Q152 / 200 · MHT CET - Medical 2026 (22 April Shift 2) · Physics
A child is standing with folded hands at the centre of the platform rotating about its central axis. The rotational kinetic energy of the system is '$K$'. The child now stretches his arms so that the moment of inertia of the system doubles. The rotational kinetic energy of the system now is '$K_1$'. The ratio $K_1 / K$ is
- A$4$
- B$\dfrac{1}{4}$
- Correct answerC$\dfrac{1}{2}$
- D$2$
Solution
By the principle of conservation of angular momentum, the angular momentum $L$ of the system remains constant since no external torque acts on it.
The rotational kinetic energy $K$ is given by $K = \dfrac{L^2}{2I}$, where $I$ is the moment of inertia.
When the child stretches his arms, the new moment of inertia becomes $I_1 = 2I$.
The new rotational kinetic energy $K_1$ is $K_1 = \dfrac{L^2}{2I_1} = \dfrac{L^2}{2(2I)} = \dfrac{1}{2} \left( \dfrac{L^2}{2I} \right) = \dfrac{K}{2}$.
Therefore, the ratio $\dfrac{K_1}{K} = \dfrac{1}{2}$.
Answer: $\dfrac{1}{2}$
The rotational kinetic energy $K$ is given by $K = \dfrac{L^2}{2I}$, where $I$ is the moment of inertia.
When the child stretches his arms, the new moment of inertia becomes $I_1 = 2I$.
The new rotational kinetic energy $K_1$ is $K_1 = \dfrac{L^2}{2I_1} = \dfrac{L^2}{2(2I)} = \dfrac{1}{2} \left( \dfrac{L^2}{2I} \right) = \dfrac{K}{2}$.
Therefore, the ratio $\dfrac{K_1}{K} = \dfrac{1}{2}$.
Answer: $\dfrac{1}{2}$
Hence, the correct answer is option: 'C'.