# Question 13de8

Jun 9, 2016

$5 A t m$

#### Explanation:

The pressure exerted by a static fluid depends only upon height $h$ of the fluid column, the density $\rho$ of the fluid, and acceleration due to gravity $g$.

The pressure in a static fluid is given by the expression

P_"static fluid" = ρgh#

1. At the surface the pressure is due to atmosphere.
2. At depth $h$ pressure is sum of pressure due to air and pressure due to water column of depth $h$
${P}_{h} = A i r + \text{pressure due to water of depth } h$
Inserting given values we obtain
$3 A t m = 1 A t m + \text{pressure due to water of depth } h$
$\implies \text{pressure due to water of depth } h = 3 - 1 = 2 A t m$
3. $\therefore \text{pressure due to water of depth } 2 h = 2 \times 2 = 4 A t m$
Now ${P}_{2 h} = A i r + \text{pressure due to water of depth } 2 h$
Inserting calculated value we obtain
${P}_{2 h} = 1 + 4 = 5 A t m$