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JEE Main 2026 (28 Jan, Shift 1) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2026 (28 Jan, Shift 1)

Q7JEE Main 2026 · 28 JAN · SHIFT 1Medium

Motion Under Gravity

Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is ______ m.

(g=10 m/s2)(g = 10\text{ m/s}^2)

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Answer: (c)

  1. Time to fall 5 m: T=2×5/10=1T = \sqrt{2 \times 5/10} = 1 s
  2. Drop 1 lands as drop 6 starts: 5 intervals in 1 s, so interval =0.2= 0.2 s.
  3. Drop 4 has fallen for 2×0.2=0.42 \times 0.2 = 0.4 s: s=12×10×0.42=0.8s = \frac{1}{2} \times 10 \times 0.4^2 = 0.8 m
  4. Height above ground =5−0.8=4.2= 5 - 0.8 = 4.2 m. So the answer is (c) 4.2 m.
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