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 $F=6 \pi \eta a v$. If this fluid is flowing through a cylindrical pipe of radius $r$, length $l$ 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

Q: 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 $F=6 \pi \eta a v$. If this fluid is flowing through a cylindrical pipe of radius $r$, length $l$ 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) From $\frac{V}{t}=k\left(\frac{p}{l}\right)^a \eta^b r^c$, we have $\left[ L ^3 T ^{-1}\right]=\left[\frac{ ML ^{-1} T ^{-2}}{ L }\right]^a\left[ ML ^{-1} T ^{-1}\right]^b[ L ]^c$ 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 $c=4$

Question

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

KVPY 2017, Diffcult

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