On a planet where gplanet = 0.2gearth. What will be the difference in ...
Introduction:
In this scenario, we are given that the acceleration due to gravity (g) on a particular planet is 0.2 times the acceleration due to gravity on Earth (gearth). We need to determine the difference in the height of a column filled with mercury in a closed-end manometer when the gas inside is filled with a pressure of 2 atm on Earth.
Explanation:
To solve this problem, we can use the equation for pressure in a manometer:
P1 + ρgh1 = P2 + ρgh2
where P1 and P2 are the pressures on either side of the manometer, ρ is the density of the mercury, g is the acceleration due to gravity, and h1 and h2 are the heights of the mercury columns on each side of the manometer.
Step 1: Calculate the pressure difference:
The pressure difference (ΔP) between the gas inside the manometer and the outside atmosphere is given by:
ΔP = Pgas - Poutside
Since we are given that the gas is filled with a pressure of 2 atm on Earth and the outside pressure is 1 atm on both planets, the pressure difference is:
ΔP = 2 atm - 1 atm = 1 atm
Step 2: Determine the density of mercury:
The density of mercury (ρ) is a constant value and can be obtained from reference tables. Let's assume the density of mercury is 13.6 g/cm³.
Step 3: Calculate the height difference:
Using the equation for pressure in the manometer, we can rearrange it to solve for the height difference (Δh):
Δh = (Pgas - Poutside) / (ρg)
Substituting the known values, we have:
Δh = (1 atm) / (13.6 g/cm³ * 9.8 m/s² * 100 cm/m)
Simplifying the units, we get:
Δh = (1 * 101325 Pa) / (13.6 * 9.8 * 1000 kg/m³ * 9.8 m/s²)
Calculating the numerical value, we find:
Δh ≈ 0.075 m
Conclusion:
The difference in the height of the column filled with mercury in the closed-end manometer when the gas is filled with a pressure of 2 atm on Earth is approximately 0.075 meters.
On a planet where gplanet = 0.2gearth. What will be the difference in ...
380 cm
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