A) 0.45 W
B) 1.32W
C) 0.66 W
D) 0.22W
Correct Answer: A
Solution :
As we know that, power transmitted \[\propto \] (amplitude)2 \[\Rightarrow \]\[\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{A_{2}^{2}}{A_{1}^{2}}\]\[\Rightarrow \]\[\frac{{{P}_{2}}}{0.20W}=\frac{{{(2\times {{10}^{-3}})}^{2}}}{(3\times {{10}^{-3}})}=\frac{9}{4}\]\[\Rightarrow \] \[\Rightarrow \]\[{{P}_{2}}=\frac{9}{4}\times 0.2=0.45W\]
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