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MCQMediumJEE 2026Circular Motion Dynamics

JEE Physics 2026 Question with Solution

In case of vertical circular motion of a particle by a thread of length rr, if the tension in the thread is zero at an angle 30∘30^\circ as shown in the figure, the velocity at the bottom point (A) of the vertical circular path is ( gg = gravitational acceleration ).

  • A

    72gr\sqrt{\dfrac{7}{2}gr}

  • B

    4gr\sqrt{4gr}

  • C

    5gr\sqrt{5gr}

  • D

    52gr\sqrt{\dfrac{5}{2}gr}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A particle moves in a vertical circle of radius rr. The tension in the thread becomes zero when the string makes an angle 30∘30^\circ with the horizontal. Find: the velocity at the bottom point AA.

At the given point, tension is zero, so gravity alone provides the centripetal force.

mv2r=mgcos⁡60∘\frac{mv^2}{r} = mg\cos 60^\circ

Therefore,

mv2r=mg2\frac{mv^2}{r} = \frac{mg}{2} v2=gr2v^2 = \frac{gr}{2}

Now use conservation of mechanical energy between the given point and the bottom point AA. The vertical height difference is

h=r(1+sin⁡30∘)=r(1+12)=3r2h = r(1 + \sin 30^\circ) = r\left(1 + \frac{1}{2}\right) = \frac{3r}{2}

So,

12mvA2=12mv2+mgh\frac{1}{2}mv_A^2 = \frac{1}{2}mv^2 + mgh

Substituting v2=gr2v^2 = \frac{gr}{2} and h=3r2h = \frac{3r}{2},

12mvA2=12m(gr2)+mg(3r2)\frac{1}{2}mv_A^2 = \frac{1}{2}m\left(\frac{gr}{2}\right) + mg\left(\frac{3r}{2}\right) 12mvA2=gr4m+3gr2m\frac{1}{2}mv_A^2 = \frac{gr}{4}m + \frac{3gr}{2}m 12mvA2=7gr4m\frac{1}{2}mv_A^2 = \frac{7gr}{4}m

Hence,

vA2=7gr2v_A^2 = \frac{7gr}{2} vA=72grv_A = \sqrt{\frac{7}{2}gr}

Therefore, the velocity at the bottom point is 72gr\sqrt{\frac{7}{2}gr}, so the correct option is A.

Common mistakes

  • Using the full weight mgmg as the centripetal force at the zero-tension point. This is wrong because only the component of gravity toward the centre contributes there. Use the appropriate component, giving mv2r=mgcos⁡60∘\frac{mv^2}{r} = mg\cos 60^\circ.

  • Taking the height difference incorrectly. This is wrong because the point is not at the top; its height above the bottom is r(1+sin⁡30∘)r(1 + \sin 30^\circ), not just rsin⁡30∘r\sin 30^\circ. Always measure vertical height from the bottom point carefully.

  • Applying energy conservation with the signs reversed. This is wrong because the particle loses gravitational potential energy while moving to the bottom, so the kinetic energy at the bottom increases. Write 12mvA2=12mv2+mgh\frac{1}{2}mv_A^2 = \frac{1}{2}mv^2 + mgh.

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