JEE Main 201610 AprilOnline

Exam: JEE Main
Year: 2016
Date: 10 April
Mode: Online
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Main 2016 · 10 April · Online · Physics · Center of Mass, Momentum and Collision

In the figure shown ABC is a uniform wire. If the center of mass of...

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

In the figure shown ABC is a uniform wire. If the center of mass of the wire lies vertically below point A , then

Options

A $1 . 85$
B $1 . 5$
C $1 . 37$
D $3$

Correct answer

C. $1 . 37$

Step-by-step solution

In the figure shown ABC is a uniform wire. If the center of mass of the wire lies vertically below point A , then

${x}_{cm} = \dfrac{x}{2} = \dfrac{( ρx )( \dfrac{x}{2} )\dfrac{1}{2} + ρy( \dfrac{y}{2} )}{ρ( x + y )}$

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

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

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