Question

Q1.if the bulk modulus of lead is 8.0 ×10^9 to Newton per metre square and initial density of the lead is 11.4 G per CC thenunder the pressure of 2.0 into 10 power 8 Newton per metre square the density of the lead is

Answer 1

Step-by-step explanation:

Let the initial volume of lead be V₁ and final volume V₂

Given,

Pressure (p) = 2 ×10⁸ N/m₂

Bulk modulus (B) = 8 × 10⁹ N/m²

If the change in volume is ΔV,

Then,

$\begin{gathered}B=\frac{pV_1}{\Delta V}\\\\\Delta V=\frac{pV_1}{B}\\\\\Delta V=\frac{2\times10^{8}V_1}{8\times10^{9}}\\\\\Delta V=\frac{V_1}{40}\end{gathered} </p><p>$

$\begin{gathered}But,\\\\\\\Delta V=V_1-V_2\\\\V_2=V_1-\Delta V\\\\V_2=V_1-\frac{V_1}{40}\\\\V_2=\frac{39V_1}{40}\end{gathered} \\ but</p><p></p><p>$

If the mass of lead is m and initial density is р₁ and final density is р₂

m = р₁V₁ = р₂V₂

$\begin{gathered}\rho_1V_1=\rho_2V_2\\\\11.4\times V_1=\rho_2\frac{39V_1}{40}\\\\\rho_2=\frac{11.4\times40}{39}\\\\\rho_2=\frac{456}{39}\\\\\rho_2=11.69\;g/cm^{2}\\ \\\rho_2=11.7\;g/cm^{2}\end{gathered} </p><p>$

Answer 2

Answer:

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