Question

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

Answer 1

Answer:

Step-by-step explanation:

Hey here is your answer

Solution:-

Height of the well = 14 m

Diameter of the well = 3 m

So, Radius of the well = 3/2 m

Volume of the earth taken out of the well = πr²h

= 22/7*(3/2)²*14

= 99 cu m

Outer radius of the embankment = R = (3/2 + 4)m = 11/2 m

Area of embankment = outer area - inner area

⇒ = πR² - πr²

= 22/7*[(11/2)² - (3/2)²]

= 22/7*[(121/4) - (9/4)]

= 22/7 × 112/4

= 88 m²

Height of the embankment = Volume/Area

= 99/88

Height of the embankment

= 1.125 m Answer

MAKE SURE IT BRAINLIEST PLEASE

Answer 2

Radius of the well (r1) = $\sf{\frac{3}{2}m,h=14m}$

Volume of earth taken out of the well = $\sf{\pi {r1}^{2} h}$

= $\sf{\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 14 = {99m}^{3} }$

Outer radius of embankment = r2 = $\sf{\frac{3}{2}+4=\frac{11}{2}m}$

$\therefore$ Area of embankment = Outer area - Inner area

$\longrightarrow$$\sf{\pi( {r2}^{2} - {r2}^{2} )}$

$\longrightarrow$$\sf{( ({ \frac{11}{2} )}^{2} - {( \frac{3}{2} )}^{2} ) {m}^{2} }$

$\longrightarrow$$\sf{\frac{22}{7} \times ( \frac{11}{2} + \frac{3}{2} )( \frac{11}{2} - \frac{3}{2} )}$

$\longrightarrow$$\sf{ \frac{22}{7} \times \frac{7}{1} \times \frac{4}{1} = {88m}^{2}}$

$\therefore$ Height of embankment = $\sf{\frac{Volume}{Area}=\frac{99}{88}}=\frac{9}{8}=1.125m$