A clock with an iron pendulum keeps correct time at $20^{\circ}$ C. How much time will it lose or gain in a day if the temperature changes to $40^{\circ}$ C? Thermal coefficient of linear expansion $α = 0.000012$ per $^{\circ}$ C.

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Solution

$Δ T = \frac{1}{2} T α Δ θ$
$Δ t = \frac{Δ T}{T^{'}}$
But $T^{'} = T$
$∴$ $Δ t = \frac{1}{2} α Δ θ (t)$
$= \frac{1}{2} \times 0.000012 \times 20 \times 24 \times 3600$
$= 10.37$ s
At higher temperature length of pendulum clock will be more. So time period will be more and it will lose the time.

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A clock with an iron pendulum keeps correct time at 20∘ C. How much time will it lose or gain in a day if the temperature changes to 40∘ C? Thermal coefficient of linear expansion α=0.000012 per∘C.

ΔT=12TαΔθΔt=ΔTT′But T′=T∴ Δt=12αΔθ(t) =12×0.000012×20×24×3600 =10.37 sAt higher temperature length of pendulum clock will be more. So time period will be more and it will lose the time.

A clock with an iron pendulum keeps correct time at $20^{\circ}$ C. How much time will it lose or gain in a day if the temperature changes to $40^{\circ}$ C? Thermal coefficient of linear expansion $α = 0.000012$ per $^{\circ}$ C.