What is the density of lead under a pressure of `2.0xx10^(8) N//m^(2) ` , if the bulk modulus of lead is ` 8.0xx10^(9) N//m^(2)` and initially the density of lead is `11.4 g//cm^(3)`?

Explanation

Here, `p=20,000 N//cm^(2)`
`= 20,000 xx 10^(4) N//m^(2) = 2 xx 10^(8) N//m^(2)`,
`B = 0.80 xx 10^(1) N/m^(2)`
As, `B=(pV)/(Delta V)`
or ` Delta V = (pV)/B = ( 2 xx 10^(8))/(0.80 xx 10^(10)) xx V = V/(40)`
`:. "Net volume" , V'=V=Delta V`
`=V-(V)/(40) = (39 V)/(40)`
Let `rho'` be the new density of lead, under the effect of pressure applied. As mass of the lead will remain the same , so,
`V rho = V'rho'`
or ` rho' = (V rho)/(V') = (V xx 11.4)/(39V//40)= (11.4 xx 40)/(39)`
`= 11.7 g//cm^(3)`.