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Correct option is B)

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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