Question:
A rod of length \(L\) is composed of two equal parts: half wood (mass \(m_w\)) and half brass (mass \(m_b\)). The moment of inertia about an axis through its centre and perpendicular to the rod is:
A rod of length \(L\) is composed of two equal parts: half wood (mass \(m_w\)) and half brass (mass \(m_b\)). The moment of inertia about an axis through its centre and perpendicular to the rod is:
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If rod is symmetric about centre, use \( \frac{1}{12}ML^2 \) directly.
Updated On: Apr 16, 2026
- \( \frac{(m_w + m_b)L^2}{6} \)
- \( \frac{(m_w + m_b)L^2}{2} \)
- \( \frac{(m_w + m_b)L^2}{12} \)
- \( \frac{(m_w + m_b)L^2}{3} \)
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The Correct Option is C
Solution and Explanation
Concept:
Moment of inertia of a rod about its centre:
\[
I = \frac{1}{12} ML^2
\]
Step 1: Treat as composite rod.
Total mass: \[ M = m_w + m_b \]
Step 2: Apply standard formula.
Even though densities differ, symmetry about centre remains. \[ I = \frac{1}{12}(m_w + m_b)L^2 \]
Step 1: Treat as composite rod.
Total mass: \[ M = m_w + m_b \]
Step 2: Apply standard formula.
Even though densities differ, symmetry about centre remains. \[ I = \frac{1}{12}(m_w + m_b)L^2 \]
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