Question

A swimmer is at 2 meter above the bottom in an 8 m deep rectangular pond which is filled to \(3/4^{\text{th}}\) of its depth. What will be the pressure on the swimmer?

• A. $100 \text{ g/cm}^2$
• B. $200 \text{ g/cm}^2$
• C. $400 \text{ g/cm}^2$
• D. $600 \text{ g/cm}^2$

Hint:

In water, every $1\text{ cm}$ of depth corresponds to a hydrostatic pressure of $1\text{ g/cm}^2$. Thus, once you find the depth of water above the swimmer is $400\text{ cm}$, the pressure is immediately $400\text{ g/cm}^2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Hydrostatic pressure ($P$) in a fluid is directly proportional to the depth of the fluid column above the point of interest.
Key Formula or Approach:
The hydrostatic pressure is given by: \[ P = \rho \cdot h \] Where: - $\rho$ is the density of water ($1 \text{ g/cm}^3$).
- $h$ is the depth of water directly above the swimmer (in $\text{cm}$).

Step 2: Detailed Explanation:
Let us analyze the dimensions given in the problem:
- Total depth of the pond = $8\text{ m}$
- The pond is filled up to $3/4^{\text{th}}$ of its depth.
Calculate the total depth of the water column ($H$): \[ H = 8 \times \frac{3}{4} = 6\text{ m} \] The swimmer is positioned $2\text{ m}$ above the bottom of the pond.
Calculate the depth of water above the swimmer ($h$): \[ h = H - 2 = 6 - 2 = 4\text{ m} = 400\text{ cm} \] Now, compute the pressure in $\text{g/cm}^2$ using the density of water ($\rho = 1 \text{ g/cm}^3$): \[ P = \rho \times h \] \[ P = 1 \text{ g/cm}^3 \times 400 \text{ cm} = 400 \text{ g/cm}^2 \] Thus, the pressure on the swimmer is $400 \text{ g/cm}^2$.

Step 3: Final Answer:
The pressure is $400 \text{ g/cm}^2$, which corresponds to Option (C).