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

Pick an option to check your answer.
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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