A man dropped a stone into a lake from a tower $500 m$ high. The sound of the splash will be heard by the man approximately after (take $g = 10 m / s^{2}$ and speed of sound in air to be $340 m / s$ )

11.5 s

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21.2 s

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10.5 s

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14.3 s

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Solution

The correct option is A $11.5 s$
Time taken by the stone to reach the surface of lake from the top of tower = $t_{1} = \sqrt{(\frac{2 h}{h}}) = \sqrt{(\frac{2 \times 500}{10})} = 10 s$ (Using $h = u t + \frac{1}{2} g t^{2}$ )
Now time taken by sound from lake to the man =
$t_{2} = \frac{h}{v} = \frac{500}{340} \approx 1.5 s$ (where $v$ is the speed of sound in air)
$\Rightarrow Total Time = t_{1} + t_{2} = 10 + 1.5 = 11.5 s$

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A man dropped a stone into a lake from a tower 500 m high. The sound of the splash will be heard by the man approximately after (take g=10 m/s2and speed of sound in air to be 340 m/s )

The correct option is A 11.5 sTime taken by the stone to reach the surface of lake from the top of tower =t1=√(2hh)=√(2×50010)=10 s(Using h=ut+12gt2) Now time taken by sound from lake to the man = t2=hv=500340≈1.5 s (where v is the speed of sound in air) ⇒Total Time=t1+t2=10+1.5=11.5 s

A man dropped a stone into a lake from a tower $500 m$ high. The sound of the splash will be heard by the man approximately after (take $g = 10 m / s^{2}$ and speed of sound in air to be $340 m / s$ )