Question

If the radius of the wire of resistance R is made n times, its mass remaining constant, then its new resistance will be

• A. $n^{2}R$
• B. $n^{4}R$
• C. $\frac{R}{n^2}$
• D. $\frac{R}{n^4}$

Answer 1

The correct option is D

$\frac{R}{n^4}$

Step 1: Given data and assumptions

Resistance of the wire is R

Assigning variables for the resistor

$L$-be the length of the resistor

$r$-be the radius of the resistor

$\rho$-resistivity of the material

$\sigma$-be the density of the material.

Step 2: Calculation of resistance

$R=\frac{\rho L}{A}=\frac{\rho L}{\pi r^2}$

Step 3: Assigning variables for the resistor new

$L\text{'}$-be the length of the resistor

$r\text{'}=nr$-be the radius of the resistor

$\rho$-resistivity of the material (Same material )

$\sigma$-be the density of the material. (Same material )

Step 4: Calculation of resistance.

$R\text{'}=\frac{\rho L\text{'}}{A\text{'}}=\frac{\rho L\text{'}}{\pi {(r\text{'})}^2}$

It is known that the mass of the material remains constant.

Then,

$\begin{array}{c}m\text{'}=m\\ \sigma AL=\sigma A\text{'}L\text{'}\\ \pi r^{2}L=\pi {\left(nr\right)}^{2}L\text{'}\\ \Rightarrow L\text{'}=\frac{L}{n^2}\end{array}$

Then the new resistance.

$\begin{array}{c}R\text{'}=\frac{\rho L\text{'}}{A\text{'}}\\ =\frac{\rho \frac{L}{n^2}}{\pi {\left(nr\right)}^2}\\ R\text{'}=\frac{\rho L}{\pi r^2}\frac{1}{n^4}\\ R\text{'}=\frac{R}{n^4}\end{array}$

Hence, the correct option is D.