Question

A pendulum clock loses $12\text{s}$ a day if the temperature is ${40}^{\circ }\text{C}$ and goes fast by $4\text{s}$ a day if the temperature is ${20}^{\circ }\text{C}$. Find the temperature in ${}^{\circ }\text{C}$ at which the clock will show correct time.

Answer 1

Let $T$ be the temperature at which the clock is correct.
Time lost per day $=(0.5\alpha$ (rise in temperature) $\times 86400$
$\Rightarrow 12=0.5\alpha (40-T)\times 86400$ ... (i)
Time gained per day $=0.5\alpha$ (drop in temperature) $\times 86400$
$\Rightarrow 4=0.5\alpha (T-20)\times 86400$ ... (ii)
Adding Eqs. (i) and (ii), we get
$32=86400\alpha (40-20)\Rightarrow \alpha =1.85\times {10}^{-5}{/}^{\circ }\text{C}$
Dividing Eq. (i) by Eq. (ii), we get
$12(T-20)=4(40-T)\Rightarrow T={25}^{\circ }\text{C}$
$\Rightarrow$ Clock shows correct time at ${25}^{\circ }\text{C}$