91Ó°ÊÓ

1. The distance of section A and B,$d_a\left(=d_b\right)=30m$
2. The total distance, D=110 m
3. The radius of pipe outside hill section is $r_A\left(=r_B\right)=2cmor0.02m$
4. The speed of water through A and B section, v=2.5 m/s
5. The time required for a dye to reach point B from point A, t=88.8 s

We can find the speed of water through the hill section from the time required for the water to travel through the hill. Then using the continuity equation, we can find the average radius of the pipe within the hill.

Formulae:

The continuity equation at two ends of a liquid flowing, $A_1{v}_1=A_2{v}_2$ (i)

where, A₁, v₁ and A₂,v₂ are the cross-sectional area and velocity of two ends 1& 2

Area of circular cross-section, $A={\mathrm{Ï€°ù}}^2$ (ii)

We know that the water flows through section A and B with speed 2.5 m/s.

So, the time required for the water to cover those sections is given by:

$\begin{array}{l}t=\frac{d_A+d_B}{v}\\ =\frac{60}{2.5}\\ =24s\end{array}$

So, 88.8-24=64.8 s is the time required for water to travel through the hill.

So, the speed of water through hill section is given by

$$\begin{gathered} v_h=\frac{d_h}{64.8} \\ =\frac{110-60}{64.8} \\ =0.772m/s \end{gathered}$$

Now, using equation (i), we get

$A_h=\frac{A_1{v}_1}{v_h}$

As from equation (iii), we know$A={\mathrm{Ï€°ù}}^2$

Hence, substituting the values, we get

$$\begin{gathered} r_h^2=\frac{r_1^2×v_1}{v_h} \\ =\frac{0.{02}^2×2.5}{0.772} \\ {r}_h=\sqrt{\frac{0.{02}^2×2.5}{0.772}} \\ =0.036m \end{gathered}$$

Therefore, the radius of the pipe in the hill section is 0.036 m or 3.6 cm.