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JEE Main 2024 (4 Apr, Shift 2) · Conservation of Energy

Conservation of Energy question from JEE Main 2024 (4 Apr, Shift 2)

Q11JEE Main 2024 · 4 APR · SHIFT 2Medium

Conservation of Energy

A body of m kgm\text{ kg} slides from rest along the curve of vertical circle from point AA to BB in frictionless path. The velocity of the body at BB is:

Figure

(given, R=14 m,g=10 m/s2R = 14\text{ m}, g = 10\text{ m/s}^2 and 2=1.4\sqrt{2} = 1.4)

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Answer: (d)

  1. A is Rsin⁡45∘R\sin 45^\circ above the centre and B is RR below it, so the drop is h=R(1+12)h = R\left(1 + \frac{1}{\sqrt{2}}\right).
  2. h=14(1+0.7)=23.8 mh = 14(1 + 0.7) = 23.8\text{ m}
  3. v=2gh=2×10×23.8=476≈21.8 m/sv = \sqrt{2gh} = \sqrt{2 \times 10 \times 23.8} = \sqrt{476} \approx 21.8\text{ m/s}
  4. So the answer is (d) 21.9 m/s.
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