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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.)
  1. 3.28×104 rad /sec
  2. 7.826×104 rad /sec
  3. 3.28×103 rad /sec
  4. 7.28×103 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=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=mgmRω²cosλ
3/5mg=mgmRω²cos0(λ=0)
3/5mg=mgmRω² (1) (cos0=1)
3/5mg=mgmRω²
mRω²=mg3/5mg
mRω²=(13/5)mg
mRω²=2/5mg
Rω²=2/5g
ω²=2g/5R
ω²=(2x10)/(5x6.4x10)(R=RadiusofEarth=6400km=6.4x10m)
ω=(6.25x10)
ω=7.8x10rad/sec
Hence,
option (B) is correct answer.

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