ADDITIONAL PROBLEMS
UNIT-04
SOLID GEOMETRY
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Prob.1 A cube of uniform mass density ρ, edge length L= 15.0cm has a mass m = 4.0kg.
[a] What is the total surface area of the cube in units of m2?
Surface area S = 6*(15.0)2 = 1350.0cm2 = 0.135 m2.
[b] What is the volume of the cube in units of m3?
V = (15.0)3 =3375cm3 = 3.375*10-3 m3.
[c] Calculate ρ in units of kg/m3 and gm/cm3.
ρ = m/V = (4.0/3.375*10-3) = 1185 kg/m3 = 4.0*103/3375=1.185 gm/cm3.
[c] What would be the mass of a cube of the same mass density as in [c] and edge length
equal to 1.5m?
Since the mass scales as the volume ( m= ρ V), and the volume of the bigger cube =
(1.5/0.15)3 =103 times the volume of the smaller cube, the mass of the bigger cube would
be 4,000.0 kg.
Prob.-2. An ice cube of edge length a = 3.0cm has a spherical air bubble of radius, r trapped
inside it. The mass of the ice cube is measured to be 26.0 gm. Determine, the radius of the
trapped air bubble. Use the mass density of ice, ρice = 103 kg/m3 and ρair = 0.
Solution:
If there were no air bubble, the mass of the ice cube would
have been equal to:
ρV =(103)(3.0x10-2)3 = 27.0*10-3 kg
Therefore the missing mass, m = (27.0 -26.0)*10-3
= 1.0*10-3kg
(This missing mass is just the mass of the missing sphere of
ice of radius r.)
ρ4π r3/3 = 1.0*10-3 kg
Prob.3 Consider a right cylindrical shell as shown in the diagram.
[a] Calculate the total surface area of the shell in units of m2.
[b] If the mass of the shell is 1.9 kg, determine the mass density (in kg/m3) of the shell’s
material.
Solution:
[a] Stot= 2π(62 – 42)+ 2π6*15 +2π4*15 = 1067.6 cm2 = 0.107 m2
[b] ρ = 1.9/V = 1.9/[ π(62 – 42)15*10-6] = 2*103kg/m3.
Prob.4 Calculate the mass of a sphere of radius, R = 18.0 cm
whose core of radius, r = 15.0 cm is filled with lead and the shell
is made out of copper. The densities of copper and lead are,
respectively, 9.0x103kg/m3 and 11.5x103kg/m3.
Solution:
M = MPb+MCu = ρPbVPb +ρCuVCu
= 11.5x103*(4π/3)(15*10-2)3 +9.0x103*(4π/3)[ (18*10-2)3 - (15*10-2)3]
= 162.5 + 92.6 = 255.1kg
Prob.5 Consider an insulated cylindrical
cable (copper conducting core + sheath) 1.0km
long and of mass 1.7x104kg. The copper
conductor is 4.0cm in diameter and surrounded by
an insulating polymer (plastic) sheathing 2cm
thick and of mass density ρ = 1500 kg/m3.
[a] Calculate the mass Mp of the plastic sheathing. Mp= ρVp= 1500π(42-­22)10-­4*103 = 5654 kg. [b] Determine ρc the mass density of the copper conductor. ρc = Mc/Vc= (17000 – 5654)/ π 4*10-­4*103 = 9.0*103kg/m3.