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At the surface of the earth, the time period of a pendulum is T1 and at a height of distance K above earth's surface is T2, then T1/T2=?
Please reply soon.
Thanks.
T is inversely proportional to sqrt(g) therefore it is directly proportional to distance from centre of earth(for points on or above earth's surface). g=GM/R2 T1/T2 =(R/(R+K)) R is radius of earth
T is proportional to 1/[(g)^1/2]
t1/t2=[g'/g]^1/2
g=GM/R^2
g'=GM/(R+K)^2
t1/t2=[R/(R+K)]
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