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Q.

A magnet oscillating in a horizontal plane has a time period of 2 sec at a place where the angle of dip is $30^{o}$ and 3 second at another place where the angle of dip is $60^{o}$ . The ratio of resultant magnetic fields at these two places is

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a

$\frac{9}{\sqrt{3}}$

b

$\frac{4}{9 \sqrt{3}}$

c

$\frac{9}{4 \sqrt{3}}$

d

$\frac{4 \sqrt{3}}{7}$

answer is C.

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Detailed Solution

$\begin{array}{l} {(B_{H})}_{1} = B_{1} \cos 30^{o} \\ {(B_{H})}_{2} = B_{2} \cos 60^{o} \\ T \propto \frac{1}{\sqrt{B_{H}}} \\ \frac{T_{1}}{T_{2}} = \sqrt{\frac{B_{H_{2}}}{B_{H_{1}}}} = \sqrt{\frac{B_{2} Cos 60^{o}}{B_{1} \cos 30^{o}}} \\ \frac{2}{3} = \sqrt{\frac{B_{2}}{B_{1}} \frac{1}{\sqrt{3}}} \\ \frac{4}{9} = \frac{B_{2}}{B_{1}} \frac{1}{\sqrt{3}} \\ \frac{B_{1}}{B_{2}} = \frac{9}{4 \sqrt{3}} \end{array}$

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