Question

The new resistance of wire of \[R\,\Omega \], whose radius is reduced half, is
A. 16 R
B. 3R
C. 2R
D. R

Answer 1

Hint: The quantity that provides the hindrance to the flow of current is known as resistance. The resistance of a conductor is dependent on the length and area of cross-section of the conductor.

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 resistivity of the material of the wire.

Complete step by step solution:
Resistance of the 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 is getting halved keeping the length constant. So,
\[{l_2} = {l_1}\] and \[{r_2} = \dfrac{{{r_1}}}{2} \\ \]
So, the ratio of the final area to the initial area of cross-section of the wire is,
\[\dfrac{{{A_2}}}{{{A_1}}} = \dfrac{{\pi r_2^2}}{{\pi r_1^2}} \\ \]
\[\Rightarrow \dfrac{{{A_2}}}{{{A_1}}} = \dfrac{{\pi {{\left( {\dfrac{{{r_1}}}{2}} \right)}^2}}}{{\pi r_1^2}} \\ \]
\[\Rightarrow \dfrac{{{A_2}}}{{{A_1}}} = \dfrac{1}{4}\]
The final volume must be equal to initial volume,
\[{A_1}{l_1} = {A_2}{l_2}\]
\[\Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{A_2}}}{{{A_1}}} = \dfrac{1}{4} \\ \]
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
\[\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\dfrac{{\rho {l_1}}}{{{A_2}}}}}{{\dfrac{{\rho {l_2}}}{{{A_2}}}}} \\ \]
\[\Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \left( {\dfrac{{{l_1}}}{{{l_2}}}} \right) \times \left( {\dfrac{{{A_2}}}{{{A_1}}}} \right) \\ \]
\[\Rightarrow \dfrac{{R\Omega }}{{{R_2}}} = \left( {\dfrac{1}{4}} \right) \times \left( {\dfrac{1}{4}} \right) \\ \]
\[\therefore {R_2} = 16R\,\Omega \]
Hence, the final resistance of the wire is 16R.

Therefore, the correct option is A.

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