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 x² = 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

  $\frac{\text{g}}{2}$
• B

  $\frac{\sqrt{\mathrm{3 }}\text{g}}{2}$
• Correct answer
  C

  $\frac{\text{g}}{\sqrt{2}}$
• D

  $\frac{\text{g}}{\sqrt{5}}$

Solution

x² = 4ay

Differentiating w.r.t. y, we get

$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2\text{a}}$

It can be considered as an inclined plane



$\therefore \text{At}\left(2\text{ a, a}\right)\text{,}\frac{\text{dy}}{\text{dx}}=1 \Rightarrow$hence $\theta =45^{\circ }$

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

hence tangential acceleration is $\text{g sin }\theta =\frac{\text{g}}{\sqrt{2}}$

Hence, the correct answer is option: 'C'.