Sound(11th And 12th > Physics ) Questions and Answers
Explanation:-
Answer: Option A. -> 6:
A
Apparent frequency due to train which is coming in is
m1=vv−vsn
Apparent frequency due to train which is leaving is
m2=vv−vsn
So the number of beats is
n1−n2=(1316−1324)320×240⇒n1−n2=6
Explanation:-
Answer: Option B. -> 36%:
B
Intensity after passing through one slab
I′=[I−20100×I]=[I−15]=415
So, intensity after passing through two slabs
I"=[I−20100×I′]=4I′5=16125∴%decrease=[(I=−1615)I]×100=36%
Explanation:-
Answer: Option B. -> 2.5 m/s:
B
Apparent frequency due to source A is
n′=v−uv×n
Apparent frequency due to source B is
n"=v+uv×n
∴n"−n′=2uv×n=10
∴u=10v2n=10×3402×680=2.5m/s
Explanation:-
Answer: Option D. -> 10:
D
When the man is approaching the factory,
n′=(v+v0v)n=(320+2320)800=(322320)800
When the man is going away from the factory,
n"=(v−v0v)n=(320−2320)800=(318320)800∴n′−n"=(320−318320)800=10Hz
Explanation:-
Answer: Option B. -> 6:
B
When the train is approaching,
n1=vv−vs×n=320320−4×243=8079×243
When the train is receding,
n2=vv+vs×n=320324×243=8081×243
Beat frequency is
n=n1−n2=80×243(179−181)=6Hz
Question 6.
A train has just completed a U-curve in a track which is a semi-circle. The engine is at the forward end of the semi-circular part of the track while the last carriage is at the rear end of the semi-circular track. The driver blows a whistle of frequency 200 Hz. Velocity of sound is 340 m/s. Then the apparent frequency as observed by a passenger in the middle of the train, when the speed of the train is 30 m/s, is
Explanation:-
Answer: Option C. -> 200 Hz:
C
The Doppler formula holds for non-collinear motion if vs and v0 are taken to be the resolved component along the line of slight, In this case, we have
v0=−vtsin45∘=−30√2m/s
vs=−vtsin45∘=30√2m/s
We have, v = 340 m/s, n = 200 Hz. The apparent frequency n’ is given by
n′=n[v−v0v−vs]=200[340+(30√2)340+(30√2)]=200Hz
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Question 7.
Two sound sources are moving in opposite directions with velocities v1 and v2(v1>v2). Both are moving away from a stationary observer. The frequency of both the sources is 900 Hz. What is the value of v1–v2 so that the beat frequency observed by the observer is 6 Hz? Speed of sound v = 300 m/s. Given that v1 and v2 ≪ v.
Explanation:-
Answer: Option B. -> 2 m/s:
B
f1=900(300300+v1)≅900(1+r1300)−1900−3v1
Similarly,
f2=900(300300+v2)=900−3v2f2−f1=6∴3(v1−v2)=6∴3(v1−v2)=6or v1−v2=2m/s
Explanation:-
Answer: Option A. -> 711 Hz:
A
As the source and the observer are approaching one another, so n’ would be larger.
f=(v+v15v−v10)600=711Hz
Explanation:-
Answer: Option A. -> 56.2 cm:
A
Apparent frequency is given by:
n′=n(vv−vs)
For a medium, frequency is inversely proportional to wavelength.
λ′=(v−vsv)λ
λ′=(320−20320)60
λ′=56.25 cm
Explanation:-
Answer: Option B. -> 30 m/s:
B
v′=vv−vsv, v"=vv+vsvv′v′′=v+vzv−vs or 65=330+v330−v11vs=330 ⇒ vs=30 m/s