JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Rotational Mechanics

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A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is

n cm

. The value of

n

is ____.

Answer: 265

Solution

Using the parallel axis theorem:

I = I_{c m} + m d^{2}

For a solid sphere:

I_{c m} = \frac{2}{5} m R^{2}

If radius of gyration is

k

, then

I = m k^{2}

m k^{2} = \frac{2}{5} m R^{2} + m d^{2}

k^{2} = \frac{2}{5} (10)^{2} + (15)^{2} = 40 + 225 = 265

Since

k = n

, we have

n = 265

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Integer answer

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