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A man slides down on a telegraphic pole ...

A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity.The frictional force between man and pole is equal to (in terms of man's weight `W`)

A

`(W)/(4)`

B

`(3W)/(4)`

C

`(W)/(2)`

D

W

Text Solution

Verified by Experts

The correct Answer is:
B
`f=w_("app")=m(g-a)`
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