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] 

At the pole, the weight is same as the actual weight.

Thus,    $100 \mathrm{N}=m\left(10{ \mathrm{ms}}^{-2}\right)\Rightarrow m=10 \mathrm{kg}$

At the equator, the apparent weight is given by

$m{g}^{\text{'}}=mg-m{\omega }^{2}R$

Also, the angular speed of an equatorial point on the earth's surface is

$\omega =(2\pi /24\times 60\times 60)$

$\Rightarrow \omega =7.27\times {10}^{-5}\mathrm{rad}/\mathrm{s}$

Hence, now,    $m{g}^{\text{'}}=100-10{\left(7.27\times {10}^{-5}\right)}^2\times 6.4\times {10}^6$

                               =  99.66N