Question:
The shift (in metre) in centre of mass when the largest possible equilateral triangular plate is removed from a uniform square plate of side \(2\,m\) with one of their sides coinciding is:
The shift (in metre) in centre of mass when the largest possible equilateral triangular plate is removed from a uniform square plate of side \(2\,m\) with one of their sides coinciding is:
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When a piece is removed from a body, treat the removed portion as negative mass in centre of mass calculations.
Updated On: Jun 18, 2026
- \(\frac{\sqrt3-4}{\sqrt3-2}\)
- \(\frac{\sqrt3-1}{4-\sqrt3}\)
- \(\frac{2-\sqrt3}{4-\sqrt3}\)
- \(\frac{\sqrt3-1}{\sqrt3-4}\)
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The Correct Option is C
Solution and Explanation
Concept:
Treat the removed triangular plate as negative mass and apply centre of mass formula.
Step 1: Area of square.
\[ A_s=2\times2=4. \] COM of square: \[ (1,1). \]
Step 2: Area of largest equilateral triangle.
Its side equals side of square: \[ a=2. \] Area \[ A_t = \frac{\sqrt3}{4}(2)^2 = \sqrt3. \] Centroid of triangle is at \[ \frac{\sqrt3}{3} \] from the base.
Step 3: Apply COM formula.
Using removed area as negative area, \[ x_{cm}=1. \] Only vertical coordinate changes. After simplification, \[ \Delta = \frac{2-\sqrt3}{4-\sqrt3}. \] Thus shift equals \[ \boxed{\frac{2-\sqrt3}{4-\sqrt3}}. \]
Step 1: Area of square.
\[ A_s=2\times2=4. \] COM of square: \[ (1,1). \]
Step 2: Area of largest equilateral triangle.
Its side equals side of square: \[ a=2. \] Area \[ A_t = \frac{\sqrt3}{4}(2)^2 = \sqrt3. \] Centroid of triangle is at \[ \frac{\sqrt3}{3} \] from the base.
Step 3: Apply COM formula.
Using removed area as negative area, \[ x_{cm}=1. \] Only vertical coordinate changes. After simplification, \[ \Delta = \frac{2-\sqrt3}{4-\sqrt3}. \] Thus shift equals \[ \boxed{\frac{2-\sqrt3}{4-\sqrt3}}. \]
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