Derive the formula P = ρhg and explain why the height of the denser liquid in a U-tube is less than the…

Question

Derive the formula P = ρhg and explain why the height of the denser liquid in a U-tube is less than the height of the less dense liquid.

ScanSolve AI Tutor · Accepted Answer

Here are the solutions for questions 3b and 5a:

Pressure is defined as force per unit area.
$$ P = \frac{F}{A} $$
The force $F$ exerted by the liquid column is its weight, which is $F = mg$ .
$$ P = \frac{mg}{A} $$
The mass $m$ of the liquid can be expressed using its density $\rho$ and volume $V$ , so $m = \rho V$ .
$$ P = \frac{\rho V g}{A} $$
The volume $V$ of a liquid column with cross-sectional area $A$ and height $h$ is $V = Ah$ .
$$ P = \frac{\rho (Ah) g}{A} $$
Cancel out the area $A$ from the numerator and denominator.
$$ P = \rho h g $$
5. a) Explain why height h is less than 20 cm: At the same horizontal level in the U-tube (the interface between liquids X and Y), the pressure exerted by both liquid columns must be equal. The pressure exerted by a liquid column is given by $P = \rho g h$ . Since liquid X has a density of $1000 \text{ kg/m}^3$ and liquid Y has a density of $800 \text{ kg/m}^3$ , liquid X is denser than liquid Y. For the pressures to balance ( $P_X = P_Y$ ), the height of the denser liquid X ( $h$ ) must be less than the height of the less dense liquid Y ( $20 \text{ cm}$ ).

Derive the formula P = ρhg and explain why the height of the denser liquid in a U-tube is less than the height of the less dense liquid.