Question

A wire made of aluminium having resistivity $\rho=2.8\times10^{-8}\,\Omega\text{ m}$ with a circular cross section and has a radius of $2\times10^{-3}$ m. A current of 5 A flows through the wire. If the voltage difference between the ends is 1 V, what is the length of the wire in meters?

• A. 50
• B. 60
• C. 90
• D. 120
• E. 110

Hint:

Always check powers of 10 carefully when working with resistivity. A small error in the exponent will lead to an answer that is off by a factor of 10 or 100.

Answer 1

The Correct Option is C

Concept:
We use Ohm's Law ($V = IR$) and the formula for resistance ($R = \rho \frac{l}{A}$). Area of circular cross-section $A = \pi r^2$.

Step 1: Calculate the required Resistance ($R$).
Given $V = 1$ V and $I = 5$ A. \[ R = \frac{V}{I} = \frac{1}{5} = 0.2 \ \Omega \]

Step 2: Calculate the cross-sectional Area ($A$).
$r = 2 \times 10^{-3}$ m. \[ A = \pi r^2 = \pi (2 \times 10^{-3})^2 = 4\pi \times 10^{-6} \text{ m}^2 \]

Step 3: Calculate the length ($l$).
\[ R = \rho \frac{l}{A} \implies l = \frac{R \times A}{\rho} \] \[ l = \frac{0.2 \times (4\pi \times 10^{-6})}{2.8 \times 10^{-8}} = \frac{0.8\pi \times 10^{-6}}{2.8 \times 10^{-8}} \] \[ l = \frac{0.8 \times 3.14159}{2.8} \times 10^2 \approx 0.897 \times 100 \approx 89.7 \text{ m} \] Rounding to the nearest option, we get 90 m.