Q.A metallic wire of length $l$ and resistance $5\Omega$ is stretched to double its length. Find its new resistivity and new resistance. Assume that there is no change in the density on stretching. OR A battery of e.m.f 10V and internal resistance $3\Omega$ is connected to a resistor R. If the current in the circuit is 0.5A. Calculate the value of R. What is the terminal voltage of the battery when the circuit is closed?

Showing the 12 most recent of 36 on this concept.

1. 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):

   $\sigma = \frac{1}{\rho}.$ …

   🔒 Start your 14-day free trial to see the full solution →
2. 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.

   …

   🔒 Start your 14-day free trial to see the full solution →
3. 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: $\rho_1:\rho_2 = 2:3$, $L_1:L_2 = 3:4$, $A_1:A_2 = 4:5$.

   …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 …

   🔒 Start your 14-day free trial to see the full solution →
5. 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:

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

   …

   🔒 Start your 14-day free trial to see the full solution →
6. 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, …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set ANNUAL1 mark
   Q.Resistivity of material of a conducting wire of length $\ell$ is $\rho$. If length of this wire is increased to $2\ell$ then what will be resistivity of material of this wire?

   ›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 …

   🔒 Start your 14-day free trial to see the full solution →
8. 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:

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

   …

   🔒 Start your 14-day free trial to see the full solution →
9. 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).

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2025Set ANNUAL1 mark
    Q.The specific resistances of copper, silver and constantan are $1.78\times10^{-6}\,\Omega$-cm, $1\times10^{-6}\,\Omega$-cm and $4.8\times10^{-6}\,\Omega$-cm respectively. Which is the best conductor and why? OR In which way should the cells be combined to get maximum current, when the external resistance is very high compared to the total internal resistance of the cells?

    ›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.

    …

    🔒 Start your 14-day free trial to see the full solution →
11. 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 …

    🔒 Start your 14-day free trial to see the full solution →
12. 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

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

    …

    🔒 Start your 14-day free trial to see the full solution →