1. Tardigrade
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  3. Physics
  4. If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [ηx ρyrz] 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

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

AIPMTAIPMT 2015Physical World, Units and Measurements

Solution:

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