Question

A cylindrical bucket,28 cm in diameter and 72 cm high, is full of water. The water is emptied into a rectangular tank, 66 cm long and 28 cm wide. Find the height of the water level of the tank​

Answer 1

hi ,

bohot Dino baad apka answer de rah hu

cylindrical bucket's diameter= 28 cm, radius=14cm

height=72 cm

rectangular tank's length =66 cm

breadth= 28 cm

as per question,

volume of water in cylindrical bucket=volume of water in rectangular tank

Πr2h=lbh

Π(14) 2∗72=66∗28∗h

h=24cm

thanks

Answer 2

Given:

• Diameter of cylindrical bucket = 28 cm.
• Height of cylindrical bucket = 72 cm
• Length of rectangular tank = 66 cm
• Width of rectangular tank = 28 cm

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To find:

• Height of bucket?

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Solution:

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∴ Radius of bucket will be, 28/2 = 14 cm respectively.

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$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{14 cm}}\put(9,17.5){\sf{72 cm\: \: }}\end{picture}$

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$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 66 cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 28 cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

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We know that,

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$\begin{gathered}\star\;{\boxed{\sf{\purple{Volume_{\;(cylinder)} = \pi r^2h}}}} \\ \end{gathered}$

$\begin{gathered}\star\;{\boxed{\sf{\pink{Volume_{\;(rectangle)} = l \times b \times h}}}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{22}{7} \times (14)^2 \times 72 = 66 \times 28 \times h\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{22}{7} \times 14 \times 14\times 72 = 66 \times 28 \times h\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf {22} \times 2\times 14\times 72 = 66 \times 28 \times h\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf {22} \times 28\times 72 = 66 \times 28 \times h\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf h = \frac{{22} \times 28\times 72 }{66 \times 28} \\ \\\end{gathered}$

$\begin{gathered}:\implies\sf h = \frac{44352 }{1848} \\ \\\end{gathered}$

$\begin{gathered}:\implies{\boxed{\sf{\pink{h = 24 \: cm}}}}\;\bigstar\\ \\\end{gathered}$

Therefore,

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• Radius of cylindrical bucket , 14 cm
• Height of rectangular tank, 24 cm