Current Electricity

JEE Main level, numerical answer, medium

67 of 380

Question 67 of 380

The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is 31.4cm31.4cm. The diameter of wire is measured as 2.4cm2.4cm. The resistivity of the given wire is measured as x×10−3Ωcmx\times {10}^{-3}\Omega cm. The value of xx is _____ .
[Take π=3.14\pi =3.14]

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Answer: 144

Since, slope of V−I graph represents the resistance, so R=tan⁡45∘=1ΩR=\tan 45^{\circ}=1\Omega
Now, resistance of wire is given by R=ρLAR=\dfrac{\rho L}{A}
So, resistivity of the wire is ρ=ARL=πr2RL=π(1.22×10−4)31.4×10−2=1.44×10−3Ωm=144×10−3Ωcm\rho =\dfrac{A}{RL}=\dfrac{\pi {r}^{2}}{RL}=\dfrac{\pi \left(1.{2}^{2}\times {10}^{-4}\right)}{31.4\times {10}^{-2}}=1.44\times {10}^{-3}\Omega m=144\times {10}^{-3}\Omega cm

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