A liquid of density 900 kg/m^3 is filled in a cylindrical tank of upper radius 0.9 m and lower radiu

Q: Why can we use Bernoulli's equation when the capillary tube is detached, but not when it is attached?

Bernoulli's equation assumes non-viscous (frictionless) flow. When the capillary tube is detached, the exit is wide and the flow is fast, making viscous losses negligible. However, when the narrow capillary tube is attached, the high surface-to-volume ratio makes viscous drag dominant, meaning Bernoulli's equation is no longer valid inside the tube.

Q: What is the difference between gauge pressure and absolute pressure in this problem?

The pressure $p$ applied at the top is a gauge pressure. Since both the top surface and the exit are exposed to the atmosphere, the atmospheric pressure $p_0$ cancels out on both sides of Bernoulli's equation, leaving only the gauge pressure $p$ and the hydrostatic pressure $\rho g H$ as the net driving force.

Q: Why did we group $\pi a^2$ and $a^2/l$ instead of solving for $a$ and $l$ separately?

The problem does not provide enough individual constraints to solve for the radius $a$ and length $l$ independently. Grouping them is a deliberate design choice to test your algebraic efficiency and dimensional awareness.

Q: Is the flow inside the capillary tube guaranteed to be streamline?

Yes, because the volume flow rate is extremely small ($8 \times 10^{-6}\text{ m}^3\text{/s}$) and the tube is very narrow, keeping the Reynolds number well below the critical threshold of 2000, which ensures laminar/streamline flow.

Q: What does the unit $\text{N-s/m}^2$ represent for viscosity?

The unit $\text{N-s/m}^2$ is equivalent to Pascal-seconds ($\text{Pa}\cdot\text{s}$), which is the SI unit for dynamic viscosity. It represents the shear stress required to produce a unit velocity gradient in the fluid.

Question

A liquid of density $900	ext{ kg/m}^3$ is filled in a cylindrical tank of upper radius $0.9	ext{ m}$ and lower radius $0.3	ext{ m}$. A capillary tube of length $l$ is attached at the bottom of the tank as shown in the figure. The capillary has outer radius $0.002	ext{ m}$ and inner radius $a$. When pressure $p$ is applied at the top of the tank volume flow rate of the liquid is $8 	imes 10^{-6}	ext{ m}^3	ext{/s}$ and if capillary tube is detached, the liquid comes out from the tank with a velocity $10	ext{ m/s}$. Determine the coefficient of viscosity of the liquid. [Given, $\pi a^2 = 10^{-6}	ext{ m}^2$ and $a^2/l = 2 	imes 10^{-6}	ext{ m}$]

Sigma Percentile · Accepted Answer

### Introduction to Fluid Dynamics and Viscosity

Imagine pouring honey versus pouring water. Honey flows slowly, clinging to the sides of the jar, resisting any attempt to make it move quickly. Water, on the other hand, splashes out effortlessly.

This fundamental difference in fluid behavior is governed by a physical property known as **viscosity**—essentially, internal friction within a fluid.

In this problem, we are challenged to determine the coefficient of viscosity of a liquid filled in a cylindrical tank.

To do this, we are presented with two beautifully contrasting scenarios: one where the fluid flows freely without friction, and another where it is restricted by a narrow capillary tube. By bridging these two scenarios, we can unlock the hidden properties of the fluid.

### Analyzing the Setup: The Two Scenarios

Let us first break down the geometry of our tank. The tank is wide at the top with a radius of $R_1 = 0.9	ext{ m}$ and tapers down to a narrow bottom of radius $R_2 = 0.3	ext{ m}$.

In **Scenario 1**, the capillary tube at the bottom is detached. The liquid flows out freely into the atmosphere with a high speed of $v_2 = 10	ext{ m/s}$. Because the exit is wide and open, we can model this flow as ideal and apply **Bernoulli's Principle** to find the total driving pressure.

In **Scenario 2**, a capillary tube of length $l$ and inner radius $a$ is attached to the bottom. This narrow tube introduces significant viscous resistance, slowing the flow to a steady volume flow rate of $Q = 8 	imes 10^{-6}	ext{ m}^3	ext{/s}$. Here, we must apply **Poiseuille's Law** to describe how viscosity restricts the flow.

