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NEET PHYSICSEasy

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10⁻⁷ m², and its resistance is measured to be 5.0 \Omega . What is the resistivity of the material at the temperature of the experiment?

A

1 × 10⁻⁷ \Omega m

B

2 × 10⁻⁷ \Omega m

C

3 × 10⁻⁷ \Omega m

D

1.6 × 10⁻⁷ \Omega m

Step-by-Step Solution

According to the NCERT text, the resistance RR of a conductor is given by R=ρlAR = \frac{\rho l}{A}, where ρ\rho is the resistivity, ll is the length, and AA is the cross-sectional area . Rearranging to solve for resistivity: ρ=RAl\rho = \frac{R A}{l}. Substituting the given values (R=5.0 ΩR = 5.0 \ \Omega, A=6.0×10−7 m2A = 6.0 \times 10^{-7} \text{ m}^2, l=15 ml = 15 \text{ m}): ρ=5.0×6.0×10−715=30×10−715=2.0×10−7 Ωm\rho = \frac{5.0 \times 6.0 \times 10^{-7}}{15} = \frac{30 \times 10^{-7}}{15} = 2.0 \times 10^{-7} \ \Omega\text{m}. This matches the solution for Exercise 3.4 .

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