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A mass of M kg is suspended by a weightl...

A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of `45^@` with the initial vertical direction is

A

`Mg (sqrt(2)-1)`

B

`Mg (sqrt(2)+1)`

C

`Mg sqrt(2)`

D

`(Mg)/(sqrt(2))`

Text Solution

Verified by Experts

The correct Answer is:
A

Work done in displacement is equal to gain in potential energy.
Work done `=F xx l sin45^(@)=(Fl)/(sqrt(2))`
Gain in potential energy
`=Mg (l-l cos 45^(@))`
`=Mgl (1-(1)/(sqrt(2))) :. (Fl)/(sqrt(2)) =(Mgl (sqrt(2)-1))/(sqrt(2))`
or `F=Mg (sqrt(2)-1)`
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