### Phase 1: Finding the Driving Pressure (Bernoulli & Continuity)

Let us analyze the free-flowing scenario first. We apply Bernoulli's equation between the top surface of the liquid (point 1) and the exit at the bottom (point 2):

$$p + p_0 + \frac{1}{2}ho v_1^2 + ho g H = p_0 + \frac{1}{2}ho v_2^2$$

Here, $p$ is the externally applied gauge pressure at the top, $p_0$ is the atmospheric pressure, and $H$ is the height of the liquid column. Since both ends are open to the atmosphere, the atmospheric pressure $p_0$ cancels out from both sides:

$$p + ho g H = \frac{1}{2}ho \left( v_2^2 - v_1^2 ight)$$

The term on the left, $\Delta p = p + ho g H$, represents the **total effective pressure** driving the fluid out of the tank. To calculate this, we need to relate the top velocity $v_1$ to the exit velocity $v_2$. We do this using the **Equation of Continuity**:

$$A_1 v_1 = A_2 v_2 \implies v_1 = \frac{A_2}{A_1} v_2 = \left(\frac{R_2}{R_1}ight)^2 v_2$$

Substituting this back into our pressure equation gives:

$$\Delta p = \frac{1}{2}ho v_2^2 \left[ 1 - \left(\frac{R_2}{R_1}ight)^4 ight]$$

Now, let us substitute the given numerical values:
- Density $ho = 900	ext{ kg/m}^3$
- Exit velocity $v_2 = 10	ext{ m/s}$
- Radii $R_1 = 0.9	ext{ m}$ and $R_2 = 0.3	ext{ m}$

$$\Delta p = \frac{1}{2} (900) (10)^2 \left[ 1 - \left(\frac{0.3}{0.9}ight)^4 ight]$$

$$\Delta p = 45000 \left[ 1 - \left(\frac{1}{3}ight)^4 ight] = 45000 \left( 1 - \frac{1}{81} ight) = 45000 \left( \frac{80}{81} ight)$$

Simplifying this fraction yields the exact driving pressure:

$$\Delta p = \frac{4}{9} 	imes 10^5	ext{ N/m}^2$$

### Phase 2: The Viscous Flow (Poiseuille's Law)

Now, let us attach the capillary tube. The flow rate $Q$ is now restricted by viscosity, and is governed by **Poiseuille's Law**:

$$Q = \frac{\pi (\Delta p) a^4}{8 \eta l}$$

We need to solve for the coefficient of viscosity $\eta$. Rearranging the equation gives:

$$\eta = \frac{\pi a^4 (\Delta p)}{8 Q l}$$

At first glance, this looks problematic because we are not given the capillary radius $a$ or length $l$ individually. However, JEE Advanced problems are famous for their elegant algebraic structures! Notice how we can group the terms:

$$\eta = \frac{(\pi a^2) (\Delta p) \left(\frac{a^2}{l}ight)}{8 Q}$$

This is a brilliant simplification because the problem explicitly provides the values of these grouped parameters:
- $\pi a^2 = 10^{-6}	ext{ m}^2$
- $\frac{a^2}{l} = 2 	imes 10^{-6}	ext{ m}$
- $Q = 8 	imes 10^{-6}	ext{ m}^3	ext{/s}$

### Phase 3: The Final Calculation

Let us substitute these values into our rearranged equation:

$$\eta = \frac{(10^{-6}) \left( \frac{4}{9} 	imes 10^5 ight) (2 	imes 10^{-6})}{8 	imes (8 	imes 10^{-6})}$$

Let us simplify the numerator first:

$$	ext{Numerator} = 10^{-6} 	imes 2 	imes 10^{-6} 	imes \frac{4}{9} 	imes 10^5 = \frac{8}{9} 	imes 10^{-7}$$

Now, let us divide by the denominator:

$$\eta = \frac{\frac{8}{9} 	imes 10^{-7}}{64 	imes 10^{-6}} = \frac{8}{9 	imes 64} 	imes 10^{-1}$$

$$\eta = \frac{1}{72} 	imes \frac{1}{10} = \frac{1}{720}	ext{ N-s/m}^2$$

Thus, the coefficient of viscosity of the liquid is exactly **$\frac{1}{720}	ext{ N-s/m}^2$** (or approximately $1.39 	imes 10^{-3}	ext{ Pa}\cdot	ext{s}$).

This elegant result shows how combining macroscopic energy conservation (Bernoulli) with microscopic viscous drag (Poiseuille) allows us to probe the fundamental properties of matter!

Visualized Solution

Visualizing the Two Flow Scenarios

Applying Bernoulli's Theorem

The Equation of Continuity

Expressing the Effective Pressure

Calculating the Numerical Value of $Δ p$

Introducing Poiseuille's Law

Rearranging for Viscosity $η$

Substituting Values and Final Calculation

The Sigma Insight: Viscosity and Stokes' Law

Introduction to Fluid Dynamics and Viscosity

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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