A solid sphere of mass 500 g and radius 5 cm is rotated about one of its diameters with an 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 .

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Accepted Answer

35. Work in SI: M=0.5 kg and R=0.05 m, so R²=2.5×10⁻³ m². Moment of inertia about a tangent, by the parallel-axis theorem: Iₜ=2/5MR²+MR²=7/5MR². Iₜ=1.4×0.5×2.5×10⁻³=1.75×10⁻³ kg m². Angular momentum about the diameter: L=2/5MR²ω=0.4×0.5×2.5×10⁻³×10=5×10⁻³ kg m² s⁻¹. Ratio: (Iₜ)/L=(1.75×10⁻³)/(5×10⁻³)=0.35=35×10⁻². x=35. The two quantities have different dimensions, so this ratio is not a physical constant — the question is simply asking for the arithmetic.

A solid sphere of mass 500 g and radius 5 cm is rotated about one of its diameters with an 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 ______.

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