Question

Figure below shows a cube of each side 15cm immersed in a tub containing water
of density 103 Kg m-3 such that its top surface is 20cm below the free surface of
water. [3]
Calculate (i) the pressure at the top of the cube
(ii) the pressure at the bottom of cube
(iii) the resultant pressure on cube.

Answer 1

AnswEr

Pressure exerted by water on the bottom of a deep dam(Hydrostatic pressure) is 120000 Pa.

Step-by-step Explanation :

GivEn :

Density of water, ϱ = 10³kg/m³

Height = 12 m

g = 10m/s²

To find :

Pressure exerted by water on the bottom of a deep dam(Hydrostatic pressure) =

SoluTion :

We know that,

$\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = \rho \times g \times h \\ \\\end{gathered} </p><p>$

Substituting the values,

$</p><p>\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = {10}^{3} \times 10 \times 12 \\ \\\end{gathered} </p><p></p><p>$

$</p><p>\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = {10}^{4} \times 12 \\ \\\end{gathered} </p><p>$

$\begin{gathered}\sf \longrightarrow \: Hydrostatic \: pressure = 10000 \times 12 \\ \\\end{gathered} </p><p>$

$\begin{gathered}\sf \therefore { \blue{Hydrostatic \: pressure = 120000 \: Pa}} \\ \\\end{gathered} </p><p>$

Hence, Pressure exerted by water on the bottom of a deep dam(Hydrostatic pressure) is 120000 Pa.

Answer 2

Given:

Length of the side of cube = 15cm

Density of water = 10³ kg/m³

The depth of water from the free surface to the cube = 20cm

To Find:

The pressure at the top of the cube.

The pressure at the bottom of the cube.

The resultant pressure on the cube.

Solution:

The depth of cube from the free surface of water = 20cm + 15cm = 35cm

The formula of pressure (P) = ρgh

And, Atmospheric pressure = $10^{5}$ Pa

i) Pressure at the top surface of the cube ($P_{1}$) = $P_{o}$ + ρgh1

⇒ $10^{5}$ + ( 10³×0.2×9.8) = 1.0196 × $10^{5}$ Pa

ii) The pressure at the bottom surface of the cube ($P_{2}$) = $P_{o}$ + ρgh2

⇒ $10^{5}$ + (10³ × 0.35×9.8) = 1.0343× $10^{5}$ Pa

iii) The resultant pressure on the cube = $P_{2} -P_{1}$

⇒ (1.0343 - 1.0196) × $10^{5}$ Pa = 1.47×10³ Pa

Hence, the pressure on the top of the cube is 1.0196×$10^{5}$ Pa and on the bottom of the cube is 1.0343×$10^{5}$ Pa. And the resultant pressure on the cube is 1.47×10³ Pa.