Question

The pressure acting on a submarine is $3\times {10}^5\text{Pa}$ 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 $1\times {10}^5\text{Pa}$, density of water is ${10}^3{\text{kg m}}^{-3}$, acceleration due to gravity $g=10{\text{m s}}^{-2}$)

• A. $\frac{200}{3}$%
• B. $\frac{5}{200}$%
• C. $\frac{200}{5}$%
• D. $\frac{3}{200}$%

Answer 1

The correct option is A $\frac{200}{3}$%

Pressure at depth $h$ is
$P=P_0+h\rho g=3\times {10}^5\text{Pa}$
$\Rightarrow h\rho g=3\times {10}^5-1\times {10}^5$
$\Rightarrow h\rho g=2\times {10}^5\text{Pa}$

As $h$ is doubled, $2h\rho g=4\times {10}^5\text{Pa}$
$\therefore$ Increased pressure, $P^{\prime }=P_0+2h\rho g$
$P^{\prime }=P_0+4\times {10}^5=5\times {10}^5\text{Pa}$

$\therefore$ % increase in pressure $=\frac{P^{\prime }-P}{P}\times 100$
$=\frac{(5-3)\times {10}^5}{3\times {10}^5}\times 100$
$=\frac{200}{3}$%

Hence, option (a) is correct.