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 :

• A. 1.85
• B. 1.37
• C. 1.5
• D. 3

Answer 1

Answer: (B)

If $CM$ lies vertically below
$A\Rightarrow$ as per choose coordinate axis in $x$ -coordinate is equal to $\frac{{\ell }_1}{2}$
$X_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$
$\frac{{\ell }_1}{2}=\frac{\left(\lambda {\ell }_1\right)\left(\frac{{\ell }_1}{4}\right)+\left(\lambda {\ell }_2\right)\left(\frac{{\ell }_2}{2}\right)}{\lambda \left({\ell }_1+{\ell }_2\right)}$
$\frac{{\ell }_1^2}{2}+\frac{{\ell }_1{\ell }_2}{2}=\frac{{\ell }_1^2}{4}+\frac{{\ell }_2^2}{2}$
$\frac{{\ell }_1^2}{4}+\frac{{\ell }_1{\ell }_2}{2}-\frac{{\ell }_2^2}{2}=0$
${\ell }_1^2+2{\ell }_1{\ell }_2-2{\ell }_2^2=0$
${\ell }_1=\frac{-2{\ell }_2\pm \sqrt{4{\ell }_1^2+4.1\left(+2{\ell }_2^2\right)}}{2}$
${\ell }_1=\frac{-2{\ell }_2+\sqrt{12{\ell }_2^2}}{2}$
${\ell }_1=\frac{-2{\ell }_2+2\sqrt{3}{\ell }_2}{2}$
${\ell }_1=(\sqrt{3}-1){\ell }_2$
$\frac{{\ell }_2}{{\ell }_1}=\frac{1}{\sqrt{3-1}}\times \frac{\sqrt{3}+1}{\sqrt{3+1}}$
$\frac{{\ell }_2}{{\ell }_1}=\frac{\sqrt{3}+1}{2}=\frac{2.732}{2}=1.366$
$\simeq 1.37$