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 $.............g / cc$
Diffcult
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We know,
$\left[\frac{1}{\rho_2}-\frac{1}{\rho_1}\right]=-\frac{P}{B} \left\{\begin{array}{l}\text { Where } \\ P=2 \times 10^8 \,N / m ^2 \\ B=8 \times 10^9 \,N / m ^2 \\ \rho_1=11.4 \,g / cc \\ \rho_2=?\end{array}\right.$
Substitute value's
$\left[\frac{1}{\rho_2}-\frac{1}{114}\right]=-\frac{2 \times 10^8}{8 \times 10^9}$
After solving, we get
$\rho_2=11.7 \,g / cc$
$\left[\frac{1}{\rho_2}-\frac{1}{\rho_1}\right]=-\frac{P}{B} \left\{\begin{array}{l}\text { Where } \\ P=2 \times 10^8 \,N / m ^2 \\ B=8 \times 10^9 \,N / m ^2 \\ \rho_1=11.4 \,g / cc \\ \rho_2=?\end{array}\right.$
Substitute value's
$\left[\frac{1}{\rho_2}-\frac{1}{114}\right]=-\frac{2 \times 10^8}{8 \times 10^9}$
After solving, we get
$\rho_2=11.7 \,g / cc$
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