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JEE Main 2025 (28 Jan, Shift 1) · Conservation of Energy

Conservation of Energy question from JEE Main 2025 (28 Jan, Shift 1)

Q8JEE Main 2025 · 28 JAN · SHIFT 1Difficult

Conservation of Energy

A bead of mass ' mm ' slides without friction on the wall of a vertical circular hoop of radius ' RR ' as shown in figure. The bead moves under the combined action of gravity and a massless spring (k) attached to the bottom of the hoop. The equilibrium length of the spring is ' RR '. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes ' RR ', would be (spring constant is ' kk ', gg is acceleration due to gravity)

Figure

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Answer: (b)

  1. Spring length becomes RR at the point where chord from the bottom =R= R: 2Rsin⁡θ2=R⇒θ=60∘2R\sin\frac{\theta}{2} = R \Rightarrow \theta = 60^\circ from the bottom.
  2. Height of that point above the bottom =R(1−cos⁡60∘)=R2= R(1 - \cos 60^\circ) = \frac{R}{2}, so the bead falls 2R−R2=3R22R - \frac{R}{2} = \frac{3R}{2}.
  3. At the top the spring (length 2R2R) is stretched by RR; at the new point it is relaxed.
  4. 12kR2+mg3R2=12mv2⇒v=3Rg+kR2m\frac{1}{2}kR^2 + mg\frac{3R}{2} = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{3Rg + \frac{kR^2}{m}}
  5. So the answer is (b).
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