The amplitude of a damped oscillator decreases to $0.9\,times$ its original magnitude in $5\,s.$ In another $10\,s$ it will decrease to $\alpha $ times its original magnitude, where $\alpha $ equals
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$A=A_{0} e^{-h t}$
$0.9 A_{0}=A_{0} e^{-k t}$
$-k t=\ln (0.9) \Rightarrow-15 k=3 \ln (0.9)$
$A=A_{0} e^{-15 k}=A_{0} e^{-\ln (0.9)^{3}}$
$=(0.9)^{3} A_{0}=0.729 A_{0}$
Hence,
option $(C)$ is correct answer.
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