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(D) From the given $V-I$ graph,the slope represents the resistance $R$ of the wire.$R = 	an(45^{\circ}) = 1 \, \Omega$.We know that resistance $R = \

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$144$

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(D) From the given $V-I$ graph,the slope represents the resistance $R$ of the wire.$R = 	an(45^{\circ}) = 1 \, \Omega$.We know that resistance $R = ho \frac{\ell}{A}$,where $\ell$ is the length and $A$ is the cross-sectional area.The area $A = \pi r^2 = \pi \left(\frac{d}{2}ight)^2 = \frac{\pi d^2}{4}$.Given $\ell = 31.4 \, cm$ and $d = 2.4 \, cm$.Substituting the values: $1 = ho \frac{31.4}{\frac{\pi 	imes (2.4)^2}{4}}$.Using $\pi = 3.14$: $1 = ho \frac{31.4}{\frac{3.14 	imes 5.76}{4}}$.$1 = ho \frac{31.4}{0.785 	imes 5.76} = ho \frac{31.4}{4.5216}$.Alternatively,$1 = ho \frac{31.4 	imes 4}{3.14 	imes 5.76} = ho \frac{10 	imes 4}{5.76} = ho \frac{40}{5.76}$.$ho = \frac{5.76}{40} = 0.144 \, \Omega \cdot cm$.Expressing this as $x 	imes 10^{-3} \, \Omega \cdot cm$,we get $ho = 144 	imes 10^{-3} \, \Omega \cdot cm$.Therefore,$x = 144$.

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