1.

A bar magnet with poles 25.0 apart and of polestrength 14.4 am rests with its cent4re on a frictionless point12 cm from its pivot the magnitude of the force is

Answer»

`15sqrt(3) N`
`7.5 SQRT(3) N`
`3.75 sqrt(3)N`
NONE of these

Solution :Anticlockwise torque = clockwise torque
`Fxx0.12 =MB sin THETA`
`F=(3.6xx0.25 xxsqrt(3)//2)/(0.12)`
`=3.75 sqrt(3)N`


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