An observer moves towards a stationary source of sound with a velocity of one-fifth of the velocity of sound…
- $5 \%$
- $10 \%$
- $20 \%$
- $25 \%$
Solution
Given, $\mathrm{v}_{\mathrm{o}}=\mathrm{v} / 5$. $\therefore \quad \mathrm{n}^{\prime}=\left(\frac{\mathrm{v}+\frac{\mathrm{v}}{5}}{\mathrm{v}}\right) \mathrm{n}=\frac{6}{5} \mathrm{n}$
Increase in apparent frequency $=\mathrm{n}^{\prime}-\mathrm{n}=\frac{6}{5} \mathrm{n}-\mathrm{n}=\frac{1}{5} \mathrm{n} \Rightarrow 20 \%$ of n
Asked in: MHT CET 2024 (03 May Shift 2)