Calculate the pressure due to a column of 0.3 m of (i) water, (ii) an oil of specificgravity 0.8 and (c) mercury of specific gravity 13.6. take density of water = 1000kg/m3. (6 marks)
Question
Calculate the pressure due to a column of 0.3 m of (i) water, (ii) an oil of specificgravity 0.8 and (c) mercury of specific gravity 13.6. take density of water = 1000kg/m3. (6 marks)
Solution
Sure, let's calculate the pressure due to a column of different liquids step by step.
(i) For water:
We know that pressure (P) = height (h) * density (ρ) * gravity (g)
Given, h = 0.3 m, ρ = 1000 kg/m^3 (for water), and g = 9.81 m/s^2 (approximate value of gravity on Earth's surface)
So, P = 0.3 m * 1000 kg/m^3 * 9.81 m/s^2 = 2943 Pa (Pascal)
(ii) For oil:
Given, the specific gravity of oil = 0.8. This means the density of oil is 0.8 times the density of water.
So, ρ = 0.8 * 1000 kg/m^3 = 800 kg/m^3
Using the same formula as above, P = 0.3 m * 800 kg/m^3 * 9.81 m/s^2 = 2355 Pa
(iii) For mercury:
Given, the specific gravity of mercury = 13.6. This means the density of mercury is 13.6 times the density of water.
So, ρ = 13.6 * 1000 kg/m^3 = 13600 kg/m^3
Using the same formula as above, P = 0.3 m * 13600 kg/m^3 * 9.81 m/s^2 = 39972 Pa
So, the pressure due to a column of 0.3 m of water is 2943 Pa, of oil is 2355 Pa, and of mercury is 39972 Pa.
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