Question

A rain drop of mass \(10g\) falls from a height of \(50m\) from rest. If the loss of energy due to air resistance is \(3J\), then the velocity of the drop on striking the ground is \((g = 10ms^{-2})\)

• A. \(10 m s^{-1}\)
• B. \(5 ms^{-1}\)
• C. \(30 ms^{-1}\)
• D. \(40 ms^{-1}\)
• E. \(20 ms^{-1}\)

Hint:

Always convert mass to kg and ensure consistent units when applying the work-energy theorem.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Use the work-energy theorem: The loss in potential energy equals the gain in kinetic energy plus the energy lost due to air resistance.

Step 2: Detailed Explanation:
Mass, \(m = 10 g = 0.01 kg\). Height, \(h = 50 m\). Initial velocity, \(u = 0\). Energy loss due to air resistance, \(W = 3 J\). Initial potential energy, \(PE_i = mgh = 0.01 \times 10 \times 50 = 5 J\). Final kinetic energy, \(KE_f = \frac{1}{2} m v^2\). By energy conservation: \[ PE_i = KE_f + \text{Loss} \] \[ 5 = \frac{1}{2} (0.01) v^2 + 3 \] \[ 5 - 3 = 0.005 v^2 \] \[ 2 = 0.005 v^2 \implies v^2 = \frac{2}{0.005} = 400 \] \[ v = \sqrt{400} = 20 \, m/s \]

Step 3: Final Answer:
The velocity on striking the ground is \(20 m s^{-1}\).