Question

If dimensions of critical velocity $v_c$ of a liquid flowing through a tube are expressed as $[{\eta }^x{\rho }^y{r}^z]$ where $\eta ,\rho$ 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

Answer: (C)

$[v_c]=[{\eta }^x{\rho }^y{r}^z]$ (given) ... (i)
Writing the dimensions of various quantities in eqn. (i), we get
$[M^{0}L{T}^{-1}]=[M{L}^{-1}{T}^{-1}{]}^x[M{L}^{-3}{T}^0{]}^y[M^{0}L{T}^0{]}^z$
$=[M^{x+y}{L}^{-x-3y+z}{T}^{-x}]$
Applying the principle of homogeneity of dimensions, we get
$x+y=0;-x-3y+z=1;-x=-1$
On solving, we get
$x=1,y=-1,z=-1$