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← IIT-JEE 2010 · Paper 1
PhysicsInteger+3Question 84 of 84

IIT-JEE 2010 · Paper 1, Physics question 84

A stationary source is emitting sound at a fixed frequency f0f_0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0f_0. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 m s⁻¹.

Answer and worked solution

Official answer: 7

Reflected frequency. A car approaching at vv acts as a moving observer and then a moving source:
f=f0c+vc−v≈f0(1+2vc)(v≪c).f=f_0\frac{c+v}{c-v}\approx f_0\left(1+\frac{2v}c\right)\quad(v\ll c).

Difference. Δff0=2Δvc=0.012\dfrac{\Delta f}{f_0}=\dfrac{2\Delta v}c=0.012, so
Δv=0.006×330=1.98 m s−1=7.13 km h−1.\Delta v=0.006\times330=1.98\ \mathrm{m\,s^{-1}}=7.13\ \mathrm{km\,h^{-1}}.

Answer: 7