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)

$1.5\, \Omega$

Answer 2

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)
$\therefore \pi \frac{D_{A}^{2}}{4} l_{A} d$
$=\pi \frac{D_{B}^{2}}{4} l_{B} d$
$\frac{l_{A}}{l_{B}}=\left(\frac{D_{B}}{D_{A}}\right)^{2} \ldots(i)$
Resistance of wire $A$ ,
$R_{A}=\frac{\rho l_{A}}{\pi r_{A}^{2}}=\frac{\rho 4 l_{A}}{\pi D_{A}^{2}} \cdots(i i)$
Resistance of wire $B$,
$R_{B}=\frac{\rho l_{B}}{\pi r_{B}^{2}}=\frac{\rho 4 l_{B}}{\pi D_{B}^{2}} \ldots(i i i)$
Divide $(i i i)$ by $(i i)$, 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 $R_{B}=\frac{R_{A}}{16}$
$=\frac{24 \text { ohm }}{16}$
$=1.5 \,ohm$