In a current carrying conductor, the ratio of the electric field and the current density at a point is called

Question

Q. In a current carrying conductor, the ratio of the electric field and the current density at a point is called (A) Resistivity (B) Conductivity (C) Resistance (D) Mobility

CBSECBSE Class XII Board 2020MCQ· 1mImportance★★★★★

✓ Free question

The ratio of electric field ( $E$ ) to current density ( $J$ ) in a conductor is defined as resistivity ( $\rho$ ), from Ohm's law in microscopic form: $E = \rho J$ . The correct option is (A).

The question tests your understanding of the microscopic form of Ohm's law — a concept that connects the macroscopic idea of resistance to the intrinsic material property. Let's build the intuition.

When current flows through a conductor, an electric field $E$ inside the material pushes the charge carriers (electrons). The resulting current density $J$ (current per unit cross-sectional area) is directly proportional to $E$ for most conductors. The constant of proportionality is the material's conductivity $\sigma$ , giving $J = \sigma E$ . Rearranging, $E = \frac{1}{\sigma} J = \rho J$ , where $\rho$ is resistivity. So $\rho = E/J$ — exactly the ratio asked.

Now, why not the other options? Let's check each:

Resistivity ( $\rho$ ): From $E = \rho J$ , we get $\rho = E/J$ . This is a material property, independent of the conductor's shape or size. It tells how strongly the material opposes current flow. This matches the given ratio.
Conductivity ( $\sigma$ ): This is the reciprocal: $\sigma = J/E$ . The question asks for $E/J$ , not $J/E$ , so conductivity is the inverse ratio.
Resistance ( $R$ ): Resistance is a macroscopic quantity: $R = V/I$ , where $V$ is potential difference across the conductor and $I$ is total current. While related to resistivity via $R = \rho L/A$ (for a uniform wire of length $L$ and area $A$ ), resistance depends on geometry, not just the local field and current density. The ratio $E/J$ is a point property, not a whole-conductor property.
Mobility ( $\mu$ ): Mobility relates drift velocity $v_d$ to electric field: $v_d = \mu E$ . It's not directly $E/J$ . Current density $J = n e v_d = n e \mu E$ , so $J/E = n e \mu$ , which is conductivity. Again, not the ratio asked.

Watch out

A common mistake is to confuse resistivity with resistance. Remember: resistivity is a material constant (like density), while resistance depends on both material and geometry. The ratio $E/J$ is a local, point-wise quantity — that's resistivity, not resistance.

Tip

To quickly recall: microscopic Ohm's law: $J = \sigma E$ or $E = \rho J$ . If the ratio is $E/J$ , it's $\rho$ (resistivity). If it's $J/E$ , it's $\sigma$ (conductivity). Think of $\rho$ as "how much field is needed per unit current density" — a measure of the material's opposition.

✓Final answer

The correct option is (A) Resistivity .

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 36 on this concept.

CBSE 2026Set DS1 mark
$Q. Write the definitions of electrical conductivity and specific conductivity.$
›Reveal solutionSolution

Conductance is the reciprocal of resistance ( $G=1/R$ ); specific conductivity is the reciprocal of resistivity ( $\sigma=1/\rho$ ).

Electrical conductivity (conductance). The conductance of a conductor is the reciprocal of its resistance and measures how easily current flows through it:

$G = \frac{1}{R}.$

Its SI unit is the siemens (S) or mho ( $\Omega^{-1}$ ).

Specific conductivity (conductivity). The conductivity of a material is the reciprocal of its specific resistance (resistivity):

$\sigma = \frac{1}{\rho}.$

It is a property of the material (independent of size/shape) and its SI unit is siemens per metre (S m $^{-1}$ ) or mho m $^{-1}$ . The two are related through the geometry of the conductor by $G = \sigma\dfrac{A}{l}$ .

✓Final answer
$G = 1/R$
CBSE 2026Set A1 markMCQ
$Q. Which factor of the following does not affect resistance of a conducting wire? (A) Length (B) Temperature (C) Material (D) Voltage applied$
›Reveal solutionSolution

R = ρL/A; ρ depends on material and temperature. Applied voltage does not change R (for an ohmic conductor).

The resistance of a wire is

$R = \rho\frac{L}{A},$

where ρ is the resistivity (a property of the material, which also varies with temperature), L is the length, and A is the cross-sectional area.

