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A second's pendulum clock havig steeel w...

A second's pendulum clock havig steeel wire is calibrated at `20^(@)`C. When temperature is increased to `30^(@)`C, then calculate how much time does the clock
`[alpha_(Steel)`=`1.2xx10^(-6^(@))C^(-1)`

A

0.3628 s

B

3.626 s

C

362.8 s

D

36.23 s

Text Solution

Verified by Experts

d) Time period of second's penulum = 2s
Change in time period,
`DeltaT = 1/2 TalphaDeltatheta`
`DeltaT = (1/2)(2)(1.2 xx 10^(-5))(306^(@)-206) = (1.2 xx 10^(-4))` s
New time period, `T^(') = T + DeltaT rArr T^(')` = 2.00012 s
time lost in a week
`DeltaT = (DeltaT)/(T^(') xx t=(1.2 xx 10^(-4))/(2.00012) x (7 xx 24 xx 3600)` = 36.23 s
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