5. A small load is suspended in the air by a spring. When this load is fully submerged in kerosene on the same spring, the magnitude of the spring's deformation remains the same. Determine the density of the material of the load. The density of kerosene $\rho=800 \mathrm{k} / \mathrm{m}^{3} .(15$ points $)$
Answer: $400 \kappa g / \mathrm{M}^{3}$

Solution. In the air, the spring is stretched: $k x=m g$ (4 points). In kerosene, the spring is compressed: $m g+k x=F_{A}=\rho_{\text {kero }} g V$ (4 points). Since the mass of the load $m=\rho_{2} V$ (3 points), we get: $\rho_{2}=\frac{\rho_{\kappa \kappa}}{2}=\frac{800}{2}=400$ kg/m ${ }^{3}$ (4 points).