A box weight 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take g = 10 ms –2 at the north pole and the radius of the earth = 6400 km) :

Solution

Physics sir · 3 steps

At the equator the spin of the Earth steals a little of g, so find that drop mω² R and subtract it.

1. Mass of box. At the pole there is no rotation effect, so mg = 196 N with g = 10 m s⁻². That gives m = 19.6 kg.

   W_(eq) = m(g - ω² R) = mg - mω² R

2. Find the drop. The Earth turns once in 24 h, so ω = 2π/86400 = 7.27 × 10⁻⁵ rad s⁻¹. Then ω² R = 5.29 × 10⁻⁹ × 6.4 × 10⁶ = 0.0338 m s⁻², and mω² R = 19.6 × 0.0338 = 0.66 N.
3. So. W_(eq) = 196 - 0.66 ≈ 195.34 N. The closest option is (c) 195.32 N.

The idea to keep. Rotation does nothing at the poles and reduces g by ω² R at the equator, about 0.034 m s⁻². Find the mass from the pole reading first, then subtract mω² R.

Answer from the NTA final answer key. Solution by PrepWiser.