Question

A plumber is hired to plumb a house that has been renovated. In doing so the length of the pipe from the water heater to the bathroom is quadrupled. In order to maintain the same conductance to the bathroom that the original pipes (of radius r) had, what radius of pipe should the plumber use compared to the original pipes?
½ r
¼ r
2r
4r

Answer 1

As we know that the conductance is given by the formula

$k = \frac{A}{\rho L}$

now here we know that

$A = \pi r^2$

L = length

now we know that conductance will remain constant while the length is quadrupled

so here we have

$k_1 = k_2$

$\frac{A_1}{\rho L_1} = \frac{A_2}{\rho L_2}$

$\frac{\pi r_1^2}{\rho L_1} = \frac{\pi r_2^2}{\rho 4L_1}$

$4 r_1^2 = r_2^2$

$r_2 = 2r_1$

so here radius becomes double

so correct answer will be

$2r$