Answer
Verified
427.8k+ 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
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Write an application to the principal requesting five class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE