Question

What pressure, in MPa, must be maintained in a
diving bell, at a depth of 1200 m, to keep out the
ocean water (S = 1.03) ?​

Answer 1

Given:

Diving bell, at a depth of 1200 m with density of ocean water as 1.03 g/cc.

To find:

Pressure in the diving bell.

Calculation:

The general expression for pressure exerted by fluid Column at a depth h is given as :

$\boxed{P = \rho \times g \times h}$

Putting available values in SI units:

$= > P = (1.03 \times 1000) \times 10 \times 1200$

$= > P = 1030 \times 10 \times 1200$

$= > P = 103 \times 12 \times {10}^{4}$

$= > P = 1236 \times {10}^{4}$

$= > P = 12.36 \times {10}^{6} \: pascal$

$= > P = 12.36 \: MPa$

So, final answer is:

$\boxed{ \sf{Pressure \: to \: be \: maintained \: in \: bell = 12.36 \: MPa }}$

Answer 2

Given :-

• Diving bell at a depth rate of 1200 m

• Density of Ocean Water = 1.03g/cc

To FinD :-

• The pressure in the diving bell

SoluTioN :-

The Question is to find the pressure in the diving bell.

The General Expression to Find The pressure exerted by a Fluid column.

$\boxed{ \bold{ \pink{p = p \times g \times h}{} }{} }{}$

Substituting the Values

$p \: = (1.03 \times 100) \times 10 \times 1200 \\ \\ 103 \times 12 \times {10}^{4} \\ \\ 1236 \times {10}^{4} \\ \\ = > 12.36 \times {10}^{6} Mpa$

$\boxed{ \bold{ \red{pressure = 12.36 \times {10}^{6}Mpa }{} }{} }{}$