Question

Two wires $A$ and $B$ of equal masses and of the same metal are taken. The diameter of the wire $A$ is half the diameter of the wire $B$. If the resistance of wire $A$ be $24\Omega$ , the resistance of $B$ will be

• A. $3\Omega$
• B. $1.5\Omega$
• C. $4.5\Omega$
• D. $6.0\Omega$

Answer 1

Answer: (B)

As both the wires $A$ and $B$ are made up of the same metal, therefore their resistivity as well their density are the same
Mass of a wire $=$ volume $\times$ Density
$\therefore$ Mass of a wire $A,M_A=\pi r_A^2{l}_{A}d$
$=\pi \frac{D_A^2}{4}{l}_{A}d$
where $d$ is the density and $D$ is the diameter and subscript $A$ for wire $A$
Mass of a wire $B,M_B=\pi r_B^2{l}_{B}d=\pi \frac{D_B^2}{4}{l}_{B}d$
$\because M_A=M_B$ (Given)
$\therefore \pi \frac{D_A^2}{4}{l}_{A}d$
$=\pi \frac{D_B^2}{4}{l}_{B}d$
$\frac{l_A}{l_B}={\left(\frac{D_B}{D_A}\right)}^2\dots (i)$
Resistance of wire $A$ ,
$R_A=\frac{\rho l_A}{\pi r_A^2}=\frac{\rho 4l_A}{\pi D_A^2}\cdots (ii)$
Resistance of wire $B$,
$R_B=\frac{\rho l_B}{\pi r_B^2}=\frac{\rho 4l_B}{\pi D_B^2}\dots (iii)$
Divide $(iii)$ by $(ii)$, we get
$\frac{R_B}{R_A}=\left(\frac{l_B}{l_A}\right){\left(\frac{D_A}{D_B}\right)}^2={\left(\frac{D_A}{D_B}\right)}^4($ Using (i))
But $D_A=\frac{1}{2}{D}_B$ (Given)
$\therefore \frac{R_B}{R_A}=\frac{1}{16}$ or $R_B=\frac{R_A}{16}$
$=\frac{24\text{ ohm }}{16}$
$=1.5ohm$