2. A column of water 40 cm high supports a 30 cm column of an unknown liquid. What is the density of the liquid? (Ans, 1.33 * 10 ^ 3 * kg * m ^ - 3)

Sure, here's how you can solve this problem:

Step 1: Understand the problem The problem is asking for the density of an unknown liquid. We know that the pressure at the bottom of a liquid column is equal to the height of the liquid column times the density of the liquid times the acceleration due to gravity. This is given by the formula P = hρg, where P is the pressure, h is the height of the liquid column, ρ is the density of the liquid, and g is the acceleration due to gravity.

Step 2: Set up the equation We know that the pressure at the bottom of the water column is equal to the pressure at the bottom of the unknown liquid column. Therefore, we can set up the following equation:

h1ρ1g = h2ρ2g

where h1 is the height of the water column, ρ1 is the density of water, g is the acceleration due to gravity, h2 is the height of the unknown liquid column, and ρ2 is the density of the unknown liquid.

Step 3: Plug in the known values We know that h1 = 40 cm = 0.4 m, ρ1 = 1000 kg/m^3 (the density of water), g = 9.8 m/s^2 (the acceleration due to gravity), and h2 = 30 cm = 0.3 m. We can plug these values into the equation:

0.4 m * 1000 kg/m^3 * 9.8 m/s^2 = 0.3 m * ρ2 * 9.8 m/s^2

Step 4: Solve for ρ2 We can cancel out the g on both sides of the equation and then solve for ρ2:

0.4 m * 1000 kg/m^3 = 0.3 m * ρ2

ρ2 = (0.4 m * 1000 kg/m^3) / 0.3 m = 1333.33 kg/m^3

So, the density of the unknown liquid is approximately 1.33 * 10^3 kg/m^3.