A) \[2.25\times {{10}^{-6}}s\]
B) \[2.5\times {{10}^{-7}}s\]
C) \[5\times {{10}^{-7}}s\]
D) \[1.125\times {{10}^{-6}}\]
Correct Answer: A
Solution :
Time period of simple pendulum, \[T=2\pi \sqrt{\frac{l}{g}}\] \[\frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}\] But at temperature\[\text{ }\!\!\theta\!\!\text{ }{{\,}^{\text{o}}}\text{C,}\] increase in length of pendulum, \[\frac{\Delta l}{l}=\alpha \Delta \theta \] \[\therefore \] \[\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \theta \] or \[\frac{\Delta \Tau }{T}=\frac{1}{2}\times 9\times {{10}^{-7}}\times (30-20)\] \[=\frac{1}{2}\times 9\times {{10}^{-7}}\times 10\] or \[\Delta \Tau =4.5\times {{10}^{-6}}\times 0.5=2.25\times {{10}^{-6}}s\]
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