Applying a voltage merely drives current through the wire — for an ohmic conductor at fixed temperature it does not alter R. Hence the applied voltage is not a factor determining resistance.

✓Final answer
$(D) Voltage applied.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Two wires that are made up of two different materials have specific resistances in the ratio 2:3, lengths 3:4 and areas 4:5. The ratio of their resistances is$
(a) 6:5
(b) 6:8
(c) 5:8
(d) 1:2
›Reveal solutionSolution

Combining the given ratios of resistivity, length and area through $R=\rho L/A$ gives $R_1:R_2 = 5:8$ .

Resistance of a wire is $R = \rho \dfrac{L}{A}$ , where $\rho$ is resistivity, $L$ is length and $A$ is area of cross-section.

Given: $\rho_1:\rho_2 = 2:3$ , $L_1:L_2 = 3:4$ , $A_1:A_2 = 4:5$ .

$\dfrac{R_1}{R_2} = \dfrac{\rho_1}{\rho_2}\cdot\dfrac{L_1}{L_2}\cdot\dfrac{A_2}{A_1} = \dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{5}{4} = \dfrac{2\times3\times5}{3\times4\times4} = \dfrac{30}{48} = \dfrac{5}{8}$

✓Final answer
$(c) 5:8.$
CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: The reciprocal of resistivity is .............$
›Reveal solutionSolution

Resistivity $\rho$ and conductivity $\sigma$ are reciprocals of each other: $\sigma = 1/\rho$ .

Resistivity ( $\rho$ ) measures how strongly a material opposes current flow. Its reciprocal, $\sigma = \dfrac{1}{\rho}$ , measures how easily a material conducts current, and is called conductivity, with SI unit siemens per metre (S/m) or $\Omega^{-1}\text{m}^{-1}$ .

✓Final answer
$The reciprocal of resistivity is conductivity.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The resistance of a wire is 'R' ohm. If it is melted and stretched to 5 times its original length, its new resistances will be$
(a) 5 R
(b) R/5
(c) 25 R
(d) R/25
›Reveal solutionSolution

Stretching a wire to 5 times its length (with volume conserved) shrinks its area by 5 times too, so resistance scales up by $5^2 = 25$ .

Original: $R = \rho \dfrac{L}{A}$ .

On melting and stretching, the volume of the metal is conserved: $A L = A' L'$ . With $L' = 5L$ :

$A' = \frac{AL}{L'} = \frac{AL}{5L} = \frac{A}{5}$

New resistance:

$R' = \rho \frac{L'}{A'} = \rho \frac{5L}{A/5} = 25 \left(\rho \frac{L}{A}\right) = 25R$

This is the standard result: for constant volume, resistance is proportional to the square of the length ( $R \propto L^2$ ), since both the increase in length and the corresponding decrease in area work in the same direction.

✓Final answer
$(c) 25 R.$
CBSE 2025Set X11 markMCQ
$Q. Resistivity of a metal wire depends on its :$
(a) area of cross-section
(b) length
(c) material
(d) volume
›Reveal solutionSolution

(c) material. Resistivity $\rho$ is an intrinsic property of the substance (and its temperature); it does not depend on the wire's dimensions. Resistance $R = \rho L/A$ depends on length and area,

✓Final answer
$\rho$
CBSE 2025Set ANNUAL1 mark
$\ell$
›Reveal solutionSolution

Resistivity is an intrinsic material property; it does not depend on the wire's length.

Resistivity $\rho$ is a property of the material of the conductor (it depends on the nature of the material and on temperature), not on the conductor's dimensions. The resistance $R = \rho L/A$ does depend on length $L$ and area $A$ — doubling the length to $2\ell$ would double the resistance (if the area stays the same) — but the resistivity $\rho$ itself stays exactly the same, since the material has not changed.

✓Final answer
$\rho$
CBSE 2025Set D1 markMCQ
$Q. Which of the following represents resistance R? ( ρ = resistivity, l = length of a material, A = cross-sectional area ) (A) ρ.(l/A) (B) ρ.(A/l) (C) l/ρA (D) lA/ρ$
›Reveal solutionSolution

Resistance of a uniform conductor is R = ρl/A.

