Question

A cylinder with radius 5.00 cm and length 20.0 cm is lowered into a tank of glucose, which has a density of 1,385 kg/m^3, the cylinder is lowered in four stages. After being lowered to a depth of 30.0 cm, the string holding the cylinder is cut. If the net force on the cylinder after the string is cut is 1.00 N, what is the density of the cylinder material?

Answer 1

The density is
`\rho = 1450 kg /m^3`

Step-by-step explanation:

From the question we are told

The radius of the cylinder is
`r = 5.00 \ cm = (5)/(100) = 0.05 \ m`

The length of the cylinder is
`l = 20.0cm = (20.0)/(100) = 0.2 \ m`

The density of the glucose is
`\rho_g = 1385 \ kg/m^3`

The depth is
`d = 30.0 \ cm = (30)/(100) = 0.30 \ m`

The net force on the string is
`F_(net) = 1.00\ N`

The net force on the string is mathematically represented as

`F_(net) = F_(wc) - F_(b)`

Here
`F_(wc)` is the force due to the weight of the cylinder

Which mathematically evaluated as

`F_(wc) = \rho V_g *g`

Where
`V_g` is the volume of liquid displaced which is equal to the volume of the cylinder which is mathematically represented as

`V_g = \pi r^2 l`

`V_g =3.142 * (0.05^2 ) * (0.200)`

`V_g = 0.00157 \ m^3`

=>

`F_(wc) = \rho * 0.00157 * g`

`F_b` is the buoyant force acting on the cylinder due to the glucose

`F_b = \rho_g * V_g * g`

`F_(b) = 1385 * 0.00157 * 9.8`

`F_(b) =21.31 N`

So substituting into formula for
`F_(net)`

`1 = \rho * (0.00157* 9.8) - 21.31`

`\rho = (22.31)/(0.00157 *9.8)`

`\rho = 1450 kg /m^3`