A) 10.3 seconds / day
B) 20.6 seconds / day
C) 5 seconds / day
D) 20 minutes / day
Correct Answer: A
Solution :
Time period \[T\propto \sqrt{l}\]Þ\[\frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}=\frac{1}{2}\alpha \Delta \theta \] Also according to thermal expansion \[l'=(1+\alpha \Delta \theta )\] \[\frac{\Delta l}{l}=\alpha +\theta \]. Hence \[\frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}=\frac{1}{2}\alpha \Delta \theta \] \[=\frac{1}{2}\times 12\times {{10}^{-6}}\times (40-20)=12\times {{10}^{-5}}\] \[\Rightarrow \Delta T=12\times {{10}^{-5}}\times 86400\] seconds / day \[T/\sqrt{n}.\]seconds / day
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