Which of the ratio of forces represents the Reynolds number $Re$?

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:

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

If $\rho$ is the density and $\eta$ is the coefficient of viscosity of a fluid which flows with a speed $v$ in a pipe of diameter $d$,the correct formula for Reynolds number $R_{e}$ is ..............

Two capillary tubes of same length each of $50 \ cm$ but of different radii $4 \ mm$ and $2 \ mm$ are connected in series. When water flows,the pressure difference between the ends of the arrangement is $P$. Then the pressure difference between the ends of the first tube is

Water from a pipe is coming at a rate of $100\, L/min$. If the radius of the pipe is $5\, cm$, the Reynolds number for the flow is of the order of: (density of water $= 1000\, kg/m^3$, coefficient of viscosity of water $= 1\, mPa\, s$)