A uniform metal rod is used as a bar pendulum | vedclass

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(D) The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgd}}$. For a uniform rod of length $L$,$T \propto \sqrt{L}$.Taking th

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$1 	imes 10^{-3} \%$

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(D) The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgd}}$. For a uniform rod of length $L$,$T \propto \sqrt{L}$.Taking the derivative,the fractional change in the time period is given by $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L}$.Since $\Delta L = L \alpha \Delta 	heta$,we have $\frac{\Delta L}{L} = \alpha \Delta 	heta$.Therefore,$\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta 	heta$.Given $\alpha = 2 	imes 10^{-6} /^{\circ}C$ and $\Delta 	heta = 10^{\circ}C$:$\frac{\Delta T}{T} = \frac{1}{2} 	imes (2 	imes 10^{-6}) 	imes 10 = 10^{-5}$.The percentage increase is $\frac{\Delta T}{T} 	imes 100 = 10^{-5} 	imes 100 = 10^{-3} \%$.

$A$ uniform metal rod is used as a bar pendulum. If the room temperature rises by $10^{\circ}C$,and the coefficient of linear expansion of the metal of the rod is $2 \times 10^{-6} /^{\circ}C$,the period of the pendulum will have a percentage increase of:

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