Question

Two wires made of same material have their length are in the ratio $1:2$ and their masses in the ratio $3:16$. The ratio of resistance of two wires is.

• A. $\frac{3}{4}$
• B. $1:2$
• C. $2:1$
• D. $4:3$

Answer 1

The correct option is D $4:3$

$\text{Given:}$

$\frac{{\ell }_1}{{\ell }_2}=\frac{1}{2},\frac{m_1}{m_2}=\frac{3}{16}$

$\text{Step 1: Defining Resistance}$

$R=\rho \frac{\ell }{A}$ ............$\left(1\right)$

Density, $d=\frac{m}{V}=\frac{m}{Al}$

$\therefore A=\frac{m}{d\ell }$

Put the value of $A$ in eq. $\left(1\right)$, we get

$R=\rho \frac{{\ell }^{2}d}{m}$ ..............$\left(2\right)$

$\text{Step 2: Ratio of resistance of two wires}$

As the two wires are made of the same material, so they will have the same density.

Let $R_1$ & $R_2$ be the resistances of the two wires, so using equation (2), we get:

$R_1=\frac{\rho {\ell }_1^{2}d}{m_1}$

& $R_2=\frac{\rho {\ell }_2^{2}d}{m_2}$

Dividing $R_1$ & $R_2$, we get:

$\frac{R_1}{R_2}={\left(\frac{{\ell }_1}{{\ell }_2}\right)}^2\times \left(\frac{m_2}{m_1}\right)$

$\Rightarrow$ $\frac{R_1}{R_2}={\left(\frac{1}{2}\right)}^2\times \left(\frac{16}{3}\right)=\frac{4}{3}$

$\therefore$ $R_1:R_2=4:3$

Hence, Option(D) is correct.