Question

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

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Answer 1

$\huge{\underline{\underline{\blue{\bf Question}}}}$

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

$\huge{ \underline{ \underline{ \green{ \bf answer}}}}$

The shape of the well will be cylindrical as shown in the figure below :

Given :

Depth ( h1 ) of well = 14m

$radius(r1) \: of \: the \: circular \: end \: of \: well = \frac{3}{2} m$

Width of embankment = 4 m

from the figure , it can be observed that from embankment will be in a cylindrical shape have other radius ,

$(r2) = 4 + \frac{3}{2} = \frac{11}{2} m$

let height of embankment be h2

volume of soil dug from well = volume of earth used to form the embankment .

$\pi \times r \frac{2}{1} \times h1 = \pi \times (r \frac{2}{2} - r \frac{2}{1} ) \times h2$

expect refer this attachment please

Answer 2

$\large\underline{\sf{Given- }}$

• A well of diameter 3m is dug 14m deep.

• The earth taken out of it has been spread evenly all around it in the shape of circular ring of width 4m to form an embankment.

$\large\underline{\sf{To\:Find - }}$

• The height of the embankment.

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$1. \: \: \: \boxed{ \sf{Volume_{(Cylinder)} = \pi \: {r}^{2} h}}$

where,

• r = radius of cylinder

• h = height of cylinder

$2. \: \: \: \boxed{ \sf{Volume_{(hollowCylinder)} = \pi \: ( {R}^{2} - {r}^{2} )h}}$

where,

• R = external radius of hollow cylinder

• r = internal radius of hollow cylinder

• h = height of hollow cylinder

$\large\underline{\sf{Solution-}}$

Dimensions of well

Diameter of well = 3 m

So,

$\sf \: Radius \: of \: well \: (r) = \dfrac{3}{2} \: m$

Height of well, h = 14 m

So,

Volume of earth dug out = Volume of cylinder of radius 3/2 m and height 14 m.

Therefore,

$\rm :\longmapsto\:Volume_{(earth \: dug \: out)} = \pi \: {r}^{2} h$

$\rm :\longmapsto\:Volume_{(earth \: dug \: out)} =\pi \: {\bigg( \dfrac{3}{2} \bigg) }^{2} \times 14$

$\rm :\longmapsto\:Volume_{(earth \: dug \: out)} =\pi \: \times \dfrac{9}{4} \times 14$

$\bf :\longmapsto\:Volume_{(earth \: dug \: out)} = \: \dfrac{63}{2} \pi \: {m}^{3} - - (1)$

Now,

Dimensions of embankment,

Inner radius of embankment, r = 3/2 m

External radius of embankment, R = r + width = 3/2+4=11/2 m

Let height of embankment be 'H' meter.

Now,

Embankment forms a hollow cylinder of inner radius 'r' and external radius 'R' having height 'H'.

So,

Volume of embankment = Volume of Hollow cylinder

$\rm :\longmapsto\:Volume_{(embankment)} = \pi \: ( {R}^{2} - {r}^{2} )H$

$\rm :\longmapsto\:Volume_{(embankment)} = \pi \: (R + r)(R - r)H$

$\rm :\longmapsto\:Volume_{(embankment)} = \pi \: \bigg( \dfrac{11}{2} + \dfrac{3}{2} \bigg) \bigg(\dfrac{11}{2} - \dfrac{3}{2} \bigg) H$

$\rm :\longmapsto\:Volume_{(embankment)} = \pi \: \times 7 \times 4 \times H$

$\bf :\longmapsto\:Volume_{(embankment)} = 28\pi \: H - - (2)$

According to statement,

The earth taken out of it has been spread even all around it in the shape of circular ring to form an embankment.

$\bf \:\therefore Volume_{(embankment)} = \:Volume_{(earth \: dug \: out)}$

$\rm :\longmapsto\:28\pi \: H = \dfrac{63}{2} \pi$

$\rm :\implies\:H = \dfrac{63}{28 \times 2} = \dfrac{9}{8} \: m$

$\bf \: Hence, \: Height_{(embankment)} \: = \: \dfrac{9}{8} \: m \:$

More information :-

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length ²+breadth ²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²