Correct Answer - A::B::D
Banking angle, `theta=tan^(-1)((v^(2))/(Rg))`
`36Km//h=10m//s`
`theta=tan^(-1)((100)/(20xx9.8))=27^(@)`
Angle of repose,
`theta_(r)=tan^(-1)(mu)=tan^(-1)(0.4)=21.8^(@)`
Sinne `thetagttheta_(r)` , vehicle cannot remain in the given position with `v=0` . At rest it will slide down. To find minimum speed, so that vehicle does not slip down, maixmum friction will act up the plane.
To find maximum speed, so that the vehicle does not skid up, maximum friction will act down the plane.
Minimum Speed
Equation of motion are,
`Ncostheta+muNsintheta=mg` ...(i)
`Nsintheta-muNcostheta=(m)/(R)v_(max)^(2)`...(ii)
Solving these two equations, we get
`v_(min)=4.2m//s`
Maximum Speed ltbr. Equations of motion are, `Ncostheta-muNsintheta=mg` ...(iii)
`Nsintheta+muNcostheta=(m)/(R)v_(max)^(2)` ...(iv)
Solving two equations, we have
`v_(max)=15m//s`