Q111 / 150 · MHT CET 2026 (20 April Shift 1) · Physics
A body is rotating about its own axis. Its rotational kinetic energy is '$x$' and its angular momentum is '$y$'. Hence its moment of inertia about its own axis is
- A$\dfrac{x}{2y}$
- B$\dfrac{y}{2x}$
- C$\dfrac{x^2}{2y}$
- Correct answerD$\dfrac{y^2}{2x}$
Solution
Rotational kinetic energy is given by $K = \dfrac{L^2}{2I}$
Substituting the given values, $K = x$ and $L = y$
$x = \dfrac{y^2}{2I}$
$I = \dfrac{y^2}{2x}$
Answer: $\dfrac{y^2}{2x}$
Substituting the given values, $K = x$ and $L = y$
$x = \dfrac{y^2}{2I}$
$I = \dfrac{y^2}{2x}$
Answer: $\dfrac{y^2}{2x}$
Hence, the correct answer is option: 'D'.