Question

A pendulum clock consists of a $1\text{mm}$ rod connected to a small, heavy bob. If it is designed to keep correct time at ${30}^{\circ }\text{C}$, how fast or slow will it go in 12 hours at ${50}^{\circ }\text{C}$? Coefficient of linear expansion of iron $=1.2\times {10}^{-5}{/}^{\circ }\text{C}$.

• A. $5.2\text{sec}$, slow
• B. $5.2\text{sec}$, fast
• C. $4\text{sec},$ fast
• D. $4\text{sec},\text{slow}$

Answer 1

The correct option is A $5.2\text{sec}$, slow

Time lost/gained due to temperature difference $\Delta \theta$ is given by
$\Delta T=T_0\times \frac{1}{2}\alpha \Delta \theta$
So, time period at temperature ${}^{\prime }{\theta }^{\prime }$ is
$T=T_0(1+\frac{\alpha \theta }{2})$
$T_{30}=T_0[1+\frac{\alpha \left(30\right)}{2}]=T_0(1+15\alpha )$
$T_{50}=T_0[1+\frac{\alpha \left(50\right)}{2}]=T_0(1+25\alpha )$
$\Rightarrow \frac{T_{50}}{T_{30}}=\frac{T_0(1+25\alpha )}{T_0(1+15\alpha )}=(1+25\alpha )(1+15\alpha {)}^{-1}$
$=(1+25\alpha )(1-15\alpha )$
$=(1+25\alpha -15\alpha -375{\alpha }^2)=(1+10\alpha )$
(neglecting square term as $\alpha$ is very small)

$\frac{T_{50}-T_{30}}{T_{50}}=\frac{T_{50}}{T_{30}}-1=(1+10\alpha )-1=10\alpha$
$=10\alpha =10\times 1.2\times {10}^{-5}=1.2\times {10}^{-4}$
$\frac{T_{50}-T_{30}}{T_{50}}$ shows fractional loss in time and this is (+) ve. Thus the clock goes slow.
Time lost in 12 hours
$\Delta t=12\times 60\times 60\times 1.2\times {10}^{-4}=5.18\approx 5.2\text{sec}$