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JEE Main 2021 (25 Jul, Shift 1) · Conservation of Energy

Conservation of Energy question from JEE Main 2021 (25 Jul, Shift 1)

Q25JEE Main 2021 · 25 JUL · SHIFT 1Medium

Conservation of Energy

A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 50 cm long. The speed of bob, when the length makes an angle of 60 ∘{}^\circ to the vertical will be (g = 10 m/s2) ____________ m/s.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. Rise of bob at 60∘60^\circ: h=L(1−cos⁡60∘)=0.5×0.5=0.25 mh = L(1 - \cos 60^\circ) = 0.5 \times 0.5 = 0.25\text{ m}.
  2. v2=u2−2gh=9−2×10×0.25v^2 = u^2 - 2gh = 9 - 2 \times 10 \times 0.25
  3. v2=4v^2 = 4
  4. v=2 m/sv = 2\text{ m/s}
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