QJEEM16Phf0bb: JEE Main 2016 Physics - Class 11 JEEM16Phf0bb Solution: In the figure shown ABC is a uniform wire. If centre of mass of …

Here   AB = x

and   BC = y

and    $\lambda$ = linear mass density.

As centre of mass is below point A, so horizontal distance of the centre of mass from B is = xcos60^o = ${x \over 2}$

$\therefore$    X_CM = ${x \over 2}$ = ${{{m_1}{x_1} + {m_2}{x_2}} \over {{m_1} + {m_2}}}$

$\Rightarrow$     ${x \over 2}$ = ${{\left( {\lambda x} \right)\left( {{x \over 2}} \right)\cos {{60}^o} + \left( {\lambda y} \right)\left( {{y \over 2}} \right)} \over {\lambda \left( {x + y} \right)}}$

$\Rightarrow$     ${x \over 2}$ = ${{{{{x^2}} \over 4} + {{{y^2}} \over 2}} \over {x + y}}$

$\Rightarrow$    x² + xy = ${{{x^2}} \over 2} + {y^2}$

$\Rightarrow$    x² + 2xy $-$ 2y² = 0

$\therefore$    x = ${{ - 2y \pm \sqrt {{{\left( {2y} \right)}^2} - 4.1\left( { - 2{y^2}} \right)} } \over {2.1}}$

=    ${{ - 2y \pm \sqrt {12{y^2}} } \over 2}$

=    $-$ y $\pm$ $\sqrt 3$ y

${x \over y} \ne - \sqrt 3 - 1$   as   ${x \over y}$ = positive.

$\therefore$     ${x \over y}$ = $\sqrt 3 - 1$

$\Rightarrow$     ${y \over x}$ = ${1 \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}$

=    ${{\sqrt 3 + 1} \over 2}$

=    ${{2.732} \over 2}$

=   1.366  $\simeq$  1.37