Question

The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then
A. The resistance will be doubled and the specific resistance will be halved.
B. The resistance will be halved and the specific resistance will remain unchanged.
C. The resistance will be halved and the specific resistance will be doubled.
D. The resistance and the specific resistance, will both remain unchanged.

Answer 1

Hint: The resistance of a material is proportional to the specific resistivity of the material and length of the material. It is inversely proportional to the area of cross-section. The specific resistance of the material is independent of the dimension of the resistor.

Formula used:
$R = \dfrac{{\rho l}}{A}$
where R is the resistance of the wire of length l and cross-sectional area A, $\rho $ is the specific resistance of the material of the wire.

Complete step by step solution:
Resistance of wire is given as \[R = \dfrac{{\rho l}}{A}\]
Let the initial length of the wire is \[{l_1}\] and the area of cross section is \[{A_1}\]. Then the initial resistance of the wire is,
\[{R_1} = \dfrac{{\rho {l_1}}}{{{A_1}}}\]
When the final length of the wire is \[{l_2}\]and the area of cross section \[{A_2}\]. Then the final resistance of the wire is,
\[{R_2} = \dfrac{{\rho {l_2}}}{{{A_2}}}\]

As it is given that the radius and the length both are getting doubled. So,
\[{l_2} = 2{l_1}\]
So, the ratio of the final area to the initial area of cross-section of the wire is,
\[\dfrac{{{A_2}}}{{{A_1}}} = \dfrac{{\pi r_2^2}}{{\pi r_1^2}}\]
\[\Rightarrow \dfrac{{{A_2}}}{{{A_1}}} = \dfrac{{\pi {{\left( {2{r_1}} \right)}^2}}}{{\pi r_1^2}}\]
\[\Rightarrow \dfrac{{{A_2}}}{{{A_1}}} = 4\]

The ratio of the final length to the initial length of the wire is,
\[\dfrac{{{l_2}}}{{{l_1}}} = 2\]
The initial resistance of the wire is given as R ohm. We need to find the final resistance of the wire. Taking the ratio of the initial resistance of the wire to the final resistance, we get
\[\dfrac{{{R_1}}}{{{R_2}}} = \left( {\dfrac{{{l_1}}}{{{l_2}}}} \right) \times \left( {\dfrac{{A2}}{{{A_1}}}} \right)\]
\[\Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{2} \times 4\]
\[\Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\dfrac{{\rho {l_1}}}{{{A_2}}}}}{{\dfrac{{\rho {l_2}}}{{{A_2}}}}}\]
\[\therefore {R_2} = \dfrac{R}{2}\]
Hence, the final resistance of the wire is getting half the initial resistance but the specific resistance of the wire remains the same as before.

Therefore, the correct option is B.

Note: If the wire is not stretched then the other dimension remains the same. If the wire is stretched then the other dimension also changes to keep the volume constant. But as in this case the length as well as the radius is changed, so we cannot assume the volume to be constant.