Question

A clock with a metallic pendulum is $5s$ fast each day at a temperature of ${15}^{\circ }C$ and $10s$ slow each day at a temperature of ${30}^{\circ }C$. Find coefficient of linear expansion for the metal.

• A. $2.6\times {10}^{-5}{{}^{\circ }C}^{-1}$
• B. $2.31\times {10}^{-5}{{}^{\circ }C}^{-1}$
• C. $4.62\times {10}^{-5}{{}^{\circ }C}^{-1}$
• D. $1.15\times {10}^{-5}{{}^{\circ }C}^{-1}$

Answer 1

Answer: (B)

$2.31\times {10}^{-5}{{}^{\circ }C}^{-1}$
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

assume @ T = $T_0$ , there is no gain and loss in time
case 1:
$\frac{\Delta t}{t}$ = gain 5 sec per day
= $\frac{5}{24\times 60\times 60}$
$\Delta T$ = $T_0$ - 15
substitute values back in eq(1)
$\frac{5}{24\times 60\times \times 60}$ = $\frac{1}{2}\times \alpha \times (T_0-15)$ eq(2)
case 2:
$\frac{\Delta t}{t}$ = loss 10 sec per day
= -$\frac{10}{24\times 60\times 60}$
$\Delta T$ = $T_0$ - 30
substitute values back in eq(1)
-$\frac{10}{24\times 60\times \times 60}$ = $\frac{1}{2}\times \alpha \times (T_0-30)$ eq(3)

$\frac{eq\left(3\right)}{eq\left(2\right)}$
-2 = $\frac{T_0-30}{T_0-15}$
$T_0-30$ = -2 $\times (T_0-15)$
$T_0-30$ = $-2T_0+30$
$3T_0$ = 60
$T_0$ = ${20}^{o}C$
substitute this value of $T_0$ in eq(2)
$\frac{5}{24\times 60\times 60}$ = $\frac{1}{2}\times \alpha \times (20-15)$
$\frac{5}{24\times 60\times 60}$ = $\frac{1}{2}\times \alpha \times 5$
solve for $\alpha$
$\alpha$ = 2.31 $\times {10}^{-5}{}^o{C}^{-1}$
so is correct.