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find T2/T1
Asked by ashutosharnold1998 | 05 Apr, 2020, 01:32: AM
Expert Answer
Acceleration due to gravity g on earth surface is given by , g = GM/R2
where G is universal gravitational constant, M is mass of earth and R is radius of earth
Acceleration due to gravity gh at a height R above earth surface is given by , gh = GM/ (R+h)2
[ gh / g ] = [ R / (R+h) ]2
Time period T of simple pendulam, T = 2π { l / g }1/2
where l is length of simple pendulam
Hence T 
1/√g or T R
1/√g or T R If T1 is period of simple pendulam on Earth surface and T2 is period of simple pendulam at a height R above earth surface
then, ( T2 / T1 ) = ( 2R ) / R = 2
Answered by Thiyagarajan K | 05 Apr, 2020, 08:55: AM
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