A pendulum clock gives correct time at $20^{\circ} C$ at a place where g = 9.800 $m s^{- 2}$ . The pendulum consists of a light steel rod connected to a heavy ball. 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 $= 12 \times 10^{- 6}^{\circ} C^{- 1}$

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Solution

$g_{1} = 9.8 m / s^{2},$

$g_{2} = 9.788 m / s^{2}$

$T_{1} = \frac{2 π \sqrt{l_{1}}}{\sqrt{g_{1}}}$

$T_{2} = \frac{2 π \sqrt{l_{2}}}{\sqrt{g_{2}}}$

$= \frac{2 π \sqrt{l_{2} (1 + α T)}}{\sqrt{g_{2}}}$

$α_{s t e e l} = 12 \times 10^{- 6} /^{\circ} C$

$Q_{1} = 20^{\circ}$

$Q_{2} = ?$

$T_{1} = T_{2}$

$\Rightarrow \frac{2 π \sqrt{l_{1}}}{\sqrt{g_{1}}} = \frac{2 π \sqrt{l_{1} (1 + α Δ T)}}{\sqrt{g_{2}}}$

$\Rightarrow \sqrt{(\frac{l_{1}}{g_{1}})} = \frac{\sqrt{l_{1} (1 + α Δ T)}}{\sqrt{g_{2}}}$

$\Rightarrow \frac{1}{9.8} = \frac{1 + 12 \times 10^{- 6} \times Δ T}{9.788}$

$\Rightarrow \frac{9.788}{9.8} = 1 + 12 \times 10^{- 6} \times Δ T$

$\Rightarrow \frac{9.788}{9.8} - 1 = 12 \times 10^{- 6} \times Δ T$

$\Rightarrow Δ T = \frac{- 0.00122}{12 \times 10^{- 6}}$

$\Rightarrow T_{2} - 20 = - 102.4$

$\Rightarrow T_{2} = - 102.4 + 20$

$= - 82.4 \approx - 82^{\circ} C$

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