The resistance of a conductor is directly proportional to its length l and inversely proportional to its cross-sectional area A, with the resistivity ρ as the constant of proportionality:

$R = \rho\,\frac{l}{A}$

Dimensionally this gives (Ω·m)(m)/(m²) = Ω, confirming the form. The other options give incorrect units.

✓Final answer
$(A) ρ\cdot(l/A).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. When the length of a copper wire is doubled keeping its area of cross-section same its resistance is :$
(a) doubled
(b) half
(c) no change
(d) four times
›Reveal solutionSolution

Resistance is directly proportional to length, so doubling the length doubles the resistance.

Resistance of a wire is given by

$R = \rho \frac{L}{A}$

where ρ is resistivity (a material constant, unchanged), L is length, and A is the cross-sectional area (kept the same here).

If $L \to 2L$ and A is unchanged:

$R' = \rho \frac{2L}{A} = 2\left(\rho\frac{L}{A}\right) = 2R$

✓Final answer
$(a) doubled — resistance is directly proportional to length.$
CBSE 2025Set ANNUAL1 mark
$1.78\times10^{-6}\,\Omega$
›Reveal solutionSolution

Comparing resistivities shows silver has the lowest value, making it the best conductor; for the alternative, cells in series maximize current when external resistance dominates.

Which is the best conductor?

The specific resistance (resistivity) $\rho$ measures how strongly a material opposes the flow of current -- the lower the resistivity, the better the conductor (since resistance $R=\rho l/A$ is directly proportional to $\rho$ ).

Given:

$\rho_{Cu} = 1.78\times10^{-6}\,\Omega\text{-cm}, \quad \rho_{Ag} = 1\times10^{-6}\,\Omega\text{-cm}, \quad \rho_{constantan} = 4.8\times10^{-6}\,\Omega\text{-cm}$

Silver has the smallest resistivity of the three, so silver is the best conductor.

✓Final answer
$n$

✓Final answer
$Connect the cells in series.$
CBSE 2024Set FS1 mark
$Q. Write the definition and dimensional formula of electrical conductivity.$
›Reveal solutionSolution

Conductivity is the reciprocal of resistivity, $\sigma=1/\rho$ , with dimensions $[M^{-1}L^{-3}T^{3}A^{2}]$ .

Definition. Electrical conductivity $\sigma$ of a material is the reciprocal of its resistivity, $\sigma=\dfrac{1}{\rho}$ . Equivalently, in the microscopic form of Ohm's law $\vec{J}=\sigma\vec{E}$ , it is the current density produced per unit electric field. Its SI unit is siemens per metre ( $\Omega^{-1}\,m^{-1}$ ).

Dimensional formula. Resistivity $\rho=\dfrac{RA}{L}$ with $[R]=[ML^{2}T^{-3}A^{-2}]$ gives

$[\rho]=\frac{[ML^{2}T^{-3}A^{-2}][L^{2}]}{[L]}=[ML^{3}T^{-3}A^{-2}].$

Hence

$[\sigma]=[\rho]^{-1}=[M^{-1}L^{-3}T^{3}A^{2}].$

✓Final answer
$\sigma=\dfrac{1}{\rho}$
CBSE 2024Set A1 markMCQ
$Q. The resistance of any wire is 500 Ω. Its electrical conductivity will be (A) 0.002 ohm^-1 (B) 0.02 ohm^-1 (C) 50 ohm^-1 (D) 500 ohm^-1$
›Reveal solutionSolution

Conductance is the reciprocal of resistance: G = 1/R = 1/500 = 0.002 Ω⁻¹.

The electrical conductance (reciprocal of resistance) is

$G = \frac{1}{R} = \frac{1}{500} = 0.002\ \Omega^{-1}\ (\text{siemens})$

(The reciprocal of resistance has unit ohm⁻¹ = siemens, matching the options; this is what the question intends.)

✓Final answer
$(A) 0.002 ohm⁻¹.$

Q.In a current carrying conductor, the ratio of the electric field and the current density at a point is called (A) Resistivity (B) Conductivity (C) Resistance (D) Mobility

Concept understanding — Resistivity

Q.In a current carrying conductor, the ratio of the electric field and the current density at a point is called (A) Resistivity (B) Conductivity (C) Resistance (D) Mobility

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