COM of Cavity & Truncated Bodies — IISER Physics MCQs with Solutions
Free IISER 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. Zero mass at the centre
D. A superposed NEGATIVE mass of the same shape ✓ Correct
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. At the original centre still
C. On the hole side of the centre
D. On the line joining the centres, on the OPPOSITE side of the disc centre from the hole ✓ Correct
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 the original centre ✓ Correct
B. Undefined
C. At R/4 from the centre
D. At R/2
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/4
B. R/8
C. R/6 (away from the hole) ✓ Correct
D. R/2
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/3
B. a√2/6 ✓ Correct
C. a/6
D. a√2/3
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. 3 kg
B. 1 kg ✓ Correct
C. 0.5 kg
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 · hard · theory
When mass is removed from a body, the COM of the remainder, the removed part, and the original body are related by:
A. All three lie on one straight line ✓ Correct
B. No fixed relation exists
C. They coincide
D. They form a right triangle
Solution: M_orig x_orig = m_rem x_rem + m_cut x_cut: the original COM divides the segment joining the other two internally in inverse mass ratio — hence collinear.
Q8 — COM of Cavity & Truncated Bodies · hard · numerical
A disc of radius 2R has a hole of radius R cut tangent to its edge (hole centre at R from the disc centre). The COM of the crescent-shaped remainder is from the disc centre at:
A. R/4
B. 2R/3
C. R/3, away from the hole ✓ Correct
D. R/2
Solution: m_hole = M/4 at distance R: shift = (M/4·R)/(3M/4) = R/3 opposite the hole — the classic lune result.