A source of sound of frequency `f_1` is placed on the ground. A detector placed at a height is released from rest on this source. The observed frequency `f(Hz)` is plotted against time `t(sec)` . The speed of sound in air is `300(m)/(s)`. Find `f_1` (`g=10(m)/(s)`).

Options

A. `0.5xx10^(3)Hz`
B. `1xx10^(3)Hz`
C. `0.25xx10^(3)`
D. `0.2xx10^(3)Hz`

Answer

B

Explanation

`f=((V+V_(0))/(V))f_(1)=f_(1)+f_(1)=(V_(0))/(V)`
`V_(0)=g t`
So, `f=f_(1)+((f_(1)g)/(V))t`
Slope of graph `=(f_(1)g)/(V)`
`(2xx10^(-3)-f_(1))/(30)=((f_(1))xx10)/(300), f_(1)=10^(3)Hz`