JEE Main 2016 · 10 Apr · Shift 1 · Physics Q5

← JEE Main 2016 · 10 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 5 of 90

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

A bent uniform wire: AB rises from B at 60° to the horizontal BC.

1.85

1.37

1.5

Answer and worked solution

Official answer: B

With B at the origin, AB = a at 60° and BC = b along x. A is at $x=\frac a2$ . The centre of mass has
$x=\frac{a\cdot\frac a4+b\cdot\frac b2}{a+b}=\frac a2\implies2b^2-2ab-a^2=0.$
So $\frac ba=\frac{1+\sqrt3}2\approx1.37$ .

Topics:Centre of mass

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JEE Main 2016 · 10 Apr · Shift 1, Physics question 5