A steel ball initially at a pressure of `10^5Pa` is heated from `20^@`C to `120^@`C keeping its volume constant. Find the final pressure inside the ball. Given that coefficient of linear expansion of steel is `1.1xx10^(-5)//^@C` and Bulk modulus of steel is `1.6xx10^(11)Nt//m^(2)`.

Explanation

On increasing temperature of ball by `100^(@)C` (from `20^(@)C` to `120^(@)C`), the thermal expansion in its volume can be given as
`DeltaV=gamma_(st)V DeltaT`
`=3alpha_(st)V DeltaT`…..1.56
Here it is given that no change is volume is allowed. This implies that the volume increment by thermal expansion is compressed elastically by external pressure. Thus elastic compression in the sphere must be equal to that given in equation(1.56). Bulk modulus of a material is defined as
`B=("increase in pressure ")/("volume strain")=(DeltaP)/(DeltAV//V)`
Here the external applied pressure to keep the volume of ball constant is given as
`DeltaP=Bxx(DeltaV)/V=B(3alpha_(st)DeltaT)`
`=1.6xx10^(11)xx3xx1.1xx10^(-5)xx100`
`=5.28xx10^(8)N//m^(2)`
`=5.28xx10^(8)Pa`
Thus this must be the excess pressure inside the ball at `120^(@)C` to keep its volume constant during heating.