A boy is located asymmetrically between two parallel walls, closer to the left wall than the right. If he claps, the time interval between successive echoes is…

Given,

• Distance of the boy from the right wall = 68 m
• Distance of the boy from the left wall = X
• Time interval between the two echoes = 0.25 s
• Speed of sound in air (v) = 340 m s^-1

Since the boy is nearer to the left wall, the echo from the left wall is heard first and that from the right wall is heard later.

Time taken by the echo from the right wall,

$\text t_2 = \frac{2 \times 68}{340} = \frac{136}{340} = 0.4 \text { s}$

Time taken by the echo from the left wall,

t₁ = t₂ − 0.25 = 0.4 − 0.25 = 0.15 s

Now,

$$\begin{aligned}\text t_1 = \frac{2\text X}{\text v} \\[1em] \Rightarrow 0.15 = \frac{2\text X}{340} \\[1em] \Rightarrow 2\text X = 0.15 \times 340 \\[1em] \Rightarrow 2\text X =51 \\[1em] \Rightarrow \text X = 25.5 \text { m}\end{aligned}$$

Hence, the value of X is 25.5 m.