Question

Out of the following, the only formula that is dimensionally acceptable, is

(1) v_e = (kη/rd)

(2) v_e = krd/η

(3) v_e = kr²d/η

(4) v_e = kη/r²d

where v_e is critical velocity, of a liquid of viscosity η , flowing in a tube of radius r. (d is the density of the liquid).

Answer 1

(1) v_e = (kη/rd)

From Newton’s law, for viscosity, namely

Hence, for this option, the dimensions of the two sides are the same. Hence this option is dimensionally acceptable. 

[It can be easily seen that the other three options are not dimensionally acceptable options].