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NCERT Exemplar · Q3

Q.A metal rod of length 10 cm10\ \text{cm} and a rectangular cross-section of 1 cm×12 cm1\ \text{cm} \times \tfrac{1}{2}\ \text{cm} is connected to a battery across opposite faces. The resistance will be

(a) maximum when the battery is connected across 1 cm × 1/2 cm faces.
(b) maximum when the battery is connected across 10 cm × 1 cm faces.
(c) maximum when the battery is connected across 10 cm × 1/2 cm faces.
(d) same irrespective of the three faces.
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✓ Free question

With R=ρ L/AR = \rho\,L/A, the metal rod gives three different resistances depending on which pair of faces the battery touches; it is maximum, R=2000 ρR = 2000\,\rho, across the two smallest (1 cm×12 cm1\ \text{cm}\times\tfrac12\ \text{cm}) faces, where the length is greatest and the area smallest.

Principle

For a uniform conductor,

R=ρ LAR = \rho\,\frac{L}{A}

where LL is the length along the current (between the connected faces) and AA is the area of those faces. A cuboid has three distinct pairs of opposite faces, so three possible (L,A)(L,A) combinations and three resistances.

The three orientations

Sides: 10 cm=0.10 m10\ \text{cm} = 0.10\ \text{m}, 1 cm=0.01 m1\ \text{cm} = 0.01\ \text{m}, 12 cm=0.005 m\tfrac12\ \text{cm} = 0.005\ \text{m}.

1. Across the 1 cm×12 cm1\ \text{cm}\times\tfrac12\ \text{cm} end faces — longest path, smallest area:

L=0.10 m,A=(0.01)(0.005)=5×10−5 m2,L = 0.10\ \text{m}, \quad A = (0.01)(0.005) = 5\times10^{-5}\ \text{m}^2,

R1=ρ 0.105×10−5=2000 ρ.R_1 = \rho\,\frac{0.10}{5\times10^{-5}} = 2000\,\rho.

2. Across the 10 cm×12 cm10\ \text{cm}\times\tfrac12\ \text{cm} faces:

L=0.01 m,A=(0.10)(0.005)=5×10−4 m2,L = 0.01\ \text{m}, \quad A = (0.10)(0.005) = 5\times10^{-4}\ \text{m}^2,

R2=ρ 0.015×10−4=20 ρ.R_2 = \rho\,\frac{0.01}{5\times10^{-4}} = 20\,\rho.

3. Across the 10 cm×1 cm10\ \text{cm}\times1\ \text{cm} faces — shortest path, largest area:

L=0.005 m,A=(0.10)(0.01)=1×10−3 m2,L = 0.005\ \text{m}, \quad A = (0.10)(0.01) = 1\times10^{-3}\ \text{m}^2,

R3=ρ 0.0051×10−3=5 ρ.R_3 = \rho\,\frac{0.005}{1\times10^{-3}} = 5\,\rho.

Comparison

R1:R2:R3=2000:20:5=400:4:1.R_1 : R_2 : R_3 = 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×12 cm1\ \text{cm}\times\tfrac12\ \text{cm} end faces — and least across the broad 10 cm×1 cm10\ \text{cm}\times1\ \text{cm} faces.

✓Final answer

The resistance is maximum, R=2000 ρR = 2000\,\rho, across the two 1 cm×12 cm1\ \text{cm}\times\tfrac12\ \text{cm} faces (minimum 5 ρ5\,\rho across the 10 cm×1 cm10\ \text{cm}\times1\ \text{cm} faces); the three values are in the ratio 400:4:1400:4:1 — option (a).

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