A person is standing at a distance of $d m$ from a wall and clapping at a rate of $4$ claps per second. The echo of his clap coincides with his clap and he hears the echo of two claps after he stops clapping. What is $d$ ? (Assume velocity of sound to be $340 m / s$ )

75 m

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85 m

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170 m

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150 m

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Solution

The correct option is B
85 m

The person is clapping at a rate of $4$ claps per second. So time taken for one clap $t$ = $\frac{1}{4}$ = $0.25 s$
Now, since he hears two claps after stopping, the echo of first would have coincided with the third clap i.e. time taken by echo to travel to the man = $2 \times t$ = $0.5 s$ .
So, distance travelled = $340 \times 0.5$ = $170 m$ .
But echo travels twice the distance between man and the wall. So, the distance between the man and the wall = $\frac{170}{2}$ = $85 m$ .

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A person is standing at a distance of d m from a wall and clapping at a rate of 4 claps per second. The echo of his clap coincides with his clap and he hears the echo of two claps after he stops clapping. What is d? (Assume velocity of sound to be 340 m/s)

A person is standing at a distance of $d m$ from a wall and clapping at a rate of $4$ claps per second. The echo of his clap coincides with his clap and he hears the echo of two claps after he stops clapping. What is $d$ ? (Assume velocity of sound to be $340 m / s$ )