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JEE Main 2021 (27 Aug, Shift 2) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2021 (27 Aug, Shift 2)

Q32JEE Main 2021 · 27 AUG · SHIFT 2Easy

Motion Under Gravity

Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.

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Answer: (d)

  1. Time to fall 9.8 m: T=2×9.8/9.8=2T = \sqrt{2 \times 9.8/9.8} = \sqrt{2} s
  2. Two intervals in TT, so drop 2 has fallen for t=22t = \frac{\sqrt{2}}{2} s, i.e. t2=0.5t^2 = 0.5 s2^2.
  3. Distance fallen =12×9.8×0.5=2.45= \frac{1}{2} \times 9.8 \times 0.5 = 2.45 m
  4. Height above floor =9.8−2.45=7.35= 9.8 - 2.45 = 7.35 m. So the answer is (d) 7.35 m.
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