A cube 5.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through and perpendicular to one face, you find that the cube weighs 6.30 N. (a) What is the density of this metal? (b) What did the cube weigh before you drilled the hole in it?

Short Answer

Expert verified

The density of the metal is, ρ=5.9×103kg/m3

The weight of the cube before the hole was drilled was, W = 7.25 N

Step by step solution

01

Identification of given data

  • The side length of the cube is,a=5.0cm1m100cm=0.05m
  • The diameter of the drill is, D=2.0cm1m100cm=0.02m
  • The weight of the cube after the hole is drilled was, W = 6.3 N
02

Step 2: Significance of density.

The density of a substance is defined as the amount of mass contained inside a unit volume.

The density can be given as,

ρ=mV………………..(1)

Where m and V are the mass and volume of the substance.

03

Determination of density of the metal

The mass of the metal can be evaluated by,

m=Wg=6.3N9.8m/s2=0.64N/m/s2=0.64kg

The volume of the cube after the drill can be evaluated as,

V=VcubeVhole=(0.05m)3π(0.01m)2(0.05m)=1.09×104m3

Now, the density of metal can be evaluated using equation (1),

ρ=0.64kg1.09×104m3=5.9×103kg/m3

Thus, the density of the metal is 5.9×103kg/m3.

04

Determination of weight of the cube after drilling hole,

The mass of the cube can be evaluated by rearranging equation (1),

m=ρVcube......................(2)

Substitute the values in equation (2),

m=5.9×103kg/m3(0.05m)3m=0.74kg

The weight of the drill hole can be evaluated by,

W=mg=(0.74kg)×9.8m/s2=7.25kgm/s21N1kgm/s2=7.25N

Thus, the weight of the cube is 7.25N

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