Animated Solution for Physics - Properties of Solids and Liquids: The pressure acting on a submarine is $3\times 10^5\text{ Pa}$ at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be (Assume that atmospheric pressure is $1\times 10^5\text{ Pa}$, density of water is $10^3{\text{ kg m}}^{-3}$, $g=10{\text{ ms}}^{-2}$)

The deep ocean is a fascinating and extreme environment. As you dive deeper, the weight of the water above you increases, leading to immense pressure. But it's not just the water pushing down on you; the Earth's atmosphere is also pressing down on the ocean's surface. This problem perfectly illustrates how these two pressures combine and how they scale with depth.

Imagine a submarine cruising at a certain depth $h$ in the ocean. The problem states that the total pressure acting on it at this depth is $3\times 10^5\text{ Pa}$. We are also given the atmospheric pressure, $p_0=1\times 10^5\text{ Pa}$.

Our goal is to find out what happens to the pressure when the submarine dives to twice its original depth, $2h$, and specifically, to calculate the percentage increase in that pressure.

To solve this, we need the fundamental equation for absolute pressure in a fluid:

Here, $p$ is the total (absolute) pressure, $p_0$ is the atmospheric pressure at the surface, $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the depth.

The term $\rho gh$ represents the gauge pressure—the pressure exerted solely by the weight of the water column above the submarine.

Let's apply our master equation to the initial state of the submarine. We know the initial pressure $p_i$ is $3\times 10^5\text{ Pa}$:

By subtracting the atmospheric pressure from the total pressure, we can isolate the gauge pressure:

This tells us that at depth $h$, the water alone exerts a pressure of $2\times 10^5\text{ Pa}$. Notice that we didn't even need to use the given values for the density of water ($\rho$) or gravity ($g$)! Treating $\rho gh$ as a single variable simplifies the math tremendously.

The New Pressure

Now, the submarine dives to a new depth of $2h$. Let's write the pressure equation for this new depth. The final pressure $p_f$ will be:

We can rearrange this to make use of the gauge pressure we just found:

Now, we simply substitute our known values back into the equation. We know $p_0=1\times 10^5\text{ Pa}$ and $\rho gh=2\times 10^5\text{ Pa}$:

So, at twice the depth, the total pressure is $5\times 10^5\text{ Pa}$. Notice that the pressure didn't simply double from $3\times 10^5$ to $6\times 10^5$. This is a common trap! The pressure doesn't double because the atmospheric pressure $p_0$ is a constant addition that doesn't scale with depth.

Percentage Increase

Finally, we need to find the percentage increase in pressure. The formula for percentage increase is:

Substituting our pressure values:

The $10^5$ terms cancel out beautifully, leaving us with:

And there we have it! The pressure increases by $\frac{200}{3}\%$, which corresponds to option (a). This problem is a fantastic reminder to always account for atmospheric pressure when dealing with absolute pressure in fluids.