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. `0.5xx10^3Hz`
B. `2xx10^3Hz`
C. `0.25xx10^3Hz`
D. `0.2xx10^3Hz`

Explanation

`f=((V+V_0)/(V))f_1=f_1+f_1(V_0)/(V)`
`V_0="gt"`
So, `f+f_1+((f_1g)/(V))t`
Slope of graph`=(f_1g)/(V)`
`((2)(10^-3)-f_1)/(30)=((f_1)(10))/(300)`
or `f_1=10^3Hz`