Embankments, Wells, and Fluid Flow Through Pipes for CBSE Class 10

In civil engineering and agricultural irrigation, geometric calculations are dynamic: excavating a well produces earth that is either spread out to raise a rectangular platform or distributed circularly around the mouth of the well to construct an embankment. Similarly, water flowing through irrigation canals and cylindrical pipelines delivers volume over time, requiring surveyors to calculate how many hectares of farmland can be irrigated within a given time window.

In CBSE Class 10 Mathematics, Chapter 12 (Surface Areas and Volumes), well-embankment problems and rate-of-flow pipe calculations represent high-frequency 4-mark and 5-mark board examination questions in Section D.

What You Will Learn

• Geometry of well excavation: The cylindrical hole and the circular ring embankment
• Formula for the height of an embankment: Equating well volume to ring volume
• The physics of fluid flow: $\text{Volume} = \text{Cross-Sectional Area} \times \text{Flow Speed} \times \text{Time}$
• Rectangular irrigation canals and standing water depth requirements
• Cylindrical pipe drainage into agricultural tanks
• Step-by-step solved CBSE board examination problems and common pitfalls

1. Well Digging and Circular Embankment Geometry

When a cylindrical well is dug:

1. The excavated earth forms a solid cylinder of volume: $\mathbf{V_{\text{earth}} = \pi r^2 h}$
2. If this excavated earth is spread evenly around the mouth of the well to a certain width $w$, it forms a hollow circular cylinder (a circular ring embankment).

                   Well Hole              Circular Embankment of Width w
                    ( r ) --------------------> ( r )--------( R )
                    |   |                                |  H |
                    | h |                                +----+
                    |   |                       Inner radius r = r_well
                    +---+                       Outer radius R = r_well + w
                 V = π r² h                     V = π(R² - r²) H

The Embankment Volume Equation:

• Inner radius of the embankment: $r = \text{Radius of the well}$.
• Outer radius of the embankment: $R = r + w$ (where $w$ is the width of the embankment).
• Area of the circular ring base: $\pi(R^2 - r^2)$.
• Let the height of the embankment be $H$. $\mathbf{\text{Volume of Embankment} = \pi(R^2 - r^2)H = \text{Volume of Earth Dug } (\pi r^2 h)}$ $\mathbf{H = \frac{r^2 h}{R^2 - r^2}}$

2. The Mechanics of Fluid Flow Through Pipes and Canals

When liquid flows through a pipe or canal of uniform cross-section at a uniform speed:

• In one second, the liquid travels a linear distance equal to the speed $v$.
• In time $t$, the length of the liquid column that flows out is: $\mathbf{L = \text{Speed} \times \text{Time} = v \times t}$

    Cross-Section Area (A)
           ( O ) =========================================>
           <---------------- Length L = Speed × Time ------>

The Master Fluid Flow Volume Equation:

$\mathbf{\text{Volume of Water Delivered} = \text{Cross-Sectional Area } (A) \times \text{Speed of Flow } (v) \times \text{Time } (t)}$

1. For a Cylindrical Pipe (Radius $r$): $\text{Volume} = (\pi r^2) \times v \times t$
2. For a Rectangular Canal (Width $b$, Depth $d$): $\text{Volume} = (b \times d) \times v \times t$

3. High-Yield Solved Board Examination Problems

Solved Example 1: The Well and Circular Embankment (NCERT Classic)

Problem: A well of diameter $3\text{ m}$ is dug $14\text{ m}$ deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width $4\text{ m}$ to form an embankment. Find the height of the embankment.

Solution:

1. Analyze Well Dimensions:
   • Diameter $d = 3\text{ m} \implies$ Radius $r = \frac{3}{2} = 1.5\text{ m}$.
   • Depth of well $h = 14\text{ m}$.
   • Volume of earth dug out: $V_{\text{earth}} = \pi r^2 h = \pi \times \left(\frac{3}{2}\right)^2 \times 14 = \pi \times \frac{9}{4} \times 14 = \frac{63\pi}{2}\text{ m}^3$
2. Analyze Embankment Dimensions:
   • Inner radius $r = 1.5\text{ m}$.
   • Width $w = 4\text{ m} \implies$ Outer radius $R = r + w = 1.5 + 4 = \mathbf{5.5\text{ m}}$.
   • Base area of the embankment ring: $\text{Area} = \pi(R^2 - r^2) = \pi(5.5^2 - 1.5^2) = \pi(5.5 - 1.5)(5.5 + 1.5) = \pi(4)(7) = \mathbf{28\pi\text{ m}^2}$
3. Equate Volumes to Find Height $H$: $\text{Volume of Embankment} = \text{Volume of Earth Dug}$ $28\pi \times H = \frac{63\pi}{2}$ $H = \frac{63}{2 \times 28} = \frac{9}{2 \times 4} = \frac{9}{8} = \mathbf{1.125\text{ m}}$
4. Therefore, <u>the height of the embankment is $1.125\text{ metres}$</u>.

Solved Example 2: Water Flow in an Irrigation Canal (NCERT Classic)

Problem: Water in a canal, $6\text{ m}$ wide and $1.5\text{ m}$ deep, is flowing with a speed of $10\text{ km/h}$. How much area will it irrigate in $30\text{ minutes}$, if $8\text{ cm}$ of standing water is needed?

Solution:

1. Analyze Canal Cross-Section:
   • Width $b = 6\text{ m}$.
   • Depth $d = 1.5\text{ m}$.
   • Cross-sectional area $A = b \times d = 6 \times 1.5 = \mathbf{9\text{ m}^2}$.
2. Calculate Length of Water Column in 30 Minutes ($0.5\text{ hour}$):
   • Speed $v = 10\text{ km/h} = 10000\text{ m/h}$.
   • Time $t = 30\text{ minutes} = 0.5\text{ hour}$.
   • Length of water column: $L = v \times t = 10000 \times 0.5 = \mathbf{5000\text{ m}}$
3. Calculate Total Volume of Water Discharged: $V = \text{Area} \times L = 9\text{ m}^2 \times 5000\text{ m} = \mathbf{45000\text{ m}^3}$
4. Calculate Irrigated Land Area:
   • Let the irrigated area be $A_{\text{field}}\text{ m}^2$.
   • Depth of standing water required $= 8\text{ cm} = \frac{8}{100}\text{ m} = 0.08\text{ m}$. $\text{Volume of Water} = \text{Area of Field} \times \text{Depth of Standing Water}$ $45000 = A_{\text{field}} \times \frac{8}{100}$ $A_{\text{field}} = \frac{45000 \times 100}{8} = 45000 \times 12.5 = \mathbf{562500\text{ m}^2}$
5. Convert to Hectares ($1\text{ hectare} = 10000\text{ m}^2$): $\text{Area} = \frac{562500}{10000} = \mathbf{56.25\text{ hectares}}$
6. Therefore, <u>the canal will irrigate $562500\text{ m}^2$ (or $56.25\text{ hectares}$) of land</u>.

4. Summary and Examination Tips

Exam Tip: In embankment problems, remember that the embankment ring does NOT cover the well opening! The well hole remains open in the center, which is why the base area is the ring $\pi(R^2 - r^2)$, not the full disk $\pi R^2$!

Common Mistake: In the irrigation problem, forgetting to convert $8 ext{ cm}$ of standing water into metres (rac{8}{100} ext{ m}). Mixing centimetres with cubic metres produces an answer off by a factor of 100!