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**Hint**We know that the ratio of isothermal bulk modulus and adiabatic bulk modulus for a monatomic gas at a given pressure is $\dfrac{3}{5}$. So the assertion is correct. We have to conform to the reason.

**Complete step by step answer**

There are different types of thermodynamic processes.

A process in which the temperature of the system is maintained throughout is called an isothermal process

In isobaric processes the pressure is maintained constant while in isochoric processes the volume is maintained constant.

If the system is insulated from the surroundings then no heat flows between the system and the surroundings and this process is adiabatic process

There are three different type of modulus in elasticity

Young’s modulus

Shear modulus

Bulk modulus

Young’s modulus: Ratio of tensile stress to tensile strain.

Shear modulus: the ratio of shear stress to shear strain.

Bulk modulus: Ratio of compressive stress to volumetric strain.

The isothermal bulk modulus of an ideal gas is the ratio of change in pressure to change in volume at constant gas temperature

$ \Rightarrow {{\text{B}}_{\text{I}}}{\text{ = }}\dfrac{{{\text{dp}}}}{{\dfrac{{{\text{dv}}}}{{\text{v}}}}}{\text{ = }}\dfrac{{{{\text{c}}_{\text{p}}}}}{{{{\text{c}}_{\text{v}}}}}{\text{ }} \to {\text{1}}$

Here, pressure and the volume are constant

\[ \Rightarrow {\text{pv = Constants }}\]

\[ \Rightarrow {\text{pv = C }}\]

Differentiating above with respect to volume,

$ \Rightarrow p.1 + v\dfrac{{dp}}{{dv}} = 0$

$ \Rightarrow p + v\dfrac{{dp}}{{dv}} = 0$

$ \Rightarrow p = - v\dfrac{{dp}}{{dv}}$

The adiabatic bulk modulus of an ideal gas is the ratio of change in pressure to change in volume when no heat exchange with surrounding.

$ \Rightarrow {B_A} = \dfrac{{dp}}{{\dfrac{{dv}}{v}}} = \dfrac{{{{\text{c}}_{\text{p}}}}}{{{{\text{c}}_{{{\text{v}}^\gamma }}}}}{\text{ }} \to {\text{2}}$

Here, the pressure, volume and the ratio of specific heats are constant

\[ \Rightarrow {\text{p}}{{\text{v}}^\gamma }{\text{ = Constants }}\]

\[ \Rightarrow {\text{p}}{{\text{v}}^\gamma }{\text{ = C}}\]

Differentiating above with respect to volume,

\[ \Rightarrow {\text{p}}{\text{.}}\gamma {{\text{v}}^{\gamma - 1}}{\text{ + }}{{\text{v}}^\gamma }{\text{.}}\dfrac{{dp}}{{dv}}{\text{ = 0}}\]

\[ \Rightarrow {\text{p}}{\text{.}}\gamma + \dfrac{{{{\text{v}}^\gamma }}}{{{{\text{v}}^{\gamma - 1}}}}{\text{.}}\dfrac{{dp}}{{dv}}{\text{ = 0}}\]

\[ \Rightarrow {\text{p}}{\text{.}}\gamma + \dfrac{{{{\text{v}}^\gamma }}}{{{{\text{v}}^\gamma }.{v^{ - 1}}}}{\text{.}}\dfrac{{dp}}{{dv}}{\text{ = 0}}\]

\[ \Rightarrow {\text{p}}{\text{.}}\gamma + {\text{v}}\dfrac{{dp}}{{dv}}{\text{ = 0}}\]

\[ \Rightarrow {\text{p}}{\text{.}}\gamma {\text{ = }} - {\text{v}}\dfrac{{dp}}{{dv}}\]

\[ \Rightarrow {\text{p}}{\text{.}}\gamma {\text{ = }} - \dfrac{{dp}}{{\dfrac{{dv}}{{\text{v}}}}}\]

\[ \Rightarrow {\text{p}}{\text{.}}\gamma {\text{ = }} - {B_I}\]

\[ \Rightarrow - v\dfrac{{dp}}{{dv}}{\text{.}}\gamma {\text{ = }} - {B_I}\]

\[ \Rightarrow - \dfrac{{{c_p}}}{{{c_v}}}{\text{.}}\gamma {\text{ = }} - {B_I}\]

\[ \Rightarrow {\text{.}}\gamma {\text{ = }}{B_I}\dfrac{{{c_v}}}{{{c_p}}}\]

From the above equation we can see that, the ratio is \[ \Rightarrow {\text{.}}\gamma {\text{ = }}{B_I}\dfrac{{{c_v}}}{{{c_p}}}\]

But in the given question the reason is given as

This ratio is equal to $\gamma = \dfrac{{{C_p}}}{{{C_v}}}$

So the reason is wrong.

The assertion is correct but the reason is wrong.

**Hence the correct answer is option (C) Assertion is correct but the reason is incorrect**

**Note**Compressibility of a liquid is the ratio of volumetric strain to the compressive stress. Bulk modulus is the ratio of compressive stress to volumetric strain. So the bulk modulus depends on the compressibility. Gases possess volume elasticity because of high compressibility. So, the magnitude of the elasticity depends on compressibility.

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