JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Rotational Mechanics

WhatsApp Email

M

and

R

be the mass and radius of a disc. A small disc of radius

R /3

is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is

\frac{4}{x} M R^{2}

. The value of x is ______.

Answer: 9

Solution

The removed disc has radius

R /3

, so its mass is

m = \frac{M}{π R ^{2}} π (\frac{R}{3})^{2} = \frac{M}{9}

, and its centre lies at

\frac{2 R}{3}

from

O

(its outer edge touches the rim). Its moment of inertia about the axis through

O

I_{hole} = \frac{1}{2} m (\frac{R}{3})^{2} + m (\frac{2 R}{3})^{2} = \frac{M R ^{2}}{162} + \frac{4 M R ^{2}}{81} = \frac{M R ^{2}}{18}

. Subtracting from the full disc,

I = \frac{1}{2} M R^{2} - \frac{M R ^{2}}{18} = \frac{8 M R ^{2}}{18} = \frac{4 M R ^{2}}{9}

. Comparing with

\frac{4}{x} M R^{2}

, the value of

x

9

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Integer answer

Practise more: all JEE Main PYQs on Rotational Mechanics →

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Rotational Mechanics

Solution

More JEE Main questions on Rotational Mechanics