An iceberg floats in seawater with 15% of its volume above the surface. If the specific weight of seawater is 10.5 kN/m^3, what is the specific weight of the iceberg (in kN/m^3)?

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Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context:Floating bodies obey Archimedes’ principle: the weight of displaced fluid equals the weight of the body. The fraction of volume submerged links the body’s specific weight to that of the fluid. This problem applies buoyancy to estimate the density (specific weight) of ice from its freeboard (portion above water). Given Data / Assumptions: Fraction above water = 15% ⇒ fraction submerged = 85% = 0.85. Specific weight of seawater, gamma_w = 10.5 kN/m^3. Hydrostatic equilibrium (no vertical acceleration). Concept / Approach:At floatation, weight of iceberg per unit volume equals weight of displaced seawater per unit volume times the submerged fraction. Therefore gamma_ice = submerged_fraction * gamma_w. The above-water percentage simply complements to 1 (or 100%). Step-by-Step Solution: Compute submerged fraction: 1 − 0.15 = 0.85.Apply equilibrium: gamma_ice = 0.85 * 10.5 kN/m^3.Calculate: gamma_ice = 8.925 kN/m^3 ≈ 8.93 kN/m^3. Verification / Alternative check: Compare with typical freshwater ice values (≈ 9.0 kN/m^3), consistent with the result given slightly denser seawater. Why Other Options Are Wrong: 12.52: exceeds seawater; object would sink.9.81: close to pure water density; would give smaller freeboard than 15% in seawater.7.83 or 6.30: too low given the observed freeboard. Common Pitfalls: Using 15% instead of 85% for the submerged fraction; confusing specific weight with specific gravity. Final Answer: 8.93

An iceberg floats in seawater with 15% of its volume above the surface. If the specific weight of seawater is 10.5 kN/m^3, what is the specific weight of the iceberg (in kN/m^3)?

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