Respuesta :
Answer:
260 g
Explanation:
Given:
- r = radius of the spherical sand = 50 micron = [tex]5\times 10^{-5}\ m[/tex]
- a = edge length of the cube = 1.0 m
- \rho = density of sand = [tex]2600\ kg/m^3[/tex]
Let n be the number of such silicon dioxide that would give the desired mass of sand.
According to the question, the surface area of all the sand particles will be equal to the surface area of the cube.
[tex]\therefore n\times \textrm{Area of a sand particle}=\textrm{Surface area of a cube }\\\Rightarrow n\times 4\pi r^2= 6a^2\\\Rightarrow n = \dfrac{6a^2}{4\pi r^2}[/tex]
Let the total mass of sand required be M.
[tex]\textrm{Total mass of sand} = \textrm{Mass of all the required sand particle}\\\Rightarrow M = n\times \textrm{mass of one sand particle}\\\Rightarrow M = n\times \textrm{Density of sand particle}\times \textrm{Volume of a sand particle}\\\Rightarrow M = n\times\rho \times \dfrac{4}{3}\pi r^3\\[/tex]
[tex]\Rightarrow M = \dfrac{6a^2}{4\pi r^2}\times\rho \times \dfrac{4}{3}\pi r^3\\\Rightarrow M = 2\times\rho \times a^2r\\\Rightarrow M = 2\times2600 \times (1)^2\times 5\times 10^{-5}\\\Rightarrow M =0.260\ kg\\\Rightarrow M =260\ g\\[/tex]
Hence, the total mass of the sand required will be equal to the 260 g.
