A clock with an iron pendulum keeps correct time at 20∘C. If temperature is changed to 40∘C, then which of the following is/are true? [Take α=0.000012/∘C]
A
Pendulum will gain 1.2×10−4 seconds per each second.
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B
Pendulum will lose 1.2×10−4 seconds per each second.
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C
Pendulum will gain 10.368 seconds/day.
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D
Pendulum will lose 10.368 seconds/day.
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Solution
The correct options are B Pendulum will lose 1.2×10−4 seconds per each second. D Pendulum will lose 10.368 seconds/day. Time period of pendulum at 20∘C T20=2π√lg where l is the length of pendulum at 20∘C. Time period at 40∘C T40=2π√l(1+αΔT)g ∴ Change in time period. T40−T20=2π√l(1+αΔT)g−2π√lg =2π√lg(√1+αΔT−1) ⇒T40−T20T20=√1+αΔT−1 Using binomial expansion, √1+αΔT=1+12αΔT ⇒T40−T20T20=12αΔT
Since the time period increases, pendulum will lose time. & Loss of time per 1 second =12×0.000012×20=1.2×10−4s Loss in 24 hr =1.2×10−4×60×60×24 =10.368s Hence options (b) and (d) are correct.