# Question 5fbd7

Jun 19, 2015

The pressure at the bottom of the second glass is $\text{2P}$.

#### Explanation:

Pressure is defined as force acting on unit area.

For the first glass, the difference between the pressure at the top, which is said to be zero, and the pressure on the bottom, which is $P$, is attributed to the weight of the water.

At the bottom of the glass, the weight of the water pushing upon the surface area of the glass causes a pressure equal to $P$. If you assume the glass to be shped like a cylinder, then youcan write

$P = \frac{m \cdot g}{S}$, where

$m$ - the mass of the water;
$S$ - the area of the cicrlec that makes up the bottom of the glass.

You can express the mass of the water by using its density and the volume of the cylinder.

$\rho = \frac{m}{V} \implies m = \rho \cdot V$

In the case of a cylinder, you have

$\left\{\begin{matrix}V = \pi \cdot {r}^{2} \cdot h \\ S = \pi \cdot {r}^{2}\end{matrix}\right.$

Therefore, the pressure at the bottom of the first glass can be written as

$P = \frac{\rho \cdot \cancel{\pi \cdot {r}^{2}} \cdot h \cdot g}{\cancel{\pi \cdot {r}^{2}}} = \rho \cdot h \cdot g$

Now for the second glass. You know that the diameter of the glass doubles, which implies that the radius doubles , and that the height double as well, compared with the first glass.

$\left\{\begin{matrix}{r}_{2} = 2 \cdot r \\ {h}_{2} = 2 \cdot h\end{matrix}\right.$

The pressure at the bottom of the second glass will be

${P}_{2} = \frac{\rho \cdot \cancel{\pi \cdot {\left(2 r\right)}^{2}} \cdot {h}_{2} \cdot g}{\cancel{\pi \cdot {\left(2 r\right)}^{2}}} = \rho \cdot {h}_{2} \cdot g$

P_2 = underbrace(rho * h * g)_(color(blue)("=P")) * 2 = color(green)("2P")#