What will be the density of lead under a pressure of 2 x 10⁸Nm^-2?Given ,normal density of lead =11.4gcm^-3 and bulk modulus of lead =8.0 x 10⁹ Nm^-2 ? Please reply soon !!!!!!!!!!!!

Here, Δp= 2 x 10⁸ N/m²
  B= 8 X 10⁹ N/m²
As B= $\frac{V\Delta p}{\Delta V}$
or ΔV= $\frac{V\Delta p}{B}$= V X $\frac{2 X {10}^8}{8 X {10}^9}$= $\frac{V}{40}$
Therefore, V'= V- ΔV= V- $\frac{V}{40}$= $\frac{39 V }{40}$
Let ρ' be the new density of lead, under the effect of pressure applied. As mass of the lead will remain the same, so Vρ= V' ρ'
or ρ'= $\frac{V\rho }{V\text{'}}=\frac{V X 11.4}{39V/40}= 11.7 g/c{m}^2$