Answer:
The density is [tex]\rho = 1450 kg /m^3[/tex]
Explanation:
From the question we are told
The radius of the cylinder is [tex]r = 5.00 \ cm = \frac{5}{100} = 0.05 \ m[/tex]
The length of the cylinder is [tex]l = 20.0cm = \frac{20.0}{100} = 0.2 \ m[/tex]
The density of the glucose is [tex]\rho_g = 1385 \ kg/m^3[/tex]
The depth is [tex]d = 30.0 \ cm = \frac{30}{100} = 0.30 \ m[/tex]
The net force on the string is [tex]F_{net} = 1.00\ N[/tex]
The net force on the string is mathematically represented as
[tex]F_{net} = F_{wc} - F_{b}[/tex]
Here [tex]F_{wc}[/tex] is the force due to the weight of the cylinder
Which mathematically evaluated as
[tex]F_{wc} = \rho V_g *g[/tex]
Where [tex]V_g[/tex] is the volume of liquid displaced which is equal to the volume of the cylinder which is mathematically represented as
[tex]V_g = \pi r^2 l[/tex]
[tex]V_g =3.142 * (0.05^2 ) * (0.200)[/tex]
[tex]V_g = 0.00157 \ m^3[/tex]
=>
[tex]F_{wc} = \rho * 0.00157 * g[/tex]
[tex]F_b[/tex] is the buoyant force acting on the cylinder due to the glucose
[tex]F_b = \rho_g * V_g * g[/tex]
[tex]F_{b} = 1385 * 0.00157 * 9.8[/tex]
[tex]F_{b} =21.31 N[/tex]
So substituting into formula for [tex]F_{net}[/tex]
[tex]1 = \rho * (0.00157* 9.8) - 21.31[/tex]
[tex]\rho = \frac{22.31}{0.00157 *9.8}[/tex]
[tex]\rho = 1450 kg /m^3[/tex]
