Question

A clock with a metal pendulum beating seconds keeps correct time at $0^{\circ }C$. If it loses $12.5s$ a day at ${25}^{\circ }C$, the coefficient of linear expansion of the metal pendulum is

• A. $\frac{1}{86400}{/}^{\circ }C$
• B. $\frac{1}{43200}{/}^{\circ }C$
• C. $\frac{1}{14400}{/}^{\circ }C$
• D. $\frac{1}{28800}{/}^{\circ }C$

Answer 1

Answer: (A)

$\frac{1}{86400}{/}^{\circ }C$

Fractional increment in time with temperature change

$\frac{\Delta t}{t}=\frac{1}{2}\alpha \Delta T$ eq(1)

where

$\frac{\Delta t}{t}$ = fractional increment

$\alpha$ = coefficient of linear expansion

$\Delta T$= change in temperature

@ $T_0=0^{o}C$ ,clocks shows correct time, and at $T={25}^{o}C$ it loses 12.5 s

$\frac{\Delta t}{t}$ = $\frac{-12.5}{24\times 60\times 60}$

$\Delta T$ = $T_0-T$ =0 - 25 = ${-25}^{0}C$

Substitute values back to eq(1)

$\frac{-12.5}{24\times 60\times 60}$ = $\frac{1}{2}\times \alpha \times (-25)$

$\frac{-12.5}{86400}$ = $-12.5\times \alpha$

$\alpha =\frac{1}{86400}{/}^{o}C$

So A is correct.Fractional increment in time with temperature change

$\frac{\Delta t}{t}=\frac{1}{2}\alpha \Delta T$ eq(1)

where

$\frac{\Delta t}{t}$ = fractional increment

$\alpha$ = coefficient of linear expansion

$\Delta T$= change in temperature

@ $T_0=0^{o}C$ ,clocks shows correct time, and at $T={25}^{o}C$ it loses 12.5 s

$\frac{\Delta t}{t}$ = $\frac{-12.5}{24\times 60\times 60}$

$\Delta T$ = $T_0-T$ =0 - 25 = ${-25}^{0}C$

Substitute values back to eq(1)

$\frac{-12.5}{24\times 60\times 60}$ = $\frac{1}{2}\times \alpha \times (-25)$

$\frac{-12.5}{86400}$ = $-12.5\times \alpha$

$\alpha =\frac{1}{86400}{/}^{o}C$

So A is correct.