Question

Two bodies rotate with kinetic energies \(E_{1}\) and \(E_{2}\), moment of inertia about their axes of rotation is \(I_{1}\) and \(I_{2}\). If \(3I_{1}=I_{2}\) and \(E_{1}=27E_{2}\) then the ratio of the angular momenta \(L_{1}\) to \(L_{2}\) is

• A. \(9:1\)
• B. \(3:1\)
• C. \(27:1\)
• D. \(1:1\)

Hint:

Express angular momentum in terms of rotational kinetic energy and moment of inertia, then take the ratio.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Rotational kinetic energy is \(E=\frac{L^{2}}{2I}\), so \(L=\sqrt{2IE}\).

Step 2: Form the ratio
\(\frac{L_{1}}{L_{2}}=\sqrt{\frac{I_{1}E_{1}}{I_{2}E_{2}}}\)

Step 3: Calculation
Given \(I_{2}=3I_{1}\) and \(E_{1}=27E_{2}\):
\(\frac{L_{1}}{L_{2}}=\sqrt{\frac{1}{3}\times27}=\sqrt{9}=3\).

Step 4: Compare the options
The ratio is \(3:1\), option (B).

Final Answer:
The ratio of L1 to L2 is 3 to 1. \[ \boxed{\text{(B) }3:1} \]