In the figure shown ABC is a uniform wire . I | vedclass

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Question

Center of mass

$\begin{array}{l}
{x_{cm}} = \frac{x}{2} = \frac{{\left( {\rho x} \right)\left( {\frac{x}{2}} \right)\frac{1}{2} + \rho y\left( {\f

Vedclass · Accepted Answer

$1.37$

Vedclass · Answer

Center of mass

$\begin{array}{l}
{x_{cm}} = \frac{x}{2} = \frac{{\left( {ho x} ight)\left( {\frac{x}{2}} ight)\frac{1}{2} + ho y\left( {\frac{y}{2}} ight)}}{{ho \left( {x + y} ight)}}\
 \Rightarrow \frac{1}{2} + \frac{y}{x} = \frac{{{y^2}}}{{{x^2}}}\
	herefore \frac{{BC}}{{AB}} = \frac{y}{x} = \frac{{1 + \sqrt 3 }}{2} = 1.37
\end{array}$

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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