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
$a+b=0 \Rightarrow-2 a-b +c=3$
$-2 a-b=-1$
Solving, we get $a=1, b=-1$ and $c=4$