Shown in the figure is a container whose top and bottom diameters are $D$ and $d$ respectively. At the bottom of the container, there is a capillary tube of outer radius $b$ and inner radius $a$. The volume flow rate in the capillary is $Q$. If the capillary is removed the liquid comes out with a velocity of $v_0$. The density of the liquid is given as $\rho$. Calculate the coefficient of viscosity $\eta$.

Step 1: Determine the pressure difference across the capillary.

The pressure difference $\Delta P$ across the capillary tube is equal to the hydrostatic pressure of the liquid column of height $h$.

$$\Delta P = \rho g h$$

Step 2: Apply Poiseuille's Law.

The volume flow rate $Q$ through the capillary of inner radius $a$ and length $L$ is given by Poiseuille's equation.

$$Q = \frac{\pi a^4 \Delta P}{8 \eta L}$$

Step 3: Express the coefficient of viscosity.

Substitute the pressure difference into the flow rate equation and solve for $\eta$.

$$Q = \frac{\pi a^4 \rho g h}{8 \eta L}$$ $$\eta = \frac{\pi a^4 \rho g h}{8 L Q}$$

Step 4: Relate the height to the efflux velocity.

Without the capillary, the efflux velocity $v_0$ is given by Torricelli's law.

$$v_0 = \sqrt{2gh}$$ $$g h = \frac{v_0^2}{2}$$

Step 5: Calculate the coefficient of viscosity.

Substitute the expression for $gh$ into the equation for $\eta$, using the capillary length $L = h$.

$$\eta = \frac{\pi a^4 \rho v_0^2}{16 L Q}$$ $$\eta = \frac{\pi a^4 \rho v_0^2}{16 h Q}$$

Answer: $\eta = \dfrac{\pi a^4 \rho v_0^2}{16 h Q}$