A pendulum clock gives correct time at 20 degree Celsius at a place where g= 9.800 m/s^2. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where g=9.788 m/s^2. At what temperature will it give correct time?coefficient of linear expansion of steel = 12x 10^-6/ degree celsius.

Dear student,

1st time

t = 2π√l/g
t = 2π√l/9.8 ------------->(1)

2nd time



t = 2π√L/G

L=l * (1 + 12x10^-6 * θ)

t = 2π√L/G
t = 2π√l*(1 + 12x10^-6 * θ)/9.788 ---------------->(2)

If it gives correct time, both "t"s of two occasions should be same
(1) = (2)

2π√l*(1 + 12x10^-6 * θ)/9.788 = 2π√l/9.8

l*(1 + 12x10^-6 * θ)/9.788 = l/9.8

1 + 12x10^-6 * θ / 9.788 = 1 / 9.8

1 + 12x10^-6 * θ = 9.788 / 9.8

1 + 12x10^-6 * θ = 0.9988

12x10^-6 * θ = -12*10^-4

10^-2 * θ = -1

θ = -100°C ( temperature difference )

Temperature - 20 = -100
Temperature = -80°C

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