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Moment of inertia of four planar rod frames about an axis in their own plane, matched to List-II
JEE Advanced 2026
Paper 1 · Question 16
Physics
Moment of inertia of planar rod frames about an axis in their own plane - each rod counts by its perpendicular distance from the axis
Easy-Moderate
Four frames of identical rods spin about an axis OCO' in their own plane under the same torque; the slower one spins up, the larger its moment of inertia. Each rod counts by its perpendicular distance from the axis: P ml^2/3 (5), Q 5ml^2/4 (1), R 2ml^2/3 (4), S ml^2/6 (2) - option A.
Open the interactive simulationSub-topics covered
- Moment of inertia as the sum of m r_perp^2; only the distance perpendicular to the axis counts
- A rod with one end on the axis at angle theta: I = (ml^2/3) sin^2 theta
- A rod parallel to the axis at distance d: I = m d^2; the parallel-axis theorem
- Rotational dynamics: the same torque gives alpha = tau / I
What it teaches
- Each rod is integrated along its length: r_perp = s sin theta, I = integral of (m/l)(s sin theta)^2 ds = (ml^2/3) sin^2 theta
- P two perpendicular rods at 45 deg to the axis: 2 x (ml^2/3)(1/2) = ml^2/3 (5)
- Q equilateral triangle, axis through C parallel to AB: 2 x (ml^2/3)(3/4) + m(3l^2/4) = 5ml^2/4 (1)
- R square about its diagonal: 4 x (ml^2/3)(1/2) = 2ml^2/3 (4); S two rods at 30 deg: 2 x (ml^2/3)(1/4) = ml^2/6 (2)
- P5 Q1 R4 S2: option A; ml^2/12 (3) belongs to no frame
Simulation ID
- ADV-2026-P1-PHY-Q16 · revision 1
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