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Question

A man has a weight 105 N on a weighing machine kept near the north pole of the earth. What will be the reduction in the weight of the man if he moves to the equator? Take g=9.8 m/s2, radius of earth (R)=6400 km, angular speed of the earth ω=7.27×105 rad/s.

A
3.62 N
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B
0.36 N
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C
10 N
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D
0 N
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Solution

The correct option is B 0.36 N
Since earth is rotating about a vertical axis passing through the poles, the effect of rotation at poles is negligible and centrifugal force Fcentrifugal=0 at the poles.
However at the equator, Fcentrifugal is maximum.

Given, the Earth is rotating with ω=7.27×105 rad/s
Fcentrifugal=mω2R, in radially outward direction at equator.

At poles:


Weight read =N=mg
m=Ng=1059.8=10.71 kg

Similarly at equator:


mg=N+mω2R
N=mgmω2R

Here, N is the weight read by the weighing machine at the equator.

N=mgmω2R=105(10.71×(7.27×105)2×(6400×1000)
N=104.64 N

Hence, difference in weight will be :
ΔN=NN=105104.64= 0.36 N

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