Question

A compass needle oscillates \(20\) times per minute at a place where the dip is \(45^\circ\) and the magnetic field is \(B_1\). The same needle oscillates \(30\) times per minute at a place where the dip is \(30^\circ\) and magnetic field is \(B_2\). Then \(B_1:B_2\) is

• A. \(9\sqrt{3}:4\sqrt{2}\)
• B. \(4\sqrt{2}:9\sqrt{3}\)
• C. \(3\sqrt{3}:2\sqrt{2}\)
• D. \(2\sqrt{2}:3\sqrt{3}\)

Hint:

For oscillations of a compass needle, use the horizontal component of earth's magnetic field: \[ B_H=B\cos\theta \] and frequency varies as \(\sqrt{B_H}\).

Answer 1

The Correct Option is D

Step 1: Use frequency relation for compass needle.
For a compass needle oscillating in earth's magnetic field, frequency is proportional to the square root of the horizontal component of magnetic field.
\[ n \propto \sqrt{B_H} \] Since,
\[ B_H = B\cos\theta \] we get
\[ n \propto \sqrt{B\cos\theta} \]

Step 2: Write the ratio of frequencies.
Given,
\[ n_1=20,\quad n_2=30 \] Dip angles are
\[ \theta_1=45^\circ,\quad \theta_2=30^\circ \] Therefore,
\[ \frac{n_1}{n_2} = \sqrt{\frac{B_1\cos45^\circ}{B_2\cos30^\circ}} \]

Step 3: Substitute values.
\[ \frac{20}{30} = \sqrt{\frac{B_1\cos45^\circ}{B_2\cos30^\circ}} \] \[ \frac{2}{3} = \sqrt{\frac{B_1\cdot \frac{1}{\sqrt{2}}}{B_2\cdot \frac{\sqrt{3}}{2}}} \]

Step 4: Square both sides.
\[ \left(\frac{2}{3}\right)^2 = \frac{B_1}{B_2}\cdot \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}} \] \[ \frac{4}{9} = \frac{B_1}{B_2}\cdot \frac{2}{\sqrt{6}} \]

Step 5: Calculate \(B_1:B_2\).
\[ \frac{B_1}{B_2} = \frac{4}{9}\cdot \frac{\sqrt{6}}{2} \] \[ \frac{B_1}{B_2} = \frac{2\sqrt{6}}{9} \] Now,
\[ \frac{2\sqrt{2}}{3\sqrt{3}} = \frac{2\sqrt{2}\cdot \sqrt{3}}{3\sqrt{3}\cdot \sqrt{3}} = \frac{2\sqrt{6}}{9} \] Thus,
\[ B_1:B_2=2\sqrt{2}:3\sqrt{3} \]

Step 6: Final conclusion.
Hence, the required ratio is
\[ \boxed{2\sqrt{2}:3\sqrt{3}} \]