Question

The pressure acting on a submarine is $3\times {{10}^{5}}Pa$ at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be:
(Assume that atmospheric pressure is $1\times {{10}^{5}}Pa$, the density of water is ${{10}^{3}}kg{{m}^{-3}}$, acceleration due to gravity g = $10m{{s}^{-1}}$)
A. $\left( \frac{200}{3} \right)%$
B. $\left( \frac{5}{200} \right)%$
C. $\left( \frac{200}{5} \right)%$
D. $\left( \frac{3}{200} \right)%$

Answer 1

Hint:We have the equation for pressure at depth using which we can solve this problem and find the pressure when the depth of the submarine is doubled. Now we have the pressure experienced by submarines at a depth and also pressure experienced by submarines when the depth is doubled. Now we can easily find the increase in pressure in terms of percentage.




Formula used:
$P={{P}_{0}}+d\rho g$
Where ‘d’ denotes the initial depth of the submarine and P is the pressure acting on the submarine at that depth. Po is the atmospheric pressure.
\[Increase\text{ }in\text{ }pressure=\frac{\Delta P}{P}\times 100%=\frac{P'-P}{P}\times 100%\]

Complete answer:
Fluid pressure is force acting per unit area on an object in a fluid. At depth the total pressure experienced by the body will be the sum of fluid pressure and atmospheric pressure.
Let ‘d’ denote the initial depth of the submarine and P be the pressure acting on the submarine at that depth. P0 is the atmospheric pressure.
Then the equation for pressure at ‘d’ is
$P={{P}_{0}}+d\rho g$
Substituting the values of density of water and acceleration due to gravity given in the question,
We get,
$
d\rho g=3\times {{10}^{5}} -(1)
$
The pressure acting on the submarine when the depth is doubled,
$
P’=Po+2d\rho g$
$P'=(1\times {{10}^{5}})+(2\times 2\times {{10}^{5}})$
$P'=5\times {{10}^{5}}Pa
$
Now,

Increase in pressure= (change in pressure / original pressure) ×100
That is,
Increase in pressure=$\dfrac{\Delta P}{P}\times 100%=\dfrac{P'-P}{P}\times 100%\]$
Increase in pressure =$\dfrac{(5-3)\times {{10}^{5}}}{3\times {{10}^{5}}}\times 100%$
Therefore, the increase in pressure in terms of percentage is
$\left( \dfrac{200}{3} \right)%
$

Therefore, the answer is: (a)





Note:Mostly what happens is that we only calculate the change in pressure and convert it to percentage. It is wrong. Actually in any question if you are asked to find the increase in some quantity ( here in question it is pressure ) , the formula is you divide the change in that quantity with the initial value and then you have to convert it to percentage.