The magnetic moments of two bar magnets of same size are in the ratio `1:2.` When they are placed one over the other with their similar poles together, then their perlod of oscillation in a magnetic field is 3s. If one of the magnets is reversed, then the period of oscillation in the same field will be q

Options

A. `sqrt3s`
B. `3sqrt3s`
C. `3 s`
D. `6 s`

Answer

To solve the problem, we need to analyze the situation of two bar magnets with different magnetic moments and how their arrangement affects the period of oscillation in a magnetic field.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The magnetic moments of the two bar magnets are in the ratio \(1:2\).
- Let the magnetic moment of the first magnet be \(M\) and the second magnet be \(2M\).
- The period of oscillation when they are placed with similar poles together is given as \(T_1 = 3 \, \text{s}\).

2. **Formula for the Period of Oscillation**:
- The period of oscillation \(T\) of a magnetic dipole in a magnetic field is given by:
\[ T = 2\pi \sqrt{\frac{I}{MB}} \]
- Here, \(I\) is the moment of inertia, \(M\) is the net magnetic moment, and \(B\) is the magnetic field strength.

3. **Finding the Net Magnetic Moment in the Initial Arrangement**:
- When the magnets are placed with similar poles together, the net magnetic moment \(M_1\) is:
\[ M_1 = M + 2M = 3M \]

4. **Finding the Period of Oscillation for the Initial Arrangement**:
- The period of oscillation can be expressed in terms of the net magnetic moment:
\[ T_1 \propto \sqrt{\frac{1}{M_1}} \implies T_1 \propto \sqrt{\frac{1}{3M}} \]
- Since \(T_1 = 3 \, \text{s}\), we can denote:
\[ T_1 = k \sqrt{\frac{1}{3M}} \quad \text{(where \(k\) is a constant)} \]

5. **Finding the Net Magnetic Moment in the Reversed Arrangement**:
- When one of the magnets is reversed, the net magnetic moment \(M_2\) becomes:
\[ M_2 = 2M - M = M \]

6. **Finding the Period of Oscillation for the Reversed Arrangement**:
- The new period of oscillation \(T_2\) will be:
\[ T_2 \propto \sqrt{\frac{1}{M_2}} \implies T_2 \propto \sqrt{\frac{1}{M}} \]
- Since we know \(T_1 \propto \sqrt{\frac{1}{3M}}\), we can relate \(T_2\) to \(T_1\):
\[ \frac{T_1}{T_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{M}{3M}} = \sqrt{\frac{1}{3}} \]
- Therefore, we can express \(T_2\) in terms of \(T_1\):
\[ T_2 = T_1 \cdot \sqrt{3} = 3 \cdot \sqrt{3} \, \text{s} \]

7. **Final Answer**:
- The period of oscillation when one of the magnets is reversed is:
\[ T_2 = 3\sqrt{3} \, \text{s} \]