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, $g=10 \,m / s ^{2}$ and radius of the earth, $R=6.4 \times 10^{3} m$ )

• A. 99.66 N
• B. 110 N
• C. 97.66 N
• D. 106 N

Answer 1

Answer: (A)

At the pole, the weight is same as the true one
Thus, $100 \,N =m\left(10 m / s ^{2}\right)$
$\Rightarrow m=10\, kg$
At the equator, the apparent weight is given by
$m g' =m g-m \omega^{2} R$
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} $
Now $m g' = 100-10\left(7.27 \times 10^{-5}\right)^{2} \times 6.4 \times 10^{3}$
$= 99.66 \,N$