Question

A clock is calibrated at a temperature of ${20}^{\circ }C$. Assume that the pendulum is a thin brass rod of negligible mass with a heavy bob attached to the end $({\alpha }_{brass}=19\times {10}^{-6}/K)$

• A. On a hot day at ${30}^{\circ }C$ the clock gains $8.2s$
• B. On a hot day at ${30}^{\circ }C$ the clock loses $8.2s$
• C. On a cold day at ${10}^{\circ }C$ the clock gains $8.2s$
• D. On a cold day at ${10}^{\circ }C$ the clock loses $8.2s$

Answer 1

Answer: (B), (C)

The correct options are
B On a hot day at ${30}^{\circ }C$ the clock loses $8.2s$
C On a cold day at ${10}^{\circ }C$ the clock gains $8.2s$

time period T = $2\Pi {\left(l/g\right)}^{1/2}$ eq(1)
where
$l$ = distance of bob from hinge
$g$ = gravitational acceleration
as temperature increases , $l$ increases so time period also increases.

option A:
on a hot day temperature increases, so Time period increases, so clock will loses time not gains.
so A is incorrect.
option B:
fractional increment in time with Temperature change
$\frac{\Delta t}{t}=\frac{1}{2}\alpha \Delta T$ eq(2)
where
$\frac{\Delta t}{t}$ = fractional increment
$\alpha$ = coefficient of linear expansion
$\Delta T$= change in temperature
@ $T_0={20}^{o}C$ ,clocks shows correct time
$T={30}^{o}C$
$\Delta T$ = $T_0-T$ =20 - 30 = ${-10}^{o}C$ = $-10K$
$\alpha$ = $19\times {10}^{-6}/K$

substitute values back to eq(2)
$\frac{\Delta t}{t}$ = $\frac{1}{2}\times (19\times {10}^{-6})\times (-10)$
$\frac{\Delta t}{t}$ = $-9.5\times {10}^{-5}$

Clock loses $9.5\times {10}^{-5}$ sec per sec. In a day it will lose
= $9.5\times {10}^{-5}\times (24\times 60\times 60)$
= $8.2secs$
B is correct.

option C:
@ $T_0={20}^{o}C$ ,clocks shows correct time
$T={10}^{o}C$
$\Delta T$ = $T_0-T$ =20 - 10 = ${10}^{o}C$ = $10K$
$\alpha$ = $19\times {10}^{-6}/K$

substitute values back to eq(2)
$\frac{\Delta t}{t}$ = $\frac{1}{2}\times (19\times {10}^{-6})\times \left(10\right)$
$\frac{\Delta t}{t}$ = $9.5\times {10}^{-5}$

Clock gains $9.5\times {10}^{-5}$ sec per sec. In a day it will lose
= $9.5\times {10}^{-5}\times (24\times 60\times 60)$
= $8.2secs$
C is correct.

option D:
on a cold day,temperature decreases , so Time period decreases as a result clock runs faster or it gains time.
so D is incorrect.