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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.

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.

Grade:

2 Answers

Chetan Mandayam Nayakar
312 Points
10 years ago

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

 

Amitabh Shrivastava
32 Points
10 years ago

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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