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 isA. `oo`B. `2sqrt(gA)`C. `sqrt(2gA)`D. `4sqrt(gA)`

A particle is projected along a horizontal field whose coefficient of

1.	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 isA. `oo`B. `2sqrt(gA)`C. `sqrt(2gA)`D. `4sqrt(gA)`
Answer» Correct Answer - C Work done against friction must equal to the initial kinetic energy. `(1)/(2)mv^(2)=underset(1)overset(oo)(int)mumgdximplies(v^(2))/(2)=Agunderset(1)overset(oo)(int)(1)/(x^(2))dx` `(v^(2))/(2)=Ag[-(1)/(x)]_(1)^(prop)` `v^(2)=2gAimpliesV=sqrt(2gA)`

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