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}$
$R \propto \frac{1}{A^2} \,\,[\because A = \pi r^2 ]$
$\Rightarrow R \propto \frac{1}{r^4}$