Question:
Two bodies A and B have their moments of inertia ' \( I \) ' and ' \( 2 I \) ' respectively about their axis of rotation. If their kinetic energy of rotation are equal then angular momentum of body A to that of body \( B \) will be in the ratio
Two bodies A and B have their moments of inertia ' \( I \) ' and ' \( 2 I \) ' respectively about their axis of rotation. If their kinetic energy of rotation are equal then angular momentum of body A to that of body \( B \) will be in the ratio
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If KE is same:
- $L \propto \sqrt{I}$
Updated On: May 4, 2026
- $1 : 2$
- \( \sqrt{2} : 1 \)
- $2 : 1$
- \( 1 : \sqrt{2} \)
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The Correct Option is D
Solution and Explanation
Concept:
Rotational kinetic energy:
\[
K = \frac{L^2}{2I}
\]
Step 1: Given condition.
\[ K_A = K_B \Rightarrow \frac{L_A^2}{2I} = \frac{L_B^2}{2(2I)} \]
Step 2: Simplify.
\[ \frac{L_A^2}{I} = \frac{L_B^2}{2I} \Rightarrow L_A^2 = \frac{L_B^2}{2} \]
Step 3: Take square root.
\[ \frac{L_A}{L_B} = \frac{1}{\sqrt{2}} \]
Step 1: Given condition.
\[ K_A = K_B \Rightarrow \frac{L_A^2}{2I} = \frac{L_B^2}{2(2I)} \]
Step 2: Simplify.
\[ \frac{L_A^2}{I} = \frac{L_B^2}{2I} \Rightarrow L_A^2 = \frac{L_B^2}{2} \]
Step 3: Take square root.
\[ \frac{L_A}{L_B} = \frac{1}{\sqrt{2}} \]
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