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Question

A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at 20C, then how fast or slow will it go in 24 hours at 40C  ?  Co-efficient of linear expansion of iron = 1.2×106/C.


A
2.08 s, slow
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B
2.08 s, fast
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C
1.04 s, slow
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D
1.04 s, fast
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Solution

The correct option is A 1.04 s, slow
Given
α=1.2×106/C.
Δθ=40C20C=20C

Time period in a pendulum is given as T=2πlg

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)
TT0=T0αΔθ2

For calculating variation of time for 24 hrsT0=86400 s

ΔT=12αΔθT0=12×α×Δθ×86400=ΔT=1.04 s

As there is increase in length due to rise of temperature, time period of pendulum increases and clock runs slow.

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