A body weighs 98 N on a spring balance on the North Pole. The same body is shifted to the equator, the weight of the body at the equator on the same scale is: (Assume g = 9.8 m/sec^{2} and radius of the earth = 6400 km)

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ISRO Refrigeration and Air Conditioning 2013

Option 3 : 97.66 N

__Concept:__

At poles, the apparent weight is equal to the true weight.

At the equator, the apparent weight gets decreased which is due to the value of acceleration due to gravity, which happens due to the centrifugal force of the earth's rotation.

The weight of the body at the equator is given by

Weight on equator (mg') **= mg – mω ^{2} R**

where, m = mass of the body,

\({\rm{\omega }} = {\rm{Angular\;velocity}} = \frac{{2{\rm{\pi }}}}{{{\rm{Time\;period\;of\;earth\;}}\left( {\rm{T}} \right)}}\;\)

R = Radius of earth (6400 km), g' = Gravitational acceleration on equator.

__Calculation:__

__Given:__

Weight = 98 N, g = 9.8 m/s^{2}, R = 6400 km = 6400 × 10^{3} m

We know, Weight = mg

\(\therefore {\rm{m}} = \frac{{{\rm{Weight}}}}{{\rm{g}}} = \frac{{98}}{{9.8}} = 10{\rm{\;kg}}\)

\(ω = \frac{{2\pi }}{T} = \frac{{2 × 3.14}}{{24 × 60 × 60}} = 7.272 × {10^{ - 5}}\) rad/sec

Weight on equator (mg'):

= mg – mω^{2}R

= (10 × 9.8) – [10 × (7.272 × 10^{-5})^{2 }× 6400 × 10^{3}]

= **97.66 N**

**The weight of the body at the equator on the same scale is 97.66 N**