Question

Masses of three wires of the same metal are in the ratio $1:2:3$ and their lengths in the ratio $3:2:1$. Electrical resistances of these wires will be in the ratio

• A. $1:2:3$
• B. $3:2:1$
• C. $1:6:27$
• D. $27:6:1$

Answer 1

Answer: (D)

$27:6:1$

Here the ratio of masses of the wires is $1:2:3$. So,
$m_1=k,m_2=2k$ and $m_3=3k$, where $k$ is the proportionality constant.
And the ratio of there lengths is $3:2:1$.
So, $l_1=3s,l_2=2s,l_3=s$, where $s$ is also a proportionality constant.

Now the resistance of a wire is given by,

$R=\rho \frac{l}{A}=\rho \frac{l^2}{V}=\rho \frac{l^{2}d}{M}$.

Using, $density\left(d\right)=\frac{Mass\left(m\right)}{Volume\left(V\right)}$. Or, $R\propto \frac{l^2}{m}$.
$\Rightarrow R_1\propto \frac{(3s{)}^2}{k}$
$\Rightarrow R_2\propto \frac{(2s{)}^2}{2k}$
$\Rightarrow R_3\propto \frac{s^2}{3k}$

Using it the ratio of their resistance is $27:6:1$.