Question 61 of 130

In the figure shown $ABC$ is a uniform wire. If the center of mass of the wire lies vertically below point $A$, then $\dfrac{BC}{AB}$ is close to

A$1.85$B$1.5$C$1.37$D$3$

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Answer: (C) $1.37$

${x}_{cm}=\dfrac{x}{2}=\dfrac{(\rho x)(\dfrac{x}{2})\dfrac{1}{2}+\rho y(\dfrac{y}{2})}{\rho (x+y)}$

$\Rightarrow \dfrac{1}{2}+\dfrac{y}{x}=\dfrac{{y}^{2}}{{x}^{2}}$

$\Rightarrow \dfrac{y}{x}=\dfrac{1+\sqrt{3}}{2}=1.37$

JEE Main Physics practice question on Center of Mass, Momentum & Collision (Centre of mass of continuous mass distribution), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.