A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to

The correct option is v³.

To find how the electrical power output of a wind-powered generator scales with the wind speed v, we can analyze the rate at which kinetic energy from the wind is intercepted by the generator's blades.

First, let the area swept by the blades of the wind generator be A, and the density of the air be ρ.

In a time interval Δt, the volume of air passing through the swept area A at a wind speed v is given by:
V = A ⋅ v ⋅ Δt

Consequently, the mass m of this volume of air is:
m = ρ ⋅ V = ρ ⋅ A ⋅ v ��� Δt

The kinetic energy (KE) of this mass of air moving at speed v is:
KE = 1 2 m v 2

Substituting the expression for mass m into the kinetic energy equation gives:
KE = 1 2 ( ρ A v Δt ) v 2 = 1 2 ρ A v 3 Δt

The total wind power input P_in intercepted by the blades is the rate of kinetic energy passing through the area per unit time:
P in = KE Δt = 1 2 ρ A v 3

Since the generator converts a fixed fraction (efficiency η) of this intercepted wind energy into electrical energy, the electrical power output P_out is:
P out = η ⋅ P in = η ⋅ 1 2 ρ A v 3

Assuming η, ρ, and A are constants, the electrical power output P_out is directly proportional to the cube of the wind speed:
P out ∝ v 3