Question

A spherical object has an outside diameter of 60.0cm . Its outer shell is composed of aluminum and is 2.80cm thick. The remainder is uniform plastic with a density of 720kg/m3 .

A) Determine the object's average density.

B) Will this object float by itself in fresh water?

Answer 1

given that ::

diameter of the sphere, d = 60 cm = 0.6 m

radius of the outer shell, r_outer = 2.8 cm = 0.028 m

(a)  the object's average density is given as :;

volume of the total sphere, V_t = 4/3 r³   { eq. 1 }

where, r = radius of the sphere = 0.3 m

inserting the values in above eq,

V_t = 4/3 (3.14) (0.3 m)³

V_t = 0.11304 m³

volume of the inside sphere, V_inside = 4/3 (r_inside)³     { eq. 2 }

where, r_inside = r - r_outside = (0.3 m - 0.028 m) = 0.2972 m

inserting the values in eq.2,

V_inside = 4/3 (3.14) (0.2972 m)³

V_inside = 0.1099 m³

total volume of the sphere, V_t = V_inside + V_outside  

or V_outside = V_t - V_inside   { eq. 3 }

inserting the values in eq.3,

  V_outside = (0.11304 m³ - 0.1099 m³)

V_outside = 0.00314 m³

we know that density of aluminium, _{al (ouside)} = 2700 kg/m³

density of sphere, _sphere = mass / volume   

_sphere = (_inside x  V_inside + _{al (ouside)} x  V_outside) / V_t     { eq. 4 }

where, _{pl (inside)} = 720 kg/m³

inserting the values in eq.4,

_sphere = [ (720 kg/m³) x (0.1099 m³) + (2700 kg/m³) x (0.00314 m³) ] / (0.11304 m³)

_sphere = (79.128 + 8.478) / 0.11304

_sphere = 154.12 kg/m³

(b)  this object will float by itself in fresh water which is given as ::

specific gravity = _sphere / _water   { eq. 5 }

where,  _water = 1000 kg/m³

inserting the values in above eq.

specific gravity = (154.12 kg/m³) / (1000 kg/m³)

specific gravity = 0.154

if the average denisty of the sphere is less than the density of water, then sphere will float.