JEE Advanced · 2006 · 09 Apr 2006 · Rotational Mechanics

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A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be

(A) $\frac{2 R}{15}$ ✓ Correct
(B) $R \frac{2}{15}$
(C) $\frac{4 R}{15}$
(D) $\frac{R}{4}$

Answer: (A)

Solution

For a solid sphere about an axis through its center:

I_{s p h er e} = \frac{2}{5} M R^{2}

For a disc of radius r about an axis through its edge and perpendicular to its plane, using the parallel axis theorem:

I_{d i sc} = I_{c m} + M d^{2} = \frac{1}{2} M r^{2} + M r^{2} = \frac{3}{2} M r^{2}

Since the mass M is conserved during recasting and the moments of inertia are equal:

\frac{2}{5} M R^{2} = \frac{3}{2} M r^{2}

Solving for r:

r^{2} = \frac{2}{5} \cdot \frac{2}{3} R^{2} = \frac{4}{15} R^{2}

r = \frac{2 R}{15}

Therefore, the answer is (A).

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option

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JEE Advanced · 2006 · 09 Apr 2006 · Rotational Mechanics

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