Physics Q46 – JEE-MAIN 2025

Home Physics Rotational MotionQuestion #454

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 _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Conceptual Principle

The moment of inertia of a body with a removed part can be found by subtracting the moment of inertia of the removed part from the moment of inertia of the complete body.

🧠 Full Solution Path

Step 1: Calculate Moment of Inertia of Complete Disc and Mass of Removed Disc✦ Active

The moment of inertia of the complete disc of mass $M$ and radius $R$ about axis AB (passing through its center O and perpendicular to its plane) is:

I_{t o t a l} = \frac{1}{2} M R^{2}

The surface mass density of the disc is $σ = \frac{M}{π R^{2}}$ . The radius of the removed disc is $r = R / 3$ . Its area is $A_{r e m o v e d} = π (R / 3)^{2} = \frac{π R^{2}}{9}$ . The mass of the removed disc is:

m = σ \times A_{r e m o v e d} = \frac{M}{π R^{2}} \times \frac{π R^{2}}{9} = \frac{M}{9}

Step 2: Calculate Moment of Inertia of Removed Disc about Axis AB○ Expand

The moment of inertia of the removed disc about its own center O' (perpendicular to its plane) is:

I_{O^{'}} = \frac{1}{2} m r^{2} = \frac{1}{2} (\frac{M}{9}) {(\frac{R}{3})}^{2} = \frac{M R^{2}}{162}

From the figure, the distance $d$ between the center O' of the removed disc and the axis AB (passing through O) is $R - R / 3 = 2 R / 3$ . Using the Parallel Axis Theorem, the moment of inertia of the removed disc about axis AB is:

I_{r e m o v e d_a b o u t_A B} = I_{O^{'}} + m d^{2} = \frac{M R^{2}}{162} + (\frac{M}{9}) {(\frac{2 R}{3})}^{2}

Simplifying the expression:

I_{r e m o v e d_a b o u t_A B} = \frac{M R^{2}}{162} + \frac{4 M R^{2}}{81} = \frac{M R^{2} + 8 M R^{2}}{162} = \frac{9 M R^{2}}{162} = \frac{1}{18} M R^{2}

💡 Teacher's Secret Hint

Ensure correct application of the parallel axis theorem and calculation of the distance 'd'.

Step 3: Calculate Moment of Inertia of Remaining Part and Determine x○ Expand

The moment of inertia of the remaining part is the difference between the moment of inertia of the complete disc and the removed disc:

I_{r e m a i n i n g} = I_{t o t a l} - I_{r e m o v e d_a b o u t_A B} = \frac{1}{2} M R^{2} - \frac{1}{18} M R^{2}

Simplifying the expression:

I_{r e m a i n i n g} = \frac{9 M R^{2} - M R^{2}}{18} = \frac{8 M R^{2}}{18} = \frac{4}{9} M R^{2}

Given that the moment of inertia of the remaining part is $\frac{4}{x} M R^{2}$ . Comparing this with our result, $\frac{4}{9} M R^{2} = \frac{4}{x} M R^{2}$ , we find:

x = 9

💡 Teacher's Secret Hint

Double-check the arithmetic when subtracting fractions.

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Physics Question 46 – JEE-MAIN 2025

Clue 1 of 3: Conceptual Principle

🧠 Full Solution Path