A clock with a metal pendulum beating seconds keeps correct time at 0- C- If it loses 12-5 seconds a day at 25- C- the coefficient of linear expansion of metal of pendulum isA- 1-86400 p e r- CB- 1-43200 p e r- CC- 1-14400 p e r- CD- 1-28800 p e r- C

The correct option is A Loss of time due to heating a pendulum is given as Δ T = 1/2αΔθ T ⇒ 12.5 = 1/2×α× (25 0)^∘ C × 86400 ⇒α = 1/86400 per ^∘ C ...

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Byju's Answer

Standard XII

Physics

Thermal Expansion

A clock with ...

Question

A clock with a metal pendulum beating seconds keeps correct time at $0^{\circ} C .$ If it loses 12.5 seconds a day at $25^{\circ} C,$ the coefficient of linear expansion of metal of pendulum is

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Solution

The correct option is A
Loss of time due to heating a pendulum is given as
$Δ T = \frac{1}{2} α Δ θ T \Rightarrow 12.5 = \frac{1}{2} \times α \times (25 - 0)^{\circ} C \times 86400 \Rightarrow α = \frac{1}{86400} p e r^{\circ} C$

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A clock with a metal pendulum beating seconds keeps correct time at 0∘C. If it loses 12.5 seconds a day at 25∘C, the coefficient of linear expansion of metal of pendulum is

The correct option is A Loss of time due to heating a pendulum is given as ΔT=12αΔθT⇒12.5=12×α×(25−0)∘C×86400⇒α=186400per∘C

A clock with a metal pendulum beating seconds keeps correct time at $0^{\circ} C .$ If it loses 12.5 seconds a day at $25^{\circ} C,$ the coefficient of linear expansion of metal of pendulum is

The correct option is A
Loss of time due to heating a pendulum is given as
$Δ T = \frac{1}{2} α Δ θ T \Rightarrow 12.5 = \frac{1}{2} \times α \times (25 - 0)^{\circ} C \times 86400 \Rightarrow α = \frac{1}{86400} p e r^{\circ} C$