Question

A wire of resistance 18 ohm is drawn until its radius reduces \[\dfrac{1}{2}th\] of its original radius then resistance of the wire is:
A. 188Ω
B. 72Ω
C. 288Ω
D. 388Ω

Answer 1

Hint:
In this question, we need to determine the resistance of a wire for the given length. For this we will use the relation between resistance, resistivity, length and the area of the wire which is given as \[R = \rho \dfrac{l}{A}\].

Complete step by step solution:
Resistance of wire \[R = 18\Omega \]
Let the initial length of the wire be \[l\]
Radius of the original wire be \[r\]
New radius after stretching be \[r'\]
New length after stretching be \[l'\]
As we know the volume on stretching is always equal, hence we can say
\[{V_1} = {V_2}\]
Where \[{V_1}\]is the volume of the original wire and \[{V_2}\]is the volume after stretching, now by we can write
\[
  {V_1} = {V_2} \\
  \pi r_1^2{h_1} = \pi r_2^2{h_2} \\
  \pi {r^2}l = \pi {\left( {\dfrac{r}{2}} \right)^2}l' \\
  l' = \dfrac{{{r^2}l}}{{\dfrac{{{r^2}}}{4}}} \\
   = 4l \\
 \]
Resistance of wire is directly proportional to its length and inversely proportional to the area of cross area of the section, which is given by the formula \[R = \rho \dfrac{l}{A}\] , where \[\rho \]is the resistivity of the material, \[l\]is the length of the material and A is the area.
We know the resistance of wire is given as \[R = \rho \dfrac{l}{A}\]
Where the resistance of wire for initial condition is given as
\[
  R = \rho \dfrac{l}{A} \\
  \rho \dfrac{l}{A} = 18\Omega - - (i) \\
 \]
Now the resistance of the wire drawn until its radius reduces \[\dfrac{1}{2}th\] of its original radius will become
\[
  R' = \rho \dfrac{{l'}}{{\pi {{\left( {r'} \right)}^2}}} \\
   = \rho \dfrac{{4l}}{{\pi {{\left( {\dfrac{r}{2}} \right)}^2}}} \\
   = \rho \dfrac{{4l}}{{\pi \dfrac{{{r^2}}}{4}}} \\
   = 16\rho \dfrac{l}{{\pi {r^2}}} \\
   = 16\rho \dfrac{l}{A} \\
   = 16R \\
 \]
Since the resistance of initial wire is 18 ohm hence the new resistance becomes
\[
  R' = 16R \\
   = 16 \times 18 \\
   = 288\Omega \\
 \]
Hence, the resistance of the wire will be 288 ohms.

Option B is correct.

Note:
When a wire is drawn to its half of the radius then the length of the wire increases since the volume of stretching remains constant. Students must note here that the volume of the wire will remain the same on compressing or elongating the length of the wire as its radius of the base will change accordingly.