___ $P a$ is the pressure at a depth of $9 m$ below the surface of the water tank. The density of water is $1000 \frac{k g}{m^{3}}$

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Solution

The expression of pressure at the bottom of a liquid of height ‘h’ and density ‘d’ is given by $ρ g h$ .
The given values are: $d = 1000 \frac{k g}{m^{3}}, h = 9 m a n d g = 10$ $\frac{m}{s^{2}}$ .
Thus, $p r e s s u r e = ρ g h$
$= 1000 \times 9 \times 10 = 90000 P a$ .

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