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

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

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

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

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

A$1,1,1$
B$1,-1,-1$
C$-1,-1,1$
D$-1,-1,-1$

AIPMT 2015, Diffcult

STD 11 Science
JEE
STD 11 - 1. units,dimensions and measurement
Physics

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