Q.A metal rod of length 10 cm and a rectangular cross-section of 1 cm×21 cm is connected to a battery across opposite faces. The resistance will be
Concept understanding — Resistivity
At a fixed temperature, a conductor's resistance is found to depend on three things: the material it is made of, its length l, and its cross-sectional area A -- specifically, R∝l and R∝1/A, combined as R=ρl/A, where ρ (rho), the RESISTIVITY (or specific resistance), is the material-dependent constant of proportionality, with SI unit the ohm-metre (Ωm). Numerically, resistivity is the resistance of a sample that has BOTH unit length and unit cross-sectional area at once. Resistivity is a property of the material itself, unlike resistance, which refers to one particular object of a given shape; conductors have resistivity of order 10−8Ωm, insulators as high as 1016Ωm, and semiconductors span the wide middle ground -- a span of some 24 orders of magnitude across ordinary materials.
Conductivity, σ=1/ρ, is resistivity's reciprocal. At the microscopic (per-point) level, resistivity relates the local electric field E inside a material to the local current density J via E=ρJ (equivalently J=σE) -- this is the point-by-point version of Ohm's law, reducing to the whole-component form R=V/I once the field and current density are integrated over an entire resistor's length and area.
Concept: Resistance and geometry, R=ρAL. The cuboid 10 cm×1 cm×21 cm has three pairs of opposite faces, each giving a different L/A:
- Across the 1 cm×21 cm end faces: L=10 cm, A=0.5 cm2 → R=ρ0.5×10−40.10=2000ρ.
- Across the 10 cm×21 cm faces: L=1 cm, A=5 cm2 → R=20ρ.
- Across the 10 cm×1 cm faces: L=21 cm, A=10 cm2 → R=5ρ.
The three values are in the ratio 2000:20:5=400:4:1. Resistance is largest when the current runs the long way through the smallest faces (greatest L, smallest A).
The resistance is maximum (R=2000ρ) when the battery is connected across the two 1 cm×21 cm faces (and minimum, 5ρ, across the 10 cm×1 cm faces) — option (a).
With R=ρL/A, the metal rod gives three different resistances depending on which pair of faces the battery touches; it is maximum, R=2000ρ, across the two smallest (1 cm×21 cm) faces, where the length is greatest and the area smallest.
Principle
For a uniform conductor,
R=ρAL
where L is the length along the current (between the connected faces) and A is the area of those faces. A cuboid has three distinct pairs of opposite faces, so three possible (L,A) combinations and three resistances.
The three orientations
Sides: 10 cm=0.10 m, 1 cm=0.01 m, 21 cm=0.005 m.
1. Across the 1 cm×21 cm end faces — longest path, smallest area:
L=0.10 m,A=(0.01)(0.005)=5×10−5 m2,
R1=ρ5×10−50.10=2000ρ.
2. Across the 10 cm×21 cm faces:
L=0.01 m,A=(0.10)(0.005)=5×10−4 m2,
R2=ρ5×10−40.01=20ρ.
3. Across the 10 cm×1 cm faces — shortest path, largest area:
L=0.005 m,A=(0.10)(0.01)=1×10−3 m2,
R3=ρ1×10−30.005=5ρ.
Comparison
R1:R2:R3=2000:20:5=400:4:1.
Resistance is greatest when the current is forced through the greatest length and smallest cross-section — the 1 cm×21 cm end faces — and least across the broad 10 cm×1 cm faces.
The resistance is maximum, R=2000ρ, across the two 1 cm×21 cm faces (minimum 5ρ across the 10 cm×1 cm faces); the three values are in the ratio 400:4:1 — option (a).
Method: Resistance Dependence on Geometry — Choosing the Current-Carrying Faces
For a conductor of given shape, this method finds how resistance changes depending on which pair of faces the current is driven across — apply it whenever a rod, block, or wire is offered in more than one possible orientation between the terminals.
Steps
Step 1: Write the resistance formula in terms of length and area
For a uniform conductor of resistivity ρ, length L (measured ALONG the direction of current flow) and cross-sectional area A (the area of the face the current enters/exits through, PERPENDICULAR to the flow):
R=ρAL
Step 2: Identify every distinct pair of current-entry faces
For a cuboid (or any solid with more than one pair of parallel faces), each pair of opposite faces defines a different possible current path: the length L for that orientation is the dimension connecting the two chosen faces, and A is the area of one of those faces (the product of the other two dimensions).
Step 3: Compute R for each orientation
Substitute the correct (L,A) pair for each orientation into Step 1's formula, keeping units consistent (convert every length to the same unit, typically metres, before multiplying or dividing).
Step 4: Compare using ratios, not absolute values
Since ρ is the same in every orientation (same material), it cancels out of any comparison: the RATIO of resistances between two orientations is simply the ratio of their respective L/A values. This lets you rank the orientations without needing to know ρ numerically.
Step 5: Applying the geometric rule
Resistance is largest for the orientation with the GREATEST length and SMALLEST cross-section (current forced the "long way" through the "thin way"), and smallest for the orientation with the shortest length and largest cross-section. Identify these directly from the object's dimensions before computing any numbers, then confirm with the Step 3 calculation.
Showing the 12 most recent of 36 on this concept.
- CBSE 2026Set DS1 markQ.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 (σ=1/ρ).
Electrical conductivity (conductance). The conductance of a conductor is the reciprocal of its resistance and measures how easily current flows through it:
G=R1.
Its SI unit is the siemens (S) or mho (Ω−1).
