Question

If dimensions of critical velocity $v_c$ of a liquid flowing through a tube are expressed as $[\eta^x?^yr^z] \,$ where $\, \eta,?$ 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)

1,-1,-1

Answer 2

$[v_c]=[\eta^x?^yr^z]$ (given) ... (i) Writing the dimensions of various quantities in eqn. (i), we get $[M^0LT^{-1}]=[ML^{-1}T^{-1}]^x[ML^{-3}T^0]^y[M^0LT^0]^z$ \hspace20mm=[M^{x+y}L^{-x-3y+z}T^{-x}] Applying the principle of homogeneity of dimensions, we get \hspace10mmx + y = 0; -x- 3y + z = 1; - x = -1 On solving, we get \hspace10mmx =1,y = -1 , z =-1