Question

Pressure at a certain depth in river water is $p_1$ and at the same depth in sea water is $p_2$. Then (Density of sea water is greater than that of river water):

• A. $p_1=p_2$
• B. $p_1>p_2$
• C. $p_1<p_2$
• D. $p_1-p_2=$ atmospheric pressure

Answer 1

The correct option is C $p_1<p_2$

The expression of the pressure at a point in fluid is given by $P=h\rho g$

where $h=$ depth of the point from the surface, $\rho =$ be the density of the fluid and $g=$ acceleration due to gravity.

Here, h and g are constant in both cases so $p_1=h{\rho }_{r}g$ and $p_2=h{\rho }_{s}g$

or $\frac{p_1}{p_2}=\frac{{\rho }_r}{{\rho }_s}$

As ${\rho }_s>{\rho }_r$ so $\frac{p_1}{p_2}<1$ or $p_1<p_2$