Question

A firecracker exploding on the surface of a lake is heard as two sounds a time interval $t$ apart by a man on a boat close to the water surface. Sound travels by a speed $v$ in the air. The distance from the exploding firecracker to the boat is:

• A. $\frac{uvt}{u+v}$
• B. $\frac{t(u+v)}{uv}$
• C. $\frac{t(u-v)}{uv}$
• D. $\frac{uvt}{u-v}$

Answer 1

The correct option is D $\frac{uvt}{u-v}$

Sound reaches the man via two media (water and air).
Time taken by sound wave in air medium is $t_{air}=\frac{x}{v_{air}}=\frac{x}{v}$
$x$ is the distance between the source and observer.
Time taken by sound wave in water medium is $t_w=\frac{x}{v_w}=\frac{x}{u}$
Here $u>v$. (Speed of sound in water is greater than speed of sound in air)
Hence $t_{air}-t_w=t\Rightarrow \frac{x}{v}-\frac{x}{u}=t\Rightarrow x(\frac{1}{v}-\frac{1}{u})=t$.
Hence $x=t\frac{uv}{u-v}$