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`        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.`
7 years ago

SAGAR SINGH - IIT DELHI
879 Points
```										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

All the best.
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Sagar Singh
B.Tech, IIT Delhi

```
7 years ago
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