A particle is projected along a horizontal field whose coefficient of friction varies as `mu=A//r^2`, where r is the distance from the origin in meters and A is a positive constant. The initial distance of the particle is `1m` from the origin and its velocity is radially outwards. The minimum initial velocity at this point so the particle never stops is

Options

A. (a) `oo`
B. (b) `2sqrt(gA)`
C. (c) `sqrt(2gA)`
D. (d) `4sqrt(gA)`

Answer

C

Explanation

Work done against friction must equal to the initial kinetic energy.
`1/2mv^2=underset1overset(oo)intmumgdximpliesv^2/2=Agunderset1overset(oo)int1/x^2dx`
`v^2/2=Ag[-1/x]_1^(oo)`
`v^2=2gAimpliesv=sqrt(2gA)`