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If the intensity of sound is 5 SI units, the amount of sound energy transmitted through a cross section of ${5 cm}^{2}$ in one second is

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25 \times 10^{- 1} W

0.25 W

25 W

25 \times 10^{- 4} W

answer is A.

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Detailed Solution

I = 5

$A = 5 {cm}^{2}$

$A = 5 \times {(10^{- 2} m)}^{2}$

$P = I \times A$

$= 5 \times 10^{- 4} \times 5$

$p = 25 \times 10^{- 4} w$

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