Question

Two bar magnets \(A\) and \(B\) are placed one over the other and are allowed to vibrate in a vibration magnetometer. They make \(20\) oscillations per minute when the similar poles of \(A\) and \(B\) are on the same side, while they make \(15\) oscillations per minute when their opposite poles lie on the same side. If \(M_A\) and \(M_B\) are the magnetic moments of \(A\) and \(B\), and \(M_A>M_B\), the ratio \(M_A : M_B\) is

• A. \(4:3\)
• B. \(25:7\)
• C. \(7:5\)
• D. \(25:16\)

Hint:

For superposed magnets: \(f \propto \sqrt{M}\). Use \(M_{eq}=M_A\pm M_B\) depending on pole orientation.

Answer 1

The Correct Option is B

Step 1: Use vibration magnetometer relation.
Time period:
\[ T = 2\pi\sqrt{\frac{I}{MB_H}} \]
So frequency \(f\) is:
\[ f \propto \sqrt{M} \]
Step 2: Effective magnetic moment.
When similar poles are on same side:
\[ M_{eq1} = M_A + M_B \]
When opposite poles are on same side:
\[ M_{eq2} = M_A - M_B \]
Step 3: Use oscillations per minute.
Oscillations per minute \(\propto f\).
So:
\[ \frac{f_1}{f_2} = \frac{20}{15} = \frac{4}{3} \]
But:
\[ \frac{f_1}{f_2} = \sqrt{\frac{M_A + M_B}{M_A - M_B}} \]
Step 4: Square both sides.
\[ \left(\frac{4}{3}\right)^2 = \frac{M_A + M_B}{M_A - M_B} \]
\[ \frac{16}{9} = \frac{M_A + M_B}{M_A - M_B} \]
Step 5: Solve for ratio.
\[ 16(M_A - M_B) = 9(M_A + M_B) \]
\[ 16M_A - 16M_B = 9M_A + 9M_B \]
\[ 7M_A = 25M_B \Rightarrow \frac{M_A}{M_B} = \frac{25}{7} \]
Final Answer: \[ \boxed{25:7} \]