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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Motion in Two Dimensions

NTA Abhyas 2020 · Physics · Motion in Two Dimensions

A bead of mass m is located on a parabolic wire with its axis...

A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equation is x^{2} = 4ay. The wire frame is fixed and the bead can slide on it without friction. The bead is released from the point y = 4a on the wire frame from rest. The tangential acceleration of the bead when it reaches the position given by y = a is :

A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and

Options

  1. Ag2\dfrac{\text{g}}{2}
  2. B3g2\dfrac{\sqrt{3} \text{g}}{2}
  3. Cg2\dfrac{\text{g}}{\sqrt{2}}
  4. Dg5\dfrac{\text{g}}{\sqrt{5}}

Correct answer

C. g2\dfrac{\text{g}}{\sqrt{2}}

Step-by-step solution

x2 = 4ay

Differentiating w.r.t. y, we get

dydx=x2aItcanbeconsideredasaninclinedplane\dfrac{\text{dy}}{\text{dx}} = \dfrac{\text{x}}{2 \text{a} It can be considered as an inclined plane}

A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and

∴At(2a, a),dydx=1⇒∴ \text{At} \left(2 \text{a, a}\right) \text{,} \dfrac{\text{dy}}{\text{dx}} = 1 ⇒ hence θ=45∘\theta = 4 {5}^{∘}

the component of weight along tangential direction is mg sin θ\theta.

hence tangential acceleration is g sinθ=g2\text{g sin} \theta = \dfrac{\text{g}}{\sqrt{2}}
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