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
606k+ 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.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

