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This is one of our many quizzes on Atmospheric pressure. Click on this to access more such competitive quizzes on Mechanics.

Atmospheric pressure Quiz 2

physics

 You decide to bake some macarons and begin to prepare the batter for it. You leave the batter to rest for some time when you notice that an air bubble is stuck at a depth of $3\,cm$ in the batter. If the absolute pressure experienced by this bubble is $101.28\,kPa$, determine the density of the batter.<br>$($Take atmospheric pressure as $101 kPa)$ 

 Arrive at an expression for the gauge pressure from the absolute pressure. Then equate the resultant expression with the theoretical definition of gauge pressure with depth to arrive at the density of the batter.  

  We know that the absolute pressure can be given in terms of atmospheric pressure and gauge pressure as:$ \\{{P}_{abs}}={{P}_{atm}}+{{P}_{gauge}}\\\Rightarrow {{P}_{gauge}}={{P}_{abs}}-{{P}_{atm}}\\ $ But we know that $P_{gauge}  = \rho_{batter} gh$ $ \\ $ Equating the above two expressions we get: $\\{{\rho }_{batter}}gh={{P}_{abs}}-{{P}_{atm}}\\\Rightarrow {{\rho }_{batter}}=\dfrac{{{P}_{abs}}-{{P}_{atm}}}{gh}\\ $ Plugging in the values from the question we get: $ \\\rho_{batter} = \dfrac{(101.28-101)\times 10^{3}}{9.8 \times 0.03}\\\therefore\rho_{batter} = 952.4\,kgm^{-3}$

 Find the global mean pressure exerted by the Earth’s atmosphere on the surface of the Earth if the total mass of the atmosphere is $5.2 \times 10^{18}\,kg$ and the radius of the earth is $6400\,km$.  

 Assume that the Earth is spherical and determine the surface area upon which the atmosphere exerts pressure. Then plug in the given values into the resultant expression and arrive at the average pressure.  

  Given that mass of the atmosphere $m_a = 5.2 \times 10^{18}kg$ and Earth radius $R = 6400km = 64 \times 10^{5}m$ $\\ $ Now, approximating the Earth to be a sphere, the pressure exerted by the atmosphere on the surface of the Earth can be given as: $\\P = \dfrac{F}{A}\\\Rightarrow P= \dfrac{m_a g}{4\pi R^2}\\ $ Plugging in the appropriate values we get: $ \\P=\dfrac{5.2\times {{10}^{18}}\times 9.8}{4\pi {{(64\times {{10}^{5}})}^{2}}}\\\therefore P=9.90\times {{10}^{8}}\,Pa$

 Suppose that you are on a motorcycle enjoying a ride with your friends when suddenly your motorcycle runs out of fuel. You decide to siphon just enough fuel from your friend’s motorcycle to ride up to the nearest fuel station. To do so, one end of a rubber tube is placed in the fuel tank of your friend’s motorcycle and the other end is fed into your motorcycle’s fuel tank such that your fuel tank is at a lower height than your friend’s. Determine the height at which your fuel tank is, if it experiences an absolute pressure of $102\,kPa$.  $($Take atmospheric pressure as 101 kPa and $rho_{fuel} = 750\,kgm^{-3})$

  Arrive at an expression for the gauge pressure from the absolute pressure. Then equate the resultant expression with the theoretical definition of gauge pressure to arrive at the suitable distance between the two fuel tanks. 

 We know that the absolute pressure can be given in terms of atmospheric pressure and gauge pressure as: $\\{{P}_{abs}}={{P}_{atm}}+{{P}_{gauge}}\\\Rightarrow {{P}_{gauge}}={{P}_{abs}}-{{P}_{atm}}\\ $ But we know that $P_{gauge}  = \rho_{fuel} gh$ $ \\ $ Equating the above two expressions we get: ${{\rho }_{fuel}}gh={{P}_{abs}}-{{P}_{atm}}\\\Rightarrow h=\dfrac{{{P}_{abs}}-{{P}_{atm}}}{{{\rho }_{fuel}}g}\\ $ Plugging in the values from the question we get: $\\h = \dfrac{(102-101)\times 10^3}{750 \times 9.8}\\\Rightarrow h= 1.36 \times 10^{-4} \times 10^3\,m\\\therefore h= 13.6\,cm$

