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MHT CET 2026 (19 May, Shift 1) · Conservation of Energy

Conservation of Energy question from MHT CET 2026 (19 May, Shift 1)

Q3MHT CET 2026 · 19 MAY · SHIFT 1Medium

Conservation of Energy

The bob of a simple pendulum of length 2m is kept at P. It is released and when it reaches Q, it losses 20% of its total energy due to air resistance. Its velocity at Q is [Take g=9.8g = 9.8 m/s2{}^{2} ]

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Answer: (c)

  1. Bob falls through h=2 mh = 2\text{ m} from P to Q; total energy (relative to Q) is mghmgh.
  2. 20% is lost, so 12mv2=0.8 mgh\frac{1}{2}mv^2 = 0.8\,mgh.
  3. v=2×0.8×9.8×2=31.36v = \sqrt{2 \times 0.8 \times 9.8 \times 2} = \sqrt{31.36}
  4. v=5.6 m/sv = 5.6\text{ m/s}. So the answer is (c).
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