Question

A well of diameter 14 m. is dug 15 m. deep. The earth taken out of it has been spread
evenly all around it in the shape of a circular ring of width 7 m. to form an embankment.
Find the height of the embankment. Smy​

Answer 1

Step-by-step explanation:

Here, a well is dug and Earth taken out of it is used to form an embankment.

Given,

Inner diameter of the well = 14m

Inner radius of the well = 14/2m = 7m

Height of the well (h) = 15m

Volume of the Earth taken out of the well = πr²h

= 22/7×(7²)×15

= 22×7×15 = 2310m³

Width = 7m

Outer radius of the embankment R = inner radius+width

Outer Radius (R) = 7+7 = 14m

The embankment is in the form of cylindrical shell, so area of embankment

Area of the embankment = Outer area - Inner area

= πR² - πr² = π(R²-r²)

= (22/7) ( 14² - 7²)

= 22/7(196-49)

= 22/7 × 147

= 22 × 21

= 462 m²

Volume of embankment= volume of earth taken out on digging the well

Area of embankment × height of embankment= volume of earth dug out

Height of embankment= volume of earth dug out/area of the embankment

Height of the embankment = 2310 / 462

Height of embankment= 5 m

Hence, the height of the embankment so formed is 5 m

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Answer 2

Answer:

$given$

$Given \: radius \: of \: the \: well = \: 7m$

$Height \: of \: the \: well = 15m$

$Width \: of \: the \: embankment = 7 m$

$Radius \: of \: the \: embankment = 7 + 7 = 14$

$Let \: h \: be \: the \: height \: of \: the \: embankment$

$Hence \: the \: volume \: of \: the \: embankment \: = \: volume \: of \: the \: well$

$That \: is,$

$\pi \: (r {}^{2} - r {}^{2} )h \: = \pi \: r {}^{2} \: h$

$(14 { }^{2} \: - 7 {}^{2} ) \: h \: = 7 { }^{2} \: (15)$

$(196 \: - 49 \: ) \: h \: = 49 \: \times 15$

$(147) \: h \: = 735$

$h \: = 735 \div 147$

$h = 5 \: m$

$Hence \: the \: height \: of \: the \: embankment \: is \: 5 \: m$

$hope \: it \: is \: helpful \: for \: u \: follow \: me \: nd \: make \:me \: brainliest$