Heat and Thermodynamics 2 Question 2
5. A pendulum clock loses $12 s$ a day if the temperature is $40^{\circ} C$ and gains $4 s$ a day if the temperature is $20^{\circ} C$. The temperature at which the clock will show correct time, and the coefficient of linear expansion $\alpha$ of the metal of the pendulum shaft are, respectively.
(2016 Main)
(a) $25^{\circ} C, \alpha=1.85 \times 10^{-5} /{ }^{\circ} C$
(b) $60^{\circ} C, \alpha=1.85 \times 10^{-4} /{ }^{\circ} C$
(c) $30^{\circ} C, \alpha=1.85 \times 10^{-3} /{ }^{\circ} C$
(d) $55^{\circ} C, \alpha=1.85 \times 10^{-2} /{ }^{\circ} C$
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Answer:
Correct Answer: 5. (a)
Solution:
$$ \begin{aligned} T _0 & =2 \pi \sqrt{\frac{L}{g}} \\ T^{\prime} & =T _0+\Delta T=2 \pi \sqrt{\frac{L+\Delta L}{g}} \\ \therefore \quad T^{\prime} & =T _0+\Delta T=2 \pi \sqrt{\frac{L(1+\alpha \Delta \theta)}{g}} \\ & ={2 \pi \sqrt{\frac{L}{g}} }(1+\alpha \Delta \theta)^{\frac{1}{2}} \approx T _0\left(1+\frac{\alpha \Delta \theta}{2}\right) \\ \therefore \quad \Delta T & =T^{\prime}-T _0=\frac{\alpha \Delta \theta T _0}{2} \cdots(i)\\ \text { or } \quad \frac{\Delta T _1}{\Delta T _2} & =\frac{\alpha \Delta \theta _1 T _0}{\alpha \theta _2 T _0} \\ \Rightarrow \quad \frac{12}{4} & =\frac{40-\theta}{\theta-20} \\ \Rightarrow \quad 3(\theta-20) & =40-\theta \\ \Rightarrow \quad 4 \theta & =100 \\ \Rightarrow \quad \theta & =25^{\circ} C \end{aligned} $$
Time gained or lost is given by
$$ \Delta T=\left(\frac{\Delta T}{T _0+\Delta T}\right) t \approx \frac{\Delta t}{T _0} t $$
From Eq. (i), $\quad \frac{\Delta T}{T _0}=\frac{\alpha \Delta \theta}{2}$
$$ \begin{aligned} \therefore & \Delta t & =\frac{\alpha(\Delta \theta) t}{2} \\ & 12 & =\frac{\alpha(40-25)(24 \times 3600)}{2} \\ \therefore & \alpha & =1.85 \times 10^{-5} /{ }^{\circ} C \end{aligned} $$
