Question

Two wires $A$ and $B$ of same material and same mass have radius $2r$ and $r$ and the length of the wire B is half the length of the wire $A$. If resistance of wire $A$ is $34\Omega$, then resistance of $B$ will be:

• A. $544\Omega$
• B. $272\Omega$
• C. $68\Omega$
• D. $17\Omega$

Answer 1

Answer: (C)

$68\Omega$

Resistance in wire, $R=\rho \frac{L}{A}=\frac{\rho L}{\pi r^2}$

$R_1=\frac{\rho L_1}{\pi r_1^2}=\frac{\rho L_1}{\pi {\left(2r\right)}^2}=\frac{1}{4}\frac{\rho L_1}{\pi r^2}=34\Omega$

$\Rightarrow \frac{\rho L_1}{\pi r^2}=34\times 4=136\Omega ......\left(1\right)$

$L_2=L_1/2$ $r_2=r$

$R_2=\rho \frac{L_2}{A_2}=\frac{\rho \frac{L_1}{2}}{\pi r_2^2}=\frac{1}{2}\frac{\rho L_1}{\pi r^2}=\frac{1}{2}\times 136=68\Omega$

Resistance of B is $68\Omega .$