what will be the density of lead under a pressure of `3xx10^8 Nm^(-2) ?` Given normal density of lead is `11.4 g//cm^3` and Bulk modulus of lead is `8.0 xx 10^9 Nm^(-2)`

Answer

`11.8 g//cm^3`

Explanation

Here, `p = 3xx10^8 Nm^(-2) , rho = 11.4 xx 10^3 kgm^(9-3)`
`B = 8.0xx 10^9 Nm^(-2)`
Let V be the volume of 1 kg of lead, then
`V = (1)/(rho) = (1)/(11.4xx10^3) = 8.77 xx 10^(-5) m^3`
Now `B = (pV)/(Delta V)`
or `Delta V = (pV)/(B) = (3xx10^8)xx(8.77xx10^(-5))/(8.0 xx 10^9)`
` =0.329 xx 10^(-5)m^3`
Now volume of 1 kg of lead
`V' = V - Delta V = 8.77 xx 10^(-5) - 0.329 xx 10^(-5)`
` = 8.441 xx 10^(-5) m^3` Now density of lead
`rho' = (1)/(V') = (1)/(8.441 xx 10^(-5)) = 11.8 xx 10^3 kg m^(-3)` = 11.8 g/c c