Question

The diameter of aluminium wire is reduced to half its original value, then its resistnace will become

• A. four times
• B. two times
• C. eight times
• D. sixteen times

Answer 1

The correct option is D sixteen times

Initial resistance of the wire $R=\frac{\rho L}{A}$

Using $L=\frac{V}{A}$ and $A=\frac{\pi d^2}{4}$ $⟹R=\frac{16\rho V}{{\pi }^2{d}^4}$

$⟹$ $\frac{R^{\prime }}{R}=\frac{d^4}{(d^{\prime }{)}^4}$ ($\because V,\rho$ are constants)

Given : $d^{\prime }=\frac{d}{2}$

$\therefore$ New resistance $R^{\prime }=2^{4}R=16R$