Question

A clock while keeps correct time at ${30}^{\circ }C$ has a pendulum rod made of brass. The number of seconds it gains (or) looses per second when the temperature falls to ${10}^{\circ }C$ is
$[\alpha$ of brass = $18\times {10}^{-6}{/}^{\circ }C]$

• A. $18\times {10}^{-6}sec$
• B. $18\times {10}^{-5}sec$
• C. 0)0018 sec
• D. 0)018 sec

Answer 1

Answer: (B)

$18\times {10}^{-5}sec$

We know, $\alpha =dl/(dt\times l)$

or, $dl=\alpha \times dt\times l$ -----(1)

Now $T=2\pi \sqrt{l/g}$ --- (2)

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

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

Using (2)

$1+dT/T=(l+dl/l{)}^{1/2}$

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

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

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

Using (1)

$dT=\left(1/2\right)\left[\right(\alpha dtl\left)/l\right]XT$

$=\left(1/2\right)\times 18\times {10}^{-6}\times \times 1$

$=18\times {10}^{-5}$