Specific conductivity (conductivity). The conductivity of a material is the reciprocal of its specific resistance (resistivity):
σ=ρ1.
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=σlA.
✓Final answerConductance G=1/R (unit S or mho); specific conductivity σ=1/ρ (unit S m−1).
- CBSE 2026Set A1 markMCQQ.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=ρAL,
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 markMCQQ.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=ρL/A gives R1:R2=5:8.
Resistance of a wire is R=ρAL, where ρ is resistivity, L is length and A is area of cross-section.
Given: ρ1:ρ2=2:3, L1:L2=3:4, A1:A2=4:5.
R2R1=ρ2ρ1⋅L2L1⋅A1A2=32×43×45=3×4×42×3×5=4830=85
✓Final answer(c) 5:8.
- CBSE 2026Set ANNUAL1 markQ.Fill in the blank: The reciprocal of resistivity is .............
›Reveal solutionSolution
Resistivity ρ and conductivity σ are reciprocals of each other: σ=1/ρ.
Resistivity (ρ) measures how strongly a material opposes current flow. Its reciprocal, σ=ρ1, measures how easily a material conducts current, and is called conductivity, with SI unit siemens per metre (S/m) or Ω−1m−1.
✓Final answerThe reciprocal of resistivity is conductivity.
- CBSE 2026Set ANNUAL1 markMCQQ.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 52=25.
Original: R=ρAL.
On melting and stretching, the volume of the metal is conserved: AL=A′L′. With L′=5L:
A′=L′AL=5LAL=5A
New resistance:
R′=ρA′L′=ρA/55L=25(ρAL)=25R
This is the standard result: for constant volume, resistance is proportional to the square of the length (R∝L2), 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 markMCQQ.Resistivity of a metal wire depends on its :(a) area of cross-section(b) length(c) material(d) volume
›Reveal solutionSolution
(c) material. Resistivity ρ is an intrinsic property of the substance (and its temperature); it does not depend on the wire's dimensions. Resistance R=ρL/A depends on length and area,
✓Final answer(c) material.
Resistivity ρ is an intrinsic property of the substance (and its temperature); it does not depend on the wire's dimensions. Resistance R=ρL/A depends on length and area, but ρ itself depends only on the material.
- CBSE 2025Set ANNUAL1 markQ.Resistivity of material of a conducting wire of length ℓ is ρ. If length of this wire is increased to 2ℓ 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 ρ 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=ρL/A does depend on length L and area A — doubling the length to 2ℓ would double the resistance (if the area stays the same) — but the resistivity ρ itself stays exactly the same, since the material has not changed.
✓Final answerResistivity remains ρ — it is unchanged, because resistivity depends only on the material, not on the wire's length. (Only the resistance would change, becoming 2R.)
- CBSE 2025Set D1 markMCQQ.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=ρAl
Dimensionally this gives (Ω·m)(m)/(m²) = Ω, confirming the form. The other options give incorrect units.
✓Final answer(A) ρ·(l/A).
- CBSE 2025Set ANNUAL1 markMCQQ.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=ρAL
where ρ is resistivity (a material constant, unchanged), L is length, and A is the cross-sectional area (kept the same here).
If L→2L and A is unchanged:
R′=ρA2L=2(ρAL)=2R
✓Final answer(a) doubled — resistance is directly proportional to length.
- CBSE 2025Set ANNUAL1 markQ.The specific resistances of copper, silver and constantan are 1.78×10−6Ω-cm, 1×10−6Ω-cm and 4.8×10−6Ω-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) ρ measures how strongly a material opposes the flow of current -- the lower the resistivity, the better the conductor (since resistance R=ρl/A is directly proportional to ρ).
Given:
ρCu=1.78×10−6Ω-cm,ρAg=1×10−6Ω-cm,ρconstantan=4.8×10−6Ω-cm
Silver has the smallest resistivity of the three, so silver is the best conductor.
✓Final answerSilver is the best conductor, because it has the lowest specific resistance (resistivity) among the three materials.
Alternative (Or):
For n identical cells, each of e.m.f. ε and internal resistance r, connected in series to an external resistance R, the current is
Iseries=R+nrnε
If instead they are connected in parallel,
Iparallel=R+r/nε
When R≫r (external resistance very large compared to the total internal resistance), the nr or r/n term in the denominator becomes negligible compared to R in both cases. In that limit the series combination delivers the full e.m.f. nε through essentially just R, giving Iseries≈nε/R, which is n times larger than Iparallel≈ε/R. Hence, when R≫r, the cells should be connected in series to get the maximum current.
✓Final answerConnect the cells in series.
- CBSE 2024Set FS1 markQ.Write the definition and dimensional formula of electrical conductivity.
›Reveal solutionSolution
Conductivity is the reciprocal of resistivity, σ=1/ρ, with dimensions [M−1L−3T3A2].
Definition. Electrical conductivity σ of a material is the reciprocal of its resistivity, σ=ρ1. Equivalently, in the microscopic form of Ohm's law J=σE, it is the current density produced per unit electric field. Its SI unit is siemens per metre (Ω−1m−1).
Dimensional formula. Resistivity ρ=LRA with [R]=[ML2T−3A−2] gives
[ρ]=[L][ML2T−3A−2][L2]=[ML3T−3A−2].
Hence
[σ]=[ρ]−1=[M−1L−3T3A2].
✓Final answerσ=ρ1, dimensional formula [M−1L−3T3A2] (SI unit Ω−1m−1).
- CBSE 2024Set A1 markMCQQ.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=R1=5001=0.002 Ω−1 (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.
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