Question

If length of resistance wire is increased by $0.5\%$ keeping the volume constant then change in resistance will be

• A. $0\%$
• B. $1\%$
• C. $0.5\%$
• D. $2\%$

Answer 1

The correct option is B

$1\%$

Step 1: Given data

Given,

length of the resistance wire is increased by $0.5\%$

Let the initial length be $l$ and final length be $l\text{'}$

$$\begin{gathered} \Rightarrow l\text{'}=l+\frac{0.5l}{100} \\ \Rightarrow l\text{'}=1.005l \end{gathered}$$

The volume of the wire remains constant.

Let $V$ be the volume of the wire, $A$ be the initial area and $A\text{'}$ be the final area

$$\begin{gathered} \Rightarrow V=lAandV=l\text{'}A\text{'} \\ \Rightarrow lA=l\text{'}A\text{'} \\ \Rightarrow lA=1.005lA\text{'} \\ \Rightarrow A=1.005A\text{'} \\ \Rightarrow A\text{'}=\frac{A}{1.005} \end{gathered}$$

Step 2: Formula and calculation

We know, the resistance of a wire is given by, $R=\frac{\rho l}{A}$

Where $\rho$ is the resistivity, $l$ is length and $A$ is area of cross-section of the wire.

$\therefore$New resistance will be

$$\begin{gathered} R\text{'}=\frac{\rho l\text{'}}{A\text{'}} \\ \Rightarrow R\text{'}=\frac{\rho \times 1.005l}{\left({\displaystyle \frac{A}{1.005}}\right)} \\ \Rightarrow R\text{'}=1.{005}^2\times \frac{\rho l}{A} \\ \Rightarrow R\text{'}=1.010025R \\ \Rightarrow R\text{'}\approx 1.01R \end{gathered}$$

Therefore, change in resistance,

$$\begin{gathered} ∆R=R\text{'}-R \\ \Rightarrow ∆R=1.01R-R \\ \Rightarrow ∆R=0.01R \end{gathered}$$

Percentage change$=\frac{R\text{'}-R}{R}\times 100=\frac{∆R}{R}\times 100=0.01\times 100=1\%$

Therefore percentage change is $1\%$

Hence, option B is the correct answer.