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System of Particles and Rotational Motion

JEE Main 2025 (Online) 22nd January Morning ShiftMCQMedium📈 Topic in 21% papers+4−1
Ques. 1JEE Main 2025, 22 January (Morning Shift)

A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above. [figure]

  1. 1732MR2
  2. 1332MR2
  3. 932MR2
  4. 732MR2

Answer

(2) 1332MR2

Solution

Remaining disc = full disc - removed disc (same axis through O).

[figure]

m= M4 (area), Ifull= 12MR2

Removed disc (parallel axis, distance R2):

I= 12m ( R2 )2+m ( R2 )2

= 38mR2= 332MR2

I= 12MR2- 332MR2= 1332MR2

Answer: (2)

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Ques. 2JEE Main 2025, 22 January (Morning Shift)

The position vectors of two 1 kg particles, (A) and (B), are given by rA= ( 1t2 i+ 2t j+ 3t k )m and rB= ( 1t i+ 2t2 j+ 3t k )m, respectively; ( 1=1 m/s2, 2=3n m/s, 3=2 m/s, 1=2 m/s, 2=-1 m/s2, 3=4p m/s), where t is time, n and p are constants. At t=1 s, | VA |= | VB | and velocities VA and VB of the particles are orthogonal to each other. At t=1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is L kg m2 s-1. The value of L is .

Ques. 3JEE Main 2025, 22 January (Evening Shift)

The torque due to the force (2 i+ j+2 k ) about the origin, acting on a particle whose position vector is ( i+ j+ k ), would be

  1. j+ k
  2. i- k
  3. i- j+ k
  4. i+ k

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