Question

Ends of two wires $A$ and $B$ having resistivity ${\rho }_A=3\times {10}^{-5}\Omega m$ and ${\rho }_B=6\times {10}^{-5}\Omega m$ of same cross section area joined in series together to form a single wire. If the resistance of the joined wire does not change with temperature, then find the ratio of their lengths $\frac{l_A}{l_B}$ , given that temperature coefficient of resistivity of wire $A$ and $B$ is ${\alpha }_A=4\times {10}^{-5}{/}^{\circ }\text{C}$ and ${\alpha }_B=-6\times {10}^{-6}{/}^{\circ }\text{C}$. Assume that mechanical dimensions do not change with temperature.

• A. $\frac{3}{7}$
• B. $\frac{10}{3}$
• C. $\frac{3}{10}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{3}{10}$

$R=R_1+R_2$
$R^{\prime }=R_1^{\prime }+R_2^{\prime }$
$R^{\prime }=R_A[1+{\alpha }_A\Delta \theta ]+R_B[1+{\alpha }_B\Delta \theta ]$
As net resistance does not change on changing temperature, so
$R_A{\alpha }_A+R_B{\alpha }_B=0$
${\rho }_A{l}_A{\alpha }_A+{\rho }_B{l}_B{\alpha }_B=0$
$\frac{l_A}{l_B}=-\frac{{\alpha }_B}{{\alpha }_A}.\frac{{\rho }_B}{{\rho }_A}$
$\frac{l_A}{l_B}=\frac{3}{10}$