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A thin magnetic iron rod of length 30 ~cm is suspended in a uniform magnetic field. Its time period of oscillation is 4 ~s . It is broken into three equal parts. The time period in seconds of oscillation of one part when suspended in the same magnetic field is

Options

  1. A1 3
  2. B2 3
  3. C3
  4. D4 3

Correct answer

D. 4 3

Step-by-step solution

Time period of magnet T=2 I M H M= magnetic moment =m l I= moment of inertia = m l^2 12 When magnet is broken into three equal parts, Then, M^ =m l 3 = M 3 I^ = m(l / 3)^2 12 = m l^2 9 12 = I 9 Now, time period T^ =2 I^ M^ H aligned & =2 I / 9 M 3 H & =2 I M H 3 & = T 3 = 4 3 sec . aligned

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