As the temperature of liquid is increased, mass & volume will

As the temperature of a liquid is increased, the mass remains the same while the volume increases and hence, the density of liquid decreases.

$\rho \propto \frac{1}{\text{v}}$

$\frac{{\rho }^{|}}{\rho }=\frac{\text{v}}{{\text{v}}^{|}}=\frac{\text{v}}{\text{v}+\text{dv}}=\frac{1}{1+\gamma \Delta \text{T}}={\rho }^{{}^{|}}=\frac{\rho }{1+\gamma \Delta \text{T}}$