A clock which keeps correct time at 20^{circ} | vedclass

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Question

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Taking the differential,the fractional change in time period is $\f

Vedclass · Accepted Answer

$10.3 \ s/	ext{day}$

Vedclass · Answer

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Taking the differential,the fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l} = \frac{1}{2} \alpha \Delta 	heta$.Given $\alpha = 12 	imes 10^{-6} /{ }^{\circ} C$ and $\Delta 	heta = 40^{\circ} C - 20^{\circ} C = 20^{\circ} C$.Substituting the values: $\frac{\Delta T}{T} = \frac{1}{2} 	imes 12 	imes 10^{-6} 	imes 20 = 120 	imes 10^{-6} = 1.2 	imes 10^{-4}$.The time lost or gained per day is $\Delta T = \frac{\Delta T}{T} 	imes T_{day}$.Since $T_{day} = 24 	imes 60 	imes 60 = 86400 \ s$,we have $\Delta T = 1.2 	imes 10^{-4} 	imes 86400 = 10.368 \ s$.Rounding to the nearest value,the clock loses $10.3 \ s/	ext{day}$.

$A$ clock which keeps correct time at $20^{\circ} C$,is subjected to $40^{\circ} C$. If the coefficient of linear expansion of the pendulum is $12 \times 10^{-6} /{ }^{\circ} C$,how much time will it gain or lose per day?

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