Physics Q46 – JEE-MAIN 2025

Home Physics Rotational MotionQuestion #453

A, B and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure. The moment of inertia of the given system about PQ axis is

\frac{x}{15} I

, where I is the moment of inertia of the disc about its diameter. The value of x is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Identify the components and axis

Recognize that the system consists of a disc, a solid sphere, and a spherical shell, and the axis of rotation is PQ.

🧠 Full Solution Path

Step 1: Identify individual moments of inertia about their centers of mass.✦ Active

Given that A is a disc, B is a solid sphere, and C is a spherical shell, all with mass $M$ and radius $R$ . The moment of inertia of the disc about its diameter is $I = \frac{1}{4} M R^{2}$ .

I_{A} = I_{d i s c, d i a m e t e r} = \frac{1}{4} M R^{2}

I_{C M, B} = I_{s o l i d s p h e r e, d i a m e t e r} = \frac{2}{5} M R^{2}

I_{C M, C} = I_{s p h e r i c a l s h e l l, d i a m e t e r} = \frac{2}{3} M R^{2}

💡 Teacher's Secret Hint

Remember the standard moment of inertia formulas for common shapes.

Step 2: Apply Parallel Axis Theorem for B and C.○ Expand

The axis PQ passes through the center of A, so $I_{A}$ is directly its moment of inertia about its diameter. For objects B and C, the axis PQ is parallel to their respective axes passing through their centers. The centers of A, B, and C form an equilateral triangle with side length $2 R$ . The axis PQ is vertical and passes through the center of A. The horizontal distance from the center of B (or C) to the axis PQ is $d = R$ . Using the Parallel Axis Theorem $I_{a x i s} = I_{C M} + M d^{2}$ :

I_{B} = I_{C M, B} + M d^{2} = \frac{2}{5} M R^{2} + M R^{2} = \frac{7}{5} M R^{2}

I_{C} = I_{C M, C} + M d^{2} = \frac{2}{3} M R^{2} + M R^{2} = \frac{5}{3} M R^{2}

💡 Teacher's Secret Hint

Carefully determine the perpendicular distance 'd' from the center of mass to the parallel axis.

Step 3: Calculate the total moment of inertia and find x.○ Expand

The total moment of inertia of the system about the PQ axis is the sum of the moments of inertia of individual objects:

I_{s y s t e m} = I_{A} + I_{B} + I_{C} = \frac{1}{4} M R^{2} + \frac{7}{5} M R^{2} + \frac{5}{3} M R^{2}

I_{s y s t e m} = M R^{2} (\frac{1}{4} + \frac{7}{5} + \frac{5}{3}) = M R^{2} (\frac{15 + 84 + 100}{60}) = \frac{199}{60} M R^{2}

We are given that $I_{s y s t e m} = \frac{x}{15} I$ , where $I = \frac{1}{4} M R^{2}$ . Substituting $I$ :

I_{s y s t e m} = \frac{x}{15} (\frac{1}{4} M R^{2}) = \frac{x}{60} M R^{2}

Equating the two expressions for $I_{s y s t e m}$ :

\frac{199}{60} M R^{2} = \frac{x}{60} M R^{2}

x = 199

💡 Teacher's Secret Hint

Ensure all terms are converted to a common denominator before summing.

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Physics Question 46 – JEE-MAIN 2025

Clue 1 of 3: Identify the components and axis

🧠 Full Solution Path