Answer to Question #125370 in Optics for Abc

Question

submarine is 20m below the surface of the sea.The pressure due to the water at the depth is P.On anotherday, the submarine is 26m below the surface of the fresh water.The density of sea is 1.3times the density of fresh water.What is the pressure due to the fresh water at a depth of 26m?

Answer 1

In general $p=\rho\cdot g\cdot h$

The pressure due to the fresh water $p_0=\rho \cdot g\cdot h_0=\rho \cdot g \cdot 26 \to \rho \cdot g=\frac{p_0}{26}$

The pressure due to the sea water $p=1.3\rho \cdot g\cdot h=1.3\rho \cdot g \cdot 20 \to \rho \cdot g=\frac{p}{1.3\cdot26}$

We have $\frac{p_0}{26}=\frac{p}{1.3\cdot26} \to p_0=p$

So, the pressure will be the same.