Question

Two copper wires whose masses are 8 and 12 gm have lengths in the ratio $3:4$. Their resistances are in the ratio.

• A. $4:9$
• B. $16:9$
• C. $27:32$
• D. $27:128$

Answer 1

The correct option is C $27:32$

Here, $m_1=8$gm$,m_2=12$gm.
And the ratio of there lengths is $3:4$.
So, $l_1=3s,l_2=4s$, where $s$ is also a proportionality constant.
Now the resistance of a wire is $R=\rho \frac{l}{A}=\rho \frac{l^2}{V}=\rho \frac{l^{2}d}{M}$.

Using, $density\left(d\right)=\frac{mass\left(m\right)}{volume\left(V\right)}$. Or, $R\propto \frac{l^2}{M}$.
$\therefore R_1\propto \frac{8^2}{3s}$
$\therefore R_1\propto \frac{64}{3s}$
$\therefore R_2\propto \frac{{12}^2}{4s}$
$\therefore R_2\propto \frac{144}{4s}$
Using it the ratio of their resistance is $27:32$.