What is the radius of the spherical cavity?

[ramalik]

An irregularly shaped chunk of concrete has a hollow spherical cavity inside. The mass of the chunk is 38 kg, and the volume enclosed by the outside surface of the chunk is 0.025 m3. What is the radius of the spherical cavity?

I think you have to find the surface area to solve for the radius and all this has to do with the density of the sphere, but I don't know how to connect it all

 

[Doc Al]

No need to mess around with surface areas, but you do need to use the density of concrete. (Look it up!) Hint: If the chunk were solid concrete, what would be its mass?

 

[Moetasim]

together

To find the radius of the spherical cavity, we can use the formula for volume of a sphere, which is V = (4/3)πr^3. We know that the volume enclosed by the outside surface of the chunk is 0.025 m3, so we can set up the equation as 0.025 = (4/3)πr^3. Solving for r, we get r = (3/4π)^(1/3) * 0.025^(1/3) = 0.059 m.

To determine the surface area of the sphere, we can use the formula A = 4πr^2. Plugging in the radius we just calculated, we get A = 4π * (0.059)^2 = 0.044 m2.

The mass of the chunk is given as 38 kg, so we can use the formula for density, which is ρ = m/V, to find the density of the sphere. Plugging in the values, we get ρ = 38/0.025 = 1520 kg/m3.

We can now use the formula for surface area of a sphere to find the surface area of the hollow sphere, which is A = 4π(R^2 - r^2), where R is the radius of the outside surface and r is the radius of the cavity. We know that the surface area of the chunk is 0.044 m2, so we can set up the equation as 0.044 = 4π(R^2 - (0.059)^2). Solving for R, we get R = (0.044/4π + (0.059)^2)^(1/2) = 0.133 m.

Therefore, the radius of the spherical cavity is 0.059 m and the radius of the outside surface is 0.133 m. The density of the sphere is 1520 kg/m3.