Dimensions: Practice Problem & Solution
Turpentine oil is flowing through a tube of length (l) and radius (r). The pressure difference between the two ends of the tube is (P). The viscosity of oil is given by \(\eta = \frac{P(r^2 - x^2)}{4vl}\) where (v) is the velocity of oil at a distance (x) from the axis of the tube. The dimensions of (eta) are: [1993]
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:
\([P] = [ML^{-1}T^{-2}]). ([r^2 - x^2] = [L^2]\). \([v] = [LT^{-1}]\). \([l] = [L]\). \([\eta] = \frac{[ML^{-1}T^{-2}][L^2]}{[LT^{-1}][L]} = \frac{[MLT^{-2}]}{[L^2T^{-1}]} = [ML^{-1}T^{-1}]\).
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