Question

Two metallic wires A and B are connected in series Wire A has length $l$ and radius r while wire B has length $2l$ and radius $2r$ If both the wires are of same material then and the ratio of the total resistance of series combination to the resistance of the wire A

• A. $\frac{3}{4}$
• B. $\frac{3}{2}$
• C. $\frac{6}{2}$
• D. $\frac{6}{5}$

Answer 1

Answer: (B)

$\frac{3}{2}$

Let the resistivity of both the wires be '$\rho$'.
The resistance of the wire A is given as $R_A=\frac{\rho l}{\pi r^2}$.

The resistance of the wire B is given as $R_B=\frac{\rho 2l}{\pi {2r}^2}$ = $\frac{\rho l}{2\pi r^2}$.

Now when the resistance are connected in seires is $R=R_A+R_B$. That is, $R=\frac{3}{2}\left(\frac{\rho l}{\pi r^2}\right)$.

So, the required ratio is $R:R_A=\frac{3}{2}\left(\frac{\rho l}{\pi r^2}\right):\frac{ql}{\pi r^2}$.

Therefore, $R:R_A=3:2$.
Hence, the ratio of the resistances in series and only wire A is 3/2.