Question

One cubic meter (1.00 m3) of aluminum has a mass of 2.70 103 kg, and the same volume of iron has a mass of 7.86 103 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 1.82 cm on an equal-arm balance.

Answer 1

1.28 cm

Step-by-step explanation:

Density (ρ) is an intensive property resulting from the quotient of mass (m) and volume (V).

The density of Al is ρAl = 2.70 × 10³ kg/m³

The density of Fe is ρFe = 7.86 × 10³ kg/m³

A Fe sphere of radius 1.82 cm has the following volume.

V = 4/3 × π × r³

V = 4/3 × π × (0.0182 cm)³

V = 2.53 × 10⁻⁵ m³

The mass corresponding to this sphere is:

2.53 × 10⁻⁵ m³ × 2.70 × 10³ kg/m³ = 0.0683 kg

The Al sphere has the same mass, so its volume is:

0.0683 kg × (1 m³/7.86 × 10³ kg) = 8.69 × 10⁻⁶ m³

The radius corresponding to an Al sphere of 8.69 × 10⁻⁶ m³ is:

V = 4/3 × π × r³

8.69 × 10⁻⁶ m³ = 4/3 × π × r³

r = 0.0128 m = 1.28 cm

Answer 2

`r_(al)=2.598cm`

Step-by-step explanation:

Density =mass/volume

`D_(al)=2.70103kg/m^(3)`

`D_(iron)=7.86103/m^(3)`

condition for balance:

`M_(iron)=M_(al)`

M=D*Volum

then:

`D_(iron)*4/3*pi*r_(ir)^(3)=D_(al)*4/3*pi*r_(al)^(3)`

`r_(al)=r_(ir)*\sqrt[3]{(D_(iron))/(D_(al))}`

`r_(al)=2.598cm`