A pendulum gives correct time at $20^{o} C$ at a place where $g = 9.8 m s^{- 2}$ . The pendulum consists of a light steel rod connected to a heavy ball. If it is taken to a different place where $g = 9.788 m s^{- 2}$ . At what temperature will it give correct time? Coefficient of linear expansion of steel is ${12 \times 10^{- 6}}^{o} C^{- 1}$ .

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Solution

We know that time period of a pendulum is given as $T = 2 π \sqrt[2]{l / g}$
Where $l$ is length of pendulum
If we change the value of $g$ then to make time period unaffected we need to change the length $l$ such that the ratio $\frac{l}{g}$
does not change.

New length $l^{'} = l (1 + α Δ t)$

So we get $\frac{l}{g} = \frac{l (1 + α Δ t)}{g^{'}}$

or $\frac{l}{9.8} = \frac{l (1 + α Δ t)}{9.788}$

Putting $α = 12 \times 10^{- 6}$
we get $Δ t = - 102^{0} C$
So it will give correct time at $20 + (- 102) = - 82^{0} C$

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