question_answer

In the figure shown/ ABC is a uniform wire. If centre of mass wire lies vertically below point A, then\[\frac{BC}{AB}\]   JEE Main Online Paper (Held On 10 April 2016)

A) 1.85                                                          

B) 1.5                

C) 3                                                                

D) 1.37                

Correct Answer: D

Solution :

If CM lies vertically below \[A\Rightarrow \]?as per choose coordinate axis in x-coordinate is equal to \[\frac{{{\ell }_{1}}}{2}\] \[{{X}_{cm}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]              \[\frac{{{\ell }_{1}}}{2}=\frac{({{\ell }_{1}})\left( \frac{{{\ell }_{1}}}{4} \right)+({{\ell }_{2}})\left( \frac{{{\ell }_{2}}}{2} \right)}{({{\ell }_{1}}+{{\ell }_{2}})}\]              \[\frac{\ell _{1}^{2}}{2}+\frac{{{\ell }_{1}}{{\ell }_{2}}}{2}=\frac{\ell _{1}^{2}}{4}+\frac{\ell _{1}^{2}}{4}+\frac{\ell _{2}^{2}}{2}\]              \[\frac{\ell _{1}^{2}}{4}+\frac{{{\ell }_{1}}+\ell  & {{ & }_{2}}}{2}-\frac{\ell _{2}^{2}}{2}=0\]              \[\ell _{1}^{2}+2{{\ell }_{1}}{{\ell }_{2}}-2\ell _{2}^{2}=0\]              \[{{\ell }_{1}}=\frac{-2{{\ell }_{2}}\pm \sqrt{4\ell _{1}^{2}+4.1(+2\ell _{2}^{2})}}{2}\]              \[{{\ell }_{1}}=\frac{-2{{\ell }_{2}}\pm \sqrt{12\ell _{2}^{2})}}{2}\]              \[{{\ell }_{1}}=\frac{-2{{\ell }_{2}}\pm 2\sqrt{3}{{\ell }_{2}}}{2}\]              \[{{\ell }_{1}}=\left( \sqrt{3}-1 \right){{\ell }_{2}}\]              \[\frac{{{\ell }_{2}}}{\ell  &  & {{ & }_{1}}}=\frac{1}{\sqrt{3-1}}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\]              \[\frac{{{\ell }_{2}}}{\ell  &  & {{ & }_{1}}}=\frac{\sqrt{3}+1}{2}=\frac{2.732}{2}=1.366\]              \[\simeq 1.37\]