A uniform cylindrical metallic conductor has an initial resistance of 12.0\,\Omega. The co… — JAMB UTME

A uniform cylindrical metallic conductor has an initial resistance of $12.0\,\Omega$ . The conductor is stretched uniformly until its length increases by $50\%$ , while maintaining constant mass and density. If a constant potential difference of $27.0\,\text{V}$ is subsequently applied across the ends of the stretched conductor, what is the electric current passing through it?

$1.0\,\text{A}$ Answer
B
$1.5\,\text{A}$
C
$2.25\,\text{A}$
D
$0.67\,\text{A}$

Answer

The electric current passing through the stretched conductor is

1.0\,\text{A}

When a metallic conductor of fixed mass and volume is stretched, increasing its length by a factor of

n = 1.5

causes its cross-sectional area to decrease by a factor of

1.5

. Because resistance is directly proportional to length and inversely proportional to cross-sectional area (

R = \rho L / A

), the new resistance becomes

n^2

times the initial resistance (

1.5^2 \times 12.0\,\Omega = 27.0\,\Omega

). By Ohm's law,

I = V / R = 27.0\,\text{V} / 27.0\,\Omega = 1.0\,\text{A}

Step-by-Step Solution

Determine the new length and cross-sectional area of the stretched conductor

L_2 = 1.5 L_1

and

A_2 = \frac{A_1}{1.5}

Increasing the length by

50\%

means

L_2 = L_1 + 0.5 L_1 = 1.5 L_1

. Since the volume

V = A \cdot L

remains constant during stretching,

A_1 L_1 = A_2 L_2

, which gives

A_2 = A_1 / 1.5

Calculate the new resistance of the conductor

R_2 = 27.0\,\Omega

Resistance is given by

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

. Substituting the new length and area gives

R_2 = \rho \frac{1.5 L_1}{A_1 / 1.5} = (1.5)^2 \rho \frac{L_1}{A_1} = 2.25 R_1 = 2.25 \times 12.0\,\Omega = 27.0\,\Omega

Apply Ohm's law to find the current

I = 1.0\,\text{A}

Using

I = \frac{V}{R_2}

, substitute

V = 27.0\,\text{V}

and

R_2 = 27.0\,\Omega

to get

I = \frac{27.0\,\text{V}}{27.0\,\Omega} = 1.0\,\text{A}

Key Concept

Resistance variation with length and cross-sectional area under constant volume constraint (

R \propto L^2

when stretched)

Estimated Time:2m 0s

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