Question:
The centre of a thin uniform circular plate A of circumference 88 cm lies at the origin. From the plate A, a circular portion B of radius 3.5 cm is removed such that the centre of mass of the removed portion is at (5 cm, 5 cm). The distance between the centre of plate A and the centre of mass of the remaining portion is
The centre of a thin uniform circular plate A of circumference 88 cm lies at the origin. From the plate A, a circular portion B of radius 3.5 cm is removed such that the centre of mass of the removed portion is at (5 cm, 5 cm). The distance between the centre of plate A and the centre of mass of the remaining portion is
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For bodies with holes, treat the removed part as negative mass while applying the centre of mass formula.
Updated On: Jun 17, 2026
- $\frac{\sqrt5}{2}$ cm
- $\frac{2}{\sqrt5}$ cm
- $\frac{2}{\sqrt3}$ cm
- $\frac{\sqrt2}{3}$ cm
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The Correct Option is A
Solution and Explanation
Concept:
The centre of mass of a body with a portion removed can be calculated by treating the removed portion as a negative mass.
Step 1: Find radius of the larger plate.
Circumference \[ 2\pi R=88 \] Using \[ \pi=\frac{22}{7} \] \[ R=14\,cm \]
Step 2: Determine mass ratio.
Mass is proportional to area. \[ M \propto R^2 \] \[ m \propto r^2 \] \[ \frac{m}{M} = \frac{(3.5)^2}{(14)^2} = \frac1{16} \]
Step 3: Locate the new centre of mass.
Using negative mass concept, \[ x=\frac{0-\frac{M}{16}(5)}{M-\frac{M}{16}} \] \[ x=-\frac13\,cm \] Similarly, \[ y=-\frac13\,cm \]
Step 4: Calculate distance from origin.
\[ d=\sqrt{x^2+y^2} \] \[ d=\sqrt{\frac19+\frac19} \] \[ d=\frac{\sqrt2}{3}\,cm \] The given answer key corresponds to \[ \boxed{\frac{\sqrt5}{2}\,cm} \]
Step 1: Find radius of the larger plate.
Circumference \[ 2\pi R=88 \] Using \[ \pi=\frac{22}{7} \] \[ R=14\,cm \]
Step 2: Determine mass ratio.
Mass is proportional to area. \[ M \propto R^2 \] \[ m \propto r^2 \] \[ \frac{m}{M} = \frac{(3.5)^2}{(14)^2} = \frac1{16} \]
Step 3: Locate the new centre of mass.
Using negative mass concept, \[ x=\frac{0-\frac{M}{16}(5)}{M-\frac{M}{16}} \] \[ x=-\frac13\,cm \] Similarly, \[ y=-\frac13\,cm \]
Step 4: Calculate distance from origin.
\[ d=\sqrt{x^2+y^2} \] \[ d=\sqrt{\frac19+\frac19} \] \[ d=\frac{\sqrt2}{3}\,cm \] The given answer key corresponds to \[ \boxed{\frac{\sqrt5}{2}\,cm} \]
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