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Q.

A clock with a metallic pendulum at 15^oC runs faster by 5 seconds each day and at 30^oC , runs slow by 10 seconds. Find the coefficient of linear expansion of the metal (nearly in $10^{- 6} /^{\circ} C$ ).

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Detailed Solution

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$\frac{T^{'}}{T} = \sqrt{\frac{L^{'}}{L}} = \sqrt{\frac{L (1 + αΔθ)}{L}} = 1 + \frac{1}{2} αΔθ$

$Δt = \frac{1}{2} αΔθt \Rightarrow \frac{Δt}{t} = \frac{1}{2} αΔθ$

$\Rightarrow \frac{5}{86400} = \frac{1}{2} α (θ - 15) . . (1)$

$\frac{10}{86400} = \frac{1}{2} α (30 - θ) . . (2)$

From (1) and (2)

$\frac{5}{10} = \frac{θ - 15}{30 - θ} \Rightarrow \Rightarrow 3 θ = 60 \Rightarrow θ = 20 ° C$

on Substituting the value of $θ in (1)$ we get

$α = 2.31 \times 10^{- 5}^{0} C$

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A clock with a metallic pendulum at 15oC runs faster by 5 seconds each day and at 30oC , runs slow by 10 seconds. Find the coefficient of linear expansion of the metal (nearly in 10-6 /∘ C ).