Question

The rate of flow of liquid in a tube of radius $r$, length $l$,whose ends are maintained at a pressure difference $p$ is $V=\frac{\pi Qp{r}^4}{\eta l}$, where $\eta$ is coefficient of the viscosity and $Q$.

• A. $8$
• B. $\frac{1}{8}$
• C. $\frac{1}{6}$
• D. $\frac{1}{16}$

Answer 1

The correct option is B $\frac{1}{8}$

The r be the radius,L be the length and P is the pressure difference.Generally the Q is represented as rate of flow of liquid but here it is taken as V, Therefore$V=\frac{\pi Qp{r}^4}{\eta l}$

As for now V=Q

Now using the Poiseuille equation for the rate of flow of liquid.

Which is $V=\pi \Delta P{r}^4/8l\eta$

Comparing with the above equation of V,

So on comparing we get the value of $Q=1/8$