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NEET PHYSICSCURRENT ELECTRICITYEasy

Question

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=6.0×107 m2A = 6.0 \times 10^{-7} \text{ m}^2, l=15 ml = 15 \text{ m}): ρ=5.0×6.0×10715=30×10715=2.0×107 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 .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from CURRENT ELECTRICITY. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSCURRENT ELECTRICITYnegligiblycurrentpassedthroughlength

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