Question

A cylindrical copper rod is reformed to thrice its original length. The resistance between its ends before the change was$R$, now its new resistance becomes

• A. $8R$
• B. $4R$
• C. $9R$
• D. $R$

Answer 1

The correct option is C

$9R$

Step 1: Given data

Initial resistance of the wire$=R$

Final resistance of the wire$=R_f$

Step 2: Assumptions

Initial length of the wire$=l_1$

Initial radius of the cross-section of the wire$=r_1$

Final length of the wire$=l_2$

Final radius of the cross-section of the wire$=r_2$

Density of the material of the wire$=\rho$

Area of the cross-section of the wire before and after stretching$=A$

Step 3: Formula used

$R=\frac{\rho l_{}}{A}$

$R=\frac{\rho l_1}{\pi {r_1}^2}$……………………………………(a)

$R_f=\frac{\rho l_2}{\pi {r_2}^2}$……………………………………(b)

Step 4: Calculation of the final resistance of the wire

Rearranging the equation (a), we get

$\rho =\frac{R\pi {r_1}^2}{l_1}$ …………………………………(c)

Rearranging the equation (b), we get

$\rho =\frac{R_f\pi {r_2}^2}{l_2}$………………………………….(d)

Comparing equations (c) and (d), we get

$\frac{l_2}{l_1}=\frac{R_f\times {r_2}^2}{R\times {r_1}^2}$ …………………………(e)

Since the area of cross-section of the wire before and after stretching remains the same, therefore we can write

$\pi {r_1}^2{l}_1=\pi {r_2}^2{l}_2$

$\frac{{r_2}^2}{{r_1}^2}=\frac{l_1}{l_2}$ …………………………….(f)

The final length of the wire is thrice the initial length, therefore we can write

$l_2=3l_1$ …………………(g)

Using equations (f) and (g) in equation (e), we get

$$\begin{gathered} \frac{3\times l_1}{l_1}=\frac{R_f}{R}\times \frac{l_1}{l_2} \\ 3=\frac{R_f}{R}\times \frac{l_1}{3\times l_1} \\ {R}_f=9R \end{gathered}$$

The final resistance of the wire is equal to nine times the initial resistance of the wire.

Hence, option (C) is correct.