JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Rotational Mechanics

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is

I_{1}

. The same rod is bent into a ring and its moment of inertia about a diameter is

I_{2}

. If

\frac{I _{1}}{I _{2}}

\frac{x π ^{2}}{3}

, then the value of x will be ______.

Answer: 8

Solution

For a uniform rod of length

ℓ

and mass

m

I_{1} = \frac{m ℓ ^{2}}{3}

When the rod is bent into a ring, the circumference equals the rod length:

ℓ = 2 π r

, so

r = \frac{ℓ}{2 π}

.

For a ring about a diameter:

I_{2} = \frac{m r ^{2}}{2} = \frac{m}{2} (\frac{ℓ}{2 π})^{2} = \frac{m ℓ ^{2}}{8 π ^{2}}

Taking the ratio:

\frac{I _{1}}{I _{2}} = \frac{\frac{m ℓ ^{2}}{3}}{\frac{m ℓ ^{2}}{8 π ^{2}}} = \frac{8 π ^{2}}{3}

Comparing with

\frac{x π ^{2}}{3}

, we get

x = 8

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer