Water flows from a tap of diameter $1.5 \ cm$ with a flow rate of $7.5 \times 10^{-5} \ m^3 \ s^{-1}$. The coefficient of viscosity of water is $10^{-3} \ Pa \cdot s$. The flow is:

We have two narrow capillary tubes $T_1$ and $T_2$. Their lengths are $l_1$ and $l_2$ and radii of cross-section are $r_1$ and $r_2$ respectively. The rate of flow of water under a pressure difference $P$ through tube $T_1$ is $8 \ cm^3/sec$. If $l_1 = 2l_2$ and $r_1 = r_2$,what will be the rate of flow when the two tubes are connected in series and the pressure difference across the combination is the same as before $(= P)$?

If water flows with a velocity of $20 \,cm \,s^{-1}$ in a pipe of radius $2 \,cm$, then the flow is (The coefficient of viscosity of water is $10^{-3} \,kg \,m^{-1} \,s^{-1}$ and density of water is $10^3 \,kg \,m^{-3}$).

The rate of flow of water in a capillary tube of length $\ell$ and radius $r$ is $V.$ The rate of flow in another capillary tube of length $2 \ell$ and radius $2 r$ for the same pressure difference would be $....V$

Two tubes of radii $r_1$ and $r_2$,and lengths $l_1$ and $l_2$,respectively,are connected in series and a liquid flows through each of them in streamline conditions. $P_1$ and $P_2$ are pressure differences across the two tubes. If $P_2 = 4P_1$ and $l_2 = \frac{l_1}{4}$,then the radius $r_2$ will be equal to:

Under a constant pressure head,the rate of flow of liquid through a capillary tube is $V$. If the length of the capillary is doubled and the diameter of the bore is halved,the rate of flow would become