A solid sphere of mass 500 g and radius 5 cm is rotated about one of its diameters with angular speed of 10 rad s⁻¹. If the moment of inertia of the sphere about its tangent is x × 10⁻² times its angular momentum about the diameter. Then the value of x will be ___________. Numerical answer type. Work it out, then open the answer.

Answer: 35. Id = 2/5MR² = 2/5(0.5)(0.05)² = 5 × 10⁻⁴ kg m ² L = Idω = 5 × 10⁻³ kg m ² /s About a tangent: It = 7/5MR² = 1.75 × 10⁻³ kg m ² It/L = 1.75 × 10⁻³5 × 10⁻³ = 0.35 = 35 × 10⁻² So x = 35.

A solid sphere of mass 500 g and radius 5 cm is rotated about one of… | JEE Main 2023 (11 Apr, Shift 1)

Q29JEE Main 2023 · 11 APR · SHIFT 1Medium

Rotational Kinetic Energy, Work & Power

A solid sphere of mass $500\text{ g}$ and radius $5\text{ cm}$ is rotated about one of its diameters with angular speed of $10\text{ rad s}^{-1}$ . 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 ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 35

$I_d = \frac{2}{5}MR^2 = \frac{2}{5}(0.5)(0.05)^2 = 5 \times 10^{-4}$ kg m $^2$
$L = I_d\omega = 5 \times 10^{-3}$ kg m $^2$ /s
About a tangent: $I_t = \frac{7}{5}MR^2 = 1.75 \times 10^{-3}$ kg m $^2$
$\frac{I_t}{L} = \frac{1.75 \times 10^{-3}}{5 \times 10^{-3}} = 0.35 = 35 \times 10^{-2}$
So $x = 35$ .

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