1.

A magnetic needle lying parallel to a magnetic field requires w units of workto turn it through 60^(@)thetorque needed to maintain the needle in this positionwill be

Answer»

w
`SQRT(3)/(2)` w
2w
`sqrt(3)`w

Solution :`W=MB COS theta rarr W=(MB)/(2) rarr 2W =MB`
`tau =MB sin 60^(@)=2w.sqrt(3)/(2)=sqrt(3)W`


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