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Chapter 12: Problem 1686

On applying an electric field of \(5 \times 10^{-8} \mathrm{Vm}^{-1}\) across a conductor, current density through it is \(2.5 \mathrm{Am}^{-2}\) The resistivity of the conductor is \(\ldots .\) (A) \(1 \times 10^{-8} \Omega \mathrm{m}\) (B) \(2 \times 10^{-8} \Omega \mathrm{m}\) (C) \(0.5 \times 10^{-8} \Omega \mathrm{m}\) (D) \(12.5 \times 10^{-8} \Omega \mathrm{m}\)

Short Answer

Expert verified
The resistivity of the conductor is (B) \(2 \times 10^{-8} \Omega \mathrm{m}\).

Step by step solution

01

Understand the relationship between electric field, current density and resistivity

We can use the following formula to relate the current density (J), electric field (E), and resistivity (ρ): \(J = \frac{E}{\rho}\) We will now use this formula to find the resistivity of the conductor.
02

Plug in the given values

The given electric field E is: \(E = 5 \times 10^{-8} \mathrm{Vm}^{-1}\) And given current density J is: \(J = 2.5 \mathrm{Am}^{-2}\) We have to find the resistivity ρ, so let's plug in the given values into the formula: \(2.5 = \frac{5 \times 10^{-8}}{\rho}\)
03

Solve for resistivity

Now, we solve for ρ: \(\rho = \frac{5 \times 10^{-8}}{2.5}\) \(\rho = 2 \times 10^{-8} \Omega \mathrm{m}\) Therefore, the resistivity of the conductor is: \(\rho = 2 \times 10^{-8} \Omega \mathrm{m}\) The correct answer is (B) \(2 \times 10^{-8} \Omega \mathrm{m}\).

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Most popular questions from this chapter

Two wires of equal diameters of resistivity's \(\rho_{1}\) and \(\rho_{2}\) are joined in series. The equivalent resistivity of the combination is.... (A) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (B) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$ (C) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (D) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$

Length of a wire of resistance \(R \Omega\) is increased to 10 times, so its resistance becomes \(1000 \Omega\), therefore \(R=\ldots .\) (The volume of the wire remains same during increase in length) (A) \(0.01 \Omega\) (B) \(0.1 \Omega\) (C) \(1 \Omega\) (D) \(10 \Omega\)

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) \(1 \times 10^{-6} \Omega \mathrm{m}\)

The drift velocity of free electrons in a conductor is \(\mathrm{V}\), when a current. \(I\) is flowing in it If both the radius and current are doubled, then drift velocity will be. (A) \((\mathrm{V} / 4)\) (B) \((\mathrm{V} / 2)\) (C) \(4 \mathrm{~V}\) (D) \(2 \mathrm{~V}\)

\(2 \mathrm{~A}\) current is obtained when a \(2 \Omega\) resistor is connected with battery having \(r \Omega\) as internal resistance \(0.5 \mathrm{~A}\) current is obtained if the above battery is connected to \(9 \Omega\) resistor. Calculate the internal resistance of the battery. (A) \(0.5 \Omega\) (B) \((1 / 3) \Omega\) (C) \((1 / 4) \Omega\) (D) \(1 \Omega\)
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