Question

A clock with a metal pendulum gains $10$ seconds each day at ${15}^{o}C$ and loses $10$ second at ${30}^{o}C$. Then temperature at which it gives correct time is:

Answer 1

Suppose $t_1=10sec$ when $T={15}^{\circ }C$
$t_2=20sec$ when $T={30}^{\circ }C$

So, $T=2\pi \sqrt{\frac{L}{z}}$

$T^{\prime }=2\pi \sqrt{\frac{L}{z}(1+\alpha \theta )}$

$=T^{\prime }(1+\frac{1}{2}\alpha \theta )$

$T^{\prime }–T=dT=\frac{1}{2}\alpha \theta T$
$△T=\frac{1}{2}\alpha \theta T$

At ${15}^{\circ },$ gain $10sec$

$10=\frac{1}{2}\alpha (t–15)T_o$---- (1)

At ${30}^{\circ }C$, $10sec$ lossed

$-10=\frac{1}{2}\alpha (t–30)T_o$---(2)

Divide (1) and (2)

$-1=\frac{t-15}{t-30},-t+30=t-15$

$t=\frac{45}{2}={22.5}^{\circ }$