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JEE Main 2025 (4 Apr, Shift 2) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2025 (4 Apr, Shift 2)

Q19JEE Main 2025 · 4 APR · SHIFT 2Medium

Angular Momentum & its Conservation

A solid sphere with uniform density and radius RR is rotating initially with constant angular velocity (ω1)(\omega_1) about its diameter. After some time during the rotation it starts losing mass at a uniform rate, with no change in its shape. The angular velocity of the sphere when its radius become R/2R/2 is xω1x\omega_1. The value of xx is _________.

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

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Answer: 32

  1. Density and shape unchanged, so M∝R3M\propto R^3: at R/2R/2, mass is M/8M/8.
  2. I=25MR2∝R5⇒I′=I32I=\frac25MR^2\propto R^5\Rightarrow I'=\frac{I}{32}
  3. Taking angular momentum conserved: Iω1=I32ωI\omega_1=\frac{I}{32}\omega
  4. ω=32ω1\omega=32\omega_1, so x=32x=32.
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