A solid sphere of mass 500 g and radius 5 cm is rotated about one of its diameter with angular speed of $10 \mathrm{rad~s}^{-1}$. If the moment of inertia of the sphere about its tangent is $x \times 10^{-2}$ times its angular momentum about the diameter. Then the value of $x$ will be ______.

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Rotational Mechanics

A solid sphere of mass 500 g and radius 5 cm is rotated about one of its diameter with angular speed of

10 rad s^{- 1}

. If the moment of inertia of the sphere about its tangent is

x \times 1 0^{- 2}

times its angular momentum about the diameter. Then the value of

x

will be ______.

Answer: 35

Solution

The moment of inertia about a diameter is

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

, and the angular momentum about it is

L = I_{d} ω = \frac{2}{5} m R^{2} ω

. By the parallel-axis theorem the moment of inertia about a tangent is

I_{t} = I_{d} + m R^{2} = \frac{7}{5} m R^{2}

. Forming the required ratio,

\frac{I _{t}}{L} = \frac{\frac{7}{5} m R ^{2}}{\frac{2}{5} m R ^{2} ω} = \frac{7}{2 ω} = \frac{7}{20} = 0.35

. Written as

x \times 1 0^{- 2}

this gives

x = 35

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Integer answer

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