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(B) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$. The change in time period due to temperature change is given by $\Delta t

Vedclass · Accepted Answer

$20$

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(B) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$. The change in time period due to temperature change is given by $\Delta t = \frac{1}{2} \alpha \Delta 	heta t$,where $\alpha$ is the coefficient of linear expansion,$\Delta 	heta$ is the change in temperature,and $t$ is the total time.Let $	heta$ be the temperature at which the clock gives the correct time. At $15 \ ^oC$,the clock is $5 \ s$ fast,so $\Delta t_1 = 5 = \frac{1}{2} \alpha (	heta - 15) 	imes (24 	imes 3600) \quad ...(1)$At $30 \ ^oC$,the clock is $10 \ s$ slow,so $\Delta t_2 = 10 = \frac{1}{2} \alpha (30 - 	heta) 	imes (24 	imes 3600) \quad ...(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{5}{10} = \frac{	heta - 15}{30 - 	heta}$$\frac{1}{2} = \frac{	heta - 15}{30 - 	heta}$$30 - 	heta = 2(	heta - 15)$$30 - 	heta = 2	heta - 30$$3	heta = 60$$	heta = 20 \ ^oC$.

$A$ pendulum clock (fitted with a small heavy bob that is connected with a metal rod) is $5 \ s$ fast each day at a temperature of $15 \ ^oC$ and $10 \ s$ slow at a temperature of $30 \ ^oC$. The temperature at which it is designed to give correct time is ........ $^oC$.

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