IIT-JEE 2009 · Paper 2 · Physics Q40

← IIT-JEE 2009 · Paper 2

PhysicsSingle correct+3 / -1Question 40 of 57

A piece of wire is bent in the shape of a parabola $y=kx^2$ ( $y$ -axis vertical) with a bead of mass $m$ on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$ -axis with a constant acceleration $a$ . The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the $y$ -axis is

\dfrac a{gk}

\dfrac a{2gk}

\dfrac{2a}{gk}

\dfrac a{4gk}

Answer and worked solution

Official answer: B

Wire frame. A pseudo-force $ma$ acts opposite to the acceleration. At equilibrium the resultant of $mg$ and $ma$ is along the normal, so the slope of the wire equals $\frac ag$ :
$\frac{dy}{dx}=2kx=\frac ag\implies x=\frac a{2gk}.$

(B)

Topics:Pseudo forces and non-inertial frames

← Previous question Next question →

IIT-JEE 2009 · Paper 2, Physics question 40