Physics

A pendulum clock keeps correct time at $$20^{o}C$$. The correction to be made during summer per day where the average temperature is $$40^{o}C$$, ($$\alpha ={ 10 }^{ -5 }/^o{ C }$$) will be:


ANSWER

$$8.64\ sec$$


SOLUTION
(D)
Formula for time period : $$T = 2\pi \sqrt{\dfrac{L }{g}} ...(1)$$ 
Actual leng with change in temperature: 
$$L=L_O(1+\alpha \Delta \theta)$$
$$\dfrac{L-L_O}{L}=\dfrac{\Delta L}{L}=\alpha \Delta \theta...(2)$$
using $$(1)$$
$$\dfrac{\Delta T}{T} =\dfrac{1}{2} \times \dfrac{\Delta L}{L}$$
 using $$(2)$$
$$\dfrac{\Delta T}{T} =\dfrac{1}{2} \alpha \Delta \theta$$

For a day  $$T= 24 hours=86400sec$$
$$\Delta T=\dfrac{1}{2} 6400\times 10^{-5}\times (40-20)$$
$$\Delta T=8.64 sec$$
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