Question

A wire is being drawn to make it thinner such that the length of the wire $I$ increases and radius $r$ decreases. Its resistance $R$ will finally be proportional to

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

Answer 1

Answer: (D)

Resistance, $R=\frac{\rho l}{A}$
If a given mass of wire is recast to increase its length or decrease its radius, then we have $m=$ volume $\times$ density $=Ald...(i)$
where, $d$ is the density
So, $A=\frac{m}{Id}$
$\therefore R=\frac{\rho l}{(\frac{m}{Id})}=\frac{\rho d}{m}\cdot l^2$
$\rho ,d$ and $m$ are constants,
$R\propto l^2$ or $\frac{R_2}{R_1}=(\frac{l_2}{l_1}{)}^2$
From above Eq. $(i)$, we have
$l=\frac{m}{A\cdot d}$
$\therefore R=\frac{\rho m}{A\cdot dA}$
$=\frac{m\rho }{d}\cdot \frac{1}{A^2}$
$R\propto \frac{1}{A^2}[\because A=\pi r^2]$
$\Rightarrow R\propto \frac{1}{r^4}$