Question

Stokes’ law states that the viscous drag force $F$ experienced by a sphere of radius $a$, moving with a speed $v$ through a fluid with coefficient of viscosity $\eta$ is given by $F=6\pi \eta a\upsilon$. If this fluid is flowing through a cylindrical pipe of radius $r$, length $I$ and pressure difference of $p$ across its two ends, then the volume of water $V$ which flows through the pipe in time $t$ can be written as $\frac{V}{t}=k{\left(\frac{p}{l}\right)}^a{\eta }^b{r}^c$ where $k$ is a dimensionless constant. Correct values of $a,b$ and $c$ are

• A. a = 1, b = - 1, c = 4
• B. a =-1, b = 1, c = 4
• C. a = 2, b = - 1, c = 3
• D. a = 1, b = - 2, c = - 4

Answer 1

Answer: (A)

From $\frac{V}{t}=k{\left(\frac{p}{l}\right)}^a{\eta }^b{r}^c$
we have
$\left[L^3{T}^{-1}\right]={\left[\frac{M{L}^{-1}{T}^{-2}}{L}\right]}^a{\left[M{L}^{-1}{T}^{-1}\right]}^b[L{]}^c$
Equating powers of $M,L$ and $T$, we get
$a+b=0\Rightarrow -2a-b+c=3$
$-2a-b=-1$
Solving, we get $a=1,b=-1$ and $c=4$