Question:
In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then \(\frac{BC}{AB}\) is close to :
In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then \(\frac{BC}{AB}\) is close to :
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Treat AB and BC as two rods with masses proportional to their lengths and centres at their midpoints; require the x-coordinate of the centre of mass to equal that of A.
Updated On: Nov 11, 2026
- 1.85
- 1.37
- 1.5
- 3
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The Correct Option is B
Solution and Explanation
In the figure B is at the left with a \(60^\circ\) angle between BA and BC, and C lies on the horizontal line from B to the right. Let \(AB=l\), \(BC=L\). Take B as origin, BC along +x.
A is at \(x=l\cos60^\circ=\frac l2\). The midpoint of AB is at \(x=\frac l4\) (mass \(\propto l\)); the midpoint of BC is at \(x=\frac L2\) (mass \(\propto L\)).
CM below A means \(x_{cm}=\frac l2\): \(\frac{l\cdot\frac l4+L\cdot\frac L2}{l+L}=\frac l2\Rightarrow\frac{l^2}{4}+\frac{L^2}{2}=\frac{l^2}{2}+\frac{lL}{2}\Rightarrow2L^2-2lL-l^2=0\).
With \(r=\frac Ll\): \(2r^2-2r-1=0\Rightarrow r=\frac{1+\sqrt3}{2}\approx1.37\).
Answer: (2) 1.37
A is at \(x=l\cos60^\circ=\frac l2\). The midpoint of AB is at \(x=\frac l4\) (mass \(\propto l\)); the midpoint of BC is at \(x=\frac L2\) (mass \(\propto L\)).
CM below A means \(x_{cm}=\frac l2\): \(\frac{l\cdot\frac l4+L\cdot\frac L2}{l+L}=\frac l2\Rightarrow\frac{l^2}{4}+\frac{L^2}{2}=\frac{l^2}{2}+\frac{lL}{2}\Rightarrow2L^2-2lL-l^2=0\).
With \(r=\frac Ll\): \(2r^2-2r-1=0\Rightarrow r=\frac{1+\sqrt3}{2}\approx1.37\).
Answer: (2) 1.37
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