Question

A bar magnet is oscillating in the earth's magnetic field with a period \(T\). When the length of the bar magnet is doubled and its mass is quadrupled, the time period is \(T_1\). The ratio of \(T_1\) to \(T\) is

• A. \(1:2\)
• B. \(4:1\)
• C. \(2:1\)
• D. \(1:4\)

Hint:

For magnet oscillation: \(T \propto \sqrt{\frac{ML^2}{L}} = \sqrt{ML}\). Track how mass and length change.

Answer 1

The Correct Option is B

Step 1: Time period of magnet.
\[ T = 2\pi \sqrt{\frac{I}{MB}} \]
where \(I\) = moment of inertia and \(M\) = magnetic moment.

Step 2: Moment of inertia dependence.
\[ I \propto ML^2 \]
Given mass becomes \(4M\) and length becomes \(2L\):
\[ I' \propto 4M \times (2L)^2 = 16ML^2 \]

Step 3: Magnetic moment dependence.
\[ \mu \propto L \]
So,
\[ \mu' = 2\mu \]

Step 4: New time period.
\[ T_1 \propto \sqrt{\frac{I'}{\mu'}} \]
\[ T_1 \propto \sqrt{\frac{16}{2}} = \sqrt{8} = 2\sqrt{2} \]

Step 5: Ratio.
\[ \frac{T_1}{T} = \frac{2\sqrt{2}}{\sqrt{1}} = 4 \]

Step 6: Final conclusion.
\[ \boxed{4:1} \] Hence, correct answer is option (B).