Question

A pendulum clock loses 12 seconds a day if the temperature is ${40}^{\circ }C$ and goes fast by 4 seconds a day if the temperature is ${20}^{\circ }C$. Find the temperature at which the clock will show correct time, and the coefficient of linear expansion of the metal of the pendulum shaft.

• A. ${26}^{\circ }C$; $1.25\times {10}^{-5}/$${}^{\circ }C$
• B. ${20}^{\circ }C$; $1.55\times {10}^{-5}/$${}^{\circ }C$
• C. ${25}^{\circ }C$;$1.85\times {10}^{-5}/$${}^{\circ }C$
• D. ${22}^{\circ }C$ ;$1.75/$${}^{\circ }C$

Answer 1

The correct option is C

${25}^{\circ }C$;$1.85\times {10}^{-5}/$${}^{\circ }C$

Let $T$ be the temperature at which the clock is correct.

Time lost per day =

$\frac{1}{2}\times a\times \left(riseintemperature\right)\times 86400s$, (see hint)

$\Rightarrow 12=\frac{1}{2}\times a\times (40-T)\times 86400.$ ...........(1)

Time gained per day -

$\frac{1}{2}\times a\times \left(riseintemperature\right)\times 86400s$,

$\Rightarrow 4=\frac{1}{2}\times a\times (T-20)\times 86400.$ ...........(2)

On adding (1) and (2), we get

$64$ = $86400\times a\times (40-20)$

$\Rightarrow$ $a$ = $1.85\times {10}^{-5}{/}^{\circ }C$.

On dividing (1) and (2), we get

$12(T-20)$ = $4(40-T)$

$\Rightarrow$ $T$ = ${25}^{\circ }C$

$\Rightarrow$ clock shows correct time at ${25}^{\circ }C$ only!