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Question

A person is standing at a latitude which makes an angle of 60 with the centre of earth. What is the % change in his actual weight here. Take reference of weight near earth's pole as actual weight. Take g=10 m/s2.

A
8%
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B
0.08%
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C
0.008%
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D
0.0008%
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Solution

The correct option is B 0.08%
Centrifugal force acting on the person at latitude θ is mω2Rcosθ
where ω= angular velocity of earth's rotation
R= radius of earth

Therefore at latitude θ:
N+mω2Rcos2θ=mg
N=mgmω2Rcos2θ

At the poles , N=mg

% change in weight=mgNmg×100
=mω2Rcos2θmg×100
=Rω2cos2θg×100
=4π2×6.4×106×1242×36002×10×4×100

% change in weight =0.08%

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