Question

The reading of a spring balance corresponds to 100 N while situated at the north pole and a body is kept on it. The weight record on the same scale, if it is shifted to the equator, is
(Take, g = 10 ms^-2 and radius of the earth, R = 6.4 X 10⁶m) [AIIMS 2015]

Moderate

Solution

At the pole, the weight is same as the actual weight.
Thus, $100 N = m (10 {ms}^{- 2}) \Rightarrow m = 10 kg$
At the equator, the apparent weight is given by
$m g^{'} = m g - m ω^{2} R$
Also, the angular speed of an equatorial point on the earth's surface is
$ω = (2 π / 24 \times 60 \times 60)$
$\Rightarrow ω = 7.27 \times 10^{- 5} rad / s$
Hence, now, $m g^{'} = 100 - 10 {(7.27 \times 10^{- 5})}^{2} \times 6.4 \times 10^{6}$
= 99.66N

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