Q52 / 60 · KVPY 2015 (SA) · Physics
Stoke's law states that the viscous drag force $F$ experience by a sphere of radius a, moving with a speed $\mathrm{v}$ through a fluid with coefficient of viscosity $\eta$, is given by $F=6 \pi n a v$
If this fluid is flowing through a cylindrical pipe of radius $\mathrm{r}$, length 1 and a pressure difference of P across its two ends, then the volume of water $\mathrm{V}$ which flows through the pipe in time $\mathrm{t}$ can be written as
$$
\frac{v}{t}=k\left(\frac{p}{l}\right)^{a} \eta^{b} r^{c}
$$
Where $\mathrm{k}$ is a dimensionless constant. Correct values of $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ are
- Correct answerA$\mathrm{a}=1, \mathrm{~b}=-1, \mathrm{c}=4$
- B$\mathrm{a}=-1, \mathrm{~b}=1, \mathrm{c}=4$
- C$\mathrm{a}=2, \mathrm{~b}=-1, \mathrm{c}=3$
- D$\mathrm{a}=1, \mathrm{~b}=-2, \mathrm{c}=-4$
Solution
$$
\begin{array}{l}
\frac{V}{t}=\left(\frac{p}{l}\right)^{a} \eta^{b} r^{c} \\
L^{3} T^{-1}=\left(M L^{-2} T^{-2}\right)^{a} L^{c}\left(M L^{-1} T^{-1}\right)^{b} \\
\mathrm{a}+\mathrm{b}=0 \\
-2 a-b+c=3 \\
-2 a-b=-1 \\
\quad c=4, a=1, b=-1
\end{array}
$$
Ans (A) is correct
Hence, the correct answer is option: 'A'.