A pendulum clock loses 12 s a day if the temperature is 40^∘ C and gains 4 s day if the tempera

step 1 the question is saying there is a pendulum and suppose length of pendulum is of the clock so len

A pendulum clock loses 12 s a day if the temperature is 40^∘...

A pendulum clock loses 12 s a day if the temperature is $40^{\circ} \mathrm{C}$ and gains $4 \mathrm{~s}$ day if the temperature is $20^{\circ} \mathrm{C}$. The temperature at which the clock will show correct time, and the co-efficient of linear expansion $(\alpha)$ of the metal of the pendulum shaft are respectively. [2016] (A) $60^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ (B) $30^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-3} /{ }^{\circ} \mathrm{C}$ (C) $55^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-2} /{ }^{\circ} \mathrm{C}$ (D) $25^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-5} /{ }^{\circ} \mathrm{C}$

A pendulum clock loses 12 s a day if the temperature is $40^{\circ} \mathrm{C}$ and gains $4 \mathrm{~s}$ day if the temperature is $20^{\circ} \mathrm{C}$. The temperature at which the clock will show correct time, and the co-efficient of linear expansion $(\alpha)$ of the metal of the pendulum shaft are respectively. [2016]
(A) $60^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
(B) $30^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
(C) $55^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-2} /{ }^{\circ} \mathrm{C}$
(D) $25^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-5} /{ }^{\circ} \mathrm{C}$

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Key Concepts

Simple Pendulum Time Period

The time period of a simple pendulum is given by T = 2??(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. This period is sensitive to changes in L; a slight increase in the pendulum's length due to thermal expansion will increase the period, causing the clock to run slower, while a decrease in length will shorten the period, making the clock run faster.

Thermal Expansion

Thermal expansion is the tendency of materials to change their dimensions in response to changes in temperature. For a pendulum, the length increases with temperature according to the coefficient of linear expansion (?), as expressed by L = L?(1 + ??T). This change in length directly influences the time period of the pendulum, thereby affecting the clock’s accuracy.

Linear Approximation and Error Analysis

For small temperature changes, the effects of thermal expansion on the pendulum’s length – and thus its time period – can be approximated using a linear relationship. This linear analysis allows one to relate small deviations in timing (expressed in seconds per day) to corresponding changes in temperature, facilitating the calculation of the temperature at which the clock runs accurately as well as the determination of the coefficient of linear expansion.

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