b. When placed into another liquid \( 14.0 \% \) of its volume is above the surface. What is the density of the liquid? Answer. Density of liquid = \( \mathrm{kg} / \mathrm{h} \).

To find the density of the liquid in which the object is placed, we can use the principle of buoyancy. According to Archimedes' principle, the weight of the liquid displaced by the object is equal to the weight of the object. Let: - \( V \) = total volume of the object - \( V_{\text{displaced}} \) = volume of the object submerged in the liquid - \( \rho_{\text{object}} \) = density of the object - \( \rho_{\text{liquid}} \) = density of the liquid - \( g \) = acceleration due to gravity (which will cancel out) Given that \( 14.0\% \) of the object's volume is above the surface, this means that \( 86.0\% \) of the object's volume is submerged. Therefore, we can express the volume submerged as: \[ V_{\text{displaced}} = 0.86 V \] The weight of the object can be expressed as: \[ \text{Weight of object} = \rho_{\text{object}} \cdot V \cdot g \] The weight of the liquid displaced is: \[ \text{Weight of liquid displaced} = \rho_{\text{liquid}} \cdot V_{\text{displaced}} \cdot g = \rho_{\text{liquid}} \cdot (0.86 V) \cdot g \] Setting the two weights equal to each other (since the object is floating): \[ \rho_{\text{object}} \cdot V \cdot g = \rho_{\text{liquid}} \cdot (0.86 V) \cdot g \] We can cancel \( V \) and \( g \) from both sides: \[ \rho_{\text{object}} = \rho_{\text{liquid}} \cdot 0.86 \] Now, we can solve for the density of the liquid: \[ \rho_{\text{liquid}} = \frac{\rho_{\text{object}}}{0.86} \] To find the density of the liquid, we need the density of the object. If you provide the density of the object, we can calculate the density of the liquid. If the density of the object is not given, we cannot determine the density of the liquid. Please provide the density of the object to proceed.