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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

An electric kettle has two coils. When one of them is switched on, the water in the kettle boils in 6 minutes. When the other coil is switched on, the water boils in 3 minutes If the two coils are connected in series the time taken to boil water in the kettle is: (A) 3 minutes (B) 6 minutes (C) 2 minutes (D) 9 minutes

Two heater wires of equal length are first connected in series and then in parallel The ratio of heat produced in the two cases is (A) \(2: 1\) (B) \(1: 2\) (C) \(4: 1\) (D) \(1: 4\)

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) \(1 \Omega\)

The effective resistance of a n number of resistors connected in parallel in \(\mathrm{x} \mathrm{ohm}\). When one of the resistors is removed, the effective resistance becomes y ohm. The resistance of the resistor that is removed is.... (A) \(\\{(\mathrm{xy}) /(\mathrm{x}+\mathrm{y})\\}\) (B) \(\\{(\mathrm{xy}) /(\mathrm{y}-\mathrm{x})\\}\) (C) \((\mathrm{y}-\mathrm{x})\) (D) \(\sqrt{x y}\)

The circuit shown in fig consists of the following \(E_{1}=6, E_{2}=2, E_{3}=3\) Volt \(\mathrm{R}_{1}=6, \mathrm{R}_{4}=3 \mathrm{Ohm}\) \(\mathrm{R}_{3}=4, \mathrm{R}_{2}=2 \mathrm{Ohm}\) $\mathrm{C}=5 \mu \mathrm{F}$$E_{1}=6 \mathrm{~V} \quad E_{2}=2 \mathrm{~V} \quad E_{3}=3 \mathrm{~V} \quad \mathrm{R}_{1}=6 \Omega$ \(R_{2}=2 \Omega \quad R_{3}=4 \Omega \quad R_{4}=3 \Omega\) The energy stored in the capacitor is. (A) \(4.8 \times 10^{-6} \mathrm{~J}\) (B) \(9.6 \times 10^{-6} \mathrm{~J}\) (C) \(1.44 \times 10^{-5} \mathrm{~J}\) (D) \(1.92 \times 10^{-5} \mathrm{~J}\)
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