A submarine experiences a pressure of 5.05 × 106 Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 × 106 Pa. Then d2 –d1 is approximately (density of water = 103 kg/m3 and acceleration due to gravity = 10 ms–2 ) :

Properties of Matter PYQ — JEE Main 2019 · 10 Apr · Shift 2 · Q44 with Solution

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Correct answer: C

For a liquid, absolute pressure at depth $d$ is

$P = P_0 + \rho g d$

where:

At depth $d_1$ :

$P_1 = P_0 + \rho g d_1 = 5.05 \times 10^6\,\text{Pa}$

At depth $d_2$ :

$P_2 = P_0 + \rho g d_2 = 8.08 \times 10^6\,\text{Pa}$

This eliminates atmospheric pressure:

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

So,

$d_2-d_1 = \frac{P_2-P_1}{\rho g}$

$P_2-P_1 = (8.08 - 5.05) \times 10^6 = 3.03 \times 10^6\,\text{Pa}$

and

$\rho g = 10^3 \times 10 = 10^4$

Hence,

$d_2-d_1 = \frac{3.03 \times 10^6}{10^4} = 3.03 \times 10^2 = 303\,\text{m}$

$303\,\text{m} \approx 300\,\text{m}$

So the correct option is C.