Question

The mass of liquid flowing per second per unit area of cross section of the tube is 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

• A. $x=y$
• B. $x=-y$
• C. $y2=x$
• D. $y=-x^2$

Answer 1

Answer: (B), (C), (D)

$y=-x^2$

We have,
$\frac{m}{At}\propto P^x{v}^y$
$\frac{m}{At}=C{P}^x{v}^y$

Using dimensional analysis:
$\left[M{L}^{-2}{T}^{-1}\right]=\left[M{L}^{-1}{T}^{-2}{]}^x\right[L{T}^{-1}{]}^y$
$\left[M{L}^{-2}{T}^{-1}\right]=\left[M^x{L}^{-x}{T}^{-2x}\right]\left[L^y{T}^{-y}\right]$
$\left[M{L}^{-2}{T}^{-1}\right]=\left[M^x{L}^{y-x}{T}^{-2x-y}\right]$

Now just equating the powers of the respective dimensions, we have:
$x=1$
$y-x=-2$
$-2x-y=-1$
$\Rightarrow y=-1$
$\therefore x=-y,y^2=x$ and $y=-x^2$
.