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Question

# A clock which keep correct time at 20∘ C is subjected to 40∘ C. If coefficient of linear expansion of the pendulum is 12×10−6/∘C. How much will it gain or lose time?

A
10.3 s/day
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B
20.6 s/day
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C
5.3 s/day
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D
30 s/day
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Solution

## The correct option is A 10.3 s/dayGiven α=1.2×10−6/∘C. Δθ=40∘C−20∘C=20∘C Time period in a pendulum is given as T=2π√lg When temperature is increased by Δθ new length is l=l0(1+αΔθ) As T is proportional to √l So new Time period T=T0√(1+αΔθ) As αΔθ <<1 applying Binomial theorem, T=T0(1+αΔθ2) T−T0=T0αΔθ2 For calculating variation of time for 24 hrs⇒T0=86400 s ΔT=12αΔθT0=12×α×Δθ×86400=⇒ΔT=1.04 s As there is increase in length due to rise of temperature, timeperiod of pendulum increases and clock runs slow

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