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Standard XII

Physics

Dimensional Analysis

The mass of t...

Question

The mass of the liquid flowing per second per unit area of cross section of the tube if proportional to $P^{x}$ and $v^{y}$ , where $P$ is the pressure difference and $v$ is the velocity then the relation between $x$ and $y$ is

x = y

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x = - y

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y^{2} = x

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y = - x^{2}

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Solution

The correct option is B $x = - y$
Given $m α P^{x} \times V^{y}$
$[{M L}^{- 2} T^{0}] = {[{M L}^{- 1} T^{- 2}]}^{x} {[M^{0} {L T}^{1}]}^{y}$
$x = 1$ } Comparing coefficients
$- 2 = - x + y$ and $- 1 = - 2 x - y$
$- 2 = - 4 x - 2 y$
$0 = (- 4 x - 2 y) - (- x + y)$
$0 = - 3 x - 3 y$
$\Rightarrow$ $3 x = - 3 y$ $\Rightarrow$ $x = - y$ .

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The mass of the liquid flowing per second per unit area of cross section of the tube if proportional to Px and vy, where P is the pressure difference and v is the velocity then the relation between x and y is

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The mass of the liquid flowing per second per unit area of cross section of the tube if proportional to $P^{x}$ and $v^{y}$ , where $P$ is the pressure difference and $v$ is the velocity then the relation between $x$ and $y$ is