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A point source emits sound waves with an average power out-put of 8.0 w .Then find intensity at 3.0 m & distance at which the sound level is 40 dB

$(a)\;0.707\;w/m^2\;,2.52\times10^{5} \;m\qquad(b)\;0.707\;w/m^2\;,2.52\times10^{3} \;m\qquad(c)\;0.707\;w/m^2\;,2.52\times10^{4} \;m\qquad(d)\;None$

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1 Answer

Answer : (c) $\;0.707\;w/m^2\;,2.52\times10^{4} \;m$
Explanation :
$I=\large\frac{P av}{4 \pi r^2}=\large\frac{80.0w}{4 \pi (3.00m)^2}=0.707w/m^2$
$10 log(\large\frac{I}{I_{0}})=40 dB$
$log I-log I_{0}=\large\frac{40}{10}=4$
$I_{0}=1.00\times 10^{-12} w/m^2$
$log I=4 + log 10^{-12}$
$log I=-8$
$I=1.00 \times 10^{-8} w/m^2$
Using this value for I in equation r = $\; \sqrt{\large\frac{Pav}{4 \pi I}}$
$=\sqrt{\large\frac{80.0 w}{4 \pi \times 1.00 \times 10^{-8} w/m^2}}$
$=2.52 \times 10^{4} m$
answered Mar 10, 2014 by yamini.v

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