Animated Solution for Physics - Properties of Solids and Liquids: A submarine experience a pressure of $5.05\times 10^6$ Pa at a depth of $d_1$ in a sea. When it goes further to a depth of $d_2$, it experiences a pressure of $8.08\times 10^6$ Pa, then $d_2-d_1$ is approximately (density of water = $10^3\text{ kg}/{\text{m}}^3$ and acceleration due to gravity = $10{\text{ ms}}^{-2}$)

Welcome to this fascinating journey into the depths of the ocean! Imagine you are inside a submarine, exploring the mysterious underwater world. As you dive deeper, you can almost feel the immense weight of the water pressing down on the hull. This problem is a classic application of fluid statics, and it beautifully demonstrates how pressure scales with depth.

When a submarine is submerged in the sea, it doesn't just face the water immediately surrounding it; it must support the weight of the entire column of water stretching all the way up to the surface, plus the weight of the Earth's atmosphere pushing down on the ocean itself.

Let's define our variables. At an initial depth $d_1$, the submarine experiences a total absolute pressure $p_1=5.05\times 10^6\text{ Pa}$. Later, it dives further down to a new depth $d_2$, where the pressure increases significantly to $p_2=8.08\times 10^6\text{ Pa}$. Our mission is to find the vertical distance it traveled, which is simply the difference in depth, $\Delta d=d_2-d_1$.

To solve this, we rely on the fundamental law of hydrostatics. The absolute pressure $p$ at any depth $h$ in an incompressible fluid of constant density $\rho$ is given by the equation:

Here, $p_0$ represents the atmospheric pressure acting on the surface of the sea. The term $\rho gh$ is the gauge pressure, which is the pressure contributed solely by the weight of the fluid column.

Let's write this equation for both depths. At the first depth $d_1$, we have:

And at the second depth $d_2$, we have:

Now, we want to find the difference in depth. The most elegant way to do this is to subtract the first equation from the second. Watch what happens to the atmospheric pressure $p_0$:

The atmospheric pressure $p_0$ completely cancels out! This is a profound physical insight: the difference in pressure between two points in a fluid depends only on the vertical distance between them and the properties of the fluid, regardless of what is happening at the surface.

We can now rearrange our equation to isolate the depth difference $\Delta d$:

It is time to plug in the numbers provided in the problem. We are given $p_1=5.05\times 10^6\text{ Pa}$, $p_2=8.08\times 10^6\text{ Pa}$, the density of seawater $\rho =10^3{\text{ kg/m}}^3$, and the acceleration due to gravity $g=10{\text{ m/s}}^2$.

Substituting these values, we get:

First, let's compute the numerator, which is the pressure difference:

Next, the denominator is simply $10^4$. Dividing the two gives:

The exact calculated difference is $303\text{ m}$. However, looking at our multiple-choice options, the question asks for an approximate value. The closest option is $300\text{ m}$, which makes perfect sense given the approximations often used in physics problems (like taking $g=10{\text{ m/s}}^2$ instead of $9.8{\text{ m/s}}^2$).

And there we have it! By understanding the linear relationship between pressure and depth, we successfully navigated the math and found the submarine's diving distance.