COM of Cavity & Truncated Bodies — JEE Main Physics MCQs with Solutions
Free JEE Main 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. At the hole's centre
B. On the hole side of the centre
C. On the line joining the centres, on the OPPOSITE side of the disc centre from the hole ✓ Correct
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. Undefined
D. At the original centre ✓ Correct
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/8
D. R/4
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/6
B. a/3
C. a√2/3
D. a√2/6 ✓ Correct
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. 2 kg
B. 0.5 kg
C. 3 kg
D. 1 kg ✓ Correct
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 · hard · numerical
From a solid sphere of radius R, a spherical cavity of radius R/2 is scooped out with its centre at R/2 from the sphere centre. The COM of the remainder is displaced from the centre by:
A. R/14 ✓ Correct
B. R/7
C. R/8
D. R/16
Solution: m_cavity = M/8 at d = R/2: shift = (M/8)(R/2)/(7M/8) = R/14, opposite the cavity.
Q8 — COM of Cavity & Truncated Bodies · medium · numerical
A square hole of side a is punched from a uniform square plate of side 2a, the hole's centre being a/2 from the plate centre. The shift of the COM of the remainder is:
A. a/8
B. a/6 ✓ Correct
C. a/4
D. a/2
Solution: Hole mass = M/4 (area ratio). Shift = (M/4)(a/2)/(3M/4) = a/6, away from the hole.