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

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Elasticity & SHM

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, T (ii) back and forth in a direction perpendicular to its plane, with a period

T_{2}

. The ratio

\frac{T _{1}}{T _{2}}

will be:

Answer: (A)

Treat the hung ring as a physical pendulum,

T = 2 π \frac{I}{m g d}

, with the pivot at the nail a distance

d = R

from the centre. Swinging in its own plane, the moment of inertia about the nail is

I_{1} = m R^{2} + m R^{2} = 2 m R^{2}

(the central value

m R^{2}

plus the parallel-axis term), so

T_{1} = 2 π \frac{2 R}{g}

. Swinging perpendicular to its plane the central moment of inertia is

\frac{1}{2} m R^{2}

, so

I_{2} = \frac{1}{2} m R^{2} + m R^{2} = \frac{3}{2} m R^{2}

and

T_{2} = 2 π \frac{3 R}{2 g}

. The ratio is

\frac{T _{1}}{T _{2}} = \frac{2 R / g}{3 R /2 g} = \frac{4}{3} = \frac{2}{3}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option