g The bottom end of a long vertical tube filled with liquid is opened in a basin exposed to air having pressure 100.8 kilo-Pascals. The top end of the tube remains closed. The maximum height of liquid in the tube that this air pressure can support is 10.5 meters. Determine the density of the liquid in units of bevelled fraction numerator k g over denominator m cubed end fraction.

979.6 kg/m³

Explanation:

We know pressure P = hρg where h = height of liquid = 10.5 m, ρ = density of liquid and g = acceleration due to gravity = 9.8 m/s²

So, density ρ = P/hg

Since P = 100.8 kPa = 100.8 × 10³ Pa

substituting the values of the variables into the equation for ρ, we have

ρ = P/hg

= 100.8 × 10³ Pa ÷ (10.5 m × 9.8 m/s²)

= 100.8 × 10³ Pa ÷ 102.9 m²/s²

= 0.9796 × 10³ kg/m³

= 979.6 kg/m³

So, the density of the liquid is 979.6 kg/m³

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