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Question: A clock which keeps correct time at \({25^ \circ }C\) has a pendulum made of a metal. The temperatur...

A clock which keeps correct time at 25C{25^ \circ }C has a pendulum made of a metal. The temperature falls to 0C{0^ \circ }C. If the coefficient of linear expansion of the metal is 1.9×105perC1.9 \times {10^{ - 5}}pe{r^ \circ }C, then find the number of seconds the clock gains per day.

Explanation

Solution

We can use the relation connecting the gain or loss of time and the change in temperature to solve this problem. The relation connecting gain or loss in time and the change in temperature is given as
Δt=12×α×Δθ×t\Delta t = \dfrac{1}{2} \times \alpha \times \Delta \theta \times t
Where, Δt\Delta t is the gain or loss in time.
α\alpha is the coefficient of linear expansion.
Δθ\Delta \theta is the change in temperature and t is the time.
Since there is an increase in temperature the value that we get will be the gain in time.

Complete step by step solution:
It is given that a clock keeps correct time at 25C{25^ \circ }C. So initial temperature θ1=25C{\theta _1} = {25^ \circ }C .
The pendulum is made of metal. We need to find the number of seconds gained per day by the clock when temperature falls to 0C{0^ \circ }C .
So final temperature θ2=0C{\theta _2} = {0^ \circ }C
So, θ1θ2=250=25C{\theta _1} - {\theta _2} = 25 - 0 = {25^ \circ }C
The relation connecting gain or loss in time and the change in temperature is given as
Δt=12×α×Δθ×t\Delta t = \dfrac{1}{2} \times \alpha \times \Delta \theta \times t
Where Δt\Delta t is the gain or loss in time.
α\alpha is the coefficient of linear expansion.
Δθ\Delta \theta is the change in temperature and t is the time.
α\alpha is given as 1.9×105perC1.9 \times {10^{ - 5}}pe{r^ \circ }C
We need to calculate the gain of time per day.
Thus t=24×60×60st = 24 \times 60 \times 60\,s
Change in temperature is Δθ=25C\Delta \theta = {25^ \circ }C
Now let us substitute all the given values in the equation.
Δt=12×19×105×25×24×60×60\Delta t = \dfrac{1}{2} \times 1 \cdot 9 \times {10^{ - 5}} \times 25 \times 24 \times 60 \times 60
Δt=20520s\therefore \Delta t = 20\cdot520\,s

This is the gain in time when the temperature falls to 0C{0^ \circ }C . We know that the time period of a pendulum increases with increase in length. When the temperature falls the length of the pendulum is decreasing so the time period becomes less. That means the time goes faster so we get a gain in time. So this value that we got is the number of seconds gained.

Note: In cases where the temperature is increasing from a low value to a high value, we will find that there will be a loss in time. Because in that case the length of the pendulum increases due to expansion of metal with temperature increase and hence the time period increases. An increase in time period means time is running slower in that case. So there will be loss of time. In our case as temperature is decreasing there is gain of time.