Question

Solve the question fast ​

Answer 1

Answer:

Given, Diameter =4 m

Radius =2 m h=21 m

V=πr

2

h=π(2)(2)(21)=84π

Volume of soil dug from well = volume of earth used to form embankment

⇒84π=π(5

2

−2

2

)h

h=

25−4

84

=

21

84

=4m

Answer 2

$\bf Given\begin{cases} & \sf{Diameter \; of\; well \; = \bf{4 \;m}} \\ & \sf{Radius \;of\; well \; = \bf{2 \;m}} \\ & \sf { Depth \; of\; earth \; (d) = \bf {21 \; m}} \\ & \sf {Width \; of \; embankment \; = \bf {3 \; m}}\end{cases}$

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$\bf { To \; find \; : }$

• Height of embankment

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$\bf {Solution \; :}$

$\sf { Volume \; of \; earth \; digged \; (v) = \; \pi r^2 d }$

$\implies \sf { v = \dfrac {22}{7} \times (2)^2 \times 21 }$

$\implies \sf {v = 264 \; m^3 }$

Now,

$\sf {Outer \: radius \: of \: embankment \: = 3 + 2 = 5 \: m}$

• Let the height of embankment = h

$\boxed {\small {\frak {\underline {\red {Volume \; of \; embankment \; = \pi (R^2 - r^2)h }}}}}$

$\small {\frak {\underline {\blue { Substituting \; the \; values :}}}}$

$\implies \sf { \dfrac {22}{7} \times (25 - 4) \times h = 264 }$

$\implies \sf { h = \dfrac { 264 \times 7}{22 \times 21}}$

$\implies \sf { h = 4 \; m}$

$\boxed {\small {\underline {\frak {\green {Therefore, \; height \; of \; embankment \; = 3 \; m}}}}}$