If the angular velocity of a body rotating about the given axis increases by 20% , then its kinetic energy of rotation will increase by

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The rotational kinetic energy of a body is given by $K = \frac{1}{2} I ω^{2}$ .

Let the initial angular velocity be $ω_{1} = ω$ .

The initial kinetic energy is $K_{1} = \frac{1}{2} I ω^{2}$ .

The angular velocity increases by $20%$ , so the new angular velocity is $ω_{2} = ω + 0.20 ω = 1.2 ω$ .

The new kinetic energy is $K_{2} = \frac{1}{2} I (1.2 ω)^{2} = 1.44 (\frac{1}{2} I ω^{2}) = 1.44 K_{1}$ .

The percentage increase in kinetic energy is $\frac{K _{2} - K _{1}}{K _{1}} \times 100% = \frac{1.44 K _{1} - K _{1}}{K _{1}} \times 100% = 0.44 \times 100% = 44%$ .

Answer: $44%$

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