The bulk modulus of water is \[2.0 \times {10^9}N/{m^2}\]. The pressure required to increase the density of water by 0.1% is:
A. \[2.0 \times {10^3}N/{m^2}\]
B. \[2.0 \times {10^6}N/{m^2}\]
C. \[2.0 \times {10^5}N/{m^2}\]
D. \[2.0 \times {10^7}N/{m^2}\]
Answer
611.1k+ views
Hint: The relative change in the volume of a body produced due to a unit compressive or tensile stress acting uniformly over its surface is known as the bulk modulus. Bulk modulus is the measure of how resistant a substance can be to a compression. It is the ratio of the infinitesimal pressure increase to the decrease of volume denoted by K or B. It is given as
\[B = - V\dfrac{{dP}}{{dV}}\]
Here V is the volume of substance and dV is the change of volume, whereas dP is the change in pressure.
Complete step by step solution:
Given the bulk modulus of water \[2.0 \times {10^9}N/{m^2}\]
Change in density is given as \[\dfrac{{\vartriangle \rho }}{\rho } = \dfrac{{0.1}}{{100}} = 0.001\]
Where density is defined as the mass per unit volume of any object which is the ratio mas the mass of the object to its volume, given as
\[\rho = \dfrac{m}{V} - - - - (i)\]
By differentiating the equation (i) with respect to V, we can write
\[
\rho = m{\left( V \right)^{ - 1}} \\
d\left( \rho \right) = md{\left( V \right)^{ - 1}} \\
d\rho = - m\dfrac{{dV}}{{{V^2}}} \\
\\
\]
Now dividing both sides of the equation we get
\[\dfrac{{d\rho }}{\rho } = - \dfrac{{dV}}{V}\]
Where \[\dfrac{{\vartriangle \rho }}{\rho } = 0.001\]
Hence we can write
\[\dfrac{{d\rho }}{\rho } = - \dfrac{{dV}}{V} = - 0.001\]
Now putting the values in bulk Modulus formula we get
\[
B = - V\dfrac{{dP}}{{dV}} \\
dP = - B\dfrac{{dV}}{V} \\
= \left( {2 \times {{10}^9}} \right)(0.001) \\
= 2 \times {10^6} \\
\]
Hence the pressure required to increase the density of water by 0.1% is \[2 \times {10^6}N/{m^2}\]
Option (B) is correct.
Note: The negative sign in the bulk modulus formula indicates that an increase in pressure is accompanied by a decrease in volume as it requires an enormous pressure to change the volume of water by a small amount.
\[B = - V\dfrac{{dP}}{{dV}}\]
Here V is the volume of substance and dV is the change of volume, whereas dP is the change in pressure.
Complete step by step solution:
Given the bulk modulus of water \[2.0 \times {10^9}N/{m^2}\]
Change in density is given as \[\dfrac{{\vartriangle \rho }}{\rho } = \dfrac{{0.1}}{{100}} = 0.001\]
Where density is defined as the mass per unit volume of any object which is the ratio mas the mass of the object to its volume, given as
\[\rho = \dfrac{m}{V} - - - - (i)\]
By differentiating the equation (i) with respect to V, we can write
\[
\rho = m{\left( V \right)^{ - 1}} \\
d\left( \rho \right) = md{\left( V \right)^{ - 1}} \\
d\rho = - m\dfrac{{dV}}{{{V^2}}} \\
\\
\]
Now dividing both sides of the equation we get
\[\dfrac{{d\rho }}{\rho } = - \dfrac{{dV}}{V}\]
Where \[\dfrac{{\vartriangle \rho }}{\rho } = 0.001\]
Hence we can write
\[\dfrac{{d\rho }}{\rho } = - \dfrac{{dV}}{V} = - 0.001\]
Now putting the values in bulk Modulus formula we get
\[
B = - V\dfrac{{dP}}{{dV}} \\
dP = - B\dfrac{{dV}}{V} \\
= \left( {2 \times {{10}^9}} \right)(0.001) \\
= 2 \times {10^6} \\
\]
Hence the pressure required to increase the density of water by 0.1% is \[2 \times {10^6}N/{m^2}\]
Option (B) is correct.
Note: The negative sign in the bulk modulus formula indicates that an increase in pressure is accompanied by a decrease in volume as it requires an enormous pressure to change the volume of water by a small amount.
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