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(B) The change in length of the rod due to an increase in temperature is given by $\Delta l = \alpha l \Delta 	heta$.Therefore,the fractional change

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$1 	imes 10^{-3}$

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(B) The change in length of the rod due to an increase in temperature is given by $\Delta l = \alpha l \Delta 	heta$.Therefore,the fractional change in length is $\frac{\Delta l}{l} = \alpha \Delta 	heta$.The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.This implies $T \propto l^{1/2}$.Differentiating both sides,we get $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}$.Substituting the expression for $\frac{\Delta l}{l}$,we get $\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta 	heta$.Given $\alpha = 2 	imes 10^{-6} {\circ}C^{-1}$ and $\Delta 	heta = 10^{\circ}C$,$\frac{\Delta T}{T} = \frac{1}{2} 	imes (2 	imes 10^{-6}) 	imes 10 = 10^{-5}$.To find the percentage change,multiply by $100$:$\frac{\Delta T}{T} 	imes 100 = 10^{-5} 	imes 100 = 10^{-3} \%$.Thus,the percentage change in the time period is $1 	imes 10^{-3} \%$.

$A$ metal rod is used as a pendulum. If the room temperature increases by $10^{\circ}C$ and the linear expansion coefficient of the metal is $2 \times 10^{-6} {\circ}C^{-1}$,then the percentage change in the time period of the pendulum is ...... $\%$

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