What will be the nature of flow of water from a circular tap,when its flow rate increases from $0.18\, L/min$ to $0.48\, L/min$? The radius of the tap is $0.5\, cm$ and the viscosity of water is $10^{-3}\, Pa\cdot s$. (Density of water: $10^{3}\, kg/m^{3}$)

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 $(= P)$?

$A$ 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?

Two capillary tubes of the same length but different radii $r_1$ and $r_2$ are fitted in parallel to the bottom of a vessel. The pressure head is $P$. What should be the radius of a single tube that can replace the two tubes so that the rate of flow is the same as before?

The Reynolds number of a flow is the ratio of

Three capillaries of length $L, L/2$ and $L/3$ are connected in series. Their radii are $r, r/2$ and $r/3$ respectively. If stream-line flow is to be maintained and the pressure difference across the first capillary is $P$,then: