___ $P a$ is the pressure at a depth of $10 m$ below the surface of the kerosene. The density of kerosene is $800 \frac{k g}{m^{3}}$ .

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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: $d = 800 \frac{k g}{m^{3}}$ , $h = 10 m a n d g = 10 \frac{m}{s^{2}}$ .
Thus, $p r e s s u r e = ρ g h$
$= 800$ $\times$ $10 \times 10 = 80000 P a$ .

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___ Pa is the pressure at a depth of 10m below the surface of the kerosene. The density of kerosene is 800kgm3.

The expression of pressure at the bottom of a liquid of height ‘h’ and density ‘d’ is given by ρgh. The given values are: d=800kgm3, h=10m and g=10ms2. Thus, pressure=ρgh =800 × 10×10=80000Pa .

___ $P a$ is the pressure at a depth of $10 m$ below the surface of the kerosene. The density of kerosene is $800 \frac{k g}{m^{3}}$ .

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: $d = 800 \frac{k g}{m^{3}}$ , $h = 10 m a n d g = 10 \frac{m}{s^{2}}$ .
Thus, $p r e s s u r e = ρ g h$
$= 800$ $\times$ $10 \times 10 = 80000 P a$ .