Question 128 of 194

A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with a time period ${T}_{1}$ and (ii) back and forth in a direction perpendicular to its plane, with a period ${T}_{2}$. The ratio $\dfrac{{T}_{1}}{{T}_{2}}$ will be :

A$\dfrac{2}{\sqrt{3}}$B$\dfrac{2}{3}$C$\dfrac{3}{\sqrt{2}}$D$\dfrac{\sqrt{2}}{3}$

Show answer and solution▾

Answer: (A) $\dfrac{2}{\sqrt{3}}$

The time period of oscillation of physical pendulum is given by
$T=2\pi \sqrt{\dfrac{I}{Mgl}}...(1)$
Where, $I$ is the moment of inertia of ring about given axis and $l$ is the distance of centre of mass of ring from the axis of rotation.
Moment of inertia of ring in its plane is given by
${I}_{1}={I}_{c}+M{d}^{2}=M{R}^{2}+M{R}^{2}=2M{R}^{2}$
Moment of inertia of ring perpendicular to its plane is given by
${I}_{2}={I}_{c}+M{d}^{2}=\dfrac{M{R}^{2}}{2}+M{R}^{2}=\dfrac{3}{2}M{R}^{2}$
Now the ratio of time period is given by
$\dfrac{{T}_{1}}{{T}_{2}}=\sqrt{\dfrac{{I}_{1}}{{I}_{2}}}$
$\Rightarrow \dfrac{{T}_{1}}{{T}_{2}}=\sqrt{\dfrac{2M{R}^{2}}{\dfrac{3}{2}M{R}^{2}}}$
$\Rightarrow \dfrac{{T}_{1}}{{T}_{2}}=\sqrt{\dfrac{4}{3}}=\dfrac{2}{\sqrt{3}}$