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JEE Main 2-April-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 2-April-2025 Paper - Shift 24 Marks
MCQ
The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :
  • A
    $\frac{1}{2} \mathrm{Mr}^{2}$
  • B
    $\frac{3}{8} \mathrm{Mr}^{2}$
  • C
    $\frac{3}{2} \mathrm{Mr}^{2}$
  • D
    $2 \mathrm{Mr}^{2}$
✓

Answer

B. $\frac{3}{8} \mathrm{Mr}^{2}$
Diameter is given as R.
$\therefore$ Radius $=\mathrm{R} / 2$
$\mathrm{I}_{\text {tan gent }}=\frac{3}{2} \mathrm{~m}\left(\frac{\mathrm{R}}{2}\right)^{2}=\frac{3}{8} \mathrm{mR}^{2}$

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