Question

1. when an object of mass 200 g is submerged in the methanol, its apparent weight is 1.052 n. when submerged in benzene, its apparent weight is 0.951 n. if the density of methanol is 0.8 g/cm³, what is the density of benzene?

Explanation:

Step1: Calculate the real - weight of the object

The mass of the object $m = 200\ g=0.2\ kg$. The real - weight of the object $W = mg$, where $g = 9.8\ m/s^{2}$. So $W=0.2\times9.8 = 1.96\ N$.

Step2: Calculate the buoyant force in methanol

The buoyant force in methanol $F_{b1}$ is given by the formula $F_{b1}=W - W_{1}$, where $W_{1}=1.052\ N$. So $F_{b1}=1.96 - 1.052=0.908\ N$.

Step3: Calculate the volume of the object using the buoyant force in methanol

According to Archimedes' principle $F_{b1}=
ho_{1}gV$, where $
ho_{1}=0.8\ g/cm^{3}=800\ kg/m^{3}$. We can solve for the volume $V$ of the object: $V=\frac{F_{b1}}{
ho_{1}g}=\frac{0.908}{800\times9.8}\ m^{3}$.

Step4: Calculate the buoyant force in benzene

The buoyant force in benzene $F_{b2}$ is given by $F_{b2}=W - W_{2}$, where $W_{2}=0.951\ N$. So $F_{b2}=1.96 - 0.951 = 1.009\ N$.

Step5: Calculate the density of benzene

Since $F_{b2}=
ho_{2}gV$, and we know $V=\frac{F_{b1}}{
ho_{1}g}$, then $
ho_{2}=\frac{F_{b2}}{gV}=\frac{F_{b2}
ho_{1}}{F_{b1}}$. Substitute $F_{b1}=0.908\ N$, $F_{b2}=1.009\ N$ and $
ho_{1}=800\ kg/m^{3}$ into the formula: $
ho_{2}=\frac{1.009\times800}{0.908}=890\ kg/m^{3}=0.89\ g/cm^{3}$.

Answer:

$0.89\ g/cm^{3}$