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At a height (h) from the surface of the ...

At a height (h) from the surface of the Earth, a simple pendulum of length 1/2 m oscillates with a frequency equal to 0.5 Hz. Then find the value of 'h' . Take, mass of the Earth, `M = 6 xx 10^(24)` kg, radius of the Earth, `R = 6400 km and pi^(2) = 10`.

Text Solution

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The frequency (n) of the given simple pendulum = 0.5 Hz.
Therefore, the time period (T) of the simple pendulum `= (1)/(n) = (1)/(0.5) = 2s`
We know, `T = 2 pi sqrt((l)/(g))`
Let the acceleration due to gravity at height (h) from the surface of the Earth `= g^(1)`
`rArr T = 2 pi sqrt((l)/(g^(1))) rArr g^(1) = 4 pi^(2) (l)/(T^(2))`
`= 4 xx 10 xx ((1)/(2))/((2)^(2)) = ms^(-2)`
`g^(1) = (GM)/((R + h)^(2))=(GM)/(g^(1))`
`rArr (64 xx 10^(5)m + h)^(2)=(6.67 xx 10^(-11) xx 6 xx 10^(24))/(5)`
`= 6.67 xx 1.2 xx 10^(13) = 80.04 xx 10^(12)`
`rArr (64 xx 10^(5)m + h) = sqrt(80.04 xx 10^(12)) approx 9 xx 10^(6)m`
`h = (9 xx 10^(6) - 6.4 xx 10^(6)) m`
`h = 2.6 xx 10^(6) m`
The position (h) of the simple pendulum from the surface of the Earth `= h = 2.6 xx 10^(6) m`.
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