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 $\theta {}^{\text{o}}\text{C,}$ increase in length of pendulum, $\frac{\Delta l}{l}=\alpha \Delta \theta$ $\therefore$ $\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \theta$ or $\frac{\Delta \mathrm{T}}{T}=\frac{1}{2}\times 9\times {10}^{-7}\times (30-20)$ $=\frac{1}{2}\times 9\times {10}^{-7}\times 10$ or $\Delta \mathrm{T}=4.5\times {10}^{-6}\times 0.5=2.25\times {10}^{-6}s$