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JEE Main 2021 (25 Jul, Shift 1) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2021 (25 Jul, Shift 1)

Q34JEE Main 2021 · 25 JUL · SHIFT 1Medium

Motion Under Gravity

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4th second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap? (Take g = 9.8 m/s2)

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Answer: (c)

  1. Droplet at 4 s has fallen 12×9.8×42=78.4\frac{1}{2} \times 9.8 \times 4^2 = 78.4 m.
  2. Next droplet (released TT s later) has fallen 78.4−34.3=44.178.4 - 34.3 = 44.1 m: 44.1=4.9(4−T)244.1 = 4.9(4 - T)^2
  3. (4−T)2=9⇒4−T=3⇒T=1(4 - T)^2 = 9 \Rightarrow 4 - T = 3 \Rightarrow T = 1 s
  4. So the answer is (c) 1 drop / second.
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