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JEE Advanced 2009Paper 2
Exam
JEE Advanced
Year
2009
Paper
Paper 2
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2009 · Paper 2 · Physics · Laws of Motion

A piece of wire is bent in the shape of a parabola y=k x^2 (y-axis...

A piece of wire is bent in the shape of a parabola y=kx2y=k x^2 (y-axis vertical) with a bead of mass mm 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 xx-axis with the constant acceleration aa. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the yy-axis isA piece of wire is bent in the shape of a parabola y=k x^2 (y-axis vertical) with a bead of mass m on it. The bead canA piece of wire is bent in the shape of a parabola y=k x^2 (y-axis vertical) with a bead of mass m on it. The bead can

Options

  1. Aagk\frac{a}{g k}
  2. Ba2gk\frac{a}{2 g k}
  3. C2agk\frac{2 a}{g k}
  4. Da4gk\frac{a}{4 g k}

Correct answer

B. a2gk\frac{a}{2 g k}

Step-by-step solution

Nsin⁡θ=mgN \sin \theta=m gNcos⁡θ=matan⁡θ=gacot⁡θ=ag=tan⁡(90∘−θ)=dydx=2kx\begin{aligned}N \cos \theta& =m a \\ \tan \theta& =\frac{g}{a}\\ \cot \theta& =\frac{a}{g}=\tan \left(90^{{}^{\circ}}-\theta\right)=\frac{d y}{d x}=2 k x \end{aligned}A piece of wire is bent in the shape of a parabola y=k x^2 (y-axis vertical) with a bead of mass m on it. The bead can\\\therefore \quad x=a2kg\frac{a}{2 k g}
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