Question

Consider a fish swimming \(5 \mathrm{m}\) below the free surface of water. The increase in the pressure exerted on the fish when it dives to a depth of \(25 \mathrm{m}\) below the free surface is \((a) 196 \mathrm{Pa}\) \((b) 5400 \mathrm{Pa}\) \((c) 30,000 \mathrm{Pa}\) \((d) 196,000 \mathrm{Pa}\) \((e) 294,000 \mathrm{Pa}\)

Short answer

Answer: (d) 196,000 Pa

Step-by-step solution

Write down the hydrostatic pressure formula

The hydrostatic pressure formula is: \(P = \rho \cdot g \cdot h\), where \(P\) is the pressure, \(\rho\) is the density of the fluid (water), \(g\) is the acceleration due to gravity, and \(h\) is the depth below the free surface.

Determine the water properties and constants

Using the standard values for the density of water and the acceleration due to gravity, we have \(\rho = 1000\,\mathrm{kg/m^3}\) and \(g = 9.81\,\mathrm{m/s^2}\).

Calculate the initial pressure on the fish

We will first find the pressure when the fish is at a depth of 5 meters. Plugging the values into the hydrostatic pressure formula, we get: \(P_1 = \rho \cdot g \cdot h_1 = 1000\,\mathrm{kg/m^3} \cdot 9.81\,\mathrm{m/s^2} \cdot 5\,\mathrm{m} = 49,050\,\mathrm{Pa}\).

Calculate the final pressure on the fish

Next, we will find the pressure when the fish is at a depth of 25 meters. Plugging the values into the hydrostatic pressure formula, we get: \(P_2 = \rho \cdot g \cdot h_2 = 1000\,\mathrm{kg/m^3} \cdot 9.81\,\mathrm{m/s^2} \cdot 25\,\mathrm{m} = 245,250\,\mathrm{Pa}\).

Calculate the increase in pressure

Now, we will find the increase in pressure by subtracting the initial pressure from the final pressure: \(\Delta P = P_2 - P_1 = 245,250\,\mathrm{Pa} - 49,050\,\mathrm{Pa} = 196,200\,\mathrm{Pa}\).

Choose the correct answer

Comparing the increase in pressure calculated in Step 5 with the options given in the exercise, we find that the correct answer is \((d) 196,000\,\mathrm{Pa}\). Note that the answer has been rounded to the nearest thousand.