The resistance of a wire of  Length (L) and the area of cross - section (A) is 0.2Ω. The resistance of a wire same material but of Length (2L) and  area of cross - section (4A) will be :

Correct option is D

Given:​

• Resistance of a wire of length $L$ and cross-sectional area $A$ is $R=0.2\Omega$.
• We need to find the resistance of a wire of the same material with length $2L$ and cross-sectional area $4A$.

Formula:
The resistance $R$ of a wire is given by the formula :

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

where $\rho$ is the resistivity of the material, $L$ is the length, and $A$ is the area of cross-section.

For the second wire, let the new resistance be R2R_2R
​$$_2$$​  for the wire with length $2L$ and area of cross-section $4A$.

Using the same formula, the resistance R2R_2R
​$$_2$$​ is:

$$R_2 = \rho \frac{2L}{4A}$$

$$R_2 = \frac{\rho \frac{L}{A}}{2}$$

                                                                           $$R_2 = \frac{R_1}{2}$$

Solution :

The original resistance $$R= \rho\frac{L}{A}= 0.2Ω.$$   

So, the resistance of the second wire will be:

                                                                       $$R_2=\frac{0.2}{2}=0.1Ω$$

The resistance of the wire with length $2L$ and cross-sectional area $4A$ will be 0.1 Ω.

The correct answer is option (d) 0.1 Ω.

R=ρLA=0.2ΩR = \rho \frac{L}{A} = 0.2 \, \Omega​​