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A body is weighed with a spring balance ...

A body is weighed with a spring balance in a train at rest, shown a weight `W`. When the train begins to move with a velocity `upsilon` around the equator from west to east and if the angular velocity of the train is `omega` then the weight shown by spring balance is

A

`W`

B

`W(1- (2 upsilon omega)/(g))`

C

`W(1- (2 upsilon omega)/(g))`

D

`W (1 - upsilon^(2)//R)`

Text Solution

Verified by Experts

The correct Answer is:
C
Observed weight of body on equator due to rotation of earth `= mg - mr omega^(2)`.
Observed weight of body due to motion of train from west to east with velocity `upsilon` (as centrifugal forces will be acting) will be
`W' = (mg - m r omega^(2)) - m upsilon^(2)//r`
`= mg - m upsilon omega - m upsilon omega = mg - 2m upsilon omega`
`= mg (1 - 2 upsilon omega//g)`
`= W (1- 2 upsilon omega//g)`
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