Question

A hollow sphere of mass '$M$' and radius '$R$' is rotating with angular frequency '$\omega$'. It suddenly stops rotating and $75$% of kinetic energy is converted to heat. If '$S$' is the specific heat of the material in $\frac{J}{kg}K$ then rise in temperature of the sphere is (M.I. of hollow sphere $=\frac{2}{3}M{R}^2$

• A. $\frac{R\omega }{4S}$
• B. $\frac{R^2{\omega }^2}{4S}$
• C. $\frac{R\omega }{2S}$
• D. $\frac{R^2{\omega }^2}{2S}$

Answer 1

Answer: (B)

$\frac{R^2{\omega }^2}{4S}$

Moment of inertia of hollow sphere $I=\frac{2}{3}M{R}^2$

Kinetic energy of the sphere $K.E=\frac{1}{2}I{w}^2$

$\therefore$ $K.E=\frac{1}{2}.\frac{2}{3}M{R}^2{w}^2=\frac{1}{3}M{R}^2{w}^2$

Heat generated $Q=\frac{3}{4}K.E$

$\therefore$ $Q=\frac{1}{4}M{R}^2{w}^2$

Let rise in the temperature be $\Delta T$.

$\therefore$ $MS\Delta T=\frac{1}{4}M{R}^2{w}^2$

$⟹$ $\Delta T=\frac{R^2{w}^2}{4S}$