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The volume of a liquid is proportional to......, given its density $\rho$, viscosity $\eta$ and $t$ the time of flow through a capillary tube of length $L$ and radius $R$, with a pressure difference $p$ across its ends
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The correct answer is:
$\frac{p R^4 t}{\eta L}$
Given, density of liquid $=\rho$
Viscosity $=\eta$
Time of flow $=t$,
Length of tube $=L$
Radius of cross-section $=R$
Pressure difference between end $s=p$
According to Poiseuille's formula, the rate of flow of liquid through a horizontal tube
$=\frac{\text { Volume of liquid flow }}{\text { Time taken }}$
i.e. $\quad Q=\frac{V}{t}=\frac{\pi}{8} \frac{p R^4}{\eta L}$
$\therefore \quad V=\frac{\pi}{8} \frac{p R^4}{\eta L} t$, (where $\frac{\pi}{8}$ is a constant.)
Hence, $\quad V \propto \frac{p R^4 t}{\eta L}$
So, the option (c) is correct, where volume of flow of liquid is proportional to $\frac{p R^4 t}{\eta L}$.
Viscosity $=\eta$
Time of flow $=t$,
Length of tube $=L$
Radius of cross-section $=R$
Pressure difference between end $s=p$
According to Poiseuille's formula, the rate of flow of liquid through a horizontal tube
$=\frac{\text { Volume of liquid flow }}{\text { Time taken }}$
i.e. $\quad Q=\frac{V}{t}=\frac{\pi}{8} \frac{p R^4}{\eta L}$
$\therefore \quad V=\frac{\pi}{8} \frac{p R^4}{\eta L} t$, (where $\frac{\pi}{8}$ is a constant.)
Hence, $\quad V \propto \frac{p R^4 t}{\eta L}$
So, the option (c) is correct, where volume of flow of liquid is proportional to $\frac{p R^4 t}{\eta L}$.
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