To find the mean pressure on the wall of a cylinder filled with a liquid up to a height [tex]\( h \)[/tex], with the density of the liquid being [tex]\( \rho \)[/tex], we can proceed as follows:
1. Identify the Known Variables:
- Height of the liquid column: [tex]\( h \)[/tex]
- Density of the liquid: [tex]\( \rho \)[/tex]
- Acceleration due to gravity: [tex]\( g \)[/tex]
2. Understand the Concept of Mean Pressure:
The mean pressure on the wall of the cylinder due to the liquid column is related to hydrostatic pressure, which varies linearly from zero at the surface (top of the liquid) to a maximum at the bottom.
3. Formula for Hydrostatic Pressure:
The hydrostatic pressure at any depth [tex]\( y \)[/tex] in the liquid is given by:
[tex]\[
P(y) = \rho g y
\][/tex]
where [tex]\( y \)[/tex] is the depth from the surface.
4. Mean Pressure Calculation:
To find the mean pressure [tex]\( P_{\text{mean}} \)[/tex], we compute the average pressure over the entire height of the liquid column. This is given by:
[tex]\[
P_{\text{mean}} = \frac{1}{h} \int_0^h \rho g y \, dy
\][/tex]
where the integral represents the area under the pressure-depth curve from [tex]\( y = 0 \)[/tex] to [tex]\( y = h \)[/tex].
5. Evaluate the Integral:
[tex]\[
\int_0^h \rho g y \, dy = \rho g \int_0^h y \, dy
\][/tex]
The integral [tex]\(\int_0^h y \, dy\)[/tex] is:
[tex]\[
\int_0^h y \, dy = \left[ \frac{y^2}{2} \right]_0^h = \frac{h^2}{2}
\][/tex]
6. Substitute the Integral Back into the Formula:
[tex]\[
P_{\text{mean}} = \frac{1}{h} \cdot (\rho g \cdot \frac{h^2}{2}) = \frac{\rho g h^2}{2h} = \frac{\rho g h}{2}
\][/tex]
7. Conclusion:
The mean pressure on the wall of the cylinder is:
[tex]\[
P_{\text{mean}} = \frac{\rho g h}{2}
\][/tex]
Thus, the correct answer from the given options is:
b. [tex]\(\frac{\rho g h}{2}\)[/tex]
