Question

A farmer connects a pipe of internal diameter 20 cm form a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 kilometer per hour, in how much time will the tank be filled?
please answer today is my maths pre board exam ​

Answer 1

Answer:

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Answer 2

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

• A pipe of internal diameter 20 cm through which water flows at the rate of 3 km per hour.

• Diameter and height of cylindrical tank is 10m and 2m.

$\large\underline{\sf{To\:Find - }}$

• Time taken to filled the tank

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

Let us consider a cylinder of radius 'r' units and height 'h' units, then volume of cylinder is given by

$\rm :\longmapsto\: \sf \: Volume_{(cylinder)} = \pi \: {r}^{2} h$

If speed of object is 'x' km per hour, it implies distance covered in 1 hour is 'x' km.

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

Let time taken to filled the cylindrical tank be 't' hours.

Now,

Given Dimensions of tank

• Diameter of cylindrical tank = 10 m

So,

• Radius of cylindrical tank, r = 5 m

And

• Height of cylindrical tank, h = 2 m

So,

• Capacity of cylindrical tank is

$\rm :\longmapsto\:Volume_{(tank)} = \pi \: {r}^{2} h$

$\rm :\longmapsto\:Volume_{(tank)} = \pi \: \times {(5)}^{2} \times 2$

$\bf :\longmapsto\: \boxed{ \bf{Volume_{(tank)} = 50\pi \: {m}^{3}}} - - - (1)$

Now,

Pipe of diameter 20 cm is connected through a canal and water flows through it at a speed of 3 km per hour.

• Diameter of cylindrical pipe = 20 cm

So,

• Radius of cylindrical pipe, R = 10 cm = 0.1 m

Speed of flow of water = 3 km per hour

It implies,

• In 1 hour, water flow is 3 km = 3000 m

• In 't' hours, water flow = 3000t m

So,

• Volume of water flows in 't' hour is given by

$\rm :\longmapsto\:Volume_{(water \: flow)} = \pi \: \times (0.1)^{2} \times 3000t$

$\rm :\longmapsto\:Volume_{(water \: flow)} = \pi \: \times \dfrac{1}{100} \times 3000t$

$\rm :\longmapsto\: \boxed{ \bf{Volume_{(water \: flow)} = 30\pi \: t \: {m}^{3} }} - - - (2)$

According to statement,

$\rm :\longmapsto\:Volume_{(water \: flow)} = Volume_{(tank)}$

$\rm :\longmapsto\:50\pi \: = 30\pi \: t$

$\rm :\longmapsto\:t \: = \: \dfrac{5}{3} \: hours = \dfrac{5}{3} \times 60 = 100 \: minutes$

$\rm :\longmapsto\:t \: = \: 1 \: \dfrac{2}{3} \: hours \: \: or \: \: 100 \: minutes$

$\boxed{\sf \: Hence, \: time \: taken \: to \: fill \: tank \: = 1 \: \dfrac{2}{3} hrs \: or \: 100 \: min}$

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Additional Information :-

$\boxed{ \sf{ \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2} h}}$

$\boxed{ \sf{ \: Volume_{(cuboid)} = lbh}}$

$\boxed{ \sf{ \: Volume_{(cube)} = {(edge)}^{3} }}$

$\boxed{ \sf{ \: CSA{(cylinder)} = 2\pi \: rh}}$

$\boxed{ \sf{ \: CSA{(cuboid)} = 2(l + b) \times h}}$

$\boxed{ \sf{ \: CSA{(cube)} = 4 \times {(edge)}^{2} }}$

$\boxed{ \sf{ \: CSA{(cone)} = \pi \: rl}}$

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

$\boxed{ \sf{ \: TSA{(cone)} = \pi \: r(l + r)}}$

$\boxed{ \sf{ \: TSA{(cuboid)} = 2(lb + bh + hl)}}$

$\boxed{ \sf{ \: TSA{(cone)} = {6(edge)}^{2} }}$