Question

A clock with a metal pendulum beating seconds keeps correct time at ${25}^{\circ }\text{C}$. If it loses $25\text{s}$ a day at ${50}^{\circ }\text{C}$, choose the correct option(s):

• A. The coefficient of linear expansion of the metal pendulum is $\frac{1}{86400}{/}^{\circ }\text{C}$.
• B. The coefficient of linear expansion of the metal pendulum is $\frac{1}{43200}{/}^{\circ }\text{C}$.
• C. The coefficient of linear expansion of the metal pendulum remains same even if the clock shows correct time at $0^{\circ }\text{C}$ and loses $25\text{s/day}$ at ${25}^{\circ }\text{C}$.
• D. The coefficient of linear expansion of the metal pendulum remains same even if clock shows correct time at $0^{\circ }\text{C}$ and loses $25\text{s/day}$ at ${50}^{\circ }\text{C}$.

Answer 1

Answer: (B), (C)

The coefficient of linear expansion of the metal pendulum remains same even if the clock shows correct time at $0^{\circ }\text{C}$ and loses $25\text{s/day}$ at ${25}^{\circ }\text{C}$.

Fractional increment in time with temperature change is given by
$\frac{\Delta t}{t}=\frac{1}{2}\alpha \Delta T......\left(1\right)$
where
$\frac{\Delta t}{t}$ = fractional increase/decrease in time
$\alpha$ = coefficient of linear expansion of pendulum
$\Delta T$ = change in temperautre

Given that, at $T_0={25}^{\circ }\text{C}$, the clock shows correct time, and at $T={50}^{\circ }\text{C}$, it loses $25\text{s}$ in a day.
$\therefore \frac{\Delta t}{t}=\frac{-25}{24\times 60\times 60}$
& $\Delta T=T_0-T=25-50=-{25}^{\circ }\text{C}$
Substituting values back into $\left(1\right)$, we get
$\frac{-25}{24\times 60\times 60}=\frac{1}{2}\times \alpha \times (-25)$
or $\alpha =\frac{1}{43200}{/}^{\circ }\text{C}$

If at $T_0=0^{\circ }\text{C}$, clock shows correct time and at $T={25}^{\circ }\text{C}$, it loses $25\text{s}$ in a day,
$\frac{\Delta t}{t}=\frac{-25}{24\times 60\times 60}$
& $\Delta T=T_0-T=-{25}^{\circ }\text{C}$
Then from $\left(1\right)$, we get
$\Rightarrow \frac{-25}{24\times 60\times 60}=\frac{1}{2}\times \alpha \times (-25)$
$\Rightarrow \alpha =\frac{1}{43200}{/}^{\circ }\text{C}$
Hence, option (c) is also correct.
But if $\Delta T=0-50=-{50}^{\circ }\text{C}$ (in option D),
$\alpha =\frac{1}{86400}{/}^{\circ }\text{C}$
i.e Option (d) is wrong.

Thus, options (b) and (c) both are correct answers.