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Chapter 12: Problem 1830

In each of the following questions, match column \(\mathrm{I}\) and column II and select the correct match out of the four given choices Column I \(\quad\) Column II (a) The unit of electrical resistivity is (p) \(\mathrm{m}^{2} \mathrm{~S}^{-1} \mathrm{~V}^{-1}\) (b) The unit of current density is (q) \(\Omega^{-1} \mathrm{~m}^{-1}\) (c) The unit of electrical conductivity is (r) \(\mathrm{Am}^{-2}\) (d) The unit of electric mobility is (s) \(\Omega \mathrm{m}\) (A) \(a-p, b-q, c-r, d-s\) (B) \(a-s, b-r, c-q, d-p\) (C) \(a-r, b-q, c-p, d-s\) (D) \(a-q, b-r, c-s, d-p\)

Short Answer

Expert verified
The short answer for this question is \(a-s, b-r, c-q, d-p\) (Option B).

Step by step solution

01

Find the units of electrical resistivity

We know that electrical resistivity (ρ) is the ratio of the electric field (E) to the current density (J). Mathematically, we have: ρ = E / J The unit of electric field (E) is Volts per meter (V/m) and the unit of current density (J) is Amperes per square meter (A/m²). Therefore, the unit of electrical resistivity (ρ) is: ρ = (V/m) / (A/m²) = Ωm
02

Find the units of current density

Current density (J) is the current (I) flowing through a conductor divided by the cross-sectional area (A) through which the current is passing. Its formula is: J = I / A The unit of current (I) is Amperes (A) and the unit of area (A) is square meters (m²). Therefore, the unit of current density (J) is: J = A / m²
03

Find the units of electrical conductivity

Electrical conductivity (σ) is the reciprocal of electrical resistivity (ρ). Its formula is: σ = 1 / ρ The unit of electrical resistivity (ρ) is Ωm, as we found in step 1. As electrical conductivity (σ) is the reciprocal of electrical resistivity (ρ), the unit of electrical conductivity (σ) is: σ = 1 / (Ωm) = S/m (Siemens per meter)
04

Find the units of electric mobility

Electric mobility (μ) is a property of a charge carrier in a material that is associated with its drift velocity (v_d) relative to an applied electric field (E). The formula for electric mobility (μ) is: μ = v_d / E The unit of drift velocity (v_d) is meters per second (m/s) and the unit of electric field (E) is volts per meter (V/m). Therefore, the unit of electric mobility (μ) is: μ = (m/s) / (V/m) = m²/(Vs)
05

Match the units and select the correct option

Now that we have the units for each term in column I, we can match them with the options in column II: (a) The unit of electrical resistivity is Ωm (s) (b) The unit of current density is A/m² (r) (c) The unit of electrical conductivity is S/m (q) (d) The unit of electric mobility is m²/(Vs) (p) Comparing these matches with the given options (A, B, C, D), we find that the correct match is: (B) \(a-s, b-r, c-q, d-p\)

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Most popular questions from this chapter

A Potentiometer wire, \(10 \mathrm{~m}\) long, has a resistance of \(40 \Omega\). It is connected in series with a resistance box and a \(2 \mathrm{v}\) storage cell If the potential gradient along the wire is $0.1 \mathrm{~m} \mathrm{v} / \mathrm{cm}$, the resistance unplugged in the box is. (A) \(260 \Omega\) (B) \(760 \Omega\) (C) \(960 \Omega\) (D) \(1060 \Omega\)

Figure, shows a network of seven resistors number 1 to 7, each equal to $1 \Omega\( connection to a \)4 \mathrm{~V}$ battery of negligible internal resistance The current I in the circuit is.... (A) \(0.5 \mathrm{~A}\) (B) \(1.5 \mathrm{~A}\) (C) \(2.0 \mathrm{~A}\) (D) \(3.5 \mathrm{~A}\)

In an experiment to measure the internal resistance of a cell by a potentiometer it is found that all the balance points at a length of $2 \mathrm{~m}$ when the cell is shunted by a 5 ohm resistance and is at a length of \(3 \mathrm{~m}\) when the cell is shunted by a 10 ohm resistance, the internal resistance of the cell is then: (A) \(1.5 \Omega\) (B) \(10 \Omega\) (C) \(15 \Omega\) (D) \(1 \Omega\)

For a cell of e.m.f \(2 \mathrm{~V}\), a balance is obtained for $50 \mathrm{~cm}\( of the potentiometer wire If the cell is shunted by a \)2 \Omega$ resistor and the balance is obtained across \(40 \mathrm{~cm}\) of the wire, then the internal resistance of the cell is. (A) \(1 \Omega\) (B) \(0.5 \Omega\) (C) \(1.2 \Omega\) (D) \(2.5 \Omega\)

The resistance of the series combination of two resistances is \(\mathrm{S}\), when they are joined in parallel the total resistance is \(\mathrm{P}\) If $S=n P\(, then the minimum possible value of \)n$ is.... (A) 4 (B) 3 (C) 2 (D) 1
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