Questions: B) 4 rho C) rho / 2 D) rho FOR 11 - 15, SHOW ALL WORK, INCLUDING ALL EQUATIONS. Each question worth 10 points each. Moments of inertia: Cylinder: I=1 / 2 mr^2 Sphere: (2/5) mr^2 Rod: (1/12) mL^2 (this L is length) FALL 2022 14. A cubical box 25.0 cm on each side is immersed in a fluid. The pressure at the top surface of the box is 109.4 x 10^3 N / m^2 and the pressure on the bottom surface is 112.0 x 10^3 N / m^2. What is the density of the fluid? FALL 2022

B) 4 rho
C) rho / 2
D) rho

FOR 11 - 15, SHOW ALL WORK, INCLUDING ALL EQUATIONS. Each question worth 10 points each.

Moments of inertia: Cylinder: I=1 / 2 mr^2 Sphere: (2/5) mr^2 Rod: (1/12) mL^2 (this L is length)

FALL 2022

14. A cubical box 25.0 cm on each side is immersed in a fluid. The pressure at the top surface of the box is 109.4 x 10^3 N / m^2 and the pressure on the bottom surface is 112.0 x 10^3 N / m^2. What is the density of the fluid?

FALL 2022

Transcript text: B) $4 \rho$ C) $\rho / 2$ D) $\rho$ FOR 11 - 15, SHOW ALL WORK, INCLUDING ALL EQUATIONS. Each question worth 10 points each. Moments of inertia: Cylinder: $\mathrm{I}=1 / 2 \mathrm{mr}^{2}$ Sphere: (2/5) $\mathrm{mr}^{2}$ Rod: (1/12) $\mathrm{mL}^{2}$ (this L is length) FALL 2022 14. A cubical box 25.0 cm on each side is immersed in a fluid. The pressure at the top surface of the box is $109.4 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}$ and the pressure on the bottom surface is $112.0 \times 10^{3}$ $\mathrm{N} / \mathrm{m}^{2}$. What is the density of the fluid? FALL 2022

Solution

Solution Steps

Step 1: Understand the Problem

We need to find the density of the fluid in which a cubical box is immersed. The pressure difference between the top and bottom surfaces of the box is given, and the side length of the cube is known.

Step 2: Identify Relevant Equations

The pressure difference in a fluid is related to the density, gravitational acceleration, and height difference by the equation:

\[ \Delta P = \rho g h \]

where:

$\Delta P$ is the pressure difference,
$\rho$ is the density of the fluid,
$g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$),
$h$ is the height difference (in this case, the side length of the cube).

Step 3: Calculate the Pressure Difference

The pressure at the top surface is $109.4 \times 10^3 \, \text{N/m}^2$ and at the bottom surface is $112.0 \times 10^3 \, \text{N/m}^2$. The pressure difference $\Delta P$ is:

\[ \Delta P = 112.0 \times 10^3 - 109.4 \times 10^3 = 2.6 \times 10^3 \, \text{N/m}^2 \]

Step 4: Solve for the Density

Using the equation $\Delta P = \rho g h$, we can solve for $\rho$:

\[ \rho = \frac{\Delta P}{g h} \]

Substitute the known values ($\Delta P = 2.6 \times 10^3 \, \text{N/m}^2$, $g = 9.81 \, \text{m/s}^2$, $h = 0.25 \, \text{m}$):

\[ \rho = \frac{2.6 \times 10^3}{9.81 \times 0.25} \]

\[ \rho = \frac{2.6 \times 10^3}{2.4525} \approx 1060.8 \, \text{kg/m}^3 \]