Question

Two wires made of copper wire are taken such that the length and area of cross-section of the second wire are twice and thrice respectively those of the first wire. If the resistance of the second wire is $12\Omega$. Find the resistance of the first wire.

• A. $3\Omega$
• B. $12\Omega$
• C. $18\Omega$
• D. $9\Omega$

Answer 1

The correct option is C

$18\Omega$

Let $l_1$ be the length of the first copper wire, $A_1$ its area of cross - section and $R_1$ its resistance :
$R_1=\rho \frac{l_1}{A_1}$........(1)
Let $L_2$ be the length of the first copper wire, $A_2$ its area of cross - section and $R_2$ its resistance :
$l_2=2l_1$
$A_2=3A_1$
$R_2=12\Omega$
$R_2=\rho \frac{l_2}{A_2}\Rightarrow 12=\rho \frac{2l_1}{3A_1}$.........(2)
Dividing (1) by (2)
$\frac{R_1}{12}=\frac{\rho l}{A_1}\times \frac{3A_1}{\rho \left(2l\right)}=\frac{3}{2}$
$\Rightarrow R_1=18\Omega$