An increase in pressure required to decreases the 200 litres volume of a liquid by 0.004% in the container is
(Bulk modulus of the liquid =2100 MPa).
A. 188 kPa
B. 8.4 kPa
C. 18.8 kPa
D. 84 kPa

Hint: A substance's bulk modulus, or K or B, is a measure of how resistant it is to compression. It's the ratio of the infinitesimal pressure rise to the volume's consequent relative reduction. Isothermal compressibility is defined as the reciprocal of the bulk modulus at a given temperature. We use this concept here.

Complete step by step solution:
The bulk modulus of a material is defined as the proportion of volumetric stress to volumetric strain when the material is deformed within the elastic limit. To put it another way, the bulk modulus is a numerical constant that is used to quantify and describe a solid or fluid elastic characteristics when pressure is applied to all surfaces. A material's bulk elastic characteristics are used to calculate how much it will compress when subjected to a certain amount of external pressure. It's crucial to find and record the ratio of pressure change to fractional volume compression in this case.
The bulk modulus K (which is generally greater than zero) can be properly described by the equation,
\[K = - V\dfrac{{dP}}{{dV}}\]
where P is pressure, V is the initial volume of the material, and\[\dfrac{{dP}}{{dV}}\] is the pressure-volume derivative.
Now in the given question it is given as
$K = 2100M{P_a} = 2100 \times {10^6}{P_a}$
Also it is given that $\dfrac{{dV}}{V} = \dfrac{{0.004}}{{100}}$
Hence upon substitution
$P = k \times \dfrac{{dV}}{V}$
\[\;P = \;2100 \times {10^6} \times \dfrac{{0.004}}{{100}}\]
Hence
$P = 84 \times {10^3}{P_a}$
Upon conversion
$1k{P_a} = 1000{P_a}$
\[ \Rightarrow P = 84{\text{ k}}{P_a}\]
Hence option D is correct

Note:
Powder diffraction under applied pressure can be used to determine the bulk modulus. It is a fluid characteristic that indicates the fluid's capacity to alter volume under pressure. Bulk modulus is a measurement of a solid's incompressibility. Furthermore, the greater the value of K for a material, the more incompressible it is by nature.