Question

If dimensions of critical velocity v_c 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

(a) -1,-1,-1

(b) 1,1,1

(c) 1,-1,-1

(d) -1,-1,1

Answer 1

Correct Option (c) 1,-1,-1

Explanation: [v_c] = [η^xρ^yr^z] (given) ..(i)

Writing the dimensions of various quantities in eqn. (i), we get

Applying the principle of homogeneity of dimensions, we get

x + y = 0;  - x - 3y + z = 1;  -x = -1

On solving, we get

x = 1, y = -1,  z = -1