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A simple pendulum made of a bob of mass $m$ and a metallic wire of negligible mass has time period 2s at T=0⁰C. If the temperature of the wire is increased and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope S. If the coefficient of linear expansion of metal is $\alpha$ then the value of S is :
Option: 1
$\alpha$

Option: 2
$\frac{\alpha }{2}$

Option: 3
$2\alpha$

Option: 4
$\frac{1}{\alpha }$

Answers (1)

best_answer

$\\ \text{Change of length of wire with temperature,} \\ \Delta \ell=\alpha \ell \Delta \theta \\ \text{Time period of pendulum at temperature} \ \theta , \\ T _{\theta}=2 \pi \sqrt{\frac{\ell+\Delta \ell}{g}} \\ =2 \pi \sqrt{\frac{\ell(1+\alpha \Delta \theta)}{g}} \\ =2 \pi \sqrt{\frac{l}{g}}(1+\alpha \Delta \theta)^{\frac{1}{2}} \\ =2 \pi \sqrt{\frac{\ell}{g}}\left(1+\frac{\Delta \ell}{\ell}\right)^{\frac{1}{2}} \\ \simeq T_{0}\left(1+\frac{\Delta \ell}{2 \ell}\right) Here \ T_{0}= \text{ time period at temperature} \ \ 0^{\circ} C \\ \therefore \text{Change in time period}, \\ \Delta T = T _{\theta}- T _{0} \\ =\frac{T_{0} \Delta \ell}{2 \ell}$

$\\ \Delta T=\frac{T_{0}(\alpha \ell \Delta \theta)}{2 \ell} \\ \\ \therefore \frac{\Delta T}{\Delta \theta}=\frac{T_{0} \alpha}{2} \\ \text{Given that} \ \ T_{0}=2 \\ \therefore \frac{\Delta T}{\Delta \theta}=\frac{2 \alpha}{2}=\alpha \\ \\\frac{\Delta T}{\Delta \theta} \text{ is the slope of} \ \Delta T \ vs \ \Delta \theta \ curve =S (given) \\ \\ \therefore S=\alpha$

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