Question

A well with 40 m internal diameter is dug 14 m deep. The soil taken out is spread all around to a width of 20 m to form a circular embankment, find the height of the embankment.

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Answer 1

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

Given that,

A well with 40 m internal diameter is dug 14 m deep.

It means,

Diameter of well, d = 40 m

So, Radius of well, r = 20 m

Height of the well, h = 14 m

So,

Volume of earth dug out = Volume of cylinder of radius 20 m and height 14 m.

We know,

Volume of cylinder of radius r and height h is given by

$\boxed{\tt{ \: \: Volume_{(cylinder)} \: = \: \pi \: {r}^{2} \: h \: \: }}$

So,

$\rm \: Volume_{(earth \: dug \: out)} = \pi \: {(20)}^{2} \times 14 \: {m}^{3} - - - (1)$

Now,

The soil taken out is spread all around to a width of 20 m to form a circular embankment.

The embankment is in the form of form of hollow cylinder having

Internal radius, r = 20 m

External Radius, R = 20 + 20 = 40 m

Let assume that Height of embankment be H m

We know,

Volume of hollow cylinder of internal radius r, external radius R and height H is given by

$\boxed{\tt{ Volume_{(hollow \: cylinder)} \: = \: \pi \: ( {R}^{2} - {r}^{2}) \: H \: \: }}$

So,

$\rm\implies \:Volume_{(embankment)} = \pi \: ( {40}^{2} - {20}^{2})H - - - (2)$

Since, Volume of earth dug out is spread to form an embankment.

$\rm\implies \:Volume_{(embankment)} = Volume_{(earth \: dug \: out)}$

So, on substituting the values from equation (1) and (2), we get

$\rm \: \pi \: ( {40}^{2} - {20}^{2})H = \pi \: {(20)}^{2} \times 14$

$\rm \: \: (40 + 20)(40 - 20)H = \:400 \times 14$

$\rm \: \: 60 \times 20 \times H = \:400 \times 14$

$\rm \: H = \dfrac{400 \times 14}{20 \times 60}$

$\rm\implies \:H = \dfrac{14}{3} \: m$

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SHORT CUT TRICK

The height of embankment can directly be evaluated as

$\boxed{\tt{ \: \: H \: = \: \frac{h {r}^{2} }{ {R}^{2} - {r}^{2} } \: }}$

where,

H is height of embankment

h is depth of well

r is radius of well

R is r + t, where t is thickness of embankment.

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ADDITIONAL INFORMATION

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