Question

A well of diameter \(3\) m is dug \(14\) m deep. The earth taken out of it is spread evenly all around it in the shape of a circular ring of width \(4\) m to form an embankment. The height of that embankment (in m) is

• A. \(1.125\)
• B. \(1.725\)
• C. \(1.825\)
• D. \(2.105\)

Hint:

In excavation problems, the dug-out volume always equals the volume of the shape formed by the earth.

Answer 1

The Correct Option is A

Concept: Volume of earth dug out = Volume of embankment formed.

Step 1: Find volume of the well.
Well is cylindrical. Diameter: \[ 3m \] Radius: \[ \frac32 m \] Depth: \[ 14m \] Volume: \[ V=\pi r^2 h \] \[ =\pi \left(\frac32\right)^2(14) \] \[ =\pi \cdot \frac94 \cdot 14 \] \[ =\frac{126\pi}{4} \] \[ =31.5\pi \]

Step 2: Find dimensions of embankment.
Inner radius: \[ \frac32 \] Width: \[ 4 \] Outer radius: \[ \frac32+4=\frac{11}{2} \] Let height of embankment be \(h\). Volume of embankment: \[ \pi(R^2-r^2)h \] \[ =\pi\left[\left(\frac{11}{2}\right)^2-\left(\frac32\right)^2\right]h \] \[ =\pi\left(\frac{121}{4}-\frac94\right)h \] \[ =\pi\left(\frac{112}{4}\right)h \] \[ =28\pi h \]

Step 3: Equate volumes.
\[ 31.5\pi=28\pi h \] \[ 31.5=28h \] \[ h=\frac{31.5}{28} \] \[ h=1.125 \] Thus, the height of embankment is: \[ \boxed{1.125} \]