MCQ
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
- A$1.85$
- B$1.5$
- ✓$1.37$
- D$3$
✓
Answer
Correct option: C.
$1.37$
c
Center of mass
Center of mass
$\begin{array}{l}
{x_{cm}} = \frac{x}{2} = \frac{{\left( {\rho x} \right)\left( {\frac{x}{2}} \right)\frac{1}{2} + \rho y\left( {\frac{y}{2}} \right)}}{{\rho \left( {x + y} \right)}}\\
\Rightarrow \frac{1}{2} + \frac{y}{x} = \frac{{{y^2}}}{{{x^2}}}\\
\therefore \frac{{BC}}{{AB}} = \frac{y}{x} = \frac{{1 + \sqrt 3 }}{2} = 1.37
\end{array}$
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