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MHT CET 2024 (3 May, Shift 1) · Vibration Magnetometer

Vibration Magnetometer question from MHT CET 2024 (3 May, Shift 1)

Q4MHT CET 2024 · 3 MAY · SHIFT 1Medium

Vibration Magnetometer

A magnetic needle of magnetic moment 6×10−2 Am26 \times 10^{-2}\text{ Am}^{2} and moment of inertia 9.6×10−5 kg m29.6 \times 10^{-5}\text{ kg m}^{2} performs simple harmonic motion in a magnetic field of 0.01 T. Time taken to complete 10 oscillations is [Take π=3.14\pi = 3.14]

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Answer: (c)

  1. T=2πIMB=2π9.6×10−56×10−2×0.01=2π0.16T = 2\pi\sqrt{\frac{I}{MB}} = 2\pi\sqrt{\frac{9.6 \times 10^{-5}}{6 \times 10^{-2} \times 0.01}} = 2\pi\sqrt{0.16}
  2. T=2×3.14×0.4=2.512 sT = 2 \times 3.14 \times 0.4 = 2.512\ \text{s}
  3. 10 oscillations: 10T=25.12 s10T = 25.12\ \text{s}
  4. So the answer is (c) 25.12 s.
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