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Question: The bulk modulus of water is \[2.0 \times {10^9}N/{m^2}\]. The pressure required to increase the den...

The bulk modulus of water is 2.0×109N/m22.0 \times {10^9}N/{m^2}. The pressure required to increase the density of water by 0.1% is:
A. 2.0×103N/m22.0 \times {10^3}N/{m^2}
B. 2.0×106N/m22.0 \times {10^6}N/{m^2}
C. 2.0×105N/m22.0 \times {10^5}N/{m^2}
D. 2.0×107N/m22.0 \times {10^7}N/{m^2}

Explanation

Solution

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=−VdPdVB = - 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 2.0×109N/m22.0 \times {10^9}N/{m^2}
Change in density is given as △ρρ=0.1100=0.001\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
ρ=mV−−−−(i)\rho = \dfrac{m}{V} - - - - (i)
By differentiating the equation (i) with respect to V, we can write

ρ=m(V)−1 d(ρ)=md(V)−1 dρ=−mdVV2   \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
dρρ=−dVV\dfrac{{d\rho }}{\rho } = - \dfrac{{dV}}{V}
Where △ρρ=0.001\dfrac{{\vartriangle \rho }}{\rho } = 0.001
Hence we can write
dρρ=−dVV=−0.001\dfrac{{d\rho }}{\rho } = - \dfrac{{dV}}{V} = - 0.001
Now putting the values in bulk Modulus formula we get

B=−VdPdV dP=−BdVV =(2×109)(0.001) =2×106  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 2×106N/m22 \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.