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 × 103 m )

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 × 103 m ) (A) 99.66 N (B) 110 N (C) 97.66 N (D) 106 N

Tardigrade Admin · Accepted Answer

At the pole, the weight is same as the true one
Thus,

100 N = m (10 m / s^{2})

\Rightarrow m = 10 k g

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 1 0^{5}

Now

m g^{'} = 100 - 10 (7.27 \times 1 0^{- 5})^{2} \times 6.4 \times 1 0^{3}

= 99.66 N

Q. The reading of a spring balance corresponds to 100N 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=10m/s2 and radius of the earth, R=6.4×103m )

99.66 N

110 N

97.66 N

106 N

Q. 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, $R = 6.4 \times 1 0^{3} m$ )