Question

12. Water is flowing at the rate of 8 m per second through a circular pipe whose internal diameteris 2 cm, into a cylindrical tank, the radius of whose base is 40 cm. Determine the increase inthe water level in 30 minutes.Hint:2222Ex 40 x 40 x h = x1x1x800 x 30 x 60. Find h.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, Water is flowing at the rate of 8 m per second through a circular pipe whose internal diameter is 2 cm.

So, it means

Radius of circular pipe, r = 1 cm

Height of circular pipe, h = 8 m = 800 cm

Now, Volume of water flows in 1 second is evaluated as

$\rm \: Volume_{(water \: flow\:in\:1\:sec)} = \pi \: {r}^{2} \: h$

$\rm \: Volume_{(water \: flow\:in\:1\:sec)} = \pi \: (1) \: (800)$

$\rm\implies \:\rm \: Volume_{(water \: flow\:in\:1\:sec)} = 800\pi$

Now, we know 1 minute = 60 sec

So, 30 minutes = 30 × 60 = 1800 sec

$\rm\implies \:\rm \: Volume_{(water \: flow\:in\:1800\:sec)} = 1800 \times 800 \: \pi \: {cm}^{3} - - (1)$

Now, Let assume that the height of water level rise in 30 minutes in cylindrical tank whose base radius, R = 40 cm is H cm

So,

$\rm \: Volume_{(water \: in \: cylindrical \: tank)} = \pi \: {(R)}^{2} \: H$

$\rm \: Volume_{(water \: in \: cylindrical \: tank)} = \pi \: {(40)}^{2} \: H$

$\rm\implies \:\rm \: Volume_{(water \: in \: cylindrical \: tank)} = 1600 \: \pi\: H \: {cm}^{3} - - (2)$

Now, According to statement,

$\rm \: Volume_{(water \: in \: cylindrical \: tank)} = Volume_{(water \: flow\:in\:1800\:sec)}$

$\rm \: 1800 \times 800 \: \pi \: = \: 1600 \: \pi \: H$

$\rm\implies \:H = 900 \: cm \: = \: 9 \: m$

$\rule{190pt}{2pt}$

Additional information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

\large\underline{\sf{Solution-}}

Solution−

Given that, Water is flowing at the rate of 8 m per second through a circular pipe whose internal diameter is 2 cm.

So, it means

Radius of circular pipe, r = 1 cm

Height of circular pipe, h = 8 m = 800 cm

Now, Volume of water flows in 1 second is evaluated as

\begin{gathered}\rm \: Volume_{(water \: flow\:in\:1\:sec)} = \pi \: {r}^{2} \: h \\ \end{gathered}

Volume

(waterflowin1sec)

=πr

2

h

\begin{gathered}\rm \: Volume_{(water \: flow\:in\:1\:sec)} = \pi \: (1) \: (800)\\ \end{gathered}

Volume

(waterflowin1sec)

=π(1)(800)

\begin{gathered}\rm\implies \:\rm \: Volume_{(water \: flow\:in\:1\:sec)} = 800\pi\\ \end{gathered}

⟹Volume

(waterflowin1sec)

=800π

Now, we know 1 minute = 60 sec

So, 30 minutes = 30 × 60 = 1800 sec

\begin{gathered}\rm\implies \:\rm \: Volume_{(water \: flow\:in\:1800\:sec)} = 1800 \times 800 \: \pi \: {cm}^{3} - - (1)\\ \end{gathered}

⟹Volume

(waterflowin1800sec)

=1800×800πcm

3

−−(1)

Now, Let assume that the height of water level rise in 30 minutes in cylindrical tank whose base radius, R = 40 cm is H cm

So,

\begin{gathered}\rm \: Volume_{(water \: in \: cylindrical \: tank)} = \pi \: {(R)}^{2} \: H \\ \end{gathered}

Volume

(waterincylindricaltank)

=π(R)

2

H

\begin{gathered}\rm \: Volume_{(water \: in \: cylindrical \: tank)} = \pi \: {(40)}^{2} \: H \\ \end{gathered}

Volume

(waterincylindricaltank)

=π(40)

2

H

\begin{gathered}\rm\implies \:\rm \: Volume_{(water \: in \: cylindrical \: tank)} = 1600 \: \pi\: H \: {cm}^{3} - - (2) \\ \end{gathered}

⟹Volume

(waterincylindricaltank)

=1600πHcm

3

−−(2)

Now, According to statement,

\begin{gathered}\rm \: Volume_{(water \: in \: cylindrical \: tank)} = Volume_{(water \: flow\:in\:1800\:sec)} \\ \end{gathered}

Volume

(waterincylindricaltank)

=Volume

(waterflowin1800sec)

\begin{gathered}\rm \: 1800 \times 800 \: \pi \: = \: 1600 \: \pi \: H \\ \end{gathered}

1800×800π=1600πH

\rm\implies \:H = 900 \: cm \: = \: 9 \: m⟹H=900cm=9m

\rule{190pt}{2pt}

Additional information :-

\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

MoreFormulae

MoreFormulae

★CSA

(cylinder)

=2πrh

★Volume

(cylinder)

=πr

2

h

★TSA

(cylinder)

=2πr(r+h)

★CSA

(cone)

=πrl

★TSA

(cone)

=πr(l+r)

★Volume

(sphere)

=

3

4

πr

3

★Volume

(cube)

=(side)

3

★CSA

(cube)

=4(side)

2

★TSA

(cube)

=6(side)

2

★Volume

(cuboid)

=lbh

★CSA

(cuboid)

=2(l+b)h

★TSA

(cuboid)

=2(lb+bh+hl)