A sound produced on the surface of a lake takes 4.5 s to reach a boat man. How much time in seconds will it take to reach a diver inside water at the same distance if speed of sound in water is 4:5 times the speed of sound in air?

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Solution

Given, $V_{w a t e r} : V_{a i r} = 4.5, t_{a i r} = 4.5 s, t_{w a t e r} = ?$

$f r o m r e l a t i o n V_{a i r} = \frac{d}{t_{a i r}} a n d V_{w a t e r} = \frac{d}{t_{w a t e r}}$

$o r V_{a i r} \times t_{a i r} {= V}_{w a t e r} \times t_{w a t e r}$

$t_{w a t e r} = \frac{V_{a i r}}{V_{w a t e r}} \times t_{a i r} = \frac{1}{4.5} \times 4.5 s = 1 s$

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A sound produced on the surface of a lake takes 4.5 s to reach a boat man. How much time in seconds will it take to reach a diver inside water at the same distance if speed of sound in water is 4:5 times the speed of sound in air?

Given, Vwater:Vair=4.5,tair=4.5s,twater=?fromrelationVair=dtair andVwater=dtwaterorVair×tair=Vwater×twatertwater=VairVwater×tair=14.5×4.5s=1s