The density of a metal at normal pressure is | vedclass

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Question

(A) The bulk modulus $B$ is defined as $B = -\frac{P}{\Delta V / V}$.Since $\Delta V = V' - V$,we have $B = -\frac{P}{(V' - V) / V} = -\frac{PV}{V' -

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$\frac{1}{1 - \frac{P}{B}}$

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(A) The bulk modulus $B$ is defined as $B = -\frac{P}{\Delta V / V}$.Since $\Delta V = V' - V$,we have $B = -\frac{P}{(V' - V) / V} = -\frac{PV}{V' - V}$.Rearranging for the change in volume: $V' - V = -\frac{PV}{B}$,which gives $V' = V(1 - \frac{P}{B})$.Density $\rho$ is defined as $\rho = \frac{m}{V}$,so $V = \frac{m}{\rho}$ and $V' = \frac{m}{\rho'}$.Substituting these into the volume equation: $\frac{m}{\rho'} = \frac{m}{\rho}(1 - \frac{P}{B})$.Therefore,$\frac{1}{\rho'} = \frac{1}{\rho}(1 - \frac{P}{B})$,which implies $\frac{\rho'}{\rho} = \frac{1}{1 - P/B}$.Since the density increases with pressure,the correct expression is $\frac{\rho'}{\rho} = (1 - P/B)^{-1} \approx 1 + P/B$ for small $P$. However,based on the standard derivation,the exact ratio is $\frac{1}{1 - P/B}$.

The density of a metal at normal pressure is $\rho$. When an additional pressure $P$ is applied,its density becomes $\rho'$. If its bulk modulus is $B$,then the ratio $\frac{\rho'}{\rho}$ is:

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