A torsional pendulum comprises a solid disc attached at its centre...

The time period of a torsional pendulum is given by:

$T = 2\pi \sqrt{\dfrac{I}{C}}$

where $I is the moment of inertia of the disc and$C$is the torsional constant of the wire.<br><br>The torsional constant of a cylindrical wire is given by:<br><br>$C = \dfrac{\pi \eta r^4}{2L}$<br><br>where$\eta$is the shear modulus,$r$is the radius, and$L$is the length of the wire. When the temperature increases by$\Delta T$, the dimensions of the wire change due to thermal expansion. The new radius and length become:<br><br>$r' = r(1 + \alpha \Delta T)$<br>$L' = L(1 + \alpha \Delta T)$<br><br>The new torsional constant$C'$is:<br><br>$C' = \dfrac{\pi \eta (r')^4}{2L'} = \dfrac{\pi \eta r^4 (1 + \alpha \Delta T)^4}{2L(1 + \alpha \Delta T)} \approx C(1 + 3\alpha \Delta T)$<br><br>Since the coefficient of linear expansion$\alpha$is given only for the thin wire, we assume the moment of inertia$I$of the disc remains constant. The new time period$T'$is:<br><br>$T' = 2\pi \sqrt{\dfrac{I}{C(1 + 3\alpha \Delta T)}} = T(1 + 3\alpha \Delta T)^{-1/2}$<br><br>Using the binomial approximation for small$\alpha \Delta T$:<br><br>$T' \approx T \left(1 - $\dfrac{3}{2}\alpha \Delta$T\right)<br><br>The magnitude of the fractional change in the time period is:<br><br>\left|\dfrac{\Delta T_{\text{period}}}{T} \right| = \dfrac{3}{2}\alpha \Delta$T$

Given values:
\alpha = 2.4 \times 10^{-5} ^\circ\textC^{-1}
$\Delta T = 45^\circ\text{C} - 5^\circ\text{C} = 40^\circ\text{C}$

Substituting the values to find the percentage change:

$\text{Percentage change} =$\dfrac{3}{2} \times ($2.4 \times 10^{-5}) \times 40 \times 100\%$

$\text{Percentage change} = 1.5 \times 9.6 \times 10^{-4} \times 100\% = 0.144\%$

Answer: $0.144\%$