Ncert Exercises & Solutions

The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as $V = \frac{π}{8} \frac{\Pr^{4}}{ηl}$ where P is the pressure difference between the two ends of the pipe and $η$ is the coefficient of the viscosity of the liquid having dimensional formula $[{ML}^{- 1} T^{- 1}]$ . Check whether the equation is dimensionally correct.

Hint: The dimensions of LHS should be equal to RHS.

Step 1: Find the dimensions of V.

The volume of a liquid flowing out per second of a pipe is given by, $V = \frac{π}{8} \frac{\Pr^{4}}{ηl}$

$Dimension of V = \frac{Dimension of volume}{Dimension of time} = \frac{[L^{3}]}{[T]} = [L^{3} T^{- 1}]$

Step 2: Find the dimensions of LHS and RHS.

( $∴$ V is the volume of liquid flowing out per second)

$\begin{array}{ll} Dimension of & P = [{ML}^{- 1} T^{- 2}] \\ Dimension of & η = [{ML}^{- 1} T^{- 1}] \\ Dimension of & l = [L] \\ Dimension of & r = [L] \end{array}$

$\begin{array}{l} Dimensions of LHS = [V] = \frac{[L^{3}]}{[T]} = [L^{3} T^{- 1}] \\ Dimensions of RHS = \frac{[{ML}^{- 1} T^{- 2}] \times [L^{4}]}{[{ML}^{- 1} T^{- 1}] \times [L]} = [L^{3} T^{- 1}] \end{array}$

As dimensions of LHS is equal to the dimensions of RHS.

Therefore, the equation is dimensionally correct.

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