Question

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 $5 \,cm$ carbon dioxide will be close to ($\ell n\, 5 \,= \,1.601, \ell n\, 2\, =\, 0.693)$.

• A. 231 s
• B. 208 s
• C. 161 s
• D. 142 s

Answer 1

Answer: (C)

$8=10e^{-\lambda\times40}$
$5=10e^{\frac{-\lambda t}{1.3}}$
In $\frac{4}{5}=-\lambda\times40$
$2\times0.693-1.601=-\lambda\times40$
$\lambda=0.005375$
In $\frac{1}{2}=-\frac{\lambda t}{1.3}$
$-0.693=-\frac{0.005375}{1.3}t$
$t=167.6$