JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Rotational Mechanics

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Solid sphere A is rotating about an axis PQ. If the radius of the sphere is 5 cm then its radius of gyration about PQ will be

x

cm. The value of x is ______.

Answer: 110

Solution

The solid sphere has moment of inertia

\frac{2}{5} m R^{2}

about an axis through its centre. The axis

P Q

runs

10 cm

from that centre, so the parallel-axis theorem gives

I_{P Q} = \frac{2}{5} m R^{2} + m d^{2}

with

R = 5 cm

and

d = 10 cm

. Writing

I_{P Q} = m k^{2}

, the radius of gyration satisfies

k^{2} = \frac{2}{5} (25) + (10)^{2} = 10 + 100 = 110

. The value of

x

110

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Integer answer

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