A fractional change in volume of oil is 1 percent when a pressure of $2 \times {10^7}{\text{N}}{{\text{m}}^{{\text{ - 2}}}}$ is applied calculate the bulk modulus and its compressibility:
A) $3 \times {10^8}{\text{N}}{{\text{m}}^{{\text{ - 2}}}},\,\,\,\,0.33 \times {10^{ - 9\,}}\,{m^2}{N^{ - 1}}$
B) $5 \times {10^9}{\text{N}}{{\text{m}}^{{\text{ - 2}}}},\,\,\,\,2 \times {10^{ - 10\,}}\,{m^2}{N^{ - 1}}$
C) $2 \times {10^9}\,{\text{N}}{{\text{m}}^{{\text{ - 2}}}},\,\,\,\,5 \times {10^{ - 10}}\,{m^2}{N^{ - 1}}$
D) $2 \times {10^9}\,{\text{N}}{{\text{m}}^{{\text{ - 2}}}},\,\,\,\,5 \times {10^{ - 9\,}}\,{m^2}{N^{ - 1}}$
Answer
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Hint Use the formula $B = \dfrac{P}{{\dfrac{{\partial V}}{V}}}$ to find the bulk modulus. Also use the formula $C = \dfrac{1}{B}$ to find compressibility.
Complete step-by-step answer:
Given that the fractional change in volume of oil is 1%.
Fractional change in volume = $\dfrac{{{\text{change}}\,{\text{in}}\,{\text{volume}}}}{{{\text{original}}\,{\text{volume}}}}$.
Or, $\dfrac{{\partial V}}{V} = 1\% = \dfrac{1}{{100}} = 0.01$
Also given that the pressure of $2 \times {10^7}{\text{N}}{{\text{m}}^{{\text{ - 2}}}}$ is applied.
Hence, P = $2 \times {10^7}{\text{N}}{{\text{m}}^{{\text{ - 2}}}}$
We need to find the bulk modulus; which is given by the following formula:
$
B = \dfrac{P}{{\dfrac{{\partial V}}{V}}} \\
\Rightarrow \dfrac{{2 \times {{10}^7}}}{{0.01}} \\
\Rightarrow 2 \times {10^9}N{m^{ - 2}} \\
$(Putting the values given in the question , we have )
Hence bulk modulus = $2 \times {10^9}\,{\text{N}}{{\text{m}}^{{\text{ - 2}}}}$.
Now, to find compressibility we use the formula,
$C = \dfrac{1}{B}$ (Where C is the compressibility and B is the bulk modulus)
Putting the values we have,
$
C = \dfrac{1}{{2 \times {{10}^9}}} \\
\Rightarrow C = 5 \times {10^{ - 10}}\,{{\text{m}}^{\text{2}}}{{\text{N}}^{{\text{ - 1}}}} \\
$
Hence the compressibility is $C = 5 \times {10^{ - 10}}\,{{\text{m}}^{\text{2}}}{{\text{N}}^{{\text{ - 1}}}}$.
Hence option C is the correct Answer.
Note i) Bulk modulus shows the rigidity of the solid. It shows how much resistance a substance is to compression.
ii) Bulk modulus of Adiabatic process = $\;\gamma P$ (Where $\gamma $ is the adiabatic constant of adiabatic process and p = pressure))
iii) Bulk modulus of Isothermal process = P ( Where P = pressure)
Complete step-by-step answer:
Given that the fractional change in volume of oil is 1%.
Fractional change in volume = $\dfrac{{{\text{change}}\,{\text{in}}\,{\text{volume}}}}{{{\text{original}}\,{\text{volume}}}}$.
Or, $\dfrac{{\partial V}}{V} = 1\% = \dfrac{1}{{100}} = 0.01$
Also given that the pressure of $2 \times {10^7}{\text{N}}{{\text{m}}^{{\text{ - 2}}}}$ is applied.
Hence, P = $2 \times {10^7}{\text{N}}{{\text{m}}^{{\text{ - 2}}}}$
We need to find the bulk modulus; which is given by the following formula:
$
B = \dfrac{P}{{\dfrac{{\partial V}}{V}}} \\
\Rightarrow \dfrac{{2 \times {{10}^7}}}{{0.01}} \\
\Rightarrow 2 \times {10^9}N{m^{ - 2}} \\
$(Putting the values given in the question , we have )
Hence bulk modulus = $2 \times {10^9}\,{\text{N}}{{\text{m}}^{{\text{ - 2}}}}$.
Now, to find compressibility we use the formula,
$C = \dfrac{1}{B}$ (Where C is the compressibility and B is the bulk modulus)
Putting the values we have,
$
C = \dfrac{1}{{2 \times {{10}^9}}} \\
\Rightarrow C = 5 \times {10^{ - 10}}\,{{\text{m}}^{\text{2}}}{{\text{N}}^{{\text{ - 1}}}} \\
$
Hence the compressibility is $C = 5 \times {10^{ - 10}}\,{{\text{m}}^{\text{2}}}{{\text{N}}^{{\text{ - 1}}}}$.
Hence option C is the correct Answer.
Note i) Bulk modulus shows the rigidity of the solid. It shows how much resistance a substance is to compression.
ii) Bulk modulus of Adiabatic process = $\;\gamma P$ (Where $\gamma $ is the adiabatic constant of adiabatic process and p = pressure))
iii) Bulk modulus of Isothermal process = P ( Where P = pressure)
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