Question

answer this...........

Answer 1

$Resistance= \frac{ \rho \: l}{a} \\ \\ \rho =resistivity\\ l=length\\ a= area$


$R=10 \: \Omega \\ \\ \rho=1.6×10^{-8} \: \Omega\ m\\ \\ required\ length=l\\ \\ d=0.5\ mm\\r=0.25 \ mm=2.5×10^{-4}\ m\\ a=\pi r^2=3.142×{(2.5×10^{-4})}^2\\a=1.964×10^{-7}\ m^2$

$l = \frac{Ra}{ \rho} \\ \\ l = \frac{10 \times 1.964 \times {10}^{ - 7} }{1.6 \times {10}^{ - 8} } \\ \\ l = 122.75 \: m$
So length required is 122.75 m
________________________________

$R \: \: \alpha \: \: \frac{1}{a} \\ \\ R \: \: \alpha \: \: \frac{1}{ {d}^{2} }$

$\text{Thus, if d is doubled,}\\ \\ \text{Resistance will becomes 1/4of the original value}\\ \\ \text{new resistance} = \frac{10}{4} = 2.5 \: \Omega$