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Question

The speed with the earth have to rotate on its axis so that a person on the equator would weight (3/5)th as much as present will be (Take the equilateral radius as 6400 km.)3.28×10−4 rad /sec7.826×10−4 rad /sec3.28×10−3 rad /sec7.28×10−3 rad /sec

A
3.28×103 rad /sec
B
7.826×104 rad /sec
C
7.28×103 rad /sec
D
3.28×104 rad /sec
Solution
Verified by Toppr

Actual weight on the equator =W=mgwhere '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−mRω²cosλ3/5mg=mg−mRω²cos0(∵λ=0)3/5mg=mg−mRω² (1) (∵cos0=1)3/5mg=mg−mRω²mRω²=mg−3/5mgmRω²=(1−3/5)mgmRω²=2/5mgRω²=2/5gω²=2g/5Rω²=(2x10)/(5x6.4x10⁶)(∵R=RadiusofEarth=6400km=6.4x10⁶m)ω=√(6.25x10⁻⁷)ω=7.8x10⁻⁴rad/secHence,option (B) is correct answer.

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