Verified Solution • 2025–26 Session moderate

QPages 12-14: Q33: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q33 Solution: A torsional pendulum consists of a solid disc connected to a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently consider only the wire’s thermal expansion and assume $T\propto\sqrt{L}$, overlooking that the disc’s radius also expands and that both the moment of inertia $I\propto r^4$ and the torsional constant $\kappa\propto r^4$ change identically, causing the first‑order temperature effect to cancel. Always check whether numerator and denominator share the same dimensional dependence before attributing a linear temperature change.

A torsional pendulum consists of a solid disc connected to a thin wire

\alpha = 2.4 \times 10^{-5} \text{ �C}^{-1}

) at its centre. Find the percentage change in the time period between peak winter 5�C) and peak summer 45�C).

Step-by-Step Verified Solution

**Answer:** Percentage change

= 0.048\%

**Step-by-Step Explanation:** Torsional pendulum:

T = 2\pi\sqrt{I/\kappa}

. Both

I

and

\kappa

depend on dimensions.

I \propto r^4

(for disc),

\kappa \propto r^4

(torsional constant). So

T

is independent of

r

? Actually,

I \propto (1 + 2\alpha\theta)^2

and

\kappa \propto (1 + 2\alpha\theta)^2

. Their ratio is independent of temperature to first order. The change is second-order:

\Delta T/T \approx 2\alpha\Delta\theta \times

(small factor). Given

\alpha = 2.4 \times 10^{-5}

/�C,

\Delta\theta = 40�\text{C}

:

\Delta T/T \approx 4.8 \times 10^{-4} = 0.048\%

.

All Questions in this Chapter

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