A clock with an iron pendulum keeps correct time at $20$${}^{\text{0}}{\text{C}}$. How much time will its loss or gain in a day if the temperature changes to $40$${}^{\text{0}}{\text{C}}$? Thermal coefficient of linear expansion α=0.000012 per${}^{\text{0}}{\text{C}}$.
Answer
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Hint: To solve this question, we use the basic theory of Thermal expansion. As we know Thermal expansion occurs when an object expands and becomes larger due to a change in the object's temperature. Similarly, in this case it will affect the operation of the clock when in a day if the temperature changes to ${\text{40}}{}^{\text{0}}{\text{C}}$. Some basic formulas are used to get our desired result in this problem.
Formula used- ${\Delta T = }\dfrac{{\text{1}}}{{\text{2}}}{\text{T}} \alpha {\Delta \theta }$
Where,
T is time period
ΔT is change in time period
Δθ is change in temperature
α is coefficient of linear expansion
Complete step-by-step answer:
Given data:
A clock which show correct time at 20°C, is now placed to 40${}^{\text{0}}{\text{C}}$
Coefficient of linear expansion of the pendulum = 12 × ${10^{ - 6}}$ per ${}^{\text{0}}{\text{C}}$.
For finding, how much will it gain or lose in time:
We have the equation,
${\Delta T = }\dfrac{{\text{1}}}{{\text{2}}}{\text{T}} \alpha {\Delta \theta }$
Where,
T is time period
ΔT is change in time period
Δθ is change in temperature
α is coefficient of linear expansion
$ \Rightarrow $${ \Delta t = }\dfrac{{{\Delta T}}}{{{{\text{T}}^{\text{'}}}}}$
But, T’ = T
$ \Rightarrow $${\Delta t = }\dfrac{{\text{1}}}{{\text{2}}}\alpha {\Delta \theta (t)}$
= $\dfrac{1}{2}$$ \times $0.000012$ \times $20$ \times $24$ \times $3600
= 10.37 s
At higher temperature the length of the pendulum clock will be increased as compared to the previous condition.
So, the time period will be more and it will lose the time.
Note: The expansion can occur in length of iron pendulum in which case it is called Linear Expansion. And If we take a square tile and then after heat it, the expansion will be on two fronts that is length and breadth, and it is called Area Expansion. Similarly, if we take a cube shape structure and heat it, all its sides expand and now the body experiences an increase in the overall volume of the structure and it is called Volume Expansion.
Formula used- ${\Delta T = }\dfrac{{\text{1}}}{{\text{2}}}{\text{T}} \alpha {\Delta \theta }$
Where,
T is time period
ΔT is change in time period
Δθ is change in temperature
α is coefficient of linear expansion
Complete step-by-step answer:
Given data:
A clock which show correct time at 20°C, is now placed to 40${}^{\text{0}}{\text{C}}$
Coefficient of linear expansion of the pendulum = 12 × ${10^{ - 6}}$ per ${}^{\text{0}}{\text{C}}$.
For finding, how much will it gain or lose in time:
We have the equation,
${\Delta T = }\dfrac{{\text{1}}}{{\text{2}}}{\text{T}} \alpha {\Delta \theta }$
Where,
T is time period
ΔT is change in time period
Δθ is change in temperature
α is coefficient of linear expansion
$ \Rightarrow $${ \Delta t = }\dfrac{{{\Delta T}}}{{{{\text{T}}^{\text{'}}}}}$
But, T’ = T
$ \Rightarrow $${\Delta t = }\dfrac{{\text{1}}}{{\text{2}}}\alpha {\Delta \theta (t)}$
= $\dfrac{1}{2}$$ \times $0.000012$ \times $20$ \times $24$ \times $3600
= 10.37 s
At higher temperature the length of the pendulum clock will be increased as compared to the previous condition.
So, the time period will be more and it will lose the time.
Note: The expansion can occur in length of iron pendulum in which case it is called Linear Expansion. And If we take a square tile and then after heat it, the expansion will be on two fronts that is length and breadth, and it is called Area Expansion. Similarly, if we take a cube shape structure and heat it, all its sides expand and now the body experiences an increase in the overall volume of the structure and it is called Volume Expansion.
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