A wire of a given material has length l and cross-sectional area A , and its resistance is 9 Ω. Another wire of the same material has length $\frac{l}{3}$ and cross-sectional area 3A. What will be the resistance of the second wire?

The correct option is 1 Ω.

Underlying Concept:

The electrical resistance of a wire depends on its dimensions and the material it is made of, according to the formula:

$R=\rho \frac{l}{A}$

Where:
• $R$ is the resistance of the wire,
• $\rho$ is the electrical resistivity of the material,
• $l$ is the length of the wire, and
• $A$ is the cross-sectional area.

Step-by-Step Derivation:

Step 1: Express the resistance of the first wire.
For the first wire, the length is $l$, the cross-sectional area is $A$, and its resistance is given as 9 Ω. Therefore:

$R_1=\rho \frac{l}{A}=9\Omega$

Step 2: Set up the formula for the second wire.
The second wire is made of the same material, which means its resistivity ($\rho$) is identical. The new dimensions are:
• New length, $l_2=\frac{l}{3}$
• New cross-sectional area, $A_2=3A$

The resistance of the second wire, $R_2$, is written as:

$R_2=\rho \frac{l_2}{A_2}$

Step 3: Substitute the new dimensions and simplify.
Substituting $l_2$ and $A_2$ into the resistance formula for the second wire:

$R_2=\rho \frac{\left(\frac{l}{3}\right)}{3A}$

Simplifying the fraction:

$R_2=\rho \frac{l}{3\times 3A}$

$R_2=\frac{1}{9}\left(\rho \frac{l}{A}\right)$

Step 4: Calculate the final resistance.
Since we know from Step 1 that $\rho \frac{l}{A}=9\Omega$, we substitute this value into the equation:

$R_2=\frac{1}{9}\times 9\Omega =1\Omega$

Thus, the resistance of the second wire is 1 Ω.