Q112 / 150 · MHT CET 2026 (18 April Shift 1) · Physics
The angular momentum of a rotating body is 'L'. When the frequency of rotating body is tripled and its kinetic energy is made one-third, the new angular momentum becomes
- A$9L$
- B$6L$
- C$\dfrac{L}{3}$
- Correct answerD$\dfrac{L}{9}$
Solution
The kinetic energy of a rotating body is given by $K = \dfrac{1}{2} L \omega$
Since $\omega = 2 \pi f$, we can write $K = \pi L f$
Therefore, the angular momentum is $L = \dfrac{K}{\pi f}$
Given that the new frequency is $f' = 3f$ and the new kinetic energy is $K' = \dfrac{K}{3}$
The new angular momentum is $L' = \dfrac{K'}{\pi f'} = \dfrac{\dfrac{K}{3}}{\pi (3f)} = \dfrac{1}{9} \dfrac{K}{\pi f} = \dfrac{L}{9}$
Answer: $\dfrac{L}{9}$
Since $\omega = 2 \pi f$, we can write $K = \pi L f$
Therefore, the angular momentum is $L = \dfrac{K}{\pi f}$
Given that the new frequency is $f' = 3f$ and the new kinetic energy is $K' = \dfrac{K}{3}$
The new angular momentum is $L' = \dfrac{K'}{\pi f'} = \dfrac{\dfrac{K}{3}}{\pi (3f)} = \dfrac{1}{9} \dfrac{K}{\pi f} = \dfrac{L}{9}$
Answer: $\dfrac{L}{9}$
Hence, the correct answer is option: 'D'.