3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 3 · AdvancedSingle correct
From a uniform circular disc of mass M (before cutting) and radius R, a circular hole of radius R/2 is cut out, the hole touching the rim of the disc (its centre is at distance R/2 from the centre O of the original disc). Find the moment of inertia of the remaining plate about an axis perpendicular to its plane passing through the centre of mass of the remaining plate.
A83MR2
B9637MR2
C3213MR2
D9641MR2
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Answer: B
Key insight: subtraction must be done about a common axis (through O), and only then may the parallel-axis theorem be used, and only through the new centre of mass. Removed mass =M/4. About O: IO=2MR2−[2(M/4)(R/2)2+4M(2R)2]=2MR2−323MR2=3213MR2. Remaining mass 3M/4; its centre of mass lies on the far side of O at x=3M/4(M/4)(R/2)=R/6. Parallel axis: Icm=IO−43M(6R)2=3213MR2−48MR2=9637MR2. Trap: stopping at 13MR2/32 (axis through O, not the CM) or adding the shift term (41MR2/96) instead of subtracting.
Exam pattern: JEE Advanced – moment of inertia of a body with a cavity, parallel-axis theorem
Question 2Level 2 · ExamSingle correct
A uniform thin square plate ABCD has mass M and side a. Find its moment of inertia about an axis lying in the plane of the plate, passing through corner A and parallel to the diagonal BD.
A127Ma2
B1213Ma2
C3Ma2
D32Ma2
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Answer: A
Key insight: for a square lamina the moment of inertia about every in-plane axis through the centre is the same, Ma2/12 (perpendicular-axis theorem with fourfold symmetry gives Ix=Iy=Idiag=21⋅6Ma2). The axis through A parallel to BD is at perpendicular distance equal to half the diagonal, a/2, from the centre. Parallel axis: I=12Ma2+M2a2=127Ma2. Traps: Ma2/3 is about an edge; 2Ma2/3 is about the perpendicular axis through a corner; 13Ma2/12 comes from using distance a instead of a/2.
Exam pattern: JEE Main/Advanced – perpendicular and parallel axis theorems for a lamina
Question 3Level 3 · AdvancedSingle correct
A plank of mass M=1 kg lies on a smooth horizontal floor. A uniform solid cylinder of mass m=2 kg and radius R rests on the plank with its axis horizontal and perpendicular to the length of the plank. A constant horizontal force F=10 N is applied at the topmost point of the cylinder, perpendicular to its axis. The cylinder rolls on the plank without slipping. The acceleration of the plank relative to the ground is
A2 m/s2 opposite to F
B2 m/s2 along F
Czero
D310 m/s2 along F
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Answer: A
Key insight: friction on the cylinder acts forward (along F), so its reaction drags the plank backward. Let a1, a2 be accelerations of the cylinder centre and plank (along F), f friction on the cylinder (along F), α the forward-rolling angular acceleration. Cylinder: F+f=ma1; torque about centre: FR−fR=21mR2α. Plank: −f=Ma2. No slip at contact: a1−αR=a2. Solving: f=3M+mFM=2 N, a2=−3M+mF=−2 m/s2, a1=6 m/s2. So the plank accelerates at 2 m/s2 opposite to F. Trap: assuming friction opposes F (giving the plank a forward acceleration) or assuming the plank stays at rest.
Exam pattern: JEE Advanced – rolling on a movable plank with constraint relation at contact
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