Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take 'g' constant with a value 10 m/s^2. The work done by the (i) gravitational force and the (ii) resistive force of air is :

1. (i) 1.25 J (ii) \(-\)8.25 J

2. (i) 100 J (ii) 8.75 J

3. (i) 10 J (ii) \(-\)8.75 J

4. (i) \(-\)10 J (ii) \(-\)8.25 J

Doubtly Asked 30 Oct 2026 NEET UG 2017 Practice test neet Save WhatsApp Report

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Answer: 3. (i) 10 J (ii) \(-\)8.75 J

Key idea: use work-energy theorem: \(W_g+W_{air}=\Delta KE\).

(i) \(W_g=mgh=0.001\times10\times1000=10\) J.

\(\Delta KE=\tfrac12(0.001)(50)^2=1.25\) J.

(ii) \(W_{air}=1.25-10=-8.75\) J (negative since resistive force opposes motion).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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