A pendulum clock loses 12;s a day if the temp | vedclass

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(C) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$. The fractional change in time period is $\frac{\Delta T}{T} = \frac{1}

Vedclass · Accepted Answer

$25^{\circ}C, \alpha = 1.85 	imes 10^{-5}/^{\circ}C$

Vedclass · Answer

(C) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$. The fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta 	heta$.For a day,$T = 24 	imes 3600 \; s$.When the clock loses $12\;s$ at $40^{\circ}C$ (where $	heta$ is the correct temperature): $\frac{12}{T} = \frac{1}{2} \alpha (40 - 	heta) \quad ...(1)$When the clock gains $4\;s$ at $20^{\circ}C$: $\frac{4}{T} = \frac{1}{2} \alpha (	heta - 20) \quad ...(2)$Dividing equation $(1)$ by $(2)$: $\frac{12}{4} = \frac{40 - 	heta}{	heta - 20} \implies 3(	heta - 20) = 40 - 	heta \implies 3	heta - 60 = 40 - 	heta \implies 4	heta = 100 \implies 	heta = 25^{\circ}C$.Substituting $	heta = 25^{\circ}C$ in equation $(2)$: $\frac{4}{24 	imes 3600} = \frac{1}{2} \alpha (25 - 20) \implies \frac{4}{86400} = \frac{1}{2} \alpha (5) \implies \alpha = \frac{8}{86400 	imes 5} = \frac{8}{432000} \approx 1.85 	imes 10^{-5}/^{\circ}C$.

$A$ pendulum clock loses $12\;s$ a day if the temperature is $40^{\circ}C$ and gains $4\;s$ a day if the temperature is $20^{\circ}C$. The temperature at which the clock will show correct time,and the coefficient of linear expansion $(\alpha)$ of the metal of the pendulum shaft are respectively:

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