Dimensions: Practice Problem & Solution
If dimension of critical velocity of liquid flowing through a tube are expressed as \(v_c \propto \eta^x \rho^y r^z\) where \(\eta\) and \(\rho\) 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: [2015-Re]
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Dimensions. Understanding the underlying formula is key to arriving at the correct answer below:
Critical velocity \(v_c = [LT^{-1}]\). Coefficient of viscosity \(\eta = [ML^{-1}T^{-1}]\). Density \(\rho = [ML^{-3}]\). Radius \(r = [L]\). Assume \(v_c \propto \eta^x \rho^y r^z\). Equating dimensions: \([LT^{-1}] = ([ML^{-1}T^{-1}])^x ([ML^{-3}])^y ([L])^z = [M^{x+y} L^{-x-3y+z} T^{-x}]\). Comparing powers: \(x+y = 0\), \(-x-3y+z = 1\), \(-x = -1\). From \(-x = -1\), we get \(x=1\). From \(x+y = 0\), we get \(1+y = 0 ⇒ y=-1\). From \(-x-3y+z = 1\), we get \(-1-3(-1)+z = 1⇒ -1+3+z=1 ⇒ 2+z=1 ⇒ z=-1\). Therefore, \(x=1, y=-1, z=-1\).
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