Question

A clock with an iron pendulum keeps correct time at ${20}^{\circ }\text{C}$. If temperature is changed to ${40}^{\circ }\text{C}$, then which of the following is/are true?
[Take $\alpha =0.000012{/}^{\circ }\text{C}$]

• A. Pendulum will gain $1.2\times {10}^{-4}$ seconds per each second.
• B. Pendulum will lose $1.2\times {10}^{-4}$ seconds per each second.
• C. Pendulum will gain $10.368$ seconds/day.
• D. Pendulum will lose $10.368$ seconds/day.

Answer 1

Answer: (B), (D)

Pendulum will lose $10.368$ seconds/day.

Time period of pendulum at ${20}^{\circ }\text{C}$
$T_{20}=2\pi \sqrt{\frac{l}{g}}$
where $l$ is the length of pendulum at ${20}^{\circ }\text{C}$.
Time period at ${40}^{\circ }\text{C}$
$T_{40}=2\pi \sqrt{\frac{l(1+\alpha \Delta T)}{g}}$
$\therefore$ Change in time period.
$T_{40}-T_{20}=2\pi \sqrt{\frac{l(1+\alpha \Delta T)}{g}}-2\pi \sqrt{\frac{l}{g}}$
$=2\pi \sqrt{\frac{l}{g}}(\sqrt{1+\alpha \Delta T}-1)$
$\Rightarrow \frac{T_{40}-T_{20}}{T_{20}}=\sqrt{1+\alpha \Delta T}-1$
Using binomial expansion, $\sqrt{1+\alpha \Delta T}=1+\frac{1}{2}\alpha \Delta T$
$\Rightarrow \frac{T_{40}-T_{20}}{T_{20}}=\frac{1}{2}\alpha \Delta T$

Since the time period increases, pendulum will lose time.
& Loss of time per $1$ second
$=\frac{1}{2}\times 0.000012\times 20=1.2\times {10}^{-4}\text{s}$
Loss in $24$ hr $=1.2\times {10}^{-4}\times 60\times 60\times 24$
$=10.368\text{s}$
Hence options (b) and (d) are correct.