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Two men are carrying a uniform bar of le...

Two men are carrying a uniform bar of length `L` on their shoulders.The bar is held horizontally such that younger man gets `((1)/(4))th` of load.Suppose the younger man is at one end of the bar.The distance of other man from the second end is ?

A

`(L)/(3)`

B

`(L)/(4)`

C

`(2L)/(3)`

D

`(3L)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
A
weight of the rod = W reaction of boy `R_(B)=(w)/(4)` Reaction of man `R_(M)=(3w)/(4)`
As the rod is in rotational equilibrium `therefore sum T=0` `R_(B) xx (L)/(2)-R_(M) xx x=0` `implies (W)/(4) xx (L)/(2)-(3 W)/(4) xx x=0 implies x=(L)/(6)` `therefore` Distance from other end.
`y=(L)/(2)-x implies y=(L)/(2)-(L)/(6)=(2L)/(6)=(L)/(3)`
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