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Question

A stone is dropped from a tower of height H metre on the upper surface of a well having water at depth d metre. If g is the acceleration due to gravity, then the time when the sound of splashing will be heard from the time of dropping the stone is [Take speed of sound in air = 330 m/s]

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Solution

The correct option is **C** ⎡⎣√2(H+d)g+(H+d330)⎤⎦ s

Let t1 be the time taken by the stone to cover the distance d+H and t2 be the time taken by the sound to come back to the dropping point by covering the distance d+H.

∴ Time when the splashed sound is heard, T=t1+t2

For downward motion of stone:

Initial velocity, u = 0 m/s

Acceleration, a = g

Distance covered, s = d + H

Using second equation of motion, s=ut+12at2⇒d+H=0×t+12g(t1)2⇒t1=√2(d+H)g....(i)

Sound travels a distance (s=d+H) to reach the dropping point of stone. Speed of sound, v=330 m/s

Time taken by sound,t2=sv=d+H330...(ii)⇒T=t1+t2=[√2(H+d)g+(H+d330)]s

Hence, the correct answer is option (c).

Let t1 be the time taken by the stone to cover the distance d+H and t2 be the time taken by the sound to come back to the dropping point by covering the distance d+H.

∴ Time when the splashed sound is heard, T=t1+t2

For downward motion of stone:

Initial velocity, u = 0 m/s

Acceleration, a = g

Distance covered, s = d + H

Using second equation of motion, s=ut+12at2⇒d+H=0×t+12g(t1)2⇒t1=√2(d+H)g....(i)

Sound travels a distance (s=d+H) to reach the dropping point of stone. Speed of sound, v=330 m/s

Time taken by sound,t2=sv=d+H330...(ii)⇒T=t1+t2=[√2(H+d)g+(H+d330)]s

Hence, the correct answer is option (c).

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