If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [$η^xρ^yr^z$] where η, ρ and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by
Answer & explanation
Correct answer: option 3
c = [$η^xρ^yr^z$]
Put dimensions of various quantities,
[M0LT-1] = [ML-1T-1]x [ML-3T0]y [M0LT0]z
= [Mx + y L- x - 3y + z T- x]
Equating power both sides, we get
x + y = 0;
- x - 3y + z = 1;
- x = - 1
On solving, we get
x = 1, y = -1, z = -1