QJEEM19Phb3a6: JEE Main 2019 Physics - Class 11 JEEM19Phb3a6 Solution: A submarine experiences a pressure of 5.05 × 10 6 Pa at a depth …

The pressure experienced by a submarine at a certain depth in the sea is given by the formula:

$P = \rho g h$

where:

• $P$ is the pressure
• $\rho$ is the density of the fluid (sea water in this case)
• $g$ is the acceleration due to gravity
• $h$ is the height (or depth in this case)

Given:

• $\rho = 10^3 \, kg/m^3$
• $g = 10 \, m/s^2$

We are looking for the difference in depth, $d_2 - d_1$ , which corresponds to the difference in pressure $\Delta P$ :

$\Delta P = P_2 - P_1 = \rho g (d_2 - d_1)$

Rearranging the above equation, we get:

$d_2 - d_1 = \frac{\Delta P}{\rho g}$

Given:

• $P_2 = 8.08 \times 10^6 \, Pa$
• $P_1 = 5.05 \times 10^6 \, Pa$

$\Delta P = P_2 - P_1 = 8.08 \times 10^6 \, Pa - 5.05 \times 10^6 \, Pa = 3.03 \times 10^6 \, Pa$

So,

$d_2 - d_1 = \frac{3.03 \times 10^6 \, Pa}{10^3 \, kg/m^3 \times 10 \, m/s^2} = 303 \, m$

The closest answer among the options provided is Option C, 300 m.