$A$ liquid flows through a pipe with a pressure difference of $P$. What should be the pressure difference required to maintain the same flow rate if the radius is doubled and the length is doubled?

In the case of a liquid, if the Reynolds number '$R_n$' is $1450$, then the flow of the liquid will

Consider a steady flow of oil in a pipeline. The cross-sectional radius of the pipeline decreases gradually as $r = r_0 e^{-\alpha x}$, where $\alpha = \frac{1}{3} \text{ m}^{-1}$ and $x$ is the distance from the pipeline inlet. If $R_1$ is the Reynolds number for a certain pipeline cross-section at a distance $x_1$ metre from the inlet and $R_2$ is for distance $(x_1 + 3)$ metre, then the ratio $\frac{R_1}{R_2}$ is

Water flows in a streamlined manner through a capillary of radius $a$,the pressure difference being $p$ and the rate of flow $Q$. If the radius is reduced to $a/2$ and the pressure is increased to $4p$,the rate of flow becomes:

If the pressure difference across a pipe of radius $r$ and length $l$ is $p$,then the volume of liquid flowing out per second is $V$. Find the correct expression for $V$.

Water is conveyed through a uniform tube of $8 \text{ cm}$ in diameter and $3140 \text{ m}$ in length at the rate of $2 \times 10^{-3} \text{ m}^3/\text{s}$. The pressure required to maintain the flow is (Viscosity of water $= 10^{-3} \text{ SI units}$):