A clock with a metal pendulum beating seconds keeps correct time at 0°C. If it loses 12.5 seconds a day at 25°C, the coefficient of linear expansion of metal of pendulum is

To find the coefficient of linear expansion of the metal pendulum, we can analyze how the time period of a pendulum changes with temperature.

The time period $T$ of a simple pendulum of length $l$ is given by the formula:
$T=2\pi \sqrt{\frac{l}{g}}$
where $g$ is the acceleration due to gravity.

When the temperature changes by $\Delta \theta$, the change in length of the pendulum due to linear expansion is given by:
$l^{\prime }=l(1+\alpha \Delta \theta )$
where $\alpha$ is the coefficient of linear expansion of the pendulum's metal.

The new time period $T^{\prime }$ becomes:
$T^{\prime }=2\pi \sqrt{\frac{l(1+\alpha \Delta \theta )}{g}}=T{(1+\alpha \Delta \theta )}^{1/2}$

Since the coefficient of linear expansion $\alpha$ is extremely small, we can use the binomial approximation ${(1+x)}^n\approx 1+nx$:
$T^{\prime }\approx T(1+\frac{1}{2}\alpha \Delta \theta )$

The fractional change in the time period is:
$\frac{\Delta T}{T}=\frac{T^{\prime }-T}{T}=\frac{1}{2}\alpha \Delta \theta$

The time lost in a total time interval $t$ is given by:
$\Delta t=t·\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \theta ·t$

We are given the following values:
Time lost in a day, $\Delta t=12.5\text{seconds}$
Total time in a day, $t=24\times 3600\text{seconds}=86400\text{seconds}$
Change in temperature, $\Delta \theta =25°\text{C}-0°\text{C}=25°\text{C}$

Substitute these values into the formula:
$12.5=\frac{1}{2}·\alpha ·25·86400$

Simplify and solve for $\alpha$:
$12.5=12.5·86400·\alpha$
$1=86400·\alpha$
$\alpha =\frac{1}{86400}\text{per}°\text{C}$

Thus, the coefficient of linear expansion of the metal of the pendulum is $1/86400\text{per}°\text{C}$.