A submarine experiences a pressure of $5.05 \times 10^6$ Pa at a depth of $\mathrm{d}_1$ in a sea. When it goes further to a depth of $\mathrm{d}_2$, it experiences a pressure of $8.08 \times 10^6 \mathrm{~Pa}$. Then $\mathrm{d}_2-\mathrm{d}_1$ is approximately (density of water $=10^3 \mathrm{~kg} / \mathrm{m}^3$ and acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ ):
Select the correct option:
A
600 m
B
500 m
C
300 m
D
400 m
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \Delta P=P_2-P_1=\rho g \Delta H \\ & 3.03 \times 10^6=10^3 \times 10 \times \Delta H \\ & \Rightarrow \Delta H \simeq 300 \mathrm{~m}\end{aligned}$
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