Question

A capillary tube of radius $r$, length $L$ is connected horizontally with a cylindrical vessel of radius $R$. A liquid of density $\rho$ and viscosity $\mu$ is filled upto height h. Flow per unit time through capillary tube is given by $\frac{V}{t}=\frac{\pi p{r}^4}{8\mu L}$ Here $p=$ pressure difference across the tube. Velocity with which level of liquid sinks in vessel

• A. $\frac{\pi \rho gh{r}^4}{8\mu L{R}^2}$
• B. $\frac{\rho gh{r}^2}{8\mu L}$
• C. $\frac{\rho gh{r}^4}{8\mu L{R}^2}$
• D. $\frac{\rho gh{r}^4}{4\mu L{R}^2}$

Answer 1

The correct option is C $\frac{\rho gh{r}^4}{8\mu L{R}^2}$

The volumetric flow rate through the capillary tube is given by,

V/t = $\frac{\pi p{r}^4}{8\mu L}$

Here, p = $\rho gh$

Now, since this is a continuous fluid, therefore, the flow rate of water flowing out would be equal to the rate of water decreasing in the cylinder.

$Ah/t=V/t$

$\pi R^{2}H/t=\frac{\pi \rho gh{r}^4}{8\mu L}$

Therefore, H/t = $\frac{\rho gh{r}^4}{8\mu L{R}^2}$