Question

A body weighs 75 gm in air, 51 gm when completely immersed in an unknown liquid and 67 gm when completely immersed in water. Find the density of the unknown liquid:

• A. 4 gm/cm\(^3\)
• B. 6 gm/cm\(^3\)
• C. 8 gm/cm\(^3\)
• D. 3 gm/cm\(^3\)

Hint:

To find the density of an unknown liquid using Archimedes' principle, use the ratio of the buoyant forces to relate the densities of the liquids.

Answer 1

The Correct Option is B

The buoyant force equals the weight loss in a liquid. The weight loss in water represents the buoyant force in water, and the weight loss in the unknown liquid represents the buoyant force in that liquid. Let \( \rho_{\text{liquid}} \) be the density of the unknown liquid. We know: \[\n\text{Loss of weight in water} = 75 - 67 = 8 \, \text{gm} \quad (\text{buoyant force in water})\n\] \[\n\text{Loss of weight in unknown liquid} = 75 - 51 = 24 \, \text{gm} \quad (\text{buoyant force in unknown liquid})\n\] Since the volume of the body is constant, Archimedes' principle allows us to state: \[\n\frac{\text{Loss of weight in unknown liquid}}{\text{Loss of weight in water}} = \frac{\rho_{\text{water}}}{\rho_{\text{liquid}}}\n\] Substituting values: \[\n\frac{24}{8} = \frac{1}{\rho_{\text{liquid}}}\n\] \[\n\rho_{\text{liquid}} = \frac{1}{3} = 6 \, \text{gm/cm}^3\n\] Therefore, the density of the unknown liquid is \( 6 \, \text{gm/cm}^3 \).