Question

A clock pendulum made of invar has a period of 0.5 s at $ 20{}^\circ $ . If the clock is used in a climate where the temperature averages $ 20{}^\circ C $ , how much time does the clock lose in each oscillation? (For inver $ \alpha =9\times {{10}^{-7}}{{/}^{o}}C $ and g = constant)

• 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}} $

Answer 1

Answer: (A)

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 $