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 = 19 " ms"^(2-)` and radius of the earth, `R = 6.4 xx 10^(6)` m)

Options

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

Answer

A

Explanation

At the pole, the weight is same as the true one
Thus, `100 N = m(10 "ms"^(-2))rArr m = 10` kg
At the equator, the apparent weight is given by
`mg'=mg - m omega^(2) R`
Also the angular speed of an equatorial point on the earth's surface is
`omega = (2pi//24 xx 60 xx60)`
`rArr omega = 7.27xx10^(5)`
Now `mg'=100-10(7.27xx10^(-5))^(2)xx6.4xx10^(6)=99.66` N