Question

The work done in tuning a magnet of magnetic moment M by angle of ${90}^o$ from the meridian, is n times the corresponding work done to turn it through an angle of ${60}^o$. The value of n is given by

• A. 2
• B. 1
• C. 0.5
• D. 0.25

Answer 1

The correct option is A 2

Given: Work done in turning ${90}^0$=$2\times$work done in turning ${60}^0$

Solution: Work done in turning ${90}^0=W_1$

Work done in turning ${60}^0=W_2$

As we know the formula $W=MB(cos{\theta }_1-cos{\theta }_2)$

Where,$cos{\theta }_1$=initial angle, $cos{\theta }_2$=final angle

In first case,${\theta }_1=0^0$,${\theta }_2={90}^0$

Now putting in he formula gives $W_1$=MB....(1)

In second case, ${\theta }_1=0^0$,${\theta }_2={60}^0$

Now putting in he formula gives$W_2=\frac{MB}{2}$....(2)

Combining eq.(1) &(2)

$W_2=\frac{W_1}{2}$

ATQ, $W_1=n\times W_2$ which gives n=2

Hence correct option is A