Question

A spherical object has a density rho. If it is compressed under high pressure to one third of its original diameter, its density will now be A spherical object has a density . If it is compressed under high pressure to one third of its original diameter, its density will now be:

a. rho/27.
b. rho/9.
c. 3rho.
d. 9rho.
e. 27rho.

Answer 1

e. 27 rho.

Step-by-step explanation:

The density of an object, assumed constant, is defined as the relationship between the mass and the volume.

If the object is compressed in such a way that the diameter is reduced to 1/3, this means, that the radius will be reduced in the same proportion.

Now, if the mass remains the same (the compression can't change it) , the volume (assumed to be a perfect sphere) will be reduced also:

V₀ = 4/3*π*r³

Vf= 4/3*π*(r/3)³ = 4/3*π*(r³/27) = V₀/27

As the volume is in in the denominator of the expression for density, this means that the new density will be equal to 27 times the original one:

ρ₀ = m/ V₀

ρ₁ = m/ V₁ = m/ (V₀/27) = 27* (m/V₀) = 27*ρ₀