 In an attempt to explain the difference in densities of various liquids you build a density tower as shown in the figure. Determine the absolute pressure acting on the bottom of the density tower if the atmospheric pressure acting on it is $101.325\,kPa$. $(\text{Take}\,\rho_{honey} = 1420\,kgm^{-3}\,\rho_{soap} = 1120\,kgm^{-3}\,\rho_{milk} = 1026\,kgm^{-3}\,\rho_{water} = 1000\,kgm^{-3}\,and\,\rho_{oil} = 960\,kgm^{-3}\,)$  $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/56805f71-0ef9-4472-9fe4-c860efbe80788172023198026677374.png"/>

  Begin by determining the gauge pressure exerted on the bottom of the tower by the differently-dense liquids constituting it. Then, account for the atmospheric pressure and find the absolute pressure.  

 The absolute pressure acting on the bottom of the density tower can be given as: $ \\P_{abs} = P_{atm} + P_{gauge}\\ $ We are given that $P_{atm} = 101.325\,kPa$ $ \\ $ The gauge pressure will be given as: $\\P_{gauge} = \rho_{honey}gh_{honey}+\rho_{soap}gh_{soap}+\rho_{milk}gh_{milk}+\rho_{water}gh_{water}+\rho_{oil}gh_{oil}\\\Rightarrow P_{gauge} = g\left((1420 \times 0.10) + (1120 \times 0.08) +(1026 \times 0.04) +(1000 \times 0.05) +(960 \times 0.07)\right)\\\Rightarrow P_{gauge} = 9.8\left(142 + 89.6 + 41.04 + 50 + 67.2 \right)\\\Rightarrow P_{gauge} = 389.84\,Pa\\ \Rightarrow P_{gauge} = 0.39\,kPa\\ $ Therefore, the absolute pressure will be: $\\P_{abs} = 101.325 + 0.39\\\therefore P_{abs} = 101.72\,kPa$

 The Burj Khalifa in Dubai is the tallest building in the world and is $828\,m$ high and is divided equally across 163 storeys. The Burj Khalifa also has an underwater aquarium that is located three storeys below the ground level extending into the sea. Determine the absolute pressure acting on the walls of this underwater aquarium. $($Take $1\,Pa = 9.87\times 10^{-6}\,atm$ and the density of seawater to be $1020\,kgm^{-3})$ 

 Begin by determining the depth at which the aquarium is located by finding the height of each storey. Find the gauge pressure at this depth and combine it with the atmospheric pressure to arrive at the absolute pressure.  

 We are given that the Burj Khalifa is $828 m$ high consisting of $163$ storeys. This means that the height of one storey would on an average be: $h = \dfrac{828}{163} = 5.1\,m$ $ \\ $ The underwater aquarium is said to be three storeys below the ground level, therefore, the depth at which it exists will be: $d = 3 \times 5.1 = 15.3\,m$ $ \\ $ Given that $\rho = 1020\,kgm^{-3}$ the gauge pressure exerted by the sea water on the aquarium at this depth will be: $\\ P_{gauge} = \rho g d\\\Rightarrow P_{gauge}  = 1020 \times 9.8 \times 15.3\\\Rightarrow P_{gauge} = 0.153 \times 10^{6}\,Pa\\\text{Converting this into atm, we get:}\\P_{gauge} = 0.153 \times 10^{6} \times 9.87 \times 10^{-6}\\\Rightarrow P_{gauge}= 1.51\,atm\\ $ Taking the atmospheric pressure to be $P_{atm} = 1\,atm\,$ the absolute pressure experienced by the aquarium can be given as: $\\P_{abs} = P_{atm} + P_{gauge}\\\Rightarrow P_{abs} = 1+1.51 \\\therefore P_{abs} = 2.51\,atm$