Question

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:

• A. $5.64sec$
• B. $6.64sec$
• C. $7.64sec$
• D. $8.64sec$

Answer 1

The correct option is D $8.64sec$

(D)

Formula for time period : $T=2\pi \sqrt{\frac{L}{g}}...\left(1\right)$

Actual leng with change in temperature:

$L=L_O(1+\alpha \Delta \theta )$

$\frac{L-L_O}{L}=\frac{\Delta L}{L}=\alpha \Delta \theta ...\left(2\right)$

using $\left(1\right)$

$\frac{\Delta T}{T}=\frac{1}{2}\times \frac{\Delta L}{L}$

using $\left(2\right)$

$\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \theta$

For a day $T=24hours=86400sec$

$\Delta T=\frac{1}{2}6400\times {10}^{-5}\times (40-20)$

$\Delta T=8.64sec$