The resistance of a wire of length L and area of cross-section A is 1.0Ω The resistance of a win of the same material but of length 4L and area of cross-section 5A will be:

Correct option is B

The correct answer is (B)  0.8 Ω

Given:

1. R is the resistance,
2. ρ\rhoρ ​is the resistivity of the material (a constant for a given material),
3. L is the length of the wire,
4. A​ is the area of cross-section.

For 1st wire: $$R_1$$​=ρ\rhoρ$$\frac{L}{A}$$​​      equation 1     

1= ρ$$\frac{L}{A}$$​ ​ 

For 2nd wire: 

1. Length = $4L$,
2. Area = $5A$,
3. Resistance $$R_2$$=
   

​$$R_2$$=ρ$$\frac{4}{5}$$   equation 2

Now from equation(1) and (2) .........

​$$R_2$$= 1×$$\frac{4}{5}$$         

$$R_2$$= 0.8 Ω

The resistance of the second wire will be  0.8 Ω0.8Ω0.8 \, \Omega​