A magnetic compass needle oscillates 30 times per minute at a place where
Question:

A magnetic compass needle oscillates 30 times per minute at a place where the dip is $45^{\circ}$, and 40 times per minute where the dip is $30^{\circ}$. If $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ are respectively the total magnetic field due to the earth and the two places, then the ratio $\mathrm{B}_{1} / \mathrm{B}_{2}$ is best given by :

1. (1) $1.8$

2. (2) $0.7$

3. (3) $3.6$

4. (4) $0.46$

Correct Option: , 4

Solution:

(4)

We have, $T=2 \pi \sqrt{\frac{I}{M B_{x}}}$

$\therefore \frac{T_{1}^{2}}{T_{2}^{2}}=\frac{B x_{2}}{B x_{1}}$

or $\left(\frac{2}{1.5}\right)^{2}=\frac{B_{2} \cos 45^{\circ}}{B_{1} \cos 30^{\circ}}=\frac{B_{2} \times 2}{\sqrt{2} \times B_{1} \times \sqrt{3}}$

$\left(\frac{4}{3}\right)^{2}=\frac{B_{2}}{B_{1}} \times \frac{2}{\sqrt{6}}$

$\therefore \quad \frac{B_{1}}{B_{2}}=\frac{9}{8 \sqrt{6}}=0.46$