Verified Solution • 2025–26 Session moderate

QPages 12-14: Q32: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q32 Solution: Show that the moment of inertia of a solid body of any shape…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently replace the radius $r$ with $r(1+\alpha\theta)$ and then write $I=I_0(1+\alpha\theta)$, forgetting that $I$ involves $r^2$; the correct substitution gives $r^2\to r^2(1+\alpha\theta)^2\approx r^2(1+2\alpha\theta)$ for small temperature changes. Remember to square the linear expansion factor and use the first‑order approximation only after expanding.

Show that the moment of inertia of a solid body of any shape changes with temperature as

I = I_0 1 + 2\alpha\theta)

, where

I_0

is the moment of inertia at 0�C and

\alpha

is the coefficient of linear expansion of the solid.

Step-by-Step Verified Solution

**Answer:**

I = I_0(1 + 2\alpha\theta)

**Step-by-Step Explanation:** For a solid body, moment of inertia

I = \int r^2 dm

. When heated, all linear dimensions scale as

r \to r(1 + \alpha\theta)

. So

r^2 \to r^2(1 + \alpha\theta)^2 \approx r^2(1 + 2\alpha\theta)

. Mass

dm

is unchanged (mass conservation). Hence

I = I_0(1 + 2\alpha\theta)

.

All Questions in this Chapter

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