JEE Main 2014 · 9 Apr · Shift 1 · Physics Q9

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PhysicsSingle correct+4 / -1Question 9 of 90

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 carbondioxide will be close to ( $\ln5=1.601$ , $\ln2=0.693$ ).

231 s

208 s

161 s

142 s

Answer and worked solution

Official answer: C

With Stokes drag, the damping constant is proportional to $\eta$ , and $A=A_0e^{-\gamma t}$ with $\gamma\propto\eta$ .

In air: $\gamma_{\text{air}}=\dfrac{\ln(10/8)}{40}=\dfrac{0.223}{40}$ .
In $\ce{CO2}$ : $\gamma=\gamma_{\text{air}}/1.3$ .

Time to halve:
$t=\frac{\ln2}{\gamma}=\frac{0.693\times1.3\times40}{0.223}\approx161\text{ s}.$

Topics:Damped and forced oscillations Viscosity and Stokes' law

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JEE Main 2014 · 9 Apr · Shift 1, Physics question 9