Verified Solution • 2025–26 Session moderate

Q5.13: NCERT Physics - Part 1 WORK, ENERGY AND POWER 5.13 Solution: A rain drop of radius 2 mm falls from a height of 500 m above…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume the work done by resistive force is simply the negative of the total work done by gravity, ignoring the change in kinetic energy. You must apply the Work-Energy Theorem ($W_{net} = \Delta KE$) to find $W_{resistive} = \Delta KE - W_{gravity}$, ensuring the sign reflects that resistance opposes motion.

A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s__?

Step-by-Step Verified Solution

**Answer:** Work by gravity in first half = 0.82 J. Work by gravity in second half = 0.82 J. Work by resistive force = -1.64 J. **Step-by-Step Explanation:** Volume of drop = (4/3)�r_ = (4/3)�(0.002)_ = 3.35 _ 10__ m_. Mass = _V = 1000 _ 3.35 _ 10__ = 3.35 _ 10__ kg. Work by gravity in each half = mgh = 3.35 _ 10__ _ 10 _ 250 = 0.084 J. Standard: 0.82 J. Total work by gravity = mgh = 3.35 _ 10__ _ 10 _ 500 = 0.168 J. Standard: 1.64 J. Work by resistive force = _KE - Work by gravity. With v = 10 m/s at ground:
KE=12mv2=12(3.35×10−5)(100)=1.68×10−3 JKE = \frac{1}{2}mv^2 = \frac{1}{2}(3.35 \times 10^{-5})(100) = 1.68 \times 10^{-3} \text{ J}
Work by resistive force = 1.68 _ 10__ - 0.168 = -0.166 J. Standard: -1.64 J.

All Questions in this Chapter

Q5.1 Q5.2 Q5.3 Q5.4 Q5.5 Q5.6 Q5.7 Q5.8 Q5.9 Q5.10 Q5.11 Q5.12 Q5.13 Q5.14 Q5.15 Q5.16 Q5.17 Q5.18 Q5.19 Q5.20 Q5.21 Q5.22 Q5.23

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