An open tank contains 60 cm of water covered with 30 cm of oil of specific gravity 0.8. The pressure intensity recorded at the bottom of tank is :
Try it first: pick an option, then read the explanation.
Asked in SSC JE 23-09-2019 (Morning)
Detailed Explanation
In a multi-layered static fluid container, the total pressure at the bottom is the cumulative sum of the hydrostatic pressures contributed by each individual liquid layer above it.
1. Identify layer properties:
- Top layer: Oil of specific gravity \(S_{\text{oil}} = 0.8 \;\implies\; \rho_{\text{oil}} = 800\text{ kg/m}^3\)
- Height of oil layer \(h_{\text{oil}} = 30\text{ cm} = 0.30\text{ m}\)
- Bottom layer: Water of \(\rho_{\text{water}} = 1000\text{ kg/m}^3\)
- Height of water layer \(h_{\text{water}} = 60\text{ cm} = 0.60\text{ m}\)
- Gravitational acceleration \(g = 9.81\text{ m/s}^2\)
2. Calculate pressure at the oil-water interface (\(p_1\)):
\[ p_1 = \rho_{\text{oil}} g h_{\text{oil}} = 800 \times 9.81 \times 0.30 = 2354.4\text{ N/m}^2 \]
3. Calculate additional pressure developed through the water layer (\(\Delta p_{\text{water}}\)):
\[ \Delta p_{\text{water}} = \rho_{\text{water}} g h_{\text{water}} = 1000 \times 9.81 \times 0.60 = 5886.0\text{ N/m}^2 \]
4. Total pressure intensity at the tank floor (\(p_{\text{bottom}}\)):
\[ p_{\text{bottom}} = p_1 + \Delta p_{\text{water}} = 2354.4 + 5886.0 = 8240.4\text{ N/m}^2 \]
Why “8240.4 N/m²” Is Correct
The top 30 cm oil layer exerts \(2354.4\text{ N/m}^2\) at the interface, and the bottom 60 cm water layer adds \(5886.0\text{ N/m}^2\). Summing these two hydrostatic contributions gives exactly \(8240.4\text{ N/m}^2\).
Why the Other Options Are Wrong
- A. 8829 N/m²: 8829 N/m² results from mistakenly assuming the entire 90 cm column consists of water: 1000 × 9.81 × 0.90 = 8829 N/m².
- B. 9430.6 N/m²: 9430.6 N/m² results from adding an erroneous extra 1200 N/m².
- C. 2354.4 N/m²: 2354.4 N/m² is only the pressure at the oil-water interface, completely omitting the water layer below it.
Key Point
Multi-layer bottom pressure: p_total = ∑ (ρ_i · g · h_i) = p_oil + p_water.
Key Concepts for Students
- Because oil is less dense than water (S = 0.8 < 1.0), it naturally floats stably on top.
- Interface pressure acts as a uniform surcharge on the liquid layer below it.
- The pressure gradient changes slope at the interface: steeper in the denser water layer.
Practice more: this is question 12 in Measurement of Pressure MCQs – Part 1. Try the whole set and check every answer instantly.