A compass needle placed at a distance r from a short magnet in tan A position showns a deflection of `60^(@)`. If the distance is increased to `r(3)^(1//3)`, then the deflection of the compass needle is:

Options

A. `10^@`
B. `20^@`
C. `30^@`
D. `40^@`

Answer

To solve the problem, we need to analyze the relationship between the deflection of the compass needle and the distance from the magnet.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- A compass needle is placed at a distance \( r \) from a short magnet, and it shows a deflection of \( 60^\circ \).
- The deflection angle \( \theta \) is related to the magnetic field produced by the magnet.

2. **Magnetic Field Relation**:
- The magnetic field \( B \) at a distance \( r \) from a short magnet is given by the formula:
\[ B \propto \frac{1}{r^3} \]
- The deflection of the compass needle is related to the tangent of the angle:
\[ \tan \theta \propto B \]
- Therefore, we can express the relationship as:
\[ \tan \theta \propto \frac{1}{r^3} \]

3. **Setting Up the Initial Condition**:
- For the initial condition, we have:
\[ \tan(60^\circ) = \sqrt{3} \]
- Thus, we can write:
\[ \tan(60^\circ) \propto \frac{1}{r^3} \implies \sqrt{3} \propto \frac{1}{r^3} \]

4. **Changing the Distance**:
- Now, the distance is increased to \( r' = r \cdot 3^{1/3} \).
- We need to find the new deflection angle \( \theta' \) when the distance is \( r' \).

5. **Applying the New Distance**:
- The new relationship becomes:
\[ \tan \theta' \propto \frac{1}{(r \cdot 3^{1/3})^3} \]
- Simplifying this:
\[ \tan \theta' \propto \frac{1}{r^3 \cdot 3} = \frac{1}{3} \cdot \frac{1}{r^3} \]

6. **Relating New and Old Angles**:
- Since \( \tan \theta \propto \frac{1}{r^3} \), we can relate the two angles:
\[ \tan \theta' = \frac{1}{3} \tan \theta \]
- Substituting \( \tan(60^\circ) = \sqrt{3} \):
\[ \tan \theta' = \frac{1}{3} \cdot \sqrt{3} = \frac{\sqrt{3}}{3} \]

7. **Finding the New Angle**:
- We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
- Therefore, we can conclude:
\[ \tan \theta' = \tan(30^\circ) \]
- Thus, \( \theta' = 30^\circ \).

### Final Answer:
The deflection of the compass needle when the distance is increased to \( r \cdot 3^{1/3} \) is \( 30^\circ \).

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