Question:
The blades of a windmill sweep out a circle of area
\[
A=2\,\text{m}^2.
\]
The wind is flowing with velocity
\[
V=6\,\text{ms}^{-1}
\]
perpendicular to the circle and the density of air is
\[
\rho=1.2\,\text{kgm}^{-3}.
\]
Then the power of the mill is
The blades of a windmill sweep out a circle of area
\[
A=2\,\text{m}^2.
\]
The wind is flowing with velocity
\[
V=6\,\text{ms}^{-1}
\]
perpendicular to the circle and the density of air is
\[
\rho=1.2\,\text{kgm}^{-3}.
\]
Then the power of the mill is
Show Hint
For wind energy problems, first find the mass flow rate:
\[
\dot m=\rho AV.
\]
Then use
\[
P=\frac12\dot mV^2
=\frac12\rho AV^3.
\]
This is the kinetic energy carried by air per second.
Updated On: Jul 29, 2026
- \(160.8\ \text{W}\)
- \(259.2\ \text{W}\)
- \(302.5\ \text{W}\)
- \(239.2\ \text{W}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
The maximum power available from wind is equal to the kinetic energy carried by air crossing the swept area per unit time.
\[
P=\frac12 \dot m V^2
\]
where
\[
\dot m=\rho AV.
\]
Step 1: Calculate the mass of air crossing the blades per second. \[ \dot m = \rho AV. \] Substituting, \[ \dot m = (1.2)(2)(6). \] \[ \dot m = 14.4\ \text{kg s}^{-1}. \]
Step 2: Calculate the power of the windmill. \[ P = \frac12 \dot m V^2. \] \[ = \frac12(14.4)(6^2). \] \[ = 7.2\times36. \] \[ = 259.2\ \text{W}. \] Therefore, \[ \boxed{P=259.2\ \text{W}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Calculate the mass of air crossing the blades per second. \[ \dot m = \rho AV. \] Substituting, \[ \dot m = (1.2)(2)(6). \] \[ \dot m = 14.4\ \text{kg s}^{-1}. \]
Step 2: Calculate the power of the windmill. \[ P = \frac12 \dot m V^2. \] \[ = \frac12(14.4)(6^2). \] \[ = 7.2\times36. \] \[ = 259.2\ \text{W}. \] Therefore, \[ \boxed{P=259.2\ \text{W}} \] \[ \boxed{\text{Answer = (B)}} \]
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