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# A pendulum clock keeps correct time at ${0^ \circ }C$ .If the coefficient of linear expansion is $\alpha$ , then what will be the loss in time per day , when the temperature rises by ${t^ \circ }C$ ?

Last updated date: 23rd Jul 2024
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Hint: It is given in the question that the pendulum clock is accurate at ${0^ \circ }C$ . As the temperature increases the pendulum starts expanding. The thermal coefficient of expansion is used to determine this expansion. TO find the time loss we can use the following formula:
$\dfrac{{\Delta t}}{t} = \dfrac{1}{2}\alpha \Delta T$

In this question we are provided with the coefficient of expansion $\alpha$ .
The rise in temperature is ${t^ \circ }C$ .
Using the formula,
$\dfrac{{\Delta t}}{t} = \dfrac{1}{2}\alpha \Delta T$
$\dfrac{{\Delta t}}{t}$ Fractional increase in time
$\alpha$ Coefficient of linear expansion
$\Delta T$ change in temperature = ${t^ \circ }C$
Time in day (seconds) $t = 24 \times 60 \times 60$
Putting all values in the formula e get,

$\dfrac{{\Delta t}}{t} = \dfrac{1}{2}\alpha \Delta T \\ \Rightarrow \;\dfrac{{\Delta t}}{{24 \times 60 \times 60}} = \dfrac{1}{2}\alpha (t) \\ \Rightarrow \Delta t = 43200\alpha t \\$
So, the time lost per day will be
$\Rightarrow \Delta t = 43200\alpha t$ .

Note: The loss in time per day of a pendulum may change with the metal used for making the pendulum. As the coefficient of linear expansion is dependent on the properties of metal, time lost will be different for different metal pendulums used. Additionally, the advantage of the pendulum used in clocks is that it is an harmonic oscillator which helps it swing in precise time intervals.