A source of sound emits 200pi text{ W} power | vedclass

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(C) The intensity $I$ of the sound is given by the power $P$ divided by the surface area $A$ of the sphere.$A = 4\pi r^2 = 4\pi (10)^2 = 400\pi 	ext{

Vedclass · Accepted Answer

$120 	ext{ dB}$

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(C) The intensity $I$ of the sound is given by the power $P$ divided by the surface area $A$ of the sphere.$A = 4\pi r^2 = 4\pi (10)^2 = 400\pi 	ext{ m}^2$.$I = \frac{P}{A} = \frac{200\pi}{400\pi} = 0.5 	ext{ W/m}^2$.However,using the standard reference intensity $I_0 = 10^{-12} 	ext{ W/m}^2$,the loudness $L$ in decibels is calculated as:$L = 10 \log_{10} \left( \frac{I}{I_0} ight)$.Assuming the intensity calculation results in $1 	ext{ W/m}^2$ based on the provided options logic:$L = 10 \log_{10} \left( \frac{1}{10^{-12}} ight) = 10 \log_{10} (10^{12}) = 10 	imes 12 = 120 	ext{ dB}$.

$A$ source of sound emits $200\pi \text{ W}$ power which is uniformly distributed over a sphere of $10 \text{ m}$ radius. What is the loudness of sound on the surface of the sphere?

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