JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Current Electricity

The current-voltage plot is a straight line passing through the origin and inclined at $45^{\circ }$, so the potential difference is numerically equal to the current and the resistance of the wire, read off as the slope $V/I$, is simply $R=1\Omega$. The resistivity then follows from $R=\rho \frac{\ell }{A}$, where the wire has length $\ell =31.4\text{cm}$ and a circular cross-section of area $A=\frac{\pi d^2}{4}$ for the measured diameter $d=2.4\text{cm}$. Taking $\pi =3.14$, $A=\frac{3.14\times (2.4{)}^2}{4}=\frac{3.14\times 5.76}{4}=4.5216{\text{cm}}^2$. Rearranging for the resistivity gives $\rho =\frac{RA}{\ell }=\frac{1\times 4.5216}{31.4}=0.144\Omega \text{cm}=144\times 10^{-3}\Omega \text{cm}$, so that $x=144$.