Two large, identical water tanks, 1 and 2, kept on the top of a building of height H, are filled wit

Q: Why does the pipe in Tank 2 increase the velocity of efflux?

The water column inside the pipe adds to the total pressure difference driving the flow. According to Bernoulli's principle, the total head is the sum of the water depth in the tank and the length of the pipe.

Q: Does the diameter of the pipe matter?

The problem states the pipe has the same internal radius as the hole. If the pipe were wider or narrower, the continuity equation would change, and we'd have to account for different velocities at the hole and the pipe exit.

Q: Why is there a negative sign in the continuity equation $-A dy/dt = av$?

The height $y$ is decreasing as time $t$ increases, so $dy/dt$ is a negative quantity. Since area and velocity are positive, we need the negative sign to balance the equation.

Q: Can we assume the velocity of the top surface is zero?

Yes, because the tanks are described as "large" and the holes as "small". This means $A \gg a$, so the velocity of the top surface ($v_{top} = \frac{a}{A}v$) is negligibly small compared to the efflux velocity.

Q: What if the pipe in Tank 2 was horizontal instead of vertical?

If the pipe were horizontal, it wouldn't add any gravitational head. The velocity of efflux would just be $\sqrt{2gy}$, exactly the same as Tank 1 (ignoring viscous losses in the pipe).

Question

Two large, identical water tanks, 1 and 2, kept on the top of a building of height H, are filled with water up to height h in each tank. Both the tanks contain an identical hole of small radius on their sides, close to their bottom. A pipe of the same internal radius as that of the hole is connected to tank 2, and the pipe ends at the ground level. When the water flows from the tanks 1 and 2 through the holes, the times taken to empty the tanks are $t_1$ and $t_2$, respectively. If $H = \left(\frac{16}{9}\right)h$, then the ratio $t_1/t_2$ is________.

Sigma Percentile · Accepted Answer

Have you ever wondered why water flows faster out of a pipe that extends downwards compared to just a hole at the bottom of a tank? This JEE Advanced problem beautifully captures this phenomenon, testing your understanding of Bernoulli's principle and fluid kinematics. Let's break it down step-by-step.

### Analyzing the Setup
We have two identical tanks, Tank 1 and Tank 2, both sitting on top of a building of height $H$. Both are filled with water to a height $h$.

Tank 1 has a simple hole at its base. Tank 2, however, has a pipe connected to its hole that runs all the way down to the ground. We need to find the ratio of the times it takes for each tank to empty, given that $H = \frac{16}{9}h$.

### The Physics of Tank 1
Let's start with the simpler case: Tank 1. As water flows out of the hole, the velocity of efflux is governed by Torricelli's Law. If the instantaneous height of the water is $y$, the velocity $v_1$ is simply:

$$v_1 = \sqrt{2gy}$$

To find the time it takes to empty, we use the equation of continuity. The rate at which the volume of water in the tank decreases must equal the rate at which water flows out of the hole. If $A$ is the cross-sectional area of the tank and $a$ is the area of the hole:

$$-A \frac{dy}{dt} = a v_1 = a\sqrt{2gy}$$

We introduce a negative sign because the height $y$ is decreasing with time. Rearranging and integrating from $y = h$ to $y = 0$:

$$\int_0^{t_1} dt = -\frac{A}{a\sqrt{2g}} \int_h^0 y^{-1/2} dy$$

Solving this integral gives us the time to empty Tank 1:

$$t_1 = \frac{2A\sqrt{h}}{a\sqrt{2g}}$$

### The Physics of Tank 2 (The Siphon Effect)
Now, let's look at Tank 2. This is where many students make a mistake. Because the pipe extends to the ground, the water column inside the pipe creates an additional "pull" or pressure difference.

Applying Bernoulli's equation between the top surface of the water in the tank and the exit of the pipe at the ground:

$$P_{atm} + \rho g (y + H) = P_{atm} + \frac{1}{2}\rho v_2^2$$

Notice that the total head driving the flow is $y + H$, not just $y$. This means the velocity of efflux at the ground is:

$$v_2 = \sqrt{2g(y+H)}$$

Because $v_2$ is always greater than $v_1$, Tank 2 will empty faster. Let's find out exactly how much faster.

### The Mathematics of Emptying Tank 2
Again, we apply the equation of continuity:

$$-A \frac{dy}{dt} = a v_2 = a\sqrt{2g(y+H)}$$

Rearranging and integrating from $y = h$ to $y = 0$:

$$\int_0^{t_2} dt = -\frac{A}{a\sqrt{2g}} \int_h^0 (y+H)^{-1/2} dy$$

The integral of $(y+H)^{-1/2}$ is $2(y+H)^{1/2}$. Evaluating the limits:

$$t_2 = \frac{2A}{a\sqrt{2g}} \left[ \sqrt{h+H} - \sqrt{H} \right]$$

### The Grand Finale: Calculating the Ratio
We now have the expressions for both $t_1$ and $t_2$. Let's find their ratio:

$$\frac{t_1}{t_2} = \frac{\frac{2A\sqrt{h}}{a\sqrt{2g}}}{\frac{2A}{a\sqrt{2g}} \left( \sqrt{h+H} - \sqrt{H} \right)}$$

The constants beautifully cancel out, leaving us with:

$$\frac{t_1}{t_2} = \frac{\sqrt{h}}{\sqrt{h+H} - \sqrt{H}}$$

The problem states that the building height $H = \frac{16}{9}h$. Let's substitute this into our ratio.

First, let's simplify the terms in the denominator:

$$\sqrt{H} = \sqrt{\frac{16}{9}h} = \frac{4}{3}\sqrt{h}$$

$$\sqrt{h+H} = \sqrt{h + \frac{16}{9}h} = \sqrt{\frac{25}{9}h} = \frac{5}{3}\sqrt{h}$$

Now, substitute these back into the ratio:

$$\frac{t_1}{t_2} = \frac{\sqrt{h}}{\frac{5}{3}\sqrt{h} - \frac{4}{3}\sqrt{h}}$$

$$\frac{t_1}{t_2} = \frac{\sqrt{h}}{\frac{1}{3}\sqrt{h}} = 3$$

**The ratio of the times is exactly 3.** Tank 1 takes three times as long to empty as Tank 2, all thanks to the extra gravitational head provided by the pipe!

Enter Numerical Value:

Visualized Solution

$System Setup$

$Velocity of Efflux (Tank 1)$

$Time to Empty Tank 1$

$Velocity of Efflux (Tank 2)$

$Time to Empty Tank 2$

$Ratio of Times$

$Substituting H = \frac{16}{9} h$

$Final Calculation$

The Sigma Insight: Flow of Fluid

Analyzing the Setup We have two identical tanks, Tank 1 and Tank 2, both sitting on top of a building of height $H$

The Physics of Tank 1 Let's start with the simpler case

The Physics of Tank 2 (The Siphon Effect) Now, let's look at Tank 2

The Mathematics of Emptying Tank 2

The Grand Finale

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity of Efflux (Tank 1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time to Empty Tank 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity of Efflux (Tank 2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time to Empty Tank 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Times

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting H=916​h

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation