Question

Two wires $\text{A}$ and $\text{B}$ made of same material and having their lengths in the ratio $6:1$ are connected in series. The potential difference across the wire are $3\text{V}$ and $2\text{V}$ respectively. If $r_A$ and $r_B$ are the radii of $\text{A}$ and $\text{B}$ respectively, then $\frac{r_B}{r_A}$ is

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

Answer 1

The correct option is B $\frac{1}{2}$

Given:
$\frac{l_A}{l_B}=\frac{6}{1};V_A=3\text{V};V_B=2\text{V}$

In series combination, same current flow through each resistance, so $V\propto R$.

$\therefore \frac{R_A}{R_B}=\frac{V_A}{V_B}=\frac{3}{2}$

$\Rightarrow \frac{\left(\rho l_A/A_A\right)}{\left(\rho l_B/A_B\right)}=\frac{3}{2}$

$\Rightarrow \frac{l_A}{l_B}\times {\left(\frac{r_B}{r_A}\right)}^2=\frac{3}{2}$

$\Rightarrow \frac{6}{1}\times {\left(\frac{r_B}{r_A}\right)}^2=\frac{3}{2}$

$\therefore \frac{r_B}{r_A}=\frac{1}{2}$

Hence, option $\left(b\right)$ is correct.