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, where the pressure in the stram 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. `(4^(1//2))g`
B. `(4^(1//3))g`
C. `(4^(1//2) - 1)g`
D. `(4^(1//3) - 1)g`

Answer

D

Explanation

`P' = 2P_(0) = 2rhoAv_(0)^(3)`
Let the new velocity of air pushed down be v
`rhoAv_(0)^(3) = 2rhoAv_(0)^(3)`
`v = 2^(1//3)v_(0)`
New thrust `F = rhoAv^(2)`
`= rhoA2^(2//3)v_(0)^(2)`
`= (mg)4^(1//3) " " [mg = rhoAv_(0)^(2)]`
`:.` Acceleration `= a = (F_("net"))/(m) = (F_("new") - F_(0))/(m)`
`= (mg4^(1//3) - mg)/(m) = (4^(1//3) - 1)`