A clock with an iron pendulum keeps correct time at 15∘C. If the room temperature rises to 20∘C, the error in seconds per day will be (coefficient of linear expansion for iron is 0.000012/∘C):
A
2.5sec
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B
2.6sec
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C
2.4sec
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D
2.2sec
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Solution
The correct option is B2.6sec The initial time period of a pendulum is Ti=2π√lig and final time period is Tf=2π√lfg
If α be the coefficient of linear expansion, then lf=li(1+αΔθ) where Δθ= change in temperature.
So, Tf=2π√li(1+αΔθ)g=Ti(1+αΔθ)1/2=Ti(1+αΔθ2) [for small value of αΔ, we can neglect the higher order term]
Or, Tf−TiTi=12αΔθ
Thus, the error in seconds per day =Tf−TiTi×86400=12αΔθ×86400=0.5(0.000012)(20−15)(86400)=2.6sec