Question

Stoke’s law states that the viscous drag force $F$ experienced by a sphere of radius $a$ , moving with a speed $v$ through a fluid and coefficient of viscosity $\eta$ , is given by $F=6\pi \eta av$ . If this fluid is flowing through a cylindrical pipe of radius $r$ , length $l$ and a pressure difference of $p$ across its two ends, the volume of water $V$ which flows through the pipe on time $t$ can be written as ${\displaystyle \frac{V}{t}}=k{\left({\displaystyle \frac{p}{l}}\right)}^a{\eta }^b{r}^c$ where, $k$ is the dimensionless constant. Correct values of $a,b\mathrm{\&}c$ are:
(A) $a=1,b=-1,c=4$
(B) $a=-1,b=1,c=4$
(C) $a=2,b=-2,c=3$
(D) $a=1,b=-2,c=-4$

Answer 1

Hint :Here, it has been asked to find the values of $a,b\mathrm{\&}c$ from the formula given above. Thus we have to calculate with the help of dimensions of each term used in the formula ${\displaystyle \frac{V}{t}}=k{\left({\displaystyle \frac{p}{l}}\right)}^a{\eta }^b{r}^c$
Where, $V$ is the volume of water, $r$ is the radius of the cylindrical pipe, $l$ is the length, $p$ is the pressure, $t$ is the time and $\eta$ is the viscosity.

Complete Step By Step Answer:
Let us first use the dimensional formulas of all the terms used in the given formula.
 ${\displaystyle \frac{V}{t}}=k{\left({\displaystyle \frac{p}{l}}\right)}^a{\eta }^b{r}^c$ ….. $(1)$
 $\left[V\right]=\left[M^0{L}^3{T}^0\right]$
 $\left[t\right]=\left[M^0{L}^0{T}^1\right]$
 $\left[p\right]={\displaystyle \frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]}}=\left[M{L}^{-1}{T}^{-2}\right]$
 $\left[l\right]=\left[L^1\right]$
 $\left[r\right]=\left[L^1\right]$
 $\left[\eta \right]={\displaystyle \frac{force}{area\times velo.gradient}}$
 $\therefore \left[\eta \right]={\displaystyle \frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]\left[T^{-1}\right]}}=\left[M{L}^{-1}{T}^{-1}\right]$
Thus, here we have written all the terms and their dimension now we have to put these values in equation $(1)$ such that:
 $eq(1)\Rightarrow {\displaystyle \frac{\left[M^0{L}^3{T}^0\right]}{\left[M^0{L}^0{T}^1\right]}}={\left[{\displaystyle \frac{\left[M{L}^{-1}{T}^{-2}\right]}{\left[L^1\right]}}\right]}^a{\left[M{L}^{-1}{T}^{-1}\right]}^b{\left[L^1\right]}^c$
 $\Rightarrow \left[M^0{L}^3{T}^{-1}\right]=\left[M^{a+b+0}{L}^{-2a-b+c}{T}^{-2a-b+0}\right]$
Now here we have to compare the powers of $[MLT]$ of right hand side with the powers of $[MLT]$ of left hand side from above equation, so we get:
 $a+b+0=0$ ….. $(i)$
 $-2a-b+c=3$ ….. $(ii)$
 $-2a-b+0=-1$ ….. $(iii)$
Now, we have to solve these equation $(i)$ , $(ii)$ and $(iii)$ to obtain the required values of $a,b\mathrm{\&}c$
Therefore, consider equations $(i)$ and $(iii)$
 $\Rightarrow a+b=-2a-b+1$
 $\Rightarrow a=1$
By using, $a=1$ in equation $(i)$ , we get
 $\Rightarrow 1+b=0$
 $\Rightarrow b=-1$
Now, put $a=1$ and $b=-1$ in equation $(ii)$ , the result is:
 $\Rightarrow -2(1)+1+c=3$
 $\Rightarrow c=4$
Thus, we calculated the values of $a=1$ , $b=-1$ and $c=4$
The correct answer is the option A.

Note :
Here, simply we have to write the dimensional formula of the given quantities in the formula in which we have to calculate the required values and then we have to put those values in that formula and compare them as we have done. Here we observed that the dimensional formula for $k$ is not written; it is because here it has been mentioned that $k$ is not having any dimension.