A simple pendulum of length 1 m is taken to height h (h = R = radius of the earth) from the earth’s surface.
The time period of small oscillation of the pendulum at that point is (g=π2ms−2)
A
2 s
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B
3 s
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C
4 s
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D
1 s
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Solution
The correct option is C 4 s At height h, g/=GM(R+h)2=GM(R+R)2=g4 ∴T=2π√lg/=2π√114×π2=2π(2π)=4s