Question

The pressure on the surface of a sphere of radius 1.2 cm is increased by 2.5×10⁷ Pa.

Calculate the resulting decrease in the volume of the sphere if it is made of lead. Bulk modulus for lead is 4.1×10¹⁰ Pa.

Answer 1

The resulting decrease in the volume is
`dV= 4.41*10^(-9) m^3`

Step-by-step explanation:

The volume of the sphere is mathematically represented as

`V_i=(4)/(3) \pi r^2`

Substituting 1.2cm =
`(1.2)/(100) = 0.012m` for radius

`V_i = (4)/(3) * 3.142 * (0.012)^3`

`= 7.23*10^(-6) m^3`

Let denote the change in volume as
`dV`

Bulk Modulus is mathematically is mathematically represented as

`\beta = (V_S)/(V_x)`

Where
`V_S` is the volumetric stress and it is mathematically evaluated as

`V_S = P* V`

substituting
`2.5*10^7Pa` for P and
`7.23*10^(-6) m^3` for V

`V_S = 2.5*10^7 * 7.23*10^(-6)`

`180.75 Pa \cdot m^3`

And
`V_x` is the volumetric strain and it is mathematically evaluated as

`V_x = (dV)/(V)`

So substituting this for
`V_x` into the Bulk Modulus equation and substituting values

`4.1 *0^10 = (180.75)/((dV)/(7.23*10^(-6)) )`

Making dV the subject

`dV = (180.75)/(4.1*10^10)`

`dV= 4.41*10^(-9) m^3`

Answer 2

Volume decreases by 4.4134x10^-9m3

Step-by-step explanation:

Detailed explanation and calculation is shown in the image below