Question

A pendulum clock having copper rod keeps correct time at ${20}^{\circ }C$. It gains $15s$ per day if cooled to $0^{\circ }C$. The coefficient of linear expansion of copper is

• A. $1.7\times {10}^{-4}{/}^{\circ }C$
• B. $1.7\times {10}^{-5}{/}^{\circ }C$
• C. $3.4\times {10}^{-4}{/}^{\circ }C$
• D. $3.4\times {10}^{-5}{/}^{\circ }C$

Answer 1

Answer: (B)

$1.7\times {10}^{-5}{/}^{\circ }C$
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
@ $T_0={20}^{o}C$ ,clocks shows correct time, and @ $T=0^{o}C$ it gains 15
$\frac{\Delta t}{t}$ = $\frac{15}{24\times 60\times 60}$
$\Delta T$ = $T_0-T$ =20 - 0 = ${20}^{0}C$

substitute values back to eq(1)
$\frac{15}{24\times 60\times 60}$ = $\frac{1}{2}\times \alpha \times \left(20\right)$
$\frac{15}{86400}$ = $10\times \alpha$
solve for $\alpha$
$\alpha =1.7\times {10}^{-5}{/}^{o}C$
so is correct.