Question

A milk tank is in the form of cylinder whose radius is 1 metre and length is 7 metre find the quantity of milk in litres that can be stored in the tank also find the area of metal sheet required to make it????​​

Answer 1

$\sf\huge\underline{Answer :-}$

Given :

Radius of the milk tank = 1 m

Length\Height of the milk tank = 7 m

To find :

Quality of milk that can be stored in the milk tank.

Area of the metal sheet required to make it.

Solution :

Let's convert the given quantities to decimetre from metres.

[Note : 1 metre = 10 decimetre]

$\sf Radius \rightarrow 1 \: m \: = \: 10 \: dcm$

$\sf Height \rightarrow 7 \: m \: = \: 70 \: dcm$

We know that,

$\boxed{\sf \star Volume \: of \: cylinder \: = \: \pi \: {r}^{2} \: h}$

$\sf\longrightarrow \dfrac{22}{7} \times 10 \times 10 \times 70 \: {dcm}^{3}$

$\sf\longrightarrow \dfrac{22}{{\cancel 7}} \times 10 \times 10 \times {\cancel {70}} \: {dcm}^{3}$

$\sf\longrightarrow (22 \times 10 \times 10 \times 10) \: {dcm}^{3}$

$\sf\longrightarrow 22000 \: {dcm}^{3}$

°.° 1 dcm³ = 1 litre

.°. 22000 dcm³ = 22000 litres.

Hence, the quantity of milk that can be stored in the cylindrical milk tank is 22000 litres.

Now,

$\boxed{\sf \star TSA \: of \: cylinder \: = \: 2 \: \pi \: r \: (r + h)}$

[Note : We'll put the quantities in metre.]

$\sf\longrightarrow 2 \times \dfrac{22}{7} \times 1 \: (1 + 7)$

$\sf\longrightarrow 2 \times \dfrac{22}{7} \times 1 \times 8$

$\sf\longrightarrow 2 \times \dfrac{22}{7} \times 8$

$\sf\longrightarrow \dfrac{352}{7}$

$\sf\longrightarrow 50.28 \: (approx)$

Hence, the area of metal sheet required to make the milk tank is 50.28 m² (approx).

Answer 2

Answer:

• Radius of cylinder = 1 metre
• Length of cylinder = 7 metre

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{1 m}}\put(6.5,17.5){\sf{7 m}}\end{picture}$

To find Quantity of milk that can be stored into it, we need to find Volume.

⠀

$\underline{\bigstar\:\textsf{Volume of the cylindrical tank :}}$

$:\implies\sf Volume=\pi r^2h\\\\\\:\implies\sf Volume=\dfrac{22}{7} \times (1 \:m)^2 \times 7 \:m\\\\\\:\implies\sf Volume = 22 \times (1 \:m)^2 \times m\\\\\\:\implies\sf Volume = 22 \times 1 \:m^2 \times m\\\\\\:\implies\sf Volume = 22 \times 1 \:m^3\\\\{\scriptsize\qquad\bf{\dag}\:\:\tt{1\:m^3=1000\:litres}} \\\\:\implies\sf Volume = 22 \times 1000\:litres\\\\\\:\implies\sf Volume = 22000\:litres$

⠀

$\therefore\:\underline{\textsf{Hence, it can hold \textbf{22000 litres} of milk}}.$

$\rule{200}{1}$

To find metal sheet required to make this tank, we need to find Total Surface Area of Tank.

⠀

$\underline{\bigstar\:\textsf{Total Surface Area of the tank :}}$

$\dashrightarrow\sf TSA=2\pi r(h+r)\\\\\\\dashrightarrow\sf TSA=2\ \times \dfrac{22}{7} \times 1 \:m \times (7 \:m+1 \:m)\\\\\\\dashrightarrow\sf TSA = 2\ \times \dfrac{22}{7} \times 1 \:m \times 8 \:m\\\\\\\dashrightarrow\sf TSA = \dfrac{352 \:m^{2} }{7}\\\\\\\dashrightarrow\sf TSA = 50.28\:m^{2}$

⠀

$\therefore\:\underline{\textsf{Metal sheet of area \textbf{50.28 m ^\text2 } is required to make it}}.$