Question

A piece of aluminium of finite length is drawn or stretched such that to reduce its diameter to one fourth its original value, its resistance will become:

• A. $256times$
• B. Four times
• C. Eight times
• D. Sixteen times

Answer 1

The correct option is A $256times$

Resistance is inversely proportional to square of diameter

Let the initial area be $A=\pi D^2/4$

Final D is D/4 So new area is $A_f=\pi \times D^2/32=A/16$

Now $L_f{A}_f=LA$ , So $F_f=16L$

$Rf=\rho \frac{L_f}{A_f}=\rho \frac{256L}{A}=256R$