Question

If the average terminal velocity of rain drop is 2 ms\(^{-1}\), then the energy transferred by rain to each square meter of the surface at a place which receives 100 cm of rain in a year is

• A. \( 1 \times 10^4 \, \text{J} \)
• B. \( 1 \times 10^3 \, \text{J} \)
• C. \( 2 \times 10^3 \, \text{J} \)
• D. \( 2 \times 10^4 \, \text{J} \)

Hint:

Energy transferred by rain per unit area = \( \frac12 \rho h v^2 \), where \( \rho \) = density of water, \( h \) = rainfall height, \( v \) = terminal velocity.

Answer 1

The Correct Option is C

Concept: The kinetic energy transferred by rain per unit area is given by \( \frac{1}{2} \rho h v^2 \), where \( \rho \) is density of water, \( h \) is height of rainfall, and \( v \) is terminal velocity.

Step 1: Identify given values. Terminal velocity \( v = 2 \, \text{m/s} \) Rainfall height \( h = 100 \, \text{cm} = 1 \, \text{m} \) Density of water \( \rho = 1000 \, \text{kg/m}^3 \)

Step 2: Recall formula for kinetic energy per unit area. Mass of rain per square meter = \( \rho \times h \times 1 = \rho h \) Kinetic energy transferred = \( \frac{1}{2} (\rho h) v^2 \)

Step 3: Substitute values. \[ E = \frac{1}{2} \times 1000 \times 1 \times (2)^2 \] \[ E = \frac{1}{2} \times 1000 \times 4 = \frac{1}{2} \times 4000 = 2000 \, \text{J} \] \[ E = 2 \times 10^3 \, \text{J} \] Correct answer: \[ E = 2 \times 10^3 \, \text{J} \quad \text{(Option C)} \]