Answer:
The gauge pressure (\(P_g\)) at a certain depth in a fluid is given by the equation:
\[ P_g = \rho \cdot g \cdot h \]
where:
- \( \rho \) is the density of the fluid,
- \( g \) is the acceleration due to gravity,
- \( h \) is the depth of the fluid.
In your case, \( P_g = 1936 \, \text{Pa} \) (gauge pressure), \( h = 25 \, \text{cm} \) (converted to meters), and \( g \) is approximately \(9.8 \, \text{m/s}^2\).
Let's solve for \( \rho \):
\[ 1936 \, \text{Pa} = \rho \cdot (9.8 \, \text{m/s}^2) \cdot 0.25 \, \text{m} \]
\[ \rho = \frac{1936}{(9.8 \times 0.25)} \]
\[ \rho \approx 800 \, \text{kg/m}^3 \]
So, the density of the unknown fluid is approximately \( 800 \, \text{kg/m}^3 \).
