With rise in temperature, density of a given body changes according to one of the following relations

The correct relation is p = ρ₀(1 - γdθ).

To understand how the density of a body changes with a rise in temperature, let us derive the expression step-by-step:

Let the initial mass of the body be $m$, the initial volume be $V_0$, and the initial density be ${\rho }_0$ at an initial temperature.

The relation between density, mass, and volume is given by:
${\rho }_0=\frac{m}{V_0}$

When the temperature of the body increases by $d\theta$, its volume expands. The new volume $V$ is given by the thermal expansion formula:
$V=V_0(1+\gamma d\theta )$
where $\gamma$ is the coefficient of volume expansion.

Since the mass $m$ of the body remains constant, the new density $\rho$ (represented as $p$ in the options) is:
$\rho =\frac{m}{V}$

Substituting the new volume expression into this density equation:
$\rho =\frac{m}{V_0(1+\gamma d\theta )}$

Since $\frac{m}{V_0}={\rho }_0$, we can rewrite this as:
$\rho =\frac{{\rho }_0}{1+\gamma d\theta }={\rho }_0{(1+\gamma d\theta )}^{-1}$

For most solids and liquids, the value of the volume expansion coefficient $\gamma$ is very small, meaning the product $\gamma d\theta \ll 1$. We can apply the binomial approximation:
${(1+x)}^n\approx 1+nx$
when $\left|x\right|\ll 1$.

Applying this approximation with $n=-1$ and $x=\gamma d\theta$, we get:
$\rho \approx {\rho }_0(1-\gamma d\theta )$

Thus, representing density as $p$ as in the given options, the relation is:
$p={\rho }_0(1-\gamma d\theta )$