Question

A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at ${20}^{\circ }C$, how fast or slow will it go in 24 hours at ${40}^{\circ }C$? Coefficient of linear expansion of iron = $1.2\times {10}^{-5}{/}^{\circ }C$.

• A. 10.4 seconds faster
• B. 10.4 minutes faster
• C. 1 minute slower
• D. 10.4 seconds slower

Answer 1

The correct option is D 10.4 seconds slower

Time period of a simple pendulum $T_t$ = $2\pi \sqrt{\frac{L_t}{g}}$.

Now length of pendulum at ${40}^{\circ }C$,
$L_{40}$ = $L_{20}(1+\alpha (40-20\left)\right)$

$\Rightarrow$ $L_{40}$ = $L_{20}$$(1+20\alpha )$

$\therefore$$T_{40}$ = $2\pi \sqrt{\frac{L_{40}}{g}}$

$T_{40}=2\pi \sqrt{\frac{L_{20}(1+20\alpha )}{g}}$

$T_{40}=2\pi \sqrt{\frac{L_{20}}{g}}(1+20\alpha {)}^{\frac{1}{2}}$

$T_{40}=2\pi \sqrt{\frac{L_{20}}{g}}(1+\frac{1}{2}(20\alpha \left)\right)$ (by binomial approximation)

$T_{40}=T_{20}(1+10\alpha )$

$\Rightarrow \frac{T_{40}}{T_{20}}$ = $1+10\alpha$

$\Rightarrow \frac{T_{40}-T_{20}}{T_{20}}$ = $10\alpha$ = $1.2\times {10}^{-4}$ seconds is the fractional loss of time.

Hence, as the temperature increases, so does the time period.

The clock slows down, in one day by $86400\times 1.2\times {10}^{-4}=10.4$ seconds.