Magnetic field of the earth is `0.3 G`. A magnet is oscillating with the rate of 5 oscillations/min. How much the magnetic field of the earth is increased, so that the number of oscillations becomes 10 per minute?

Answer

To solve the problem, we need to understand the relationship between the frequency of oscillation of a magnet and the magnetic field strength. The frequency \( f \) of oscillation is related to the magnetic field \( B \) by the following relationship:

\[ f \propto \sqrt{B} \]

This means that if we double the frequency, the magnetic field must increase by a factor of four (since \( (2f)^2 = 4f^2 \)).

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Initial frequency \( f_1 = 5 \) oscillations/min
- Initial magnetic field \( B_H = 0.3 \) G

2. **Determine Final Conditions**:
- Final frequency \( f_2 = 10 \) oscillations/min

3. **Calculate the Ratio of Frequencies**:
\[ \frac{f_2}{f_1} = \frac{10}{5} = 2 \]

4. **Relate Frequency to Magnetic Field**:
Since frequency is proportional to the square root of the magnetic field:
\[ \frac{f_2}{f_1} = \sqrt{\frac{B'_H}{B_H}} \]
where \( B'_H \) is the new magnetic field strength.

5. **Square Both Sides**:
\[ \left(\frac{f_2}{f_1}\right)^2 = \frac{B'_H}{B_H} \]
Substituting the values:
\[ 2^2 = \frac{B'_H}{0.3} \]
\[ 4 = \frac{B'_H}{0.3} \]

6. **Solve for \( B'_H \)**:
\[ B'_H = 4 \times 0.3 = 1.2 \text{ G} \]

7. **Calculate the Increase in Magnetic Field**:
The increase in the magnetic field \( \Delta B_H \) is given by:
\[ \Delta B_H = B'_H - B_H = 1.2 - 0.3 = 0.9 \text{ G} \]

### Final Answer:
The magnetic field of the earth needs to be increased by **0.9 G**.

---