A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45° and 40 times per minute where the dip is 30°. If B$$_1$$ and B$$_2$$ are the net magnetic fields due to the earth at the two places respectively, then the ratio B$$_1$$/B$$_2$$ is approximately equal to

We begin by recalling the theory of the small oscillations of a magnetic (compass) needle. When the needle is displaced slightly in the horizontal plane, the restoring torque is $$\tau = M B_H \sin\theta \approx M B_H \theta$$ for small $$\theta,$$ where $$M$$ is the magnetic moment and $$B_H$$ is the horizontal component of the earth’s magnetic field. For a body executing small-angle simple harmonic motion, the angular frequency is given by the standard formula

$$\omega = \sqrt{\dfrac{\text{restoring torque per unit angle}}{\text{moment of inertia}}} = \sqrt{\dfrac{M B_H}{I}}.$$

The ordinary (linear) frequency is $$f = \dfrac{\omega}{2\pi},$$ so we have

$$f = \dfrac{1}{2\pi}\sqrt{\dfrac{M}{I}}\,\sqrt{B_H}.$$

Everything except $$B_H$$ is constant for the same needle; hence

$$f \propto \sqrt{B_H} \quad\Longrightarrow\quad B_H \propto f^2.$$

Now we convert the given “oscillations per minute” to “oscillations per second” (hertz):

At the first place, the needle makes 30 oscillations in 60 s, so

$$f_1 = \dfrac{30}{60} = 0.50\ \text{Hz}.$$

At the second place, the needle makes 40 oscillations in 60 s, so

$$f_2 = \dfrac{40}{60} = 0.667\ \text{Hz}.$$

Using $$B_H \propto f^2,$$ we obtain

$$\dfrac{B_{H1}}{B_{H2}} = \left(\dfrac{f_1}{f_2}\right)^2 = \left(\dfrac{0.50}{0.667}\right)^2 = \left(0.75\right)^2 = 0.5625.$$

However, the question asks for the ratio of the total earth’s magnetic fields $$B_1$$ and $$B_2,$$ not just their horizontal components. The total field is related to its horizontal component by the well-known relation

$$B_H = B\cos\delta,$$

where $$\delta$$ is the dip (angle of inclination). Thus

$$B = \dfrac{B_H}{\cos\delta}.$$

So, for the two places,

$$\dfrac{B_1}{B_2} = \dfrac{B_{H1}/\cos\delta_1}{B_{H2}/\cos\delta_2} = \dfrac{B_{H1}}{B_{H2}}\;\dfrac{\cos\delta_2}{\cos\delta_1}.$$

The dips are $$\delta_1 = 45^\circ$$ and $$\delta_2 = 30^\circ.$$ We substitute their cosines:

$$\cos45^\circ = \dfrac{1}{\sqrt2} \approx 0.7071,\qquad \cos30^\circ = \dfrac{\sqrt3}{2} \approx 0.8660.$$

Therefore,

$$\dfrac{\cos\delta_2}{\cos\delta_1} = \dfrac{0.8660}{0.7071} \approx 1.2247.$$

Multiplying this with the earlier ratio of horizontal components, we get

$$\dfrac{B_1}{B_2} = 0.5625 \times 1.2247 \approx 0.688.$$

This value is very close to 0.7. Among the given options, the one that matches is 0.7.

Hence, the correct answer is Option D.