Question

A ring is hung on a nail. It can oscillate, without slipping or sliding

$\left(i\right)$ In its plane with a time period $T_1$ and,

$\left(ii\right)$ Back and forth in a direction perpendicular to its plane, with a period$T_2$.

The ratio $\frac{T_1}{T_2}$ will be:

• A. $\frac{3}{\sqrt{2}}$
• B. $\frac{\sqrt{2}}{3}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{2}{3}$

Answer 1

The correct option is C

$\frac{2}{\sqrt{3}}$

Step 1. Given Data,

A ring is hung on a nail oscillate, without slipping or sliding

$\left(i\right)$ In its plane with a time period $T_1$ and,

$\left(ii\right)$ Back and forth in a direction perpendicular to its plane, with a period$T_2$.

Let the radius of ring is $R$ and mass is $M$.

Step 2. We have to find the ratio of $\frac{T_1}{T_2}$,

We know that,

$T=2\mathrm{\pi }\sqrt{\frac{I}{Mgd}}$ (where, $T$ is time period, $I$ is moment of Inertia, $M$ is mass,

$g$ is gravity and $d$ is distance .)

Now, in its plane with a time period $T_1$,

$$\begin{gathered} T_1=2\mathrm{\pi }\sqrt{\frac{I_1}{Mgd}} \\ \Rightarrow {\mathrm{T}}_1=2\mathrm{\pi }\sqrt{\frac{2{\mathrm{MR}}^2}{\mathrm{Mgd}}} \end{gathered}$$

Now, back and forth in a direction perpendicular to its plane, with a period$T_2$

$$\begin{gathered} T_2=2\mathrm{\pi }\sqrt{\frac{I_2}{Mgd}} \\ \Rightarrow {\mathrm{T}}_2=2\mathrm{\pi }\sqrt{\frac{\left({\displaystyle \frac{M{R}^2}{2}}+{\mathrm{MR}}^2\right)}{\mathrm{Mgd}}} \\ \Rightarrow {\mathrm{T}}_2=2\mathrm{\pi }\sqrt{\frac{\left({\displaystyle \frac{3{\mathrm{MR}}^2}{2}}\right)}{\mathrm{Mgd}}} \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{T_1}{T_2}=\frac{2\mathrm{\pi }\sqrt{\frac{2{\mathrm{MR}}^2}{\mathrm{Mgd}}}}{2\mathrm{\pi }\sqrt{\frac{\left({\displaystyle \frac{3{\mathrm{MR}}^2}{2}}\right)}{\mathrm{Mgd}}}} \\ \Rightarrow \frac{T_1}{T_2}=\frac{\sqrt{2{\mathrm{MR}}^2}}{\sqrt{\left(\frac{3{\mathrm{MR}}^2}{2}\right)}} \\ \Rightarrow \frac{T_1}{T_2}=\frac{\sqrt{4{\mathrm{MR}}^2}}{\sqrt{3{\mathrm{MR}}^2}} \\ \Rightarrow \frac{T_1}{T_2}=\frac{2}{\sqrt{3}} \end{gathered}$$

Hence, option C is correct.