Question

A seconds pendulum clock has a steel wire. The clock shows correct time at ${25}^{0}C$. How much time does the clock loose or gain, in one week, when the temperature is increased to ${35}^{0}C$?
($\alpha steel=1.2\times {10}^{-5}{/}^{0}C$)

• A. 321.5 s
• B. 3.828 s
• C. 82.35 s
• D. 36.28 s

Answer 1

The correct option is D 36.28 s

According to the question,

We know that the loss ad gain in time period of any pendulum is $\frac{\Delta T}{T}$

$=\frac{1}{2}\alpha \Delta \theta$ (Where $\Delta T$ is gain or loss of time is

$T=7$ ,week $=7\times 86400$ seconds $=604800$ seconds, and $\Delta \theta$ is change in term).

Now, putting all the values from the question, we get,

$\frac{\Delta T}{604800}=\frac{1}{2}\times 1.2\times {10}^{-5}\times \left(35-25\right)$

$\frac{\Delta T}{604800}=\frac{1}{2}\times 1.2\times {10}^{-5}\times 10$

$\Delta T=36.25$ seconds.

Hence, as the temperature increases the length of steel increases and its time period is also increases.

So, it will become slow by $36.28$ seconds.