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A pendulum clock keeps correct time at 20 deg C at a location where g = 9.800 m/s². The bob hangs from a light steel rod. The clock is moved to another location where g = 9.788 m/s². At what temperature will it again keep correct time? (Linear expansion coefficient of steel = 12 * 10⁻⁶ per deg C.)

1. 122 deg C
2. 71 deg C
3. -69 deg C
4. -82 deg C

Correct answer: -82 deg C

Solution

Keeping T = 2*pi*sqrt(L/g) fixed requires DeltaL/L = (g'-g)/g; this is negative, so the rod must contract (cool) by about 102 deg C, giving roughly -82 deg C.

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