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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2016
  • Hard
  • 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 

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Answer & Solution

Correct Answer

(C) \(1.37\)

Step-by-step Solution

Detailed explanation

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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