The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $10\, cm$ to $8\, cm$ in $40\, seconds$ . Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$ . The time in which amplitude of this pendulum will reduce from $10\, cm$ to $5\, cm$ in carbon dioxide will be close to ..... $s$ $(ln\, 5 = 1.601,ln\, 2 = 0 .693)$
JEE MAIN 2014, Medium
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As we know,
$x=x_{0} e^{-b t/ 2 m}$
From question,
$8=10 \mathrm{e}^{-\frac{40 \mathrm{b}}{2 \mathrm{m}}}$ $...(i)$
Similarly, $5=10 e^{-\frac{b t}{2 m}}$ $...(ii)$
Solving eqns $(i)$ and $(ii)$ we get
$t \cong 142 s$
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