Actual weight on the equator =W=mg
where 'm' is the mass and 'g' is the gravity.
According to the given condition,
Weight on the equator =W′=mg′
Weight on the equator =W′=3/5mg ........... (1)
We know that,λ=0 at the equator.
Now,
mg′=mg−mRω²cosλ
3/5mg=mg−mRω²cos0(∵λ=0)
3/5mg=mg−mRω² (1) (∵cos0=1)
3/5mg=mg−mRω²
mRω²=mg−3/5mg
mRω²=(1−3/5)mg
mRω²=2/5mg
Rω²=2/5g
ω²=2g/5R
ω²=(2x10)/(5x6.4x10⁶)(∵R=RadiusofEarth=6400km=6.4x10⁶m)
ω=√(6.25x10⁻⁷)
ω=7.8x10⁻⁴rad/sec
Hence,
option (B) is correct answer.