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Physics Question on thermal properties of matter

A pendulum clock loses 12 s12\,s a day if the temperature is 40∘C40^{\circ}C and gains 4 s4\,s a day if the temperature is 20∘C20^{\circ}C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α)(\alpha) of the metal of the pendulum shaft are respectively :

A

25∘C;α=1.85×10−5/∘C25^{\circ} C ; \alpha = 1.85 \times 10^{-5} / {^{\circ}C}

B

60∘C;α=1.85×10−4/∘C60^{\circ} C ; \alpha = 1.85 \times 10^{-4} / {^{\circ}C}

C

30∘C;α=1.85×10−3/∘C30^{\circ} C ; \alpha = 1.85 \times 10^{-3} / {^{\circ}C}

D

55∘C;α=1.85×10−2/∘C55^{\circ} C ; \alpha = 1.85 \times 10^{-2} / {^{\circ}C}

Answer

25∘C;α=1.85×10−5/∘C25^{\circ} C ; \alpha = 1.85 \times 10^{-5} / {^{\circ}C}

Explanation

Solution

Time loss or gain is given by
Δt=(ΔTT)t=12α.Δθ.t\Delta t = \left(\frac{\Delta T}{T}\right)t = \frac{1}{2} \alpha. \Delta\theta.t
∴ 12=12α(40−θ0)×1d\therefore \, 12 = \frac{1}{2} \alpha\left(40 - \theta_{0}\right) \times1d .....(1)
4=12α(θ0−20)×1d4 = \frac{1}{2} \alpha\left( \theta_{0} - 20\right)\times1d .....(2)
(1)(2)\frac{(1)}{(2)} gives
3=(40−θ0)(θ0−20)3 = \frac{\left(40 - \theta_{0}\right)}{\left(\theta_{0} - 20\right)}
Solving θ0=25∘C\theta_{0} = 25^{\circ} C and putting in (2)
α=85×29×60×60=\alpha = \frac{8}{5 \times 29\times 60\times 60} =