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What is the density of lead under the pressure of 2.0\times 10^{8}N/m^{2}, if the bulk modulus of lead is 8.0\times 10^{9}N/m^{2} and initially the density of lead is 11.4gm/cm^{3}?

Option: 1

11.69gm/cm^{3}


Option: 2

12gm/cm^{3}


Option: 3

40gm/cm^{3}


Option: 4

13.4gm/cm^{3}


Answers (1)

best_answer

\text { The changed density, } \rho^{\prime}=\frac{\rho}{1-\frac{d P}{B}}

Substituting the values, we have

\rho^{\prime}=\frac{11.4}{1-\frac{2.0 \times 10^8}{8.0 \times 10^9}}

\rho^{\prime}=\frac{11.4 \times 8.0 \times 10^9}{8.0 \times 10^9-2.0 \times 10^8}

= 11.69gm/cm^{3}

 

Posted by

Irshad Anwar

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