Question

Work done in turning a magnet of magnetic moment M by
an angle of 90º from the mgnetic meridian is n times the
corresponding work done to turn through an angle of 60º,
where n is
(a) 1/2 (b) 2 (c) 1/4 (d) 1

Answer 1

Answer:

W=MB(cos0−cos90)=MB

W1=MB(cos0−cos60)=0.5MB

n=2

Explanation:

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Answer 2

✪ Question :-

Work done in turning a magnet of magnetic moment M by  an angle of 90º from the mgnetic meridian is n times the  corresponding work done to turn through an angle of 60º,  where n is

(a) 1/2 (b) 2 (c) 1/4 (d) 1

✪ Solution :-

(b) 2

✪ Explanation :-

Angle of magnet (θ) = 90° and 60°. Work done in turning the magnet through 90° :-

$\sf{W=MB(cos\ 0^\circ-cos\ 90^\circ)}\\\\\sf{\implies W=MB(1-0)}\\\\\sf{\implies W=MB}\:\:\:\: \cdots (i)$

$\rule{130}1$

Again,

$\sf{W'=MB(cos\ 0^\circ -cos\ 60^\circ)}\\\\\sf{\implies W'=MB(1-\dfrac{1}{2} )}\\\\\sf{\implies W'=MB(\dfrac{1}{2} )}\\\\\sf{\implies W'=\dfrac{MB}{2} }\:\:\:\: \cdots (ii)$

From (i) and (ii), we get :-

MB = W = 2W'

⇒W = 2W'

or n = 2

Therefore, the answer is (b) 2.