Hydrostatics with specific gravity — an oil (specific gravity 0.8) has an atmospheric pressure on its surface of 0.1 kg/cm². Find the pressure at 2.5 m depth expressed as head of water.

Introduction / Context:
Converting pressure in mixed units into an equivalent water head is a common hydrostatics task. Specific gravity links the pressure due to a column of one fluid to an equivalent height of water that would exert the same pressure at the reference point.

Given Data / Assumptions:

• Specific gravity of oil S = 0.8 (rho_oil = 0.8 * rho_water).
• Depth in oil h_oil = 2.5 m.
• Atmospheric pressure at surface p_atm = 0.1 kg/cm².
• 1 kgf/cm² ≈ 10 m of water head (consistent engineering approximation).

Concept / Approach:

Total pressure head at depth equals atmospheric head plus the oil hydrostatic head converted to water head. Oil head, when expressed as equivalent water head, is S * h_oil because p = rho * g * h and rho_oil = S * rho_water.

Step-by-Step Solution:

Verification / Alternative check (if short method exists):

Direct pressure: p = p_atm + rho_oil * g * h_oil; divide by rho_water * g to convert to water head → gives 1 + 2.0 = 3.0 m, consistent.

Why Other Options Are Wrong:

1 m ignores hydrostatic head; 2 m ignores atmospheric contribution; 3.5 and 4 m overestimate the oil contribution.

Common Pitfalls (misconceptions, mistakes):

Forgetting to add atmospheric head when asked for absolute pressure head; using 1 kg/cm² as exactly 9.81 m instead of the standard 10 m approximation in such problems (difference is minor here).

Final Answer:

3 metres of water