Water Resources·MCQ explained

A pressure gauge fitted on the side of a tank filled with liquid reads 50 kPa and 100 kPa at height of 5 m and 10 m. What is the approximate density of the liquid (in kg/m³)? (take g = 10)

Try it first: pick an option, then read the explanation.

Correct answer: B. 1000

Asked in SSC JE 25-09-2019 (Morning)

Detailed Explanation

According to the hydrostatic law, the rate of increase of pressure between two vertical points in a liquid column is directly proportional to the fluid's mass density and the vertical separation distance.

1. Identify the recorded gauge values:

  • Pressure at upper gauge: \(p_1 = 50\text{ kPa} = 50,000\text{ Pa}\)
  • Pressure at lower gauge: \(p_2 = 100\text{ kPa} = 100,000\text{ Pa}\)
  • Pressure difference: \(\Delta p = p_2 - p_1 = 100,000 - 50,000 = 50,000\text{ Pa}\)
  • Vertical distance between gauges: \(\Delta h = 10\text{ m} - 5\text{ m} = 5\text{ m}\)
  • Acceleration due to gravity: \(g = 10\text{ m/s}^2\)

2. Apply the hydrostatic pressure-depth relation:

\[ \Delta p = \rho \cdot g \cdot \Delta h \]

3. Solve for fluid density \(\rho\):

\[ \rho = \frac{\Delta p}{g \cdot \Delta h} = \frac{50,000\text{ Pa}}{10\text{ m/s}^2 \times 5\text{ m}} = \frac{50,000}{50} = 1000\text{ kg/m}^3 \]

Why “1000” Is Correct

The pressure increases by \(50\text{ kPa}\) over a vertical depth increase of \(5\text{ m}\). Dividing the pressure gradient of \(10\text{ kPa/m} = 10,000\text{ N/m}^3\) by \(g = 10\text{ m/s}^2\) yields a liquid density of exactly \(1000\text{ kg/m}^3\) (water).

Liquid Free SurfaceG₁P₁ = 50 kPa (5 m)G₂P₂ = 100 kPa (10 m)5 m10 mΔh = 10 - 5 = 5 mΔP = 100 - 50 = 50 kPaΔP = ρgΔh ⇒ 50,000 = ρ × 10 × 5 ⇒ ρ = 1000 kg/m³
Tank with pressure gauges at two elevations: Δp = 50 kPa over Δh = 5 m gives ρ = 1000 kg/m³.

Why the Other Options Are Wrong

  • A. 5000: 5000 kg/m³ results from omitting the 5 m distance and dividing 50,000 by 10.
  • C. 10: 10 kg/m³ results from failing to convert kPa to Pa (50 / 50 = 1).
  • D. 100: 100 kg/m³ results from a power-of-ten arithmetic error.

Key Point

Liquid density from two gauge elevations: ρ = Δp / (g · Δh).

Key Concepts for Students

  • In fluid hydrostatics, the vertical pressure gradient is uniform: dp/dh = ρg.
  • Two pressure gauges mounted at known elevation differences can function as an in-situ densitometer.
  • 1000 kg/m³ is the standard density of pure water at 4°C.

Practice more: this is question 14 in Measurement of Pressure MCQs – Part 1. Try the whole set and check every answer instantly.