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### IIT-JEE 2010 Paper 1 Offline

Numerical

A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. 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 ms$$-$$1.

## Explanation

Let car B be the observer (moving towards S).

The frequency observed is

$${f_1} = {f_0}\left( {{{c + v} \over c}} \right)$$

When sound gets reflected, the frequency observed by source S is

$${f_2} = {f_1}\left( {{c \over {c - v}}} \right)$$

where v is the speed of car and c is the speed of sound. Therefore,

$${f_2} = {f_0}\left( {{{c + v} \over {c - v}}} \right)$$

Now, $$d{f_x} = {f_0}\left[ {{{(c - v)dv - (c + v)( - dv)} \over {{{(c - v)}^2}}}} \right]$$

$$= {{2{f_0}c\,dv} \over {{{(c - v)}^2}}}$$

That is,

$${{2{f_0}c\,dv} \over {{{(c - v)}^2}}} = \left( {{{1.2} \over {100}}} \right){f_0}$$

$$\Rightarrow dv = {{1.2} \over {100}} \times {{{{(c - v)}^2}} \over {2c}}$$

Since, v << c, we get c $$-$$ v $$\simeq$$ c.

Therefore,

$$dv = {{1.2} \over {100}} \times {c \over 2} = 1.98$$ m/s

$$= 1.98 \times {{18} \over 5}$$ km/h = 7 km/h.

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