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TS EAMCET · Physics · Thermal Properties of Matter

The moment of inertia \(I\) of uniform rod about a perpendicular bisector increases to \(I+\Delta I\), if the temperature is increased slightly by \(\Delta T\). If the coefficient of linear expansion is \(\alpha\), then \(\frac{\Delta I}{I}\) is \[ \left(\text { Assume } \frac{\Delta T}{T} \ll 1\right) \]

  1. A \(\alpha \Delta T\)
  2. B \(2 \alpha \Delta T\)
  3. C \(3 \alpha \Delta T\)
  4. D \(4 \alpha \Delta T\)
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Answer & Solution

Correct Answer

(B) \(2 \alpha \Delta T\)

Step-by-step Solution

Detailed explanation

Let initial moment of inertia of rod about perpendicular bisector, \(I=\frac{1}{12} M L^2\) where, \(M=\) mass of rod and \(\quad L=\) initial length of rod. After increasing temperature by \(\Delta T\). Now, moment of inertia,…
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