A metallic wire has a resistance of 12.0\,\Omega at 0\,^\circ\text{C} and a temperature co… — JAMB UTME

A metallic wire has a resistance of $12.0\,\Omega$ at $0\,^\circ\text{C}$ and a temperature coefficient of resistance of $4.0 \times 10^{-3}\,^\circ\text{C}^{-1}$ . The wire is uniformly stretched until its length increases by $25\%$ . Assuming the density and total volume of the wire remain constant during stretching, what is the resistance of the stretched wire at $50\,^\circ\text{C}$ ?

Answer: 22.5 Ω

Answer

The resistance of the stretched wire at

50\,^\circ\text{C}

22.5\,\Omega

Stretching a wire by

25\%

increases its length by a factor of

1.25

and reduces its cross-sectional area by a factor of

1.25

(since volume is conserved). The resistance at

0\,^\circ\text{C}

scales as

(1.25)^2 = 1.5625

, giving

18.75\,\Omega

. Accounting for the temperature increase to

50\,^\circ\text{C}

via

R(T) = R'_0(1 + \alpha T)

yields

18.75 \times (1 + 4.0 \times 10^{-3} \times 50) = 18.75 \times 1.20 = 22.5\,\Omega

Step-by-Step Solution

Calculate the resistance of the wire at

0\,^\circ\text{C}

after uniform stretching.

R'_0 = 18.75\,\Omega

Uniform stretching by

25\%

increases length to

L' = 1.25 L_0

. Volume conservation (

V = A L

) requires area to decrease to

A' = A_0 / 1.25

. Since

R = \rho L / A

R'_0 = R_0 (L'/L_0)^2 = 12.0 \times (1.25)^2 = 18.75\,\Omega

Apply the temperature coefficient formula to calculate resistance at

50\,^\circ\text{C}

R(50) = 22.5\,\Omega

Using

R(T) = R'_0 (1 + \alpha T)

, substitute

R'_0 = 18.75\,\Omega

\alpha = 4.0 \times 10^{-3}\,^\circ\text{C}^{-1}

, and

T = 50\,^\circ\text{C}

to find

R(50) = 18.75 \times (1 + 0.20) = 22.5\,\Omega

Key Concept

Combined effects of dimensional deformation and temperature on electrical resistance

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