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

$T_t$ = $2\pi \sqrt{\frac{L_t}{g}}$.

Now, $I_{40}$ = $I_{20}(1+\alpha (40-20\left)\right)$

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

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

= $2\pi \sqrt{l_{20}(1+20\alpha )g}$

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

= $2\pi \sqrt{\frac{L_{20}}{g}}(1+\frac{1}{2}\times 20\alpha )$ (by binomial approximation, as 2 is very small)

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

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

$\Rightarrow \frac{T_{40}-T_{20}}{T_{20}}$ = $10\alpha$ = $1.2\times {10}^{-4}$ 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 \(86400~\times~(1.2 × 10^{-6})\) seconds = $10.4$ seconds.