A police car with a siren of frequency $8 \mathrm{kHz}$ is moving with uniform velocity $36 \mathrm{~km} /…
- $8.50 \mathrm{kHz}$
- $8.25 \mathrm{kHz}$
- $7.75 \mathrm{kHz}$
- $7.50 \mathrm{kHz}$
Solution

Apparent frequency of sound heard by car driver (observer) reflected from the building will be $ \begin{aligned} f^{\prime} & =f\left(\frac{v+v_0}{v-v_s}\right) \\ & =8\left(\frac{320+10}{320-10}\right) \\ & =8.5 \mathrm{kHz} \end{aligned} $ $\therefore$ Correct option is (a). Analysis of Question (i) Question is simple. (ii) Driver will listen two sounds, direct and reflected. Direct sound will be of $8 \mathrm{kHz}$ as driver has no relative motion with the car. But reflected sound is of increased frequency because driver and image of car both are approaching towards each other
Asked in: JEE Advanced 2011 (Paper 1)