Answer
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Hint: Bulk modulus is the ability of a material to withstand volumetric changes due to compressive forces applied to it. The bulk modulus of a material can be defined as the ratio of the volumetric stress of a material to its volumetric strain.
Formula used:
$\Delta {{P = - K}}\dfrac{{\Delta {{V}}}}{{{V}}}$
Complete step by step solution:
The bulk modulus of a material is a measure of how resistant the particular material is to compression. It can also be called the incompressibility of a material. Bulk modulus is defined as the proportion of volumetric stress of a material related to its volumetric strain when the deformation of the material is within its elastic limit. The symbol used for Bulk modulus is ${{K}}$ and the dimensions are of force per unit area and the S.I unit of Bulk modulus is ${{N/}}{{{m}}^{ - 2}}$.
Bulk modulus of a material is
${{K = - }}$$\Delta {{P}}\dfrac{{{V}}}{{\Delta {{V}}}}$,
where $\Delta {{P}}$is the change in pressure of force applied per unit area,
$\Delta {{V}}$ is the change in volume of the material due to compression and
$V$ is the initial volume of the material. Therefore,
$\Rightarrow \Delta {{P = - K}}\dfrac{{\Delta {{V}}}}{{{V}}}$
Since the decrease of the water sample is given as $0.01\% $
$\Rightarrow \Delta {{V = }} - \dfrac{{0.01}}{{100}} \times {{V}}$
$ \Rightarrow \dfrac{{\Delta {{V}}}}{{{V}}} = - 0.0001$
On putting the values in the equation, we get:
$\Rightarrow \Delta {{P = 0}}{{.0001}} \times {{2}}{{.1}} \times {{1}}{{{0}}^9}{{N/}}{{{m}}^2}$
$ \Rightarrow \Delta {{P = 2}}{{.1}} \times {{1}}{{{0}}^5}{{N/}}{{{m}}^2}$
Therefore, the increase in pressure required to decrease the volume of the water sample by $0.01\% $ is $2.1 \times {10^5}{{N/}}{{{m}}^2}$.
Note: Bulk modulus should not be confused with young’s modulus or shear modulus. Young’s modulus is the ratio of tensile stress to tensile strain and shear modulus is the ratio of shear stress to shear strain, while bulk modulus is the ratio of tensile stress to tensile strain.
Formula used:
$\Delta {{P = - K}}\dfrac{{\Delta {{V}}}}{{{V}}}$
Complete step by step solution:
The bulk modulus of a material is a measure of how resistant the particular material is to compression. It can also be called the incompressibility of a material. Bulk modulus is defined as the proportion of volumetric stress of a material related to its volumetric strain when the deformation of the material is within its elastic limit. The symbol used for Bulk modulus is ${{K}}$ and the dimensions are of force per unit area and the S.I unit of Bulk modulus is ${{N/}}{{{m}}^{ - 2}}$.
Bulk modulus of a material is
${{K = - }}$$\Delta {{P}}\dfrac{{{V}}}{{\Delta {{V}}}}$,
where $\Delta {{P}}$is the change in pressure of force applied per unit area,
$\Delta {{V}}$ is the change in volume of the material due to compression and
$V$ is the initial volume of the material. Therefore,
$\Rightarrow \Delta {{P = - K}}\dfrac{{\Delta {{V}}}}{{{V}}}$
Since the decrease of the water sample is given as $0.01\% $
$\Rightarrow \Delta {{V = }} - \dfrac{{0.01}}{{100}} \times {{V}}$
$ \Rightarrow \dfrac{{\Delta {{V}}}}{{{V}}} = - 0.0001$
On putting the values in the equation, we get:
$\Rightarrow \Delta {{P = 0}}{{.0001}} \times {{2}}{{.1}} \times {{1}}{{{0}}^9}{{N/}}{{{m}}^2}$
$ \Rightarrow \Delta {{P = 2}}{{.1}} \times {{1}}{{{0}}^5}{{N/}}{{{m}}^2}$
Therefore, the increase in pressure required to decrease the volume of the water sample by $0.01\% $ is $2.1 \times {10^5}{{N/}}{{{m}}^2}$.
Note: Bulk modulus should not be confused with young’s modulus or shear modulus. Young’s modulus is the ratio of tensile stress to tensile strain and shear modulus is the ratio of shear stress to shear strain, while bulk modulus is the ratio of tensile stress to tensile strain.
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