Answer
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Hint The stress applied to a body is directly proportional to the strain induced in it. If the stain is volumetric, we get bulk modulus after removing the proportionality sign. From this equation we need to find the change in volume of the solid copper cube.
Step by step solution
We are given the length of the edge of the solid copper cube, $L = 10cm = 0.1m$ .
This cube is subjected to a pressure of 7 M Pa. $ \Rightarrow $ Pressure, $P = 7MPa = 7 \times {10^6}Pa{\text{ or N}}{{\text{m}}^{{\text{ - 2}}}}$ .
The volume of this cube will be, $V = {L^3}$
$
\Rightarrow V = {\left( {0.1} \right)^3} \\
\Rightarrow V = {10^{ - 3}}{m^3} \\
$
The value of bulk modulus of copper is also given, \[ \Rightarrow K = 140GPa = 140 \times {10^9}Pa{\text{ or N}}{{\text{m}}^{{\text{ - 2}}}}\].
Now we will use the expression for bulk modulus, K and substitute all the given values.
\[
\Rightarrow K = \dfrac{P}{{\dfrac{{\Delta V}}{V}}} \\
\Rightarrow 140 \times {10^9} = \dfrac{{7 \times {{10}^6}}}{{\dfrac{{\Delta V}}{{{{10}^{ - 3}}}}}} \\
\Rightarrow 140 \times {10^9} = \dfrac{{7 \times {{10}^6} \times {{10}^{ - 3}}}}{{\Delta V}} \\
\Rightarrow \Delta V = \dfrac{{7 \times {{10}^6} \times {{10}^{ - 3}}}}{{140 \times {{10}^9}}} \\
\Rightarrow \Delta V = 5 \times {10^{ - 8}}{m^3} \\
\Rightarrow \Delta V = 5 \times {10^{ - 8}} \times {\left( {{{10}^2}} \right)^3}c{m^3} \\
\Rightarrow \Delta V = 5 \times {10^{ - 2}}c{m^3} \\
\]
Therefore, option (A) is correct.
Note The physical significance of bulk modulus is that it is a measure of how resistant a substance is to compression when pressure is applied on it. In other words, bulk modulus is a measure of rigidity of a substance. Rigidity and bulk modulus of a material are inversely related to each other.
Step by step solution
We are given the length of the edge of the solid copper cube, $L = 10cm = 0.1m$ .
This cube is subjected to a pressure of 7 M Pa. $ \Rightarrow $ Pressure, $P = 7MPa = 7 \times {10^6}Pa{\text{ or N}}{{\text{m}}^{{\text{ - 2}}}}$ .
The volume of this cube will be, $V = {L^3}$
$
\Rightarrow V = {\left( {0.1} \right)^3} \\
\Rightarrow V = {10^{ - 3}}{m^3} \\
$
The value of bulk modulus of copper is also given, \[ \Rightarrow K = 140GPa = 140 \times {10^9}Pa{\text{ or N}}{{\text{m}}^{{\text{ - 2}}}}\].
Now we will use the expression for bulk modulus, K and substitute all the given values.
\[
\Rightarrow K = \dfrac{P}{{\dfrac{{\Delta V}}{V}}} \\
\Rightarrow 140 \times {10^9} = \dfrac{{7 \times {{10}^6}}}{{\dfrac{{\Delta V}}{{{{10}^{ - 3}}}}}} \\
\Rightarrow 140 \times {10^9} = \dfrac{{7 \times {{10}^6} \times {{10}^{ - 3}}}}{{\Delta V}} \\
\Rightarrow \Delta V = \dfrac{{7 \times {{10}^6} \times {{10}^{ - 3}}}}{{140 \times {{10}^9}}} \\
\Rightarrow \Delta V = 5 \times {10^{ - 8}}{m^3} \\
\Rightarrow \Delta V = 5 \times {10^{ - 8}} \times {\left( {{{10}^2}} \right)^3}c{m^3} \\
\Rightarrow \Delta V = 5 \times {10^{ - 2}}c{m^3} \\
\]
Therefore, option (A) is correct.
Note The physical significance of bulk modulus is that it is a measure of how resistant a substance is to compression when pressure is applied on it. In other words, bulk modulus is a measure of rigidity of a substance. Rigidity and bulk modulus of a material are inversely related to each other.
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