JEE Advanced 2024 — Paper 1 — PHY-III solution

JEE Advanced2024Paper 1PHY-III

A source ( $S$ ) of sound has frequency $240$ Hz. When the observer ( $O$ ) and the source move towards each other at a speed $v$ with respect to the ground (as shown in Case 1 in the figure), the observer measures the frequency of the sound to be $288$ Hz. However, when the observer and the source move away from each other at the same speed $v$ with respect to the ground (as shown in Case 2 in the figure), the observer measures the frequency of sound to be $n$ Hz. The value of $n$ is _______.

Your answer

Solution

Let $c$ be the speed of sound. Doppler formula:

Case 1 (approach): $288 = 240 \cdot \frac{c + v}{c - v}$ .

Case 2 (separation): $f_{2} = 240 \cdot \frac{c - v}{c + v}$ .

Multiplying the two:

$288 \cdot f_{2} = 240 \cdot 240 = 57600$

$\Rightarrow f_{2} = \frac{57600}{288} = 200$ Hz.

$∴ n = 200$ .

← Back to full paper

Practice more PHY-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →