The work done in turning a magnet of magnetic moment M by an angle of 90- from the meridian is n times the corresponding work done to turn it through an angle of 60-A- n -1B- n -1-2C- n -2D- n -1-4

The correct option is C n=2W_1=MB[cos0^∘ cos90^∘] =MB[1 0] W_1=MB ...(1) in second case W_2=MB[cos 0^∘ cos60^∘] W_2=MB/2 ....(2) (1) ÷(2) W_1/W_2=2 n=2 ...

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Byju's Answer

Standard XII

Physics

Magnetic Field

The work done...

Question

The work done in turning a magnet of magnetic moment M by an angle of $90^{\circ}$ from the meridian is n times the corresponding work done to turn it through an angle of $60^{\circ}$ .

n = \frac{1}{2}

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n = 2

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n = \frac{1}{4}

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n = 1

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Solution

The correct option is B $n = 2$
$W_{1} = M B [c o s 0^{\circ} - c o s 90^{\circ}]$
$= M B [1 - 0]$
$W_{1} = M B$ ...(1)
in second case
$W_{2} = M B [c o s 0^{\circ} - c o s 60^{\circ}]$
$W_{2} = \frac{M B}{2}$ ....(2)
$(1) \div (2)$
$\frac{W_{1}}{W_{2}} = 2$
$n = 2$

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The work done in turning a magnet of magnetic moment M by an angle of 90∘ from the meridian is n times the corresponding work done to turn it through an angle of 60∘.

The correct option is B n=2W1=MB[cos0∘−cos90∘] =MB[1−0] W1=MB ...(1) in second case W2=MB[cos0∘−cos60∘] W2=MB2 ....(2) (1)÷(2) W1W2=2 n=2

The work done in turning a magnet of magnetic moment M by an angle of $90^{\circ}$ from the meridian is n times the corresponding work done to turn it through an angle of $60^{\circ}$ .

The correct option is B $n = 2$
$W_{1} = M B [c o s 0^{\circ} - c o s 90^{\circ}]$
$= M B [1 - 0]$
$W_{1} = M B$ ...(1)
in second case
$W_{2} = M B [c o s 0^{\circ} - c o s 60^{\circ}]$
$W_{2} = \frac{M B}{2}$ ....(2)
$(1) \div (2)$
$\frac{W_{1}}{W_{2}} = 2$
$n = 2$