Answer to Question #148579 in Physics for Nevil

Question

A solid weighs 0.850 kgf in air and 0.575 kgf in water. Find the density of the solid in SI units. Take density of water as 1000 kgm^–3

Answer 1

$mg-\rho_1\cdot g\cdot V=0.85\cdot9.8=8.33\to \rho gV-\rho_1\cdot g\cdot V=8.33$

$mg-\rho_2\cdot g\cdot V=0.575\cdot9.8=5.64\to \rho gV-\rho_1\cdot g\cdot V=5.64$

$\frac{}{}$ $\frac{\rho -\rho_1}{\rho -\rho_1}=\frac{8.33}{5.64}\to\rho=\frac{8.33\rho_2-5.64\rho_1}{8.33-5.64}=$

$=\frac{8.33\cdot1000-5.64\cdot1.29}{8.33-5.64}=3094(kg/m^3)$