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Physics · Motion in one Dimension

JEE Main 2026 — 28 January, Morning Shift — Question 28

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)\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)

  1. Option A:

    2.5

  2. Option B:

    4

  3. Option C:

    4.2

    Correct
  4. Option D:

    3.8

Answer: C

Step-by-step solution

Time to reach ground =2 h g=2×510=1sec=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 5}{10}}=1 \mathrm{sec} Five drops per second Time between each drop =0.2sec=0.2 \mathrm{sec}. Time of fall for 4th 4^{\text {th }} drop is 1−0.6=0.4sec1-0.6=0.4 \mathrm{sec} Height fall of 4th 4^{\text {th }} drop is =12×10×0.42=0.8 m=\frac{1}{2} \times 10 \times 0.4^{2}=0.8 \mathrm{~m} Height from ground =5−0.8=4.2 m=5-0.8=4.2 \mathrm{~m}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Morning Shift
Subject
Physics
Chapter
Motion in one Dimension
Topic
Motion Under Gravity

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