91Ó°ÊÓ

• The width ofthe stream 1 is${\mathrm{b}}_1=8.2\text{m}$.
• The depth ofthe stream 1 is${\mathrm{h}}_1=3.4\text{m}$.
• The speed ofthe current 1 is${\mathrm{v}}_1=2.3\text{m/s}$.
• The width of stream 2 is${\mathrm{b}}_2=6.\text{8m}$.
• The depth of stream 2 is${\mathrm{h}}_2=3.2\text{m}$.
• The speed ofthe current 2 is${\mathrm{v}}_2=2.6\text{m/s}$.
• The width ofthe river stream is$b=10.5\text{m}$.
• The speed of the river current is $\mathrm{v}=\text{2.9″¾/²õ}$.

For ideal fluid, the rate of flow of volume is constant. The rate of flow of fluid is equal to the product of the area of the tube at any point and the speed of the fluid at that point.

The flow within any tube of flow obeys the equation of continuity:

Here is the volume flow rate, is the cross-sectional area of the tube of flow at any point, and is the speed of the fluid at that point.

Use the equation of continuity to find the depth of the river.

The formula is as follows:

${\mathrm{A}}_1{\mathrm{v}}_1={\mathrm{A}}_2{\mathrm{v}}_2$

Here, A is the corresponding area and v is the corresponding velocity.

Let, the depth of river beh.

Using the equation of continuity and substituting the values of area and cross section, we can find height.

$\begin{array}{c}{\mathrm{A}}_1{\mathrm{v}}_1+{\mathrm{A}}_2{\mathrm{v}}_2=\mathrm{Av}\\ (8.2{\text{m}}^2×3.4)\left(2.3\text{m/s}\right)+(6.8{\text{m}}^2×3.2)\left(2.6\text{m/s}\right)=(10.5\text{m}×\mathrm{h})\left(2.9\text{m/s}\right)\\ \mathrm{h}=\frac{64.124+56.576}{30.45}\\ \mathrm{h}=3.96\text{″¾}\\ ≈4.0\text{″¾}\end{array}$

Hence, the depth of the river is$4.0m$