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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Taking the natural logarithm on both sides: $\ln T = \ln(2\pi) + \f

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$\frac{1}{2} \alpha T \Delta 	heta$

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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Taking the natural logarithm on both sides: $\ln T = \ln(2\pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln g$.Differentiating both sides,we get: $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}$.We know that the linear expansion of the length is given by $\Delta l = l \alpha \Delta 	heta$,which implies $\frac{\Delta l}{l} = \alpha \Delta 	heta$.Substituting this into the expression for the change in time period: $\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta 	heta$.Therefore,the change in time period is $\Delta T = \frac{1}{2} \alpha T \Delta 	heta$.

If the temperature of a pendulum with time period $T$ is increased by $\Delta \theta$,the change in the time period of the pendulum is .......

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