C. Tripling the length and reducing the radius by a factor of 2 is the change to a pipe would increase the conductance by a factor of 12.
<u>Explanation:</u>
As we know that the resistance is directly proportional to the length of the pipe and it is inversely proportional to the cross sectional area of the pipe.
So it is represented as,
R∝ l /A [ area is radius square]
So k is the proportionality constant used.
R = kl/A
Conductance is the inverse of resistance, so it is given as,
C= 1/R.
R₁ = kl₁ / A₁
R₂ = kl₂/A₂
R₂/R₁ = 1/12 [∵ conductance is the inverse of resistance]
= l₂A₁ / l₁A₂
If we chose l₁/l₂= 3 and A₂/A₁= 4 So R₂/R₁= 1/3×4 = 1/12
So tripling the length and reducing the radius by a factor of 2 would increase the conductance by a factor of 12.
