Question:
A container is filled to a height of 20 cm with water. A 30 cm thick layer of oil with specific gravity 0.8 floats on the top of water. If the density of water is 1000 kg/m\(^3\) and atmospheric pressure is \(1 \times 10^5\) Pa, then the total pressure at the bottom of the container is:
[Acceleration due to gravity = 10 m/s\(^2\)]
A container is filled to a height of 20 cm with water. A 30 cm thick layer of oil with specific gravity 0.8 floats on the top of water. If the density of water is 1000 kg/m\(^3\) and atmospheric pressure is \(1 \times 10^5\) Pa, then the total pressure at the bottom of the container is:
[Acceleration due to gravity = 10 m/s\(^2\)]
[Acceleration due to gravity = 10 m/s\(^2\)]
Show Hint
Total pressure at a depth in a multi-layer fluid is the sum of atmospheric pressure and hydrostatic pressures of each layer: \(P = P_\text{atm} + \sum_i \rho_i g h_i\).
Updated On: Jun 19, 2026
- \(1.044 \times 10^5\) Pa
- \(1.24 \times 10^5\) Pa
- \(1.062 \times 10^5\) Pa
- \(1.15 \times 10^5\) Pa
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the problem.
We have two fluid layers: water at the bottom (\(h_w = 0.2~\text{m}\)) and oil on top (\(h_o = 0.3~\text{m}\)), with oil specific gravity \(SG = 0.8 \Rightarrow \rho_o = 0.8 \times 1000 = 800~\text{kg/m}^3\). Atmospheric pressure \(P_\text{atm} = 1 \times 10^5~\text{Pa}\).
Step 2: Pressure due to water.
\[ P_w = \rho_w g h_w = 1000 \cdot 10 \cdot 0.2 = 2000~\text{Pa} \]
Step 3: Pressure due to oil.
\[ P_o = \rho_o g h_o = 800 \cdot 10 \cdot 0.3 = 2400~\text{Pa} \]
Step 4: Total pressure at bottom.
\[ P_\text{total} = P_\text{atm} + P_w + P_o = 1 \times 10^5 + 2000 + 2400 = 1.044 \times 10^5~\text{Pa} \]
Step 5: Conclusion.
Hence, the total pressure at the bottom of the container is \(1.044 \times 10^5\) Pa.
We have two fluid layers: water at the bottom (\(h_w = 0.2~\text{m}\)) and oil on top (\(h_o = 0.3~\text{m}\)), with oil specific gravity \(SG = 0.8 \Rightarrow \rho_o = 0.8 \times 1000 = 800~\text{kg/m}^3\). Atmospheric pressure \(P_\text{atm} = 1 \times 10^5~\text{Pa}\).
Step 2: Pressure due to water.
\[ P_w = \rho_w g h_w = 1000 \cdot 10 \cdot 0.2 = 2000~\text{Pa} \]
Step 3: Pressure due to oil.
\[ P_o = \rho_o g h_o = 800 \cdot 10 \cdot 0.3 = 2400~\text{Pa} \]
Step 4: Total pressure at bottom.
\[ P_\text{total} = P_\text{atm} + P_w + P_o = 1 \times 10^5 + 2000 + 2400 = 1.044 \times 10^5~\text{Pa} \]
Step 5: Conclusion.
Hence, the total pressure at the bottom of the container is \(1.044 \times 10^5\) Pa.
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