A block of mass is placed on a surface with a vertical cross section given by `y=x^3/6`. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is:

Options

A. `(1)/(3)m`
B. `(1)/(2) m`
C. `(1)/(6)m`
D. `(2)/(3)m`

Answer

C

Explanation

Block is under limiting friction, so `mg sin theta =mumg cos theta rArr mu = tan theta….(i)`
Equation of the surface `y = (x^(3))/(6)`
`mu mg cos theta`
Slope of the curved surface
`tantheta=(dy)/(dx)=(x^(2))/(2)...(ii)`
`mu = (x^(2))/(2)`
`mu = (x^(2))/(2)=0.5`
`rArrx^(2)=1` or `x =1`
So `y = (x^(3))/(6) =((1)^(2))/(6) = (1)/(6)`
.