Question

Answer the which is given in attachment.

Answer 1

We have to find a dimensionally consistent relation. Let us see what quantities we have:



On the LHS we have Volume of liquid flowing per second.


It is here given as a symbol of V.
So

$[\, V\, ] = \frac{[ \, Volume \, ]}{[ \, Time \, ]} \\ \\ \\ \implies [ \, V \, ] = \frac{[ \, L^3 \, ]}{[ \, T \, ]} \\ \\ \\ \implies [ \, V \, ] = [ \, L^3 \, T^{-1} \, ]$


Other quantities are:


-> Coefficient of Viscosity

Viscosity comes under the topic of Fluid Mechanics. Basically, as a liquid flows, there is some resistance to the flow, as layers of the liquid slide against each other.


The Viscous Force is given by:

$F = - \eta A \, \frac{dv}{dx}$

Here, 

$F$ = Viscous Force
$\eta$ = Coefficient of viscosity
$A$ = Area of contact of liquid surface
$v$ = velocity
$x$ = Length of flowing liquid perpendicular to direction of flow

$\frac{dv}{dx}$ = Velocity Gradient. This shows us how velocity of liquid changes at different layers


$F = ma \\ \\ Unit \, \, of \, \, F = kg \, m/s^2 \\ \\ \implies [ \, F \, ] = [ \, M \, L \, T^{-2} \, ]$


We have:

$[tex]F = -\eta A \frac{dv}{dx} \\ \\ \\ \implies [\, F \, ] = [ \, \eta \, ] \, \, [ \, A \, ] \, \, \frac{[ \, v \, ]}{[\, x \, ]} \\ \\ \\ \implies [ \, \eta \, ] = \frac{[\, F \, ] \, \, [ \, x \, ]}{[ \, A \, ] \, \, [ \, v \, ]} \\ \\ \\ \implies [ \, \eta\, ] = \frac{[ \, M \, L \, T^{-2} \, ] \, \, [ \, L \, ]}{[ \, L^2 \, ] \, \, [\, L \, T^{-1} \, ]} \\ \\ \\ \implies [ \, \eta \, ] = [ \, M \, L^{-1} \, T^{-1} \, ]$


-> Tube of radius r

Clearly, dimensions of r are [ L ]


-> Tube of length l

Again, dimensions are [ L ]



-> Pressure Difference P


Now,

$P = \frac{Force}{Area} \\ \\ \\ \implies [ \, P \, ] = \frac{[ \, F \, ]}{[\, A \, ]} \\ \\ \\ \implies [ \, P \, ] = \frac{[ \, M \, L \, T^{-2} \, ]}{[\, L^2 \, ]} \\ \\ \\ \implies [ \, P \, ] = [ \, M \, L^{-1} \, T^{-2} \, ]$


Now we check the options:

The first one is:

$V = \frac{\pi P r^4}{8 \eta l}$

We know that the dimensions of LHS are $[ \, L^3 \, T^{-1} \, ]$

Now, we check dimensions of RHS:


$\frac{[ \, \pi P r^4 \, ]}{[ \, 8\eta l \,]} \\ \\ \\ = \frac{[\, P \, r^4 \, ]}{[ \, \eta l \, ]} \\ \\ \\ = \frac{[\, M \, L^{-1} \, T^{-2} \, ] \, \, [ \, L^4\, ]}{[ \, M \, L^{-1} \, T^{-1} \, ] \, \, [ \, L \, ]} \\ \\ \\ = [ \, L^3 \, T^{-1} \, ] \\ \\ \\ = \text{Dimensions of LHS}$


Thus, the first option itself is the dimensionally consistent relation. 

So, the answer is Option (1).

$\mathbb{H} \mathfrak{ope \, \, it\, \, helps} \\ \mathbb{P} \mathfrak{urva} \\ \mathbb{BRAINLY \, \, COMMUNITY}$