Answer

Verified

452.4k+ views

Hint- For solving such questions draw a rough diagram for better understanding. Find out the total volume of earth dug first of all in order to find the volume of embankment and finally height of the embankment.

The shape of the well be cylindrical as shown in the given figure drawn

Given depth ${h_1}$ of the well $ = 14m$

Radius $\left( {{r_1}} \right)$ of the circular end of the well $ = \dfrac{3}{2}m$

Width of the embankment $ = 4m$

As show in the figure the embankment will also be in shape of hollow cylinder so, its outer radius is

$\left( {{r_2}} \right) = 4 + \dfrac{3}{2} = \dfrac{{11}}{2}m$

Let the height of the embankment be ${h_2}$.

So the volume of the soil dug from the well

= volume of the earth used to form the embankment

As, Volume of the soil dug

= volume of the cylinder

\[

= \pi r_1^2{h_1} \\

= \pi \times {\left( {\dfrac{3}{2}} \right)^2} \times 14 \\

= \pi \times \dfrac{9}{4} \times 14 \\

\] ---- (1)

Also volume of the embankment

=volume of hollow cylinder

\[

= \pi \times \left( {r_2^2 - r_1^2} \right) \times {h_2} \\

= \pi \times \left( {{{\left( {\dfrac{{11}}{2}} \right)}^2} - {{\left( {\dfrac{3}{2}} \right)}^2}} \right) \times {h_2} \\

= \pi \times \left( {\left( {\dfrac{{121}}{4}} \right) - \left( {\dfrac{9}{4}} \right)} \right) \times {h_2} \\

= \pi \times \left( {\dfrac{{112}}{4}} \right) \times {h_2} \\

\] --- (2)

From equation (1) and (2), comparing to find the value of ${h_2}$.

$

\Rightarrow \pi \times \left( {\dfrac{9}{4}} \right) \times 14 = \pi \times \left( {\dfrac{{112}}{4}} \right) \times {h_2} \\

\Rightarrow \left( {\dfrac{9}{4}} \right) \times 14 = \left( {\dfrac{{112}}{4}} \right) \times {h_2} \\

\Rightarrow {h_2} = \dfrac{9}{4} \times 14 \times \dfrac{4}{{112}} \\

\Rightarrow {h_2} = \dfrac{9}{8} \\

\Rightarrow {h_2} = 1.125m \\

$

Hence the height of the embankment will be $1.125m$.

Note- Figures are the most important part of questions containing these types of practical problems. Formulas of volume of cylinder, hollow cylinder and others are very useful and must be remembered. In order to solve problems of real life or practical type, try to relate it with some geometrical figures in order to solve the problem easily and fast.

The shape of the well be cylindrical as shown in the given figure drawn

Given depth ${h_1}$ of the well $ = 14m$

Radius $\left( {{r_1}} \right)$ of the circular end of the well $ = \dfrac{3}{2}m$

Width of the embankment $ = 4m$

As show in the figure the embankment will also be in shape of hollow cylinder so, its outer radius is

$\left( {{r_2}} \right) = 4 + \dfrac{3}{2} = \dfrac{{11}}{2}m$

Let the height of the embankment be ${h_2}$.

So the volume of the soil dug from the well

= volume of the earth used to form the embankment

As, Volume of the soil dug

= volume of the cylinder

\[

= \pi r_1^2{h_1} \\

= \pi \times {\left( {\dfrac{3}{2}} \right)^2} \times 14 \\

= \pi \times \dfrac{9}{4} \times 14 \\

\] ---- (1)

Also volume of the embankment

=volume of hollow cylinder

\[

= \pi \times \left( {r_2^2 - r_1^2} \right) \times {h_2} \\

= \pi \times \left( {{{\left( {\dfrac{{11}}{2}} \right)}^2} - {{\left( {\dfrac{3}{2}} \right)}^2}} \right) \times {h_2} \\

= \pi \times \left( {\left( {\dfrac{{121}}{4}} \right) - \left( {\dfrac{9}{4}} \right)} \right) \times {h_2} \\

= \pi \times \left( {\dfrac{{112}}{4}} \right) \times {h_2} \\

\] --- (2)

From equation (1) and (2), comparing to find the value of ${h_2}$.

$

\Rightarrow \pi \times \left( {\dfrac{9}{4}} \right) \times 14 = \pi \times \left( {\dfrac{{112}}{4}} \right) \times {h_2} \\

\Rightarrow \left( {\dfrac{9}{4}} \right) \times 14 = \left( {\dfrac{{112}}{4}} \right) \times {h_2} \\

\Rightarrow {h_2} = \dfrac{9}{4} \times 14 \times \dfrac{4}{{112}} \\

\Rightarrow {h_2} = \dfrac{9}{8} \\

\Rightarrow {h_2} = 1.125m \\

$

Hence the height of the embankment will be $1.125m$.

Note- Figures are the most important part of questions containing these types of practical problems. Formulas of volume of cylinder, hollow cylinder and others are very useful and must be remembered. In order to solve problems of real life or practical type, try to relate it with some geometrical figures in order to solve the problem easily and fast.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Guru Purnima speech in English in 100 words class 7 english CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Select the word that is correctly spelled a Twelveth class 10 english CBSE