Question

A well with 10 m inside diameter is dug 14 m deep. Earth taken out of it is spread all around to a width of 5 m to form an embankment. Find the height of the Embankment.​

Answer 1

Given :-

• The diameter of well = 10m
• The height of well = 14m
• embankment spread = 5m

To Find :-

• The height of the Embankment = ?

Solution :-

• To calculate the height of Embankment at first we have to find the volume of well. As we know that the shape of well is in cylindrical form with diameter 10 m and height is 14m. Then embankment is spread all around the well with height of 5m. Simply by applying formula to calculate it's height.

Calculation begins :-

⇒ Radius of well = 10/2 = 5m

⇒ Height of well = 14m

⇒ Radius of inner embankment = 5m

⇒ Radius of outer embankment = 10m

Formula Used for volume :-

⇒ Volume of well = π r² h

⇒ Volume of Embankment = (Top area of circle of embankment - Bottom area of circle embankment) * h

• Outer radius (r) = 10 and Inner radius(R) = 5m

⇒Volume of well = Volume of Embankment

⇒ π r² h = h*(π r² - π R²)

⇒π × 5² × 14 = h × π(10² - 5²)

⇒ 5² × 14 = h × (100 - 25)

⇒ 25 × 14 = h × 75

⇒ h × 3 = 14

⇒ h = 14/3

⇒ h = 4.6 m

Hence,

• The height of Embankment = 4.6m

Answer 2

Answer:

4.66 m (approx.)

Step-by-step explanation:

$\blue{ \large \frak{ \underline{Given: - }} }$

• Inside Diameter is 10 m

• Height of the earth dug out is 14 m

$\blue{ \large \: \frak{ \underline{To \: Find : - }}}$

• Height of the Embankment.

$\blue{ \large \frak{ \underline{Solution: - }}}$

$\bf \: Radius \: = \frac{10}{2} = 5 \: cm$

Now,

$\sf \: Volume_{(Earth \: dug \: out)} =\pi {r}^{2} h \\ \\ ➥ \: \sf \: Volume_{(Earth \: dug \: out)} = \frac{22}{7} \times 5 \times 5 \times 14 \\ \\➥ \bf Volume_{(Earth \: dug \: out)} = 1100 \: {cm}^{3}$

And,

$\bf \: Outer \: radius = 5 + 5 = 10 \: cm$

Also,

$\sf \: Area \: of \: embankment = \pi(R {}^{2} - {r}^{2} ) \\ ➯ \sf \: \pi( {R} + r)( R - r) \\➯ \sf \: \frac{22}{7} \times (10 + 5) \times (10 - 5) \\➯ \sf \frac{22}{7} \times 15 \times 5 \\ ➯\bf \frac{22}{7} \times 75 \: {m}^{2}$

So,

$\sf \: Height \: of \: embankment \\ = \rm \frac{Volume \: of \: the \: earth \: dug - out}{Area \: of \: the \: embankment} \\ \\ \\ ✏ \sf \: \frac{1100}{ \frac{22}{7}× 75 } \\ \\ ✏ \sf \: \frac{7 \times 1100}{22 \times 75} \\ \\ ✏ \large \color{grey} \underline{ \boxed{ \frak{ \green{\: 4.66 \: m\: (approx.)}}}}$

So,

$\tiny \underline{ \therefore \sf\color{purple} \: Height _{(Embankment) \: is \: 4.66 \: m \: (approximately).}}$

▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