Question

A uniform wire of resistance R is uniformly compressed along its length, until its radius becomes \[n\] times the original radius. Now, the resistance of the wire becomes
A. $\dfrac{R}{n} \\ $
B. $\dfrac{R}{{{n^4}}} \\ $
C. $\dfrac{R}{{{n^2}}} \\ $
D. $nR$

Answer 1

Hint:To solve this question we will first find the value of length after the wire is compressed until its radius becomes \[n\] times the given radius of the wire. For this we will use the concept of conservation of volume on expansion or compression of a given wire. Then we will substitute the known values in the formula of resistance.

Formula Used:
Resistance, \[R = \dfrac{{\rho l}}{A}\]
where $\rho $ is the resistivity, $l$ is the length and $A$ is the area of the cross section of the conductor (here, wire).
Volume, $V = A \times l$ and $A = \pi {r^2}$ where $r$ is the radius of the wire.

Complete step by step solution:
Let initial length, radius and resistance of the wire be $l,\,r,\,R$ respectively.
Let the final length, radius and resistance of the wire be ${l_2},\,{r_2},\,{R_2}$ respectively.
Given: ${r_2} = nr$ ...(1)
To find the final length of the wire we will use conservation of volume concept and equate the volume of the wire in the two given situations.
Initially, $V = A \times l$
Also, $A = \pi {r^2}$
Therefore, $V = \pi {r^2} \times l$ ...(2)
Finally, $V = \pi {\left( {nr} \right)^2} \times {l_2} \\ $.....(from equation (1)) ...(3)
Equating equations (2) and (3), we get,
$\pi {r^2} \times l = \pi {\left( {nr} \right)^2} \times {l_2} \\ $
Simplifying this, we get, ${l_2} = \dfrac{l}{{{n^2}}}$ ...(4)
Now, we know that, \[R = \dfrac{{\rho l}}{A}\]
This implies that the initial resistance is,
\[R = \dfrac{{\rho l}}{{\pi {r^2}}} \\ \] ...(5)
And the final resistance is
\[{R_2} = \dfrac{{\rho {l_2}}}{{\pi {{\left( {nr} \right)}^2}}} \\ \] ...(6)
Dividing equation (5) and (6), we get;
\[\dfrac{{{R_2}}}{R} = \dfrac{{\left( {\dfrac{{\rho {l_2}}}{{\pi {{\left( {nr} \right)}^2}}}} \right)}}{{\left( {\dfrac{{\rho l}}{{\pi {r^2}}}} \right)}} \\ \]
Since resistivity is the property of the material (or the conductor) and has the same value for a given material at a given temperature. Therefore,
\[\dfrac{{{R_2}}}{R} = \dfrac{{\left( {\dfrac{{{l_2}}}{{{n^2}}}} \right)}}{l}\]
Substituting equation (4) in the above expression, we get,
\[\dfrac{{{R_2}}}{R} = \dfrac{{\left( {\dfrac{l}{{{n^2} \times {n^2}}}} \right)}}{l} \\ \]
Simplifying this,
\[\dfrac{{{R_2}}}{R} = \dfrac{1}{{{n^4}}} \\ \]
Thus, ${R_2} = \dfrac{1}{{{n^4}}}R$

Hence, option B is the correct answer.

Note: This question can be done with one more approach. Rather than dividing the expressions of the initial and final resistances, we can find the value of resistivity in both the cases and equate the two expressions obtained. This can be done because the value of resistivity is the same for a given material at a given temperature.