Answer
Which is the average radius of the pipe in the hill
$r_{2} = 0.03599 m$
Work Step by Step
According to the equation of continuity for the flow of ideal fluid.
$A_{1}V_{1} = A_{2}V_{2}$
$V_{2} =\frac{A_{1}V_{1} }{A_{2}}$
as we know $A = πr^{2}$
$V_{2} =\frac{ πr_{1}^{2}V_{1} }{ πr_{2}^{2}}$
$V_{2} =\frac{ r_{1}^{2} V_{1} }{ r_{2}^{2}}$
$r_{2} = \sqrt \frac{ r_{1}^{2} V_{1} }{ V_{2}}$ (1)
in order to calculate the velocity at midpoint we use the fowling relation
$V_{2} =\frac{\Delta x }{\Delta t}$
$V_{2} = \frac{50 m}{88.8 s-\frac{60 m}{2,5 m/s}}$
$V_{2} = \frac{50 m}{64.8 s}$
$V_{2} = 0.772 m/s$
by putting the value of $V_{2} $ in equation (1)
$r_{2} = \sqrt \frac{ r_{1}^{2} V_{1} }{ V_{2}}$
$r_{2} = \sqrt \frac{( (0.02 m)^{2}\times (2.5 m/s) }{ 0.772 m/s}$
$r_{2} = 0.03599 m$
Which is the average radius of the pipe in the hill
$r_{2} = 0.03599 m$
