Question

A well with 10 m inside diameter is dug 14 m deep. Earth taken out of it is spread all around to a width of 5 m to form an embankment. Find the height of embankment​

Answer 1

Well with 10m diameter is dug 14m deep.

Cylindrical radius=

$\frac{10}{2} = 5m$

Volume= πr²h

$\frac{22}{7} \times 5 \times 5 \times 14$

=1100 cubic metre

Is taken out and spread width of 5m.

R1=5+5= 10m

r= 5m

V=(R²-r²)h

1100=

$\frac{22}{7} ( {10}^{2} - {5}^{2})h$

350=(100−25)h

$h = \frac{350}{75} = \frac{14}{3}$

⇒h=4.66m

Hence, the answer is 4.66m.

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Answer 2

Given:

• A well with 10 m inside diameter is dug 14 m deep.
• Earth taken out of it is spread all around to a width of t m to form an embankment.

To find:

Height of the embankment.

Solution:

According to the question:

Volume of the earth dug out = π r$^{2}$h m$^{3}$

= 22/7 × 5 × 5 × 14 m$^{3}$ = 1100 m$^{3}$

Area of the embankment (shaded region) = π (R$^{2}$ – r$^{2}$)

$\implies$ Area of the embankment or the shaded region = π (10$^{2}$ – 5$^{2}$) m$^{2}$ = 22/7 × 75 m$^{2}$

Height is unknown. So, let the height be h. Then,

Volume of the hollow cylinder = Volume of earth dugout.

$\implies$ π(R$^{2}$ – r$^{2}$)h = 1100

$\implies \: \frac{22}{7} (100 - 25) \times h = 1100 \\ \\ \: \implies h = \frac{1100 \times 7}{22 \times 75} m = \frac{14}{3} m \\ \\ = 4.66 \: m$

So, height of the embankment = Volume of the earth dugout ÷ Area of the embankment.

$\implies$ Height of the embankment

$= \frac{1100}{ \frac{22}{7} \times 75} \\ \\ = \frac{7 \times 1100}{22 \times 75} \\ \\ = 4.66 \: m$

Thus, height of the embankment = 4.66 m