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JEE Main 2022 · Physics · Kinematics and Laws of Motion · hard

A small ball tied to a string of length L moves in a vertical circle. At the lowest point its speed is u=√(3.5 gL). The string goes slack at some…

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A small ball tied to a string of length LL moves in a vertical circle. At the lowest point its speed is u=3.5 gLu=\sqrt{3.5\,gL}. The string goes slack at some point and the ball then moves as a projectile. The maximum height reached by the ball above the lowest point of the circle is
  1. A. 3L2\dfrac{3L}{2}
  2. B. 27L16\dfrac{27L}{16}
  3. C. 19L16\dfrac{19L}{16}
  4. D. 7L4\dfrac{7L}{4}
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Answer: (B) 27L16\dfrac{27L}{16}

Let the string go slack at angle φ\varphi above the horizontal through the centre. Slack means T=0T=0, so v2=gLsin⁡φv^{2}=gL\sin\varphi. Energy: v2=u2−2gL(1+sin⁡φ)v^{2}=u^{2}-2gL(1+\sin\varphi). So 3.5−2−2sin⁡φ=sin⁡φ3.5-2-2\sin\varphi=\sin\varphi, giving sin⁡φ=12\sin\varphi=\frac12, v2=gL2v^{2}=\frac{gL}{2}. The slack point is at height L(1+sin⁡φ)=3L2L(1+\sin\varphi)=\frac{3L}{2}. The velocity is tangential, so its vertical component is vcos⁡φv\cos\varphi. Further rise =v2cos⁡2φ2g=(gL/2)(3/4)2g=3L16=\frac{v^{2}\cos^{2}\varphi}{2g}=\frac{(gL/2)(3/4)}{2g}=\frac{3L}{16}. Total =3L2+3L16=27L16=\frac{3L}{2}+\frac{3L}{16}=\frac{27L}{16}.
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