Respuesta :
Answer:
[tex]\Delta V_1=2\times \Delta V_2[/tex]
Explanation:
The bulk modulus is a constant that describes how resistant a substance is to compression.
It is defined as the ratio between increase in pressure and the resulting decrease in a volume of the material.
It is given by a formula :
[tex]Bulk\,\,modulus\,\,(K)=\frac{volumetric\,\,stress}{volumetric\,\,strain}[/tex]
OR
[tex]K=\frac{\Delta P}{(\frac{\Delta V}{V}) }[/tex]
where:
[tex]\Delta V[/tex] & [tex]\Delta P[/tex] are the change in volume and change in pressure respectively.
V= original volume
According to the given:
[tex]K_2=2K_1[/tex]
[tex]V_1=V_2[/tex]
[tex]\Delta P_1=\Delta P_2[/tex]
So,
[tex]K_2=\frac{\Delta P_2}{(\frac{\Delta V_2}{V_2}) }[/tex]
[tex]K_2=\frac{\Delta P_2\times V_2}{\Delta V_2}[/tex].................................(1)
&
[tex]K_1=\frac{\Delta P_1}{(\frac{\Delta V_1}{V_1}) }[/tex]
[tex]K_1=\frac{\Delta P_1\times V_1}{\Delta V_1}[/tex]..................................(2)
From the given conditions we compare equations (1) & (2):
[tex]K_2=2K_1[/tex]
[tex]\frac{\Delta P_2\times V_2}{\Delta V_2}=2\times \frac{\Delta P_1\times V_1}{\Delta V_1}[/tex]
cancelling the equal terms
[tex]\frac{1}{\Delta V_2} =\frac{2}{\Delta V_1}[/tex]
[tex]\Delta V_1=2\times \Delta V_2[/tex]
The material in first case undergoes twice the volume reduction than that of the material in first case under the given conditions.
