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:

$8.64\ sec$

##### 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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Single Correct Medium Published on 18th 08, 2020
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