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JEE Main 2024 (30 Jan, Shift 2) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2024 (30 Jan, Shift 2)

Q31JEE Main 2024 · 30 JAN · SHIFT 2Medium

Angular Momentum & its Conservation

Two discs of moment of inertia I1=4 kg m2I_1 = 4\text{ kg m}^{2} and I2=2 kg m2I_2 = 2\text{ kg m}^{2}, about their central axes & normal to their planes, rotating with angular speeds 10 rad/s10\text{ rad/s} & 4 rad/s4\text{ rad/s} respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is _________ J.

Numerical answer type. Work it out, then open the answer.

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Answer: 24

  1. Loss on coupling: ΔK=12I1I2I1+I2(ω1−ω2)2\Delta K = \frac{1}{2}\frac{I_1I_2}{I_1+I_2}(\omega_1-\omega_2)^2.
  2. I1I2I1+I2=4×26=43 kg m2\frac{I_1I_2}{I_1+I_2} = \frac{4\times 2}{6} = \frac{4}{3}\ \text{kg m}^2, ω1−ω2=6 rad/s\omega_1-\omega_2 = 6\ \text{rad/s}.
  3. ΔK=12⋅43⋅36=24 J\Delta K = \frac{1}{2}\cdot\frac{4}{3}\cdot 36 = 24\ \text{J}
  4. Answer: 24 J
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