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T h e t i m e p e r i o d o f o s c i l l a t i o n s o f m a g n e t T = 2 π \sqrt{\frac{I}{MH}} where, I = moment of inertia of magnet M = pole strength \times L and H = horizontal component of Earth' s magnetic field . I = \frac{{mL}^{2}}{12} Here m is mass of magnet When the three equal parts of magnet are placed on one another with their like poles together, then Moment of inertia of one part will be I' = \frac{(m / 3) {(L / 3)}^{2}}{12} So moment of Inertia of 3 parts will be I'_{3} = \frac{(m / 3) {(L / 3)}^{2}}{12} \times 3 = \frac{1}{12} \frac{{mL}^{2}}{9} = \frac{I}{9} New magnetic moment M' = Pole strength \times \frac{L}{3} \times 3 = M Hence T' = 2 π \sqrt{\frac{I'}{M' H}} = 2 π \sqrt{\frac{I / 9}{MH}} = \frac{T}{3} As given T = 2 s T' = \frac{2}{3} s

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