An automobile enters a turn of radius `R`. If the road is banked at an angle of `45^(@)` and the coefficient of friction is 1, the minimum and maximum speed with which the automobile can negotiate the turn without skidding is `:`

Options

A. `sqrt((rg)/(2))and sqrt(rg)`
B. `sqrt(rg)/(2)and sqrt(rg)`
C. `sqrt(rg)/(2)and 2sqrt(rg)`
D. 0 and inifinite

Answer

D

Explanation

`F.B.D.`for minimum speed `( w.r.t ` automobile `) :`

`Sigmaf_(y)=N- mg cos theta -(mv^(2))/(R) sin theta =0`
`Sigmaf_(x)=(mv^(2))/(R) cos theta + mu N - mg sin theta =0`
`rArr ( mv^(2))/(R)cos theta+ mu ( mg cos theta + ( mv^(2))/(R) sin theta )- mg sin theta =0`
`rArr v^(2)=((mu R g cos theta - Rg sin theta ))/((cos theta + mu sing theta ))`
for `theta =45^(@)` and `mu =1:`
`v_(min)=(Rg-Rg)/(1+1)=0`
`F.B.D` for maximum speed `( w.r.t.` automobile)

`Sigma f_(x)=(mv^(2))/(R)cos theta -mg sin theta - mu ( mg cos theta +( mv^(2))/(R) sin theta )=0`
for `theta 45^(@) ` and `mu =1`
`v_(max)=oo(` infinite `)`