A parabolic bow1 with its bottom at origin has the shape `y = (x^(2))/(20)` where `x` and `y` are in metre The maximum height at which a small mass m can be placed on the bowl without slipping is (coeff of static friction `0.5`
.

Options

A. `1.25Nm`
B. `2.5m`
C. `1.0m`
D. `4.0m`

Answer

a

Explanation

As `y =(x^(2))/(20) , (dy)/(dx) =(2x)/(20)`
Angle of repose `theta_(r)` is such that
`mu = tan theta_(r) = (dy)/(dx) = (2x)/(20) = (x)/(10)`
`x = 10 mu = 10 xx 0.5 = 5m`
Corresponding value of `y = (x^(2))/(20) = (5^(2))/(20) = 1.25m` .