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 loss in weight in water is the buoyant force due to the displacement of water, and the loss in weight in the unknown liquid is the buoyant force due to the displacement of the unknown liquid. Let \( \rho_{\text{liquid}} \) be the density of the unknown liquid. We know: \[ \text{Loss of weight in water} = 75 - 67 = 8 \, \text{gm} \quad (\text{buoyant force in water}) \] \[ \text{Loss of weight in unknown liquid} = 75 - 51 = 24 \, \text{gm} \quad (\text{buoyant force in unknown liquid}) \] The volume of the body is the same in both liquids, so using Archimedes' principle: \[ \frac{\text{Loss of weight in unknown liquid}}{\text{Loss of weight in water}} = \frac{\rho_{\text{water}}}{\rho_{\text{liquid}}} \] Substitute the values: \[ \frac{24}{8} = \frac{1}{\rho_{\text{liquid}}} \] \[ \rho_{\text{liquid}} = \frac{1}{3} = 6 \, \text{gm/cm}^3 \] Thus, the density of the unknown liquid is \( 6 \, \text{gm/cm}^3 \).