A wind power generators converts wind energy into electrical energy.Asume that the generator converts fixed fraction of wind energy intercpted by its blades into electrical energy.For wind speed v, the electrical output will be proportional to

1.v 2.v² 3.v³ 4.v⁴

ans - v³

Dear Manoj, According to Bernouli's Theorm the net trust experienced propotional to the velocities difference above and below the wing of the areoplane Thus P1-P2= (1/2) *d *(V2^2-V1^2) And net force experienced Fnet= A * (1/2) *d *(V2^2-V1^2).....Eq 1 Where A= Total area of the of the wings and d = density of air and V1 and V2 are velocities above and below the wings. Now the shape of the wind turbines is similar to that of the aeroplane . Thus, the power output P = Fnet*Velocity of the wind = A * (1/2) *d *V^2 * V (Since we can assume that the wind mill has V2=0 as maximum amount of the air flow takes place in the bulged part of the wings as shown in the figure thus giving V2 = V (velocity of the wind) This implies Output power is proportional to V^3 where all the terms in the expression[A*(1/2)*d] can be assumed to be constant Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best STUDENT NAME !!! Regards, Askiitians Experts Adapa Bharath

Consider a circular plane of unit area perpendicular to the wind velocity. Wind crossing this plane in unit time would be trapped in the cylinder having this plane as its base and length V. (volume of this cylinder = Vx(unit area) = V)

=> Energy crossing the circular plane per unit time = Power of wind = ((1/2)xdxV²)xV

(since K.E. of wind per unit volume = ((1/2)xdxV² and other forms of energy carried by wind are independent of V.)

=> Power of generator would be proportional to V³.

Let ρ be the density of the air and A be the cross-section area of the blades. Consider a cylindrical volume V with cross-sectional area AA and length x . The mass of air contained in volume V is m=ρAx and its kinetic energy is,K=12mv2 = 12ρAxv2.The time taken by this volume to intersect the blades is Δt=x/v. The air transfers its entire kinetic energy to the blades (assuming 100% efficiency ,or ,as given in the question , a fraction that is fixed constant). Thus, the power generated is,P=KΔt=12ρAv3.Clearly it is proportional to v3.