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Chapter 12: Problem 1707

A wire \(50 \mathrm{~cm}\) long and \(1 \mathrm{~mm}^{2}\) in cross-section carries a current of \(4 \mathrm{~A}\) when connected to a \(2 \mathrm{~V}\) battery. The resistivity of the wire is: (A) \(2 \times 10^{-7} \Omega \mathrm{m}\) (B) \(5 \times 10^{-7} \Omega \mathrm{m}\) (C) \(4 \times 10^{-6} \Omega \mathrm{m}\) (D) \(1 \times 10^{-6} \Omega \mathrm{m}\)

Short Answer

Expert verified
The resistivity of the wire is \(1\times10^{-6} \Omega m\) (Option D).

Step by step solution

01

Calculate the resistance using Ohm's law

Ohm's law states that the voltage is equal to the product of current (I) and resistance (R): \[V = IR\] We are given the current (I = 4 A) and the voltage (V = 2 V). We will use Ohm's law to find the resistance (R): \[R = \frac{V}{I} = \frac{2}{4}\]
02

Find the resistance value

By solving for R, we find its value: \[R = \frac{2}{4} = 0.5 \Omega\]
03

Use the resistance formula to find resistivity

The formula for the resistance in terms of resistivity (ρ), length (L), and cross-sectional area (A) is: \[R = \frac{\rho L}{A}\] We are given the length (L = 50 cm = 0.5 m) and cross-sectional area (A = \(1 mm^2 = 1\times10^{-6} m^2\)). We will use the observed resistance (R = 0.5 Ω) and the formula to find the resistivity (ρ): \[\rho = \frac{RA}{L} = \frac{0.5 \times 1\times10^{-6}}{0.5}\]
04

Calculate the resistivity

By solving for ρ, we find its value: \[\rho = \frac{0.5 \times 1\times10^{-6}}{0.5} = 1\times10^{-6} \Omega m\] The resistivity of the wire is therefore \(1\times10^{-6} \Omega m\), which corresponds to option (D).

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Most popular questions from this chapter

How many dry cells, each of emf \(1.5 \mathrm{~V}\) and internal resistance $0.5 \Omega\(, much be joined in series with a resistor of \)20 \Omega$ to give a current of \(0.6 \mathrm{~A}\) in the circuit? (A) 2 (B) 8 (C) 10 (D) 12

The length of a potentiometer wire is \(600 \mathrm{~cm}\) and it carries a current of \(40 \mathrm{~m} \mathrm{~A}\) for cell of emf \(2 \mathrm{~V}\) and internal resistance \(10 \Omega\), the null point is found to be at $500 \mathrm{~cm}$ on connecting a voltmeter across the cell, the balancing length is decreased by \(10 \mathrm{~cm}\) The resistance of the voltmeter is (A) \(500 \Omega\) (B) \(290 \Omega\) (C) \(49 \Omega\) (D) \(20 \Omega\)

Three identical resistors connected in series with a battery, together dissipate \(10 \mathrm{~W}\) of power. What will be the power dissipated, if the same resistors are connected in paralle1 across the same battery? (A) \(60 \mathrm{~W}\) (B) \(30 \mathrm{~W}\) (C) \(90 \mathrm{~W}\) (D) \(120 \mathrm{~W}\)

Resistance of a wire at \(50^{\circ} \mathrm{C}\) is \(5 \Omega\), and at \(100^{\circ} \mathrm{C}\) it is \(6 \Omega\) find its resistance at $0^{\circ} \mathrm{C}$ (A) \(4 \Omega\) (B) \(3 \Omega\) (C) \(2 \Omega\) (D) \(1 \Omega\)

A potentiometer wire of length \(1 \mathrm{~m}\) and resistance \(10 \Omega\) is connected in series with a cell of e.m.f \(2 \mathrm{~V}\) with internal resistance \(1 \Omega\) and a resistance box of a resistance \(R\) if potential difference between ends of the wire is \(1 \mathrm{~V}\) the value of \(R\) is. (A) \(4.5 \Omega\) (B) \(9 \Omega\) (C) \(15 \Omega\) (D) \(20 \Omega\)
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