If dimensions of critical velocity νc of a liquid flowing through a tube are expressed as ηxρyrz 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

Question

If dimensions of critical velocity νc of a liquid flowing through a tube are expressed as ηxρyrz 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

Infinity Learn · Accepted Answer

$$vc=$$ηxρ$$yr2$$   (given)                         . . . .(i)Writing$$ the dimensions of various quantities in eqn. $$(i), we $$getM0LT-1=ML-1$$ $$T-1xML-3$$ $$T0yM0LT0z=Mx+y$$ $$L-x-3y+z$$ $$T-xApplying$$ the principle of homogeneity of dimensions, we $$getx+y=0$$ ;$$-x-3$$ $$y+z=1$$ ;$$-x=-1$$ On solving, we get $$x=1$$, $$y=-1$$, $$z=-1$$

If dimensions of critical velocity $ν_{c}$ of a liquid flowing through a tube are expressed as $[η^{x} ρ^{y} r^{z}]$ 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

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material