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Physics Question 28 – JEE-MAIN 2026

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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 5 m/s. The work done by the unknown resistive force is _______ J. (take g=10 m/s2)

The problem involves gravity and an unknown resistive force, both doing work on the raindrop.

🥷
Ninja Strategy•Energy Dissipation Estimation

Estimate the work done by gravity and the final kinetic energy. Since the final kinetic energy is much smaller than the work done by gravity, the resistive force must have dissipated almost all the gravitational potential energy, meaning its work should be close to -10 J.

Step 1: Convert units and identify initial/final states✦ Active

Given mass m=1 g=1×10−3 kg. Initial velocity u=0 m/s. Height h=1 km=1000 m. Final velocity v=5 m/s. Acceleration due to gravity g=10 m/s2.

Step 2: Calculate work done by gravity and change in kinetic energy○ Expand

The work done by gravity is Wg=mgh. The change in kinetic energy is ΔKE=12mv2−12mu2.

Wg=(1×10−3 kg)(10 m/s2)(1000 m)=10 J
ΔKE=12(1×10−3 kg)(5 m/s)2−0=12(1×10−3)(25)=0.0125 J
💡 Teacher's Secret Hint

Ensure all units are consistent (SI) before calculation.

Step 3: Apply Work-Energy Theorem to find work done by resistive force○ Expand

According to the Work-Energy Theorem, the net work done is the sum of work done by gravity and the resistive force, and it equals the change in kinetic energy: Wnet=Wg+Wr=ΔKE.

10 J+Wr=0.0125 J
Wr=0.0125−10=−9.9875 J

Rounding to two decimal places, Wr≈−9.99 J. The closest option is −9.98 J.

💡 Teacher's Secret Hint

Remember that resistive forces always do negative work, which is consistent with the calculated value.

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