The air of density `rho` and moving with a velocity `v` strikes perpendicular the inclined surface of area `A` and of a wedge kept on a horizontal surface. The mass of the wedge is `m`. Assuming the collisions to be perfectly inelastic, the minimum value of the coefficient of friction between the wedge and the ground so that the wedge does not move is (Assume mass of particles of air is negligible)

Options

A. `(rho Av^(2)sin theta)/(mg+rhoAv^(2) cos theta)`
B. `tan theta`
C. `(rho A v^(2))/(mg) tan theta`
D. `(rho Av^(2))/(mg+rhoAv^(2) cos theta)`

Answer

A

Explanation

`N = rho a v^(2) cos theta + m g, mu(N) = f`
`mu(rho a v^(2) cos theta + mg) = rho a v^(2) sin theta`
`mu = (rho a v^(2) sin theta)/(rho a v^(2) cos theta + mg)`