A bar magnet oscillates with a frequency of 10 oscillations per minute . When another bar magnet is placed on its axis at a small distance , it oscillates at 14 oscillations per minute . Now , the second bar magnet is turned so that poles are instantaneous , keeping the location same . The new frequency of oscillation will be

Options

A. 2 vibrations/min
B. 4 vibrations/min
C. 10 vibrations/min
D. 14 vibrations/min

Explanation

To solve the problem step by step, we need to analyze the situation of the two bar magnets and how their interaction affects the frequency of oscillation.

### Step 1: Understand the initial conditions
The first bar magnet oscillates with a frequency of 10 oscillations per minute (F1 = 10 oscillations/min).

### Step 2: Convert frequency to time period
The time period (T1) is the reciprocal of frequency:
\[ T_1 = \frac{1}{F_1} = \frac{1}{10 \text{ oscillations/min}} \]
To convert this to seconds, we note that 1 minute = 60 seconds:
\[ T_1 = \frac{60 \text{ seconds}}{10} = 6 \text{ seconds} \]

### Step 3: Analyze the effect of the second bar magnet
When the second bar magnet is placed on the axis of the first one, the frequency increases to 14 oscillations per minute (F2 = 14 oscillations/min).

### Step 4: Convert the new frequency to time period
The time period (T2) is:
\[ T_2 = \frac{1}{F_2} = \frac{1}{14} \text{ oscillations/min} \]
Converting this to seconds:
\[ T_2 = \frac{60 \text{ seconds}}{14} \approx 4.29 \text{ seconds} \]

### Step 5: Relate time periods and magnetic fields
The time period of oscillation of a magnet in a magnetic field is given by:
\[ T \propto \frac{1}{\sqrt{B}} \]
Thus, we can write the ratio of the time periods in terms of the magnetic fields:
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_2}{B_1}} \]
Where \( B_1 \) is the magnetic field due to the Earth's magnetic field and \( B_2 \) is the combined magnetic field when the second magnet is placed.

### Step 6: Substitute the time periods
Substituting the values we found:
\[ \frac{6}{4.29} = \sqrt{\frac{B_H + B}{B_H}} \]
Squaring both sides gives:
\[ \left(\frac{6}{4.29}\right)^2 = \frac{B_H + B}{B_H} \]
Let \( x = \frac{B}{B_H} \):
\[ \left(\frac{6}{4.29}\right)^2 = 1 + x \]

### Step 7: Solve for x
Calculating \( \left(\frac{6}{4.29}\right)^2 \):
\[ \left(\frac{6}{4.29}\right)^2 \approx 1.67 \]
Thus,
\[ 1.67 = 1 + x \implies x \approx 0.67 \]
This means:
\[ \frac{B}{B_H} \approx 0.67 \implies B \approx 0.67 B_H \]

### Step 8: Determine the new frequency when the second magnet's poles are reversed
When the second magnet's poles are reversed, the magnetic field becomes \( B_H - B \):
\[ B_3 = B_H - B = B_H - 0.67 B_H = 0.33 B_H \]

### Step 9: Relate the new time period to the original
Using the relationship:
\[ \frac{T_3}{T_1} = \sqrt{\frac{B_1}{B_3}} \]
Substituting \( B_1 \) and \( B_3 \):
\[ \frac{T_3}{6} = \sqrt{\frac{B_H}{0.33 B_H}} = \sqrt{\frac{1}{0.33}} \]
Calculating gives:
\[ T_3 = 6 \sqrt{3} \approx 10.39 \text{ seconds} \]

### Step 10: Calculate the new frequency
The new frequency \( F_3 \) is:
\[ F_3 = \frac{1}{T_3} \approx \frac{1}{10.39} \text{ oscillations/seconds} \]
Converting to oscillations per minute:
\[ F_3 \approx \frac{60}{10.39} \approx 5.77 \text{ oscillations/min} \]

### Final Answer
The new frequency of oscillation will be approximately **5.77 oscillations per minute**.