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QPages 11-12: Q8: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 11-12: Q8 Solution: If the temperature of a uniform rod is slightly increased by ,…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often forget that the moment of inertia depends on the square of the length ($L^2$), leading them to incorrectly assume a linear increase of $alpha I Delta t$. You must apply the binomial approximation to $(1+alpha Delta t)^2$ to correctly identify the factor of 2 in the change.

If the temperature of a uniform rod is slightly increased by Δt\Delta t, its moment of inertia II about a perpendicular bisector increases by

Multiple Choice Options

  • A zero
  • B αIΔt\alpha I \Delta t
  • C 2αIΔt2\alpha I \Delta t
  • D 3αIΔt3\alpha I \Delta t

Step-by-Step Verified Solution

**Answer:** (c) 2_I_t. **Step-by-Step Explanation:** For a uniform rod, the moment of inertia about a perpendicular bisector is I = ML_/12. When temperature increases by _t, L increases by _L = L__t. The new moment of inertia I' = M(L+_L)_/12 � I(1 + 2__t). Hence, the increase is 2_I_t.

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