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 $g=10m{s}^{-2}$)

• 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

The correct option is A

$\left(\frac{200}{3}\right)\%$

Step 1: Given data

The pressure acting on a submarine$\left(P\right)=3\times {10}^{5}Pa$

Atmospheric pressure$\left(P_{\circ }\right)=1\times {10}^{5}Pa$

Acceleration due to gravity$\left(g\right)=10m{s}^{-2}$

Density of water=${10}^{3}kg{m}^{-3}$

Step 2: Formula used

The pressure at depth $h$ can be computed using the following formula:

$$\begin{gathered} P=P_{\circ }+h\rho g----\left(1\right) \\ whereh=heightand\rho =density \end{gathered}$$

Step 3: Determine the value of $h\rho g$

Substitute $P=3\times {10}^{5}Pa$ and $P_{\circ }=1\times {10}^{5}Pa$ in equation $\left(1\right)$,

$\begin{array}{rcl}\left(3\times {10}^5\right)& =& \left(1\times {10}^5\right)+h\rho g\\ h\rho g& =& 2\times {10}^5----\left(2\right)\end{array}$

Step 4: Compute the increased pressured

The value of $h\rho g$ when the height gets doubled,

$\begin{array}{rcl}2h\rho g& =& 4\times {10}^5----\left(3\right)\end{array}$

So, the pressure increased after increasing the height is,

$$\begin{gathered} P^{\text{'}}=P_{\circ }+2h\rho g \\ {P}^{\text{'}}=\left(1\times {10}^5\right)+\left(4\times {10}^5\right) \\ {P}^{\text{'}}=5\times {10}^{5}Pa \end{gathered}$$

Step 5: Compute percentage increase in pressure

The percent increase in pressure is,

$\begin{array}{rcl}\%increase& =& \frac{increasedpressure\left(P^{\text{'}}\right)-initialpressure\left(P\right)}{initialpressure}\times 100\\ & =& \frac{\left(5\times {10}^5\right)-\left(3\times {10}^5\right)}{\left(3\times {10}^5\right)}\times 100\\ & =& \left(\frac{200}{3}\right)\%\end{array}$

Hence, option A is the correct answer.