Question

Se sumerge en agua un cubo de un material que tiene una densidad de 0,2Kg/lt, y una

Arista de 9cm. Calcula: La parte del cubo que flota y la parte que se hunde.

a. 18% y 21%

b. 20% y 25%

c. 20% y 26.8%

d. 25% y 31%

Answer 1

The percentage of the cube that floats = 80%

The percentage of the cube that sinks =20%

Step-by-step explanation:

Given that the density of the cube material,
`\rho= 0.2 kg/l`

Length of the sides of the cube,
`a= 9 cm`

So, the volume of the cube,
`v=a^3=(9cm)^3=729 cm^3`

As
`1000cm^3= 1 l,`

so the volume of the cube
`v= 729/1000 l = 0.729 l.`

Hence, the mass of the cube,
`m=\fho v = 0.2 * 0.729 = 0.1458 kg`.

The gravitational force acting on the cube in the downward direction,

`F_c= mg` [g is the acceleration due to gravity]

`\Rightarrow F_g=0.1458g` Newtons ...(i)

At equilibrium condition, let x cm of the cube sinks in the water.

So, the volume of the water replaces,

`v_w=9* 9* x = 81x cm^3.`

Or,
`v_w= 81 x/ 1000 l = 0.081x l`

The density of the water,
`\rho_w= 1 kg/l`

So, the mass of water replaced,

`m_w= \rho_w v_w = 1* 0.081x= 0.081x kg.`

Now, the buoyancy force acting on the cube in the upward direction,

`F_b = m_w * g` [ g is the acceleration due to gravity]

`\Rightarrow F_b= 0.081x* g` Newtons ...(ii)

At equilibrium condition, the gravitational force acting on the cube in the downward direction is equal to the buoyancy force acting on the cube in the upward direction, so by from equations (i) and (ii), we have,

`F_g= F_b\\\\\Rightarrow 0.1458g = 0.081x* g \\\\\Rightarrow 0.081x=0.1458 \\\\\Rightarrow x= 0.1458/0.081 = 1.8 cm`.

So, the part of the cube that sinks = 1.8 cm

As the remaining portion of the cube is floating, so the part of the cube that floats
`= 9-1.8= 7.2 cm`.

The percentage of the cube that floats

=
`\frac {7.2}{9}* 100 = 80\%`

The percentage of the cube that sinks

`= \frac {1.8}{9}* 100 = 20\%`

Please note that none of the given options are correct.