Question

A well of diameter 3 m is dug 21 m deep. The earth taken out of it is spread evenly all around it to a width of 4 m to form an embankment, find the height of embankment.​

Answer 1

Answer:

Answer:

0.5235 m.

Step-by-step explanation:

Diameter of well = 3 m

Radius of well = Diameter/2 = 3/2 =1.5 m

Depth of well = 21 m

Well is in the shape of cylinder

Volume of well = \pi r^{2} hπr

2

h

= 3.14 \times 1.5 ^{2} \times 21=3.14×1.5

2

×21

= 148.365 m^3=148.365m

3

Radius of embankment = Radius of well + width

= 1.5 + 8

= 9.5 m

Since , Embankment is in the shape of cylinder .

So, volume of embankment = \pi r^{2} hπr

2

h

= 3.14 \times 9.5^{2} \times h=3.14×9.5

2

×h

= 283.385h=283.385h

Since the earth is taken out from the well is used to make embankment .

So, volume will be equal.

\Rightarrow 148.365 =283.385h⇒148.365=283.385h

\Rightarrow \frac{148.365}{283.385}=h⇒

283.385

148.365

=h

\Rightarrow 0.5235=h⇒0.5235=h

Hence the height of the embankment is 0.5235 m.

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Answer 2

QUESTION:-

A well of diameter 3 m is dug 21 m deep. The earth taken out of it is spread evenly all around it to a width of 4 m to form an embankment, find the height of embankment.

ANSWER:-

Let

• radius of the well → r
• height of the well → h

Then

• r = 3/2 m
• h = 21 m

Let r1 be the internal radius and R the external radius of the embankment and let h1 be its height.

then , r1 = radius of the well = r = 3/2 m

and R = internal radius + width of embankment

$⟹( \dfrac{3}{2} + 4)m$

$⟹ \dfrac{11}{2} m$

Volume of earth spread = Volume of earth dug Volume of embankment = Volume of earth dug

⟹ π(R² - r1²)h1 = πr²h

⟹ (R² - r1²)h1 = r²h

$⟹ [( { \dfrac{11}{2} )}^{2} -( { \dfrac{3}{2}) }^{2} ]h1 = [( { \dfrac{3}{2} )}^{2} \times 21 ]m$

$⟹ ( { \dfrac{11}{2} } + { \dfrac{3}{2} } )( { \dfrac{11}{2} } - { \dfrac{3}{2} } )h1 = ( { \dfrac{9}{4} }\times 21)m$

$⟹ ( \dfrac{14}{2})( \dfrac{8}{2} )h1 = ( { \dfrac{9}{4} }\times 21)m$

$⟹(7)(4)h1 = ( { \dfrac{9}{4} }\times 21)m$

$⟹ h1 = ( { \dfrac{9}{4} }\times 21 \times \dfrac{1}{7} \times \dfrac{1}{4} )m$

$⟹ \dfrac{27}{16} m$

$⟹ 1.69m \: (approx.)$