The helicopter has a mass `m` and maintains its height by imparting a downward momentum to a column of air defined by the slipstream boundary as shown in figure. The propeller blades can project a downward air speed `v_0`, where the pressure in the stream below the blades is atmospheric and the radius of the circular cross section of the slipstream is `r`. Neglect any rotational energy of the air, the temperature rise due to air friction and any change in air density `rho`.
If the power is doubled, the acceleration of the helicopter is :-

Options

A. `(mg)/(2r) sqrt(mg)/(pi rho)`
B. `(mg)/(r ) sqrt(mg)/(pi rho)`
C. `(2 mg)/(r ) sqrt(mg)/(pi rho)`
D. `(mg)/(r ) sqrt(mg)/(2pi rho)`

Answer

A

Explanation

Thrust `= F = rhoAv^(2) " " [A = pir^(2)]`
at equilibrium `rhoAv^(2) = mg`
Power `P = Fv = rhoAv^(3)`
`= (mg)/(r ) sqrt((mg)/(p pi))`