Question

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)?

Answer 1

Final density of lead = 11.69 g/cm³

Given:

Bulk modulus of lead is 8.0 x $10^9$  N/m²

Pressure =  2 x $10^8$  N/m²

Density of lead is 11.4 g/cm³ = 11.4 x $10^3$  N/m³

To find:

Density of lead under a pressure.

Formula used:

Bulk modulus = $\frac{\Delta p }{\frac{\Delta \rho}{ \rho }}$

Where $\Delta p$ = Change in pressure.

           $\Delta \rho$ = Change in density.

Explanation:

Density of lead is 11.4 x $10^3$  N/m³

Bulk modulus = $\frac{\Delta p }{\frac{\Delta \rho}{ \rho }}$

           $\Delta \rho$ = $\frac{2 \times 10^8\times 11.4 \times 10^3}{8\times 10^9 }$ = 0.05 x $10^3$  N/m³ = 0.05 g/cm³

So final density of lead = initial density + Change in density

                                      = 11.4 + 0.05

                                      = 11.69 g/cm³

Final density of lead = 11.69 g/cm³

To learn more....

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