Question

A submarine experiences a pressure of $5.05\times {10}^6\text{Pa}$ at depth of $d_1$ in a sea. When it goes further to a depth of $d_2$, it experiences a pressure of $8.08\times {10}^6\text{Pa}$. Then $d_2-d_1$ is approximately

(density of water $={10}^3{\text{kg/m}}^3$ and acceleration due to gravity $=10{\text{ms}}^{-2})$

• A. $600\text{m}$
• B. $400\text{m}$
• C. $300\text{m}$
• D. $500\text{m}$

Answer 1

The correct option is C $300\text{m}$

Given:
Pressure at depth of $\left(d_1\right),P_1=5.05\times {10}^6\text{Pa}$
Pressure at depth of $\left(d_2\right),P_2=8.08\times {10}^6\text{Pa}$.
Density of water $={10}^3{\text{kg/m}}^3$

Now, absolute pressure at some depth in sea,
$P_1=P_0+\rho g{d}_1$
$P_2=P_0+\rho g{d}_2$
Pressure difference,

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

$\Rightarrow 3.03\times {10}^6={10}^3\times 10\times \Delta d$

$\Rightarrow \Delta d\simeq 300\text{m}$

Hence option $\left(A\right)$ is correct.