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- 3.28×10−4 rad /sec
- 7.826×10−4 rad /sec
- 3.28×10−3 rad /sec
- 7.28×10−3 rad /sec

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Solution

Verified by Toppr

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.

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