Q197 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
A boy stands $x\ \text{m}$ away from a high wall and claps his hands at a steady rate so that each clap coincides with the echo of the one before. If he makes $60$ claps in $2$ minutes and the velocity of sound is $340\ \text{m/s}$, then the distance $x$ from the wall is
- A$170\ \text{m}$
- Correct answerB$340\ \text{m}$
- C$255\ \text{m}$
- D$510\ \text{m}$
Solution
Time taken for $60$ claps is $2$ minutes $= 120\ \text{s}$.
Time interval between two consecutive claps is $T = \dfrac{120}{60} = 2\ \text{s}$.
Let the distance of the boy from the wall be $x$. The sound travels a distance of $2x$ (to the wall and back) to form an echo.
Time taken for the echo to return is $t = \dfrac{2x}{v}$.
Since each clap coincides with the echo of the previous one, the time for the echo to return equals the time interval between claps, so $t = T$.
$\dfrac{2x}{340} = 2$
$2x = 680$
$x = 340\ \text{m}$
Answer: $340\ \text{m}$
Time interval between two consecutive claps is $T = \dfrac{120}{60} = 2\ \text{s}$.
Let the distance of the boy from the wall be $x$. The sound travels a distance of $2x$ (to the wall and back) to form an echo.
Time taken for the echo to return is $t = \dfrac{2x}{v}$.
Since each clap coincides with the echo of the previous one, the time for the echo to return equals the time interval between claps, so $t = T$.
$\dfrac{2x}{340} = 2$
$2x = 680$
$x = 340\ \text{m}$
Answer: $340\ \text{m}$
Hence, the correct answer is option: 'B'.