Question

A pendulum clock runs fast by 5 seconds per day at ${20}^{\circ }C$ and goes slow by 10 seconds per day at ${35}^{\circ }C$. It shows correct time at a temperature of :

• A. ${27.5}^{\circ }C$
• B. ${25}^{\circ }C$
• C. ${30}^{\circ }C$
• D. ${33}^{\circ }C$

Answer 1

Answer: (B)

${25}^{\circ }C$

We know, $\alpha =dl/(dt\times l)$ [dl = length increase, dt = temperature increase]
Or, $dl=\alpha \times dt\times l$ ---- (1)

Now. $T=2\pi \sqrt{l/g}$ ---(2) [T = time period]

$T+dT=2\pi \sqrt{(l+dl)/g}$

or, $T(1+dT/T)=2\pi \sqrt{l/g}\sqrt{(1+dl/l)}$ ---- (3)

Comparing with (2), we get,
$1+dT/T=(1+dl/l{)}^{1/2}$

$=1+\left(1/2\right)dl/l$

$dT/T=\left(1/2\right)dl/l$

$dT=\left(1/2\right)dl/l\times T$

using (1)
$dT=\left(1/2\right)(\alpha \times dt\times l\times T)/l$

Or, $dT=\left(1/2\right)(\alpha \times dt\times T)$

In case 1, the clock goes first by 5 sec at ${20}^0$ C

In case 2, the clock goes slow by 10 sec at ${35}^0$ C

So, $10=\left(1/2\right)\alpha (35-t)\times 86400$

and $5=\left(1/2\right)\alpha (t-20)\times 86400$

Solving these two equations we get, t = ${25}^0$C