Q127 / 160 · TS EAMCET 2014 · Physics

Wires $A$ and $B$ have resistivities $\rho_A$ and $\rho_B$, $\left(\rho_B=2 \rho_A\right)$ and have lengths $l_A$ and $l_B$. If the diameter of the wire $B$ is twice that of $A$ and the two wires have same resistance, then $\frac{l_B}{l_A}$ is

  • Correct answer
    A
    $2$
  • B
    $1$
  • C
    $\frac{1}{2}$
  • D
    $\frac{1}{4}$

Solution

The given, $$ \rho_B=2 \rho_A, R_A=R_B=R $$ Let, $r_A=r$ $$ r_B=2 r $$ According to formula, $$ \begin{aligned} R_A & =\rho_A \cdot \frac{I_A}{\pi r_A^2} \\ R_B & =\rho_B \cdot \frac{I_B}{\pi r_B^2} \\ \because \quad R_A & =R_B \\ \rho_A \cdot \frac{I_A}{\pi r_A^2} & =\rho_B \cdot \frac{I_B}{\pi r_B^2} \end{aligned} $$ $\begin{aligned} \frac{I_B}{I_A} & =\frac{r_B^2}{r_A^2} \cdot \frac{\rho_A}{\rho_B} \\ & =\frac{(2 r)^2}{r^2} \cdot \frac{\rho_A}{2 \rho_B} \\ & =\frac{4 r^2}{r^2} \cdot \frac{\rho_A}{2 \rho_A} \\ & =\frac{4}{2}=2\end{aligned}$

Hence, the correct answer is option: 'A'.

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