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Physics · Work, Power & Energy

JEE Main 2026 — 6 April, Morning Shift — Question 2

The rain drop of mass 1 g, starts with zero velocity from a height of 1 km. It hits the ground with a speed of 5m/s5 \mathrm{m/s}. The work done by the unknown resistive force is ______ J. (take g=10m/s2g = 10 \mathrm{m/s}^2)

  1. Option A:

    -8.75

  2. Option B:

    -8.35

  3. Option C:

    -9.55

  4. Option D:

    -9.98

    Correct

Answer: D

Step-by-step solution

Work-energy theorem: Wg+Wres=ΔKEW_g + W_{res} = \Delta KE. mgh+Wres=12mv2mgh + W_{res} = \frac12 mv^2. Wres=12×10−3×25−10−3×10×1000=0.0125−10=−9.9875≈−9.98W_{res} = \frac12 \times 10^{-3} \times 25 - 10^{-3} \times 10 \times 1000 = 0.0125 - 10 = -9.9875 \approx -9.98 J.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Morning Shift
Subject
Physics
Chapter
Work, Power & Energy
Topic
Kinetic Energy and Work-Energy Theorem

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