The pressure on the sample is 4.375 × 10⁷ Pa

<h3>Further explanation</h3>
Let's recall Hydrostatic Pressure formula as follows:


<em>where:</em>
<em>P = hydrosatic pressure ( Pa )</em>
<em>ρ = density of fluid ( kg/m³ )</em>
<em>g = gravitational acceleration ( m/s² )</em>
<em>h = height of a column of liquid ( m )</em>
<em>F = force ( N )</em>
<em>A = cross-sectional area ( m² )</em>
Let us now tackle the problem!

<u>Given:</u>
diameter of large cylinder = D₁ = 10.0 cm
diameter of small cylinder = D₂ = 2.0 cm
area of the sample = A = 4.0 cm² = 4.0 × 10⁻⁴ m²
magnitude of force = F = 350 N
<u>Asked:</u>
pressure on the sample = P = ?
<u>Solution:</u>
<em>Firstly , we will find the force acting on the small cylinder:</em>







<em>Next, we could find the force acting on the large cylinder:</em>








<em>Finally, we could calculate the pressure on the sample:</em>




<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Pressure