 18 \times {5 \over {18}}m{s^{ - 1}} = 5m{s^{ - 1}}

Frequency of source,

υ0{\upsilon _0}

= 1392 Hz Speed of sound, v = 343 m s –1 The frequency of the honk heard by the motorcyclist is

v′=υ0(v+vmv+vC)=1392(343+10343+5)v' = {\upsilon _0}\left( {{{v + {v_m}} \over {v + {v_C}}}} \right) = 1392\left( {{{343 + 10} \over {343 + 5}}} \right)
=1392×353348=1412 Hz= {{1392 \times 353} \over {348}} = 1412\,Hz
Q24
If n 1 , n 2 and n 3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by
A 1n=1n1+1n2+1n3{1 \over n} = {1 \over {{n_1}}} + {1 \over {{n_2}}} + {1 \over {{n_3}}}
B 1n=1n1+1n2+1n3{1 \over {\sqrt n }} = {1 \over {\sqrt {{n_1}} }} + {1 \over {\sqrt {{n_2}} }} + {1 \over {\sqrt {{n_3}} }}
C n=n1+n2+n3\sqrt n = \sqrt {{n_1}} + \sqrt {{n_2}} + \sqrt {{n_3}}
D n == n 1 + n 2 + n 3
Correct Answer
Option A
Solution
n=12lTmn = {1 \over {2l}}\sqrt {{T \over m}}
⇒n∝1l⇒nl=constant, K\Rightarrow n \propto {1 \over l} \Rightarrow nl = constant,\,K

∴\therefore

n1l1=K,n2l2=K,n3l3=K{n_1}{l_1} = K,{n_2}{l_2} = K,{n_3}{l_3} = K

Also,

l=l1+l2+l3l = {l_1} + {l_2} + {l_3}
⇒Kn=Kn1+Kn2+Kn3\Rightarrow {K \over n} = {K \over {{n_1}}} + {K \over {{n_2}}} + {K \over {{n_3}}}
⇒1n=1n1+1n2+1n3\Rightarrow {1 \over n} = {1 \over {{n_1}}} + {1 \over {{n_2}}} + {1 \over {{n_3}}}
Q25
The length of the wire between two ends of a sonometer is 100 cm. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio 1 : 3 : 5.
A 150023cm,50023cm{{1500} \over {23}}cm,{{500} \over {23}}cm
B 150023cm,{{1500} \over {23}}cm, 30023cm{{300} \over {23}}cm
C 30023cm,150023cm{{300} \over {23}}cm,{{1500} \over {23}}cm
D 150023cm,200023cm{{1500} \over {23}}cm,{{2000} \over {23}}cm
Correct Answer
Option D
Solution

From formula,

f=1xTmf = {1 \over x}\sqrt {{T \over m}}
⇒1f∝l\Rightarrow {1 \over f} \propto l
∴l1:l2:l3=1f1:1f2:1f3\therefore {l_1}:{l_2}:{l_3} = {1 \over {{f_1}}}:{1 \over {{f_2}}}:{1 \over {{f_3}}}

= f 2 f 3 : f 1 f 3 : f 1 f 2 [Given: f 1 : f 2 : f 3 = 1 : 3 : 5] = 15 : 5 : 3 Therefore the positions of two bridges below the wire are

15×10015+5+3cm{{15 \times 100} \over {15 + 5 + 3}}cm

and

15×100+5×10015+5+3cm{{15 \times 100 + 5 \times 100} \over {15 + 5 + 3}}cm
⇒150023cm,  200023cm\Rightarrow {{1500} \over {23}}cm,\,\,{{2000} \over {23}}cm
Q26
Two sources P and Q produce notes of frequency 660 Hz. each. A listener moves from P to Q with a speed of 1 ms −-1 . If the speed of sound is 330 m/s, then the number of beats heard by the listener per second will be
A 4
B 8
C 2
D zero
Correct Answer
Option A
Solution
Δff=vC{{\Delta f} \over f} = {v \over C}
⇒(Beats)/2f=vC\Rightarrow {{\left( {Beats} \right)/2} \over f} = {v \over C}
⇒Beats=2fvC=4\Rightarrow Beats = {{2fv} \over C} = 4
Q27
A source of unknown frequency gives 4 beats/s when sounded with a source of known frquency 250 Hz. The second harmonic of the source of unknown frequency gives five beats per second, when sounded with a source of frequency 513 Hz. The unknown frequency is
A 240 Hz
B 260 Hz
C 254 Hz
D 246 Hz
Correct Answer
Option C
Solution

Let

υ\upsilon

be frequency of the unknown source. As it gives 4 beats per second when sounded with a source of frequency 250 Hz,

∴υ=250±4=246 Hz or 254 Hz\therefore \upsilon = 250 \pm 4 = 246\,Hz\,or\,254\,Hz

Second harmonic of this unknown source = 492 Hz or 508 Hz which gives 5 beats per second, when sounded with a source of frequency 513 Hz.

Therefore unknown frequency,

υ\upsilon

= 254 Hz.

Q28
If we study the vibration of a pipe open at both ends. then the following statement is not true.
A All harmonics of the fundamental frequency will be generated.
B Pressure change will be maximum at both ends.
C Open end will be antinode.
D Odd harmonics of the fundamental frequency will be generated.
Correct Answer
Option B
Solution

Pressure change will be minimum at both ends.

Q29
A wave travelling in the + ve x-direction having displacement along y-direction as 1 m, wavelength 2π\pi m and frequency of 1π{1 \over \pi } Hz is represented by
A y = sin(10π\pi x −- 20π\pi t)
B y = sin(2π\pi x + 2π\pi t)
C y = sin(x −- 2t)
D y == sin(2π\pi x −- 2π\pi t)
Correct Answer
Option C
Solution

The standard equation of a wave travelling along +ve x-direction is given by

y=Asin(kx−ωt){\rm{y = Asin(kx - }}\omega {\rm{t)}}

where A = Amplitude of the wave k = angular wave number ω\omega = angular frequency of the wave Given: A = 1 m,

λ=2πm,υ=1πHz\lambda = {\rm{2}}\pi m,\upsilon = {1 \over \pi }Hz

As

k=2πλ=2π2π=1k = {{2\pi } \over \lambda } = {{2\pi } \over {2\pi }} = 1
ω=2πυ=2π×1π=2\omega = 2\pi \upsilon = 2\pi \times {1 \over \pi } = 2

∴\therefore The equation of the given wave is y = 1 sin (1x – 2t) = sin(x – 2t)

Q30
A train moving at a speed of 220 m s −-1 towards a stationary object, emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is (Speed of sound in air is 330 m s −-1 )
A 3500 Hz
B 4000 Hz
C 5000 Hz
D 3000 Hz
Correct Answer
Option C
Solution

Here, Speed of the train, v T = 220 ms –1 Speed of sound in air, v = 330 ms –1 The frequency of the echo detected by the driver of the train is

υ′=υ(v+vTv−vT)=1000(330+220330−220)\upsilon ' = \upsilon \left( {{{v + {v_T}} \over {v - {v_T}}}} \right) = 1000\left( {{{330 + 220} \over {330 - 220}}} \right)
=1000×550110=5000 Hz= 1000 \times {{550} \over {110}} = 5000\,Hz
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