Question

A ground is 100 m long and 80 m wide. A well 40 m long, 20 m wide
and 18 m deep is dug and the amount of the soil taken out are spreaded
over rest of the ground. By how much height of ground will rise up
?​

Answer 1

Answer :-

$\:\\\large{\underbrace{\underline{\sf{What\;the\;Question\;Says\;:-}}}}$

Here the concept of Volume of Cylinder and Volume of Cuboid has been used. We see that the ground is in the shape of cuboid. Also the well is in the shape of Cuboid. We know that whatever quantity of soil will be taken out will be equal to volume of well, since well is to be emptied full. Then that soil is to be spread all over the field equally. So the volume of ground after spreading soil will be equal to the volume of soil taken out that is volume of well. Without height the ground will remain as a rectangular plane but when its raised to some height by spreading soil, we can find out its volume.

Let's do it !!

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★ Formula Used :-

$\:\\\large{\boxed{\sf{\odot\;\;\:Volume\;of\;Cuboid\;=\;\bf{Length\:times\:Breadth\:times\:Height}}}}$

$\:\\\large{\boxed{\sf{\odot\;\;\:Volume\;of\;Field\;=\;\bf{Volume\;of\;Well_{(soil\;taken\;out)}}}}}$

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★ Question :-

A ground is 100 m long and 80 m wide. A well 40 m long, 20 m wide and 18 m deep is dug and the amount of the soil taken out are spread over rest of the ground. By how much height of ground will rise up ?

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★ Solution :-

Given,

» Length of ground = 100 m

» Breadth of ground = 80 m

» Length of Well = 40 m

» Breadth of Well = 20 m

» Height of Well = 18 m

• Let the height to which the ground is to be raised be 'h'.

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~ For the Volume of Ground :-

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:Volume\;of\;Cuboid_{(Ground)}\;=\;\bf{Length\:\times\:Breadth\:\times\:Height}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:Volume\;of\;Cuboid_{(Ground)}\;=\;\bf{100\:\times\:80\:\times\:h}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:Volume\;of\;Cuboid_{(Ground)}\;=\;\bf{8000\;\times\:h\;\;=\;\;\underline{\underline{8000\:h\;\;m^{3}}}}}}$

$\:\\\large{\boxed{\boxed{\tt{Volume\;\;of\;\;Ground\;\;=\;\;\bf{8000\:h\;\;m^{3}}}}}}$

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~ For the Volume of Well that is soil taken out :-

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:Volume\;of\;Cuboid_{(Well)}\;=\;\bf{Length\:\times\:Breadth\:\times\:Height}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:Volume\;of\;Cuboid_{(Well)}\;=\;\bf{40\:\times\:20\:\times\:18\:\;=\;\:\underline{\underline{14400\;\;m^{3}}}}}}$

$\:\\\large{\boxed{\boxed{\tt{Volume\;\;of\;\;Well\;\;=\;\;\bf{14400\:h\;\;m^{3}}}}}}$

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~ For the height by which ground is raised :-

$\:\\\large{\sf{:\Longrightarrow\;\;\:Volume\;of\;Field\;=\;\bf{Volume\;of\;Well_{(soil\;taken\;out)}}}}$

$\:\\\large{\sf{:\Longrightarrow\;\;\:8000\:h\;=\;\bf{14400}}}$

$\:\\\large{\sf{:\Longrightarrow\;\;\:h\;=\;\bf{\dfrac{\cancel{14400}}{\cancel{8000}}\;\:=\;\;\underline{\underline{1.8\;\;m}}}}}$

$\:\\\large{\underline{\underline{\mapsto\;\;\:\rm{Thus,\;height\;at\;which\;the\;ground\;is\;raised\;is\;\;\boxed{\bf{1.8\;\;m}}}}}}$

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★ More formulas to know :-

$\:\\\sf{\leadsto\;\;\;Volume\;of\;Cylinder\;=\;\pi r^{2}h}$

$\:\\\sf{\leadsto\;\;\;Volume\;of\;Cube\;=\;(Side)^{3}}$

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$\:\\\sf{\leadsto\;\;\;Volume\;of\;Cuboid\;=\;Length\:\times\:Breadth\:\times\:Height}$

Figure :-

$\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf x\:cm}\put(7.7,6.3){\sf y\:cm}\put(11.3,7.45){\sf z\:cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

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$\:\\\sf{\leadsto\;\;\;Volume\;of\;Sphere\;=\;\dfrac{4}{3}\:\times\:\pi r^{3}}$

$\:\\\sf{\leadsto\;\;\;Volume\;of\;Hemisphere\;=\;\dfrac{2}{3}\:\times\:\pi r^{3}}$

$\:\\\sf{\leadsto\;\;\;Volume\;of\;Cone\;=\;\dfrac{1}{3}\:\times\:\pi r^{2}h}$

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Answer 2

Given,

A ground is 100 m long and 80 m wide. A well 40 m long, 20 m wide  and 18 m deep is dug and the amount of the soil taken out are spread over rest of the ground.

To find :

By how much height of ground will rise up .

Solution :

Length of ground = 100 m

Width of ground = 80 m

Again, length of well = 40 m

Width of well = 20 m

Depth of well = 18 m

∴ Volume of well = 40 * 20 * 18

∴ Volume of well = 14400 m³

∵ Volume of earth dugout = Volume of well

∴ Volume of earth dugout = 14400 m³

Now let the rise of height of ground be 'h' m

∴ Volume of ground = Volume of earth dugout

⇒ 100 * 80 * h = 14400

⇒ 8000h = 14400

⇒ h = 14400/8000

⇒ h = 1.8 m

∴ Height of ground will rise by 1.8 m.