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 $\frac{\ell_{1}}{2}$
$X _{ cm }=\frac{ m _{1} x _{1}+ m _{2} x _{2}}{ m _{1}+ m _{2}}$
$\frac{\ell_{1}}{2}=\frac{\left(\lambda \ell_{1}\right)\left(\frac{\ell_{1}}{4}\right)+\left(\lambda \ell_{2}\right)\left(\frac{\ell_{2}}{2}\right)}{\lambda\left(\ell_{1}+\ell_{2}\right)}$
$\frac{\ell_{1}^{2}}{2}+\frac{\ell_{1} \ell_{2}}{2}=\frac{\ell_{1}^{2}}{4}+\frac{\ell_{2}^{2}}{2}$
$\frac{\ell_{1}^{2}}{4}+\frac{\ell_{1} \ell_{2}}{2}-\frac{\ell_{2}^{2}}{2}=0$
$\ell_{1}^{2}+2 \ell_{1} \ell_{2}-2 \ell_{2}^{2}=0$
$\ell_{1}=\frac{-2 \ell_{2} \pm \sqrt{4 \ell_{1}^{2}+4.1\left(+2 \ell_{2}^{2}\right)}}{2}$
$\ell_{1}=\frac{-2 \ell_{2}+\sqrt{12 \ell_{2}^{2}}}{2}$
$\ell_{1}=\frac{-2 \ell_{2}+2 \sqrt{3} \ell_{2}}{2}$
$\ell_{1}=(\sqrt{3}-1) \ell_{2}$
$\frac{\ell_{2}}{\ell_{1}}=\frac{1}{\sqrt{3-1}} \times \frac{\sqrt{3}+1}{\sqrt{3+1}}$
$\frac{\ell_{2}}{\ell_{1}}=\frac{\sqrt{3}+1}{2}=\frac{2.732}{2}=1.366$
$\simeq 1.37$