JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Rotational Mechanics

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Moment of inertia about an axis AB for a rod of mass 40 kg and length 3 m is same as that of a solid sphere of mass of 10 kg and radius R about an axis parallel to AB axis with separation of 3 m as shown in figure below. The value of R is given as

\frac{α}{2}

. The value of

α

is ______.

Answer: 15

Solution

The axis

A B

is perpendicular to the rod through one end, so

I_{r o d} = \frac{M _{1} L ^{2}}{3} = \frac{40 \times 3 ^{2}}{3} = 120 kg \cdot m^{2}

. For the sphere about a parallel axis

3

m from its centre, the parallel-axis theorem gives

I_{s p h er e} = \frac{2}{5} M_{2} R^{2} + M_{2} L^{2} = \frac{2}{5} \times 10 \times R^{2} + 10 \times 3^{2} = 4 R^{2} + 90

. Equating the two moments of inertia,

120 = 4 R^{2} + 90

, so

R^{2} = \frac{30}{4} = \frac{15}{2}

and

R = \frac{15}{2}

. Comparing with

R = \frac{α}{2}

, the value of

α

15

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Integer answer

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JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Rotational Mechanics

Solution

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