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Q.

From a height of $2 m$ a drop of water of radius $2 \times 10^{- 3} m$ fall and produces a sound. The sound produced can be heard upto a distance of $20 m$ . If the gravitational energy is converted into sound energy is $0.5 s$ , then calculate the intensity at a distance of $20 m$ .

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a

$2 \times 10^{- 7} W / m^{2}$

b

$2.6 \times 10^{- 6} W / m^{2}$

c

$2.6 \times 10^{- 7} W / m^{2}$

d

$3 \times 10^{- 7} W / m^{2}$

answer is C.

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Detailed Solution

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$P . E = m g h; Δ P . E = m g h = (\frac{4}{3} π r^{3}) ρ h g P o w e r = \frac{4}{3} \frac{π r^{3} ρ g h}{t} I \times A = \frac{4}{3} \frac{π r^{3} ρ g h}{t} \Rightarrow I = \frac{4}{3} \frac{π r^{3} ρ g h}{A \times t} I = \frac{4}{3} \frac{π r^{3} ρ g h}{4 \times r^{2} t} = \frac{r ρ g h}{t}$

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