A metallic wire of uniform area of cross-section has a resistance R, resistivity ρ, and power rating P at V volts. The wire is uniformly stretched to reduce the radius to half the original radius. The values of resistance, resistivity, and power rating at V volts are now denoted by R′, ρ′, and P′ respectively. The corresponding values are correctly re lated as _______.

The correct option is:
$\mathrm{\rho \prime }=\rho$, $\mathrm{R\prime }=16R$, $\mathrm{P\prime }=\frac{1}{16}P$

Let's understand the step-by-step physical principles and derivations that lead to this conclusion.

1. Effect on Resistivity ($\rho$):
Resistivity is an intrinsic property of a material. It depends only on the nature of the material of the wire and its temperature, not on its dimensions (length, radius, or cross-sectional area). Since the wire is only stretched and the material remains the same, its resistivity does not change. Therefore, we have:
$\mathrm{\rho \prime }=\rho$

2. Effect on Resistance ($R$):
The resistance of a wire is given by the formula:
$R=\rho \frac{l}{A}=\rho \frac{l}{\pi r^2}$
where $l$ is the length and $r$ is the radius of the wire.

When the wire is stretched, its volume ($V$) remains constant. Let the initial volume be $V=A·l=\pi r^{2}l$, and the final volume be $\mathrm{V\prime }=\mathrm{A\prime }·\mathrm{l\prime }=\pi {\mathrm{r\prime }}^2\mathrm{l\prime }$.
Since volume is conserved:
$\pi r^{2}l=\pi {\mathrm{r\prime }}^2\mathrm{l\prime }$

We are given that the radius is reduced to half of its original value, i.e., $\mathrm{r\prime }=\frac{r}{2}$. Substituting this into the volume conservation equation:
$r^{2}l={\left(\frac{r}{2}\right)}^2\mathrm{l\prime }$
$r^{2}l=\frac{r^2}{4}\mathrm{l\prime }$
$\mathrm{l\prime }=4l$

Now, we can express the new resistance $\mathrm{R\prime }$ in terms of the new length and radius:
$\mathrm{R\prime }=\mathrm{\rho \prime }\frac{\mathrm{l\prime }}{\pi {\mathrm{r\prime }}^2}$
Substituting $\mathrm{\rho \prime }=\rho$, $\mathrm{l\prime }=4l$, and $\mathrm{r\prime }=\frac{r}{2}$:
$\mathrm{R\prime }=\rho \frac{4l}{\pi {\left(\frac{r}{2}\right)}^2}$
$\mathrm{R\prime }=\rho \frac{4l}{\pi \frac{r^2}{4}}$
$\mathrm{R\prime }=16\left(\rho \frac{l}{\pi r^2}\right)$
$\mathrm{R\prime }=16R$

3. Effect on Power Rating ($P$) at constant voltage $V$:
The power rating of a wire connected across a voltage $V$ is given by:
$P=\frac{V^2}{R}$

The new power rating $\mathrm{P\prime }$ at the same voltage $V$ is:
$\mathrm{P\prime }=\frac{V^2}{\mathrm{R\prime }}$

Substituting the new resistance value $\mathrm{R\prime }=16R$ into the power equation:
$\mathrm{P\prime }=\frac{V^2}{16R}$
$\mathrm{P\prime }=\frac{1}{16}\left(\frac{V^2}{R}\right)$
$\mathrm{P\prime }=\frac{1}{16}P$

Conclusion:
Combining the three results, we get:
$\mathrm{\rho \prime }=\rho$, $\mathrm{R\prime }=16R$, and $\mathrm{P\prime }=\frac{1}{16}P$