Answer
a) $R=\dfrac{kL}{d^2}$
b) $ k \approx 0.002917$
c) $137$ohms
d)$\dfrac{3}{4}$
Work Step by Step
a) As per the given problem, we have:
$R=\dfrac{kL}{d^2}$
Here, $k$ is defined as constant of proportionality
b) From part (a), we have $R=\dfrac{kL}{d^2}$
Plug the given values in the above expression, we get
$140=k\dfrac{1.2}{(0.005)^2} \implies k \approx 0.002917$
c) From part (a), we have $R=k\dfrac{kL}{d^2}$
Then, $R=(0.002917)\dfrac{3}{(0.008)^2} \approx 137$ohms
d) From part (a), we have $R'=k\dfrac{kL'}{d'^2}$
Then, $R'=(0.002917)\dfrac{3L}{(2d)^2} =\dfrac{3}{4}R$
