The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2 s . The magnet - Physics MCQ AIEEE 2004 2026 | ExamTest.live | ExamTest.live

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The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2 s . The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be:

A $2 s$
B $2/3 s$
C $23 s$
D $2/ 3 s$

Solution & Step-by-step Explanation

Time period T = 2 π \frac{I}{MB} .Initially, I = \frac{1}{12} m L^{2} and M = m_{p} L (where m_{p} is pole strength).When cut into 3 parts:Length of each part L^{'} = L /3, Mass m^{'} = m /3, Magnetic moment M^{'} = M /3 .Combination of 3 parts (stacked):Total Moment of Inertia I_{n e w} = 3 \times (\frac{1}{12} m^{'} (L^{'})^{2}) = 3 \times \frac{1}{12} (\frac{m}{3}) (\frac{L}{3})^{2} = \frac{1}{27} I .Total Magnetic Moment M_{n e w} = 3 \times M^{'} = 3 \times \frac{M}{3} = M .New time period T^{'} : T^{'} = 2 π \frac{I /27}{MB} = \frac{1}{27} (2 π \frac{I}{MB}) = \frac{T}{3 3} \dots Wait, recalculating stack . Actually, if the parts are stacked, the total mass is m, but length is L /3 . I_{co mb} = \frac{1}{12} m (L /3)^{2} = I /9 . M_{co mb} = 3 \times (m_{p} \times L /3) = M . T^{'} = 2 π \frac{I /9}{MB} = \frac{1}{3} T = \frac{2}{3} s

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The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2 s . The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be:

A

2 s

B

2/3 s

C

23 s

D

2/ 3 s

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