A thin rectangular magnet suspended freely has a period of oscillation equal to T . Now it is broken into two equal halves (each having half of the or - Physics MCQ AIEEE 2003 2026 | ExamTest.live

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A thin rectangular magnet suspended freely has a period of oscillation equal to T . Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T^{'}, the ratio T^{'} / T is:

Solution & Step-by-step Explanation

The time period of a magnet is T = 2 π \frac{I}{MB}, where I is the moment of inertia and M is the magnetic dipole moment.When broken into two equal halves:New mass m^{'} = m /2 and new length l^{'} = l /2 .New moment of inertia I^{'} = \frac{1}{12} m^{'} (l^{'})^{2} = \frac{1}{12} (\frac{m}{2}) (\frac{l}{2})^{2} = \frac{I}{8} .New magnetic moment M^{'} = pole strength \times \frac{l}{2} = \frac{M}{2} .New time period T^{'} : T^{'} = 2 π \frac{I /8}{( M /2 ) B} = 2 π \frac{I}{4 MB} = \frac{1}{2} (2 π \frac{I}{MB}) = \frac{T}{2} Thus, T^{'} / T = 1/2 .

Try it yourself before checking the explanation above.

A thin rectangular magnet suspended freely has a period of oscillation equal to T . Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T^{'}, the ratio T^{'} / T is:

\frac{1}{2 2}

1/2

2

1/4

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