JEE Physics 2023 Question with Solution

Given: The compass needle makes $20$ oscillations per minute at dip $30^\circ$ and $30$ oscillations per minute at dip $60^\circ$.

Find: The value of $x$ if

$$\frac{B_1}{B_2} = \frac{4}{\sqrt{x}}$$

where $B_1$ and $B_2$ are the total magnetic fields at the two places.

For a compass needle,

$$f = \frac{1}{2\pi} \sqrt{\frac{MB_H}{I}}$$

where $M$ is the magnetic moment, $B_H$ is the horizontal component of Earth's magnetic field, and $I$ is the moment of inertia.

Since $M$ and $I$ are constant,

$$f \propto \sqrt{B_H}$$

So,

$$f^2 \propto B_H$$

Also, the horizontal component is

$$B_H = B \cos \delta$$

Therefore,

$$f^2 \propto B \cos \delta$$

For the two places,

$$\frac{f_1^2}{f_2^2} = \frac{B_1 \cos \delta_1}{B_2 \cos \delta_2}$$

Substituting $f_1 = 20$, $f_2 = 30$, $\delta_1 = 30^\circ$, and $\delta_2 = 60^\circ$,

$$\frac{20^2}{30^2} = \frac{B_1 \cos 30^\circ}{B_2 \cos 60^\circ}$$ $$\frac{400}{900} = \frac{B_1 \left(\sqrt{3}/2\right)}{B_2 \left(1/2\right)}$$ $$\frac{4}{9} = \frac{B_1}{B_2} \sqrt{3}$$ $$\frac{B_1}{B_2} = \frac{4}{9\sqrt{3}}$$

Given that

$$\frac{B_1}{B_2} = \frac{4}{\sqrt{x}}$$

Equating,

$$\frac{4}{\sqrt{x}} = \frac{4}{9\sqrt{3}}$$ $$\sqrt{x} = 9\sqrt{3}$$ $$x = \left(9\sqrt{3}\right)^2 = 81 \times 3 = 243$$

Therefore, the value of $x$ is $243$.

Given: Frequency of oscillation depends on the horizontal component of Earth's magnetic field.

Find: The numerical value of $x$.

The key idea is that a compass oscillates due to the horizontal component $B_H$, not the total field directly. Hence,

$$f^2 \propto B_H = B \cos \delta$$

So the ratio becomes

$$\frac{f_1^2}{f_2^2} = \frac{B_1 \cos 30^\circ}{B_2 \cos 60^\circ}$$

Now,

$$\frac{20^2}{30^2} = \frac{B_1}{B_2} \cdot \frac{\sqrt{3}/2}{1/2}$$ $$\frac{4}{9} = \frac{B_1}{B_2} \sqrt{3}$$ $$\frac{B_1}{B_2} = \frac{4}{9\sqrt{3}} = \frac{4}{\sqrt{x}}$$

Thus,

$$\sqrt{x} = 9\sqrt{3}$$ $$x = 243$$

Therefore, the required numerical answer is $243$.