Answer:
The resulting decrease in the volume is [tex]dV= 4.41*10^{-9} m^3[/tex]
Explanation:
The volume of the sphere is mathematically represented as
[tex]V_i=\frac{4}{3} \pi r^2[/tex]
Substituting 1.2cm = [tex]\frac{1.2}{100} = 0.012m[/tex] for radius
[tex]V_i = \frac{4}{3} * 3.142 * (0.012)^3[/tex]
[tex]= 7.23*10^{-6} m^3[/tex]
Let denote the change in volume as [tex]dV[/tex]
Bulk Modulus is mathematically is mathematically represented as
[tex]\beta = \frac{V_S}{V_x}[/tex]
Where [tex]V_S[/tex] is the volumetric stress and it is mathematically evaluated as
[tex]V_S = P* V[/tex]
substituting [tex]2.5*10^7Pa[/tex] for P and [tex]7.23*10^{-6} m^3[/tex] for V
[tex]V_S = 2.5*10^7 * 7.23*10^{-6}[/tex]
[tex]180.75 Pa \cdot m^3[/tex]
And [tex]V_x[/tex] is the volumetric strain and it is mathematically evaluated as
[tex]V_x = \frac{dV}{V}[/tex]
So substituting this for [tex]V_x[/tex] into the Bulk Modulus equation and substituting values
[tex]4.1 *0^10 = \frac{180.75}{\frac{dV}{7.23*10^{-6}} }[/tex]
Making dV the subject
[tex]dV = \frac{180.75}{4.1*10^10}[/tex]
[tex]dV= 4.41*10^{-9} m^3[/tex]
