A metallic body of material with density of 8 | vedclass

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(C) Let $v_2$ be the volume of the material and $v_1$ be the volume of the cavity.Given density of material $\rho = 8000\ kg/m^3$ and density of water

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(C) Let $v_2$ be the volume of the material and $v_1$ be the volume of the cavity.Given density of material $ho = 8000\ kg/m^3$ and density of water $ho_w = 1000\ kg/m^3$.Mass in air: $M = ho v_2 = 10.0\ kg$.Apparent mass in water: $M' = M - 	ext{buoyant force} = ho v_2 - ho_w(v_1 + v_2) = 7.5\ kg$.Substituting $M = 10.0\ kg$ into the second equation:$10.0 - ho_w(v_1 + v_2) = 7.5 \implies ho_w(v_1 + v_2) = 2.5\ kg$.Now,take the ratio of the two equations:$\frac{ho_w(v_1 + v_2)}{ho v_2} = \frac{2.5}{10.0} = \frac{1}{4}$.Substitute the densities:$\frac{1000}{8000} \left( \frac{v_1}{v_2} + 1 ight) = \frac{1}{4}$.$\frac{1}{8} \left( \frac{v_1}{v_2} + 1 ight) = \frac{1}{4}$.$\frac{v_1}{v_2} + 1 = 2$.$\frac{v_1}{v_2} = 1$.

$A$ metallic body of material with density of $8000\ kg/m^3$ has a cavity inside. $A$ spring balance shows its mass to be $10.0\ kg$ in air and $7.5\ kg$ when immersed in water. The ratio of the volume of the cavity to the volume of the material of the body must be

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