Question

A seconds pendulum clock has a steel wire. The clock shows correct time at ${25}^{\circ }\text{C}$. How much time does the clock lose or gain in one week, when the temperature is increased to ${35}^{\circ }\text{C}$ ? $({\alpha }_{steel}=1.2\times {10}^{-5}{/}^{\circ }\text{C})$

• A. $5.18\text{s}$ gain
• B. $5.18\text{s}$ lose
• C. $36.28\text{s}$ gain
• D. $36.28\text{s}$ lose

Answer 1

The correct option is D $36.28\text{s}$ lose

Time period $T=2\pi \sqrt{\frac{l}{g}}$
length changes as temperature increases

$l_{35}=l_{25}(1+\alpha \Delta \theta )$

$T_{35}=2\pi \sqrt{\frac{l_{35}}{g}}$

$T_{25}=2\pi \sqrt{\frac{l_{25}}{g}}$

So, $T_{35}=2\pi \sqrt{\frac{l_{25}(1+\alpha \Delta \theta )}{g}}$

$T_{35}=T_{25}(1+\alpha \Delta \theta {)}^{\frac{1}{2}}$

for $\alpha \Delta \theta <<1$

$T_{35}=T_{25}(1+\frac{1}{2}\alpha \Delta \theta )$

$T_{35}=T_{25}+I_{25}\times \frac{1}{2}\alpha \Delta \theta$

$T_{35}-T_{25}=\Delta T=T_{25}\times \frac{1}{2}\alpha \Delta \theta =7\times 24\times 60\times 60\times 0.5\times 1.2\times {10}^{-5}\times 10=36.28s$