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Step-by-Step Solution
Step 1: Understand the given information
The submarine experiences a total pressure of $3 \times 10^5 \text{ Pa}$ at some depth. Atmospheric pressure is given as $1 \times 10^5 \text{ Pa}$, density of water is $1000 \,\text{kg/m}^3$, and $g=10 \,\text{m/s}^2$. We assume that the total pressure on the submarine at a certain depth $h$ in water is:
$P = P_0 + \rho g h, \\
\text{where } P_0 \text{ is the atmospheric pressure}, \, \rho \text{ is the density of water}, \, g \text{ is gravitational acceleration, and } h \text{ is the depth.}$
Step 2: Calculate the gauge pressure at the initial depth
The total pressure at depth $h$ is $3 \times 10^5 \,\text{Pa}.$ Since atmospheric pressure is $1 \times 10^5 \,\text{Pa}$, the additional pressure due to the water column (gauge pressure) is:
$ \rho g h = P - P_0 = 3 \times 10^5 - 1 \times 10^5 = 2 \times 10^5 \text{ Pa}.$
Step 3: Find the pressure at doubled depth
If the depth is doubled (i.e., new depth $2h$), the new gauge pressure due to water will be:
$ \rho g \cdot (2h) = 2 \times (\rho g h) = 2 \times (2 \times 10^5 \text{ Pa}) = 4 \times 10^5 \text{ Pa}.$
Hence, the total pressure at this doubled depth is:
$ P' = P_0 + 4 \times 10^5 = 1 \times 10^5 + 4 \times 10^5 = 5 \times 10^5 \text{ Pa}.$
Step 4: Calculate the percentage increase in pressure
The original pressure was $P = 3 \times 10^5 \text{ Pa}$. The new pressure is $P' = 5 \times 10^5 \text{ Pa}$. The percentage increase is:
$ \text{% increase} = \frac{P' - P}{P} \times 100
= \frac{5 \times 10^5 - 3 \times 10^5}{3 \times 10^5} \times 100
= \frac{2 \times 10^5}{3 \times 10^5} \times 100
= \frac{200}{3}\%. $
Answer
The percentage increase in the pressure acting on the submarine is $ \frac{200}{3}\%.$
