COM of Cavity & Truncated Bodies — NEET Physics MCQs with Solutions
Free NEET Physics COM of Cavity & Truncated Bodies MCQs with step-by-step solutions (8 questions). Part of Centre of Mass, Momentum & Collisions. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — COM of Cavity & Truncated Bodies · easy · theory
To find the COM of a body with a cavity, the standard method is to treat the cavity as:
A. A positive extra mass
B. A point mass at the edge
C. A superposed NEGATIVE mass of the same shape ✓ Correct
D. Zero mass at the centre
Solution: Full body (positive) + cavity region (negative mass) reproduces the actual object; apply the usual COM formula with the negative term.
Q2 — COM of Cavity & Truncated Bodies · medium · theory
A circular hole is punched in a uniform disc, away from the centre. The COM of the remaining lamina lies:
A. On the hole side of the centre
B. On the line joining the centres, on the OPPOSITE side of the disc centre from the hole ✓ Correct
C. At the hole's centre
D. At the original centre still
Solution: Removing mass on one side pushes the balance point the other way along the line of centres.
Q3 — COM of Cavity & Truncated Bodies · easy · numerical
From a uniform disc of radius R a concentric hole of radius R/2 is cut. The COM of the ring-like remainder is:
A. At R/4 from the centre
B. At R/2
C. At the original centre ✓ Correct
D. Undefined
Solution: Concentric removal keeps full symmetry — COM stays at the centre.
Q4 — COM of Cavity & Truncated Bodies · medium · numerical
From a uniform disc of radius R, a circular hole of radius R/2 is cut with its centre at R/2 from the disc centre. The COM of the remainder shifts from the centre by:
A. R/6 (away from the hole) ✓ Correct
B. R/2
C. R/4
D. R/8
Solution: Shift = (m_hole·d)/(M − m_hole) = [(M/4)(R/2)]/(3M/4) = R/6, opposite to the hole.
Q5 — COM of Cavity & Truncated Bodies · hard · numerical
From a uniform square plate of side 2a, one quadrant (an a×a square) is removed. The COM of the remaining plate shifts from the centre by a distance of:
A. a√2/6 ✓ Correct
B. a/3
C. a√2/3
D. a/6
Solution: The removed quarter (mass M/4) had its centre at (a/2, a/2) — a distance a√2/2 from the plate centre. Shift = (M/4)(a√2/2)/(3M/4) = a√2/6, directed away from the removed corner.
Q6 — COM of Cavity & Truncated Bodies · medium · numerical
A disc of radius R has mass 9 kg; a hole of radius R/3 is drilled with centre at 2R/3 from the disc centre. The mass removed is:
A. 0.5 kg
B. 3 kg
C. 1 kg ✓ Correct
D. 2 kg
Solution: Mass ∝ area: m = 9 × (R/3)²/R² = 9/9 = 1 kg — the first step of every cavity problem.
Q7 — COM of Cavity & Truncated Bodies · easy · numerical
A uniform metre stick is cut at the 40 cm mark and the shorter piece is discarded. The COM of the remaining 60 cm piece (from its own left end at 40 cm) is at the stick mark:
A. 75 cm
B. 60 cm
C. 70 cm ✓ Correct
D. 80 cm
Solution: The 60 cm piece spans 40–100 cm; its midpoint is at 70 cm.
Q8 — COM of Cavity & Truncated Bodies · medium · numerical
From a disc of radius 6 cm, a hole of radius 2 cm is cut with its centre 3 cm from the disc centre. The COM shift of the remainder is:
A. 1/3 cm
B. 3/8 cm ✓ Correct
C. 3/4 cm
D. 1 cm
Solution: m_hole/M = 4/36 = 1/9; shift = (1/9 × 3)/(8/9) = 3/8 cm away from the hole.