If the Bulk modulus of lead is 8.0 times 10^9 | vedclass

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(D) The Bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.Since density $\rho = \frac{m}{V}$,we have $V = \frac{m}{\rho}$.Taking the

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$11.7$

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(D) The Bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.Since density $ho = \frac{m}{V}$,we have $V = \frac{m}{ho}$.Taking the derivative,$\Delta V = -\frac{m}{ho^2} \Delta ho$.Thus,$\frac{\Delta V}{V} = -\frac{\Delta ho}{ho}$.Substituting this into the Bulk modulus formula: $B = \frac{\Delta P}{\Delta ho / ho}$,which gives $\frac{\Delta ho}{ho} = \frac{\Delta P}{B}$.Given: $B = 8.0 	imes 10^9 \, N/m^2$,$\Delta P = 2.0 	imes 10^8 \, N/m^2$,and $ho_1 = 11.4 \, g/cc$.$\Delta ho = ho_1 	imes \frac{\Delta P}{B} = 11.4 	imes \frac{2.0 	imes 10^8}{8.0 	imes 10^9} = 11.4 	imes 0.025 = 0.285 \, g/cc$.The final density $ho_2 = ho_1 + \Delta ho = 11.4 + 0.285 = 11.685 \, g/cc \approx 11.7 \, g/cc$.

If the Bulk modulus of lead is $8.0 \times 10^9 \, N/m^2$ and the initial density of the lead is $11.4 \, g/cc$,then under the pressure of $2.0 \times 10^8 \, N/m^2$,the density of the lead is ............. $g/cc$.

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