Question

1)Selvi’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44 m × 95cm. The overhead tank has its radius 60 cm and height 95 cm. Find the height of the water left in the sump after the overhead tank has been completely filled with water from the sump which had been full. Compare the capacity of the tank with that of the sump.
(Use π = 3.14)

2)An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet as shown in the figure. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold.
(Take π = 22/7)
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Answer 1

1. ❆ ϲοиϲєρτ :-

In this question, we will take the help of the branch of maths, "Volume". We have to find out the water left in the sump after the transfer of its water when it was full into a tank. For this we will take the help of "Volume" as well as some basic maths operations.

$\sf \dag \: \red {Given :-}$

For the sump, which is of cuboidal shape,

• Length = 1.57 m
• Width = 1.44 m
• Height = 95 cm = 0.95 m

Now we will find out the volume of sump, using the formula,

$\boxed{ \orange \bigstar \: \sf \green {volume \: of \: cuboid \: = length \: \times breadth \: \times height \: (F.1)}}$

Putting the values in the formula, we get,

$\sf \longrightarrow \: volume \: of \: sump \: =1.4 \times 1.57 \times 0.95 \\ \sf \longrightarrow \: volume \: of \: sump \: = \: 2.14776 {m}^{3} \: \: \: \: \: \: \: \: \: \: \: \:$

For the tank, which is of cylindrical shape,

• Radius = 60 cm = 0.6 m
• Height = 95 cm = 0.95 cm

Now we will find out the volume of cylindrical tank, using the formula,

$\boxed { \sf \orange\bigstar \: \green{volume \: of \: cylinder \: = 2 \pi {r}^{2} h}}$

Putting the values, we get,

$\sf \longrightarrow \: volume \: of \: tank \: = 2 \times 3.14 \times ( {0.6}^{2} ) \times 0.95 \\ \sf \longrightarrow \: volume \: of \: tank \: = 1.074 {m}^{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Therefore,

$\sf \longrightarrow \: volume \: of \: water \: = volume \: of \: sump \: - volume \: of \: water \: \\ \sf \longrightarrow \: volume \: of \: water \: =2.147 {m}^{3} - 1.074 \: {m}^{3} = 1.073 \: {m}^{3} \: \: \: \: \: \: \\ \sf \longrightarrow \: volume \: of \: water \: =1.073 \: {m}^{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

❆ ϲοиϲєρτ :-

Now we have to find out the height of the water left in the sump after the transfer of its water into overhead tank. Here, again we will use the concept of "Volume". We have already found out the volume of water. And we are provided with length and width of the water ( i.e. same to that of the tank ). So, its very easy for us to calculate the height of water. So, let's do it.

Given (for the water) :-

• Volume = 1.073 m³
• Width = 1.44 m
• Length = 1.57 m

Now, we have to find out the height. So again we will use the formula of volume of a cuboid i.e. F. 1. Therefore,

$\sf \leadsto \: volume \: of \: water \: = height \: \times lenght \: \times breadth \: \\ \sf \leadsto \: 1.073 \: {m}^{3} = h \times 1.57 \times 1.44 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \leadsto \:h = \frac{ \cancel{1.073 \: {m}^{3}} }{ \cancel{1.57 \times 1.44}} = 0.475 \: m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \pink{ \therefore \: height \: of \: water \: = \: 0.475 \: m}$

❆ Concept :-

Now we have been asked to compare the capacity of the tank with that of the sump. We know that capacity means how much of matter can be stored in them i.e. their volume. So, we can say that,

$\boxed{\sf \bigstar \: \frac{capacity \: of \: tank}{capacity \: of \: sump} = \frac{volume \: of \: tank}{volume \: of \: sump} }$

Now we will simply put the values to get the required answer. Let's do it!

$\sf \leadsto\: \frac{capacity \: of \: tank}{capacity \: of \: sump} = \frac{volume \: of \: tank}{volume \: of \: sump} \\ \sf \leadsto \: \frac{1.074 \: {m}^{3} }{2.147 \: {m}^{3} } = 0.49 = 0.5 = \frac{1}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Therefore, the capacity of sump is double than that of the tank.

Answer 2

➦ Given :

➟The sump has dimensions

➻ 1.57 m × 1.44 m × 95cm

• ➻Length =1.57 m
• ➻ Breadth= 1.44 m
• ➻Height =95cm

➟ The overhead tank has

• ➻ radius 60 cm
• ➻ height 95 cm.

➦ To Find

➻ The height of the water left in the sump

➻ Compare the capacity of the tank with that of the sump.

➦ Explanation of step 1

⇛ The volume of water in the overhead tank is equals to the volume of the water removed from the sump.

⇛ We have to find Volume of the overhead tank (cylinder)

➦ Formula used

➲ Volume of the overhead tank (cylinder)

• ➻π r²h

⇛Place the given value in this formula

➻π (radius) ²x height

➳π (60 cm) ² x 95 cm.

➳π (0.6 m)²x 0.95 m

➪3.14 x 0.6 x 0.6 x 0.95 m³

➦ Explanation of step 2

⇛ We have to calculate now Volume of the water in the sump when full

➦ Formula used

➲ Volume of the water in the sump when full

• ➻ l × b × h

➲ Where

➳ l means length =1.57 m

➳ b means breadth =1.44 m

➳ h means height = 95 cm

⇛Place the given value in this formula

➻length × breadth × height

➻1.57 m x 1.44 m x 95 cm

➻1.57 m x 1.44 mx 0.95 m ³

➦ Explanation of step 3

⇛ We have to find the volume of the water left or remained in the sump after filling the overhead tank

➦ Formula used

➻ Volume of the water in the sump when full - (minus)Volume of the overhead tank (cylinder)

➲ Where

➻ Volume of the water in the sump when full

• (1.57 x 1.44 x 0.95) m³

➻ Volume of the overhead tank (cylinder)

(3.14 x 0.6 x 0.6 x 0.95) m²

⇛Place the given value in this formula

(1.57 x 1.44 x 0.95) m³- (3.14 x 0.6 x 0.6 x 0.95) m²

➻ (1.57 x 0.6×0.6 x 0.95×2 )m³

➲ Now

➻ After the overhead tank is filled completely then the height of water in the sump be h meter,

➲ Then

➦ Formula used

➻The volume of water left in the sump ÷ ( Divided ) by length × breadth

➲ Where

➻ length × breadth

• 1.57 × 1.44

➻The volume of water left in the sump

(1.57 x 0.6×0.6 x 0.95×2 )m³

⇛Place the given value in this formula

➻ (1.57 x 0.6×0.6 x 0.95×2 ) /1.57 × 1.44 m

⇛Here 1.57 get cancelled out

➻( 0.6×0.6 x 0.95×2) / 1.44 m

➻0.684 ÷ 1.44 m

• ➻0.47 m

⇛In order to convert meter into centimeter multiply it by 100

⇛ 0.475 × 100

➻ 47.5 cm

➲ Hence

⇛ Therefore, height of water left in the sump after filling the overhead tank is 47.5 cm.

➲Now

⇛ We have to compare capacity of tank with that of sump

➦ Formula used

⇛ Capacity of tank / Capacity of sump

➲ Where

• Volume of the overhead tank (cylinder)

⇛(3.14 x 0.6 x 0.6 x 0.95) m²

• Volume of water when sump is full

⇛(0.57 × 1.44 ×0.95)

⇛Place the given value in this formula

➲ (3.14 x 0.6 x 0.6 x 0.95) / (0.57 × 1.44 ×0.95)

⇛ After solving this we get

➲ 1/2 as answer

➦Therefore

⇛ Hence, the capacity of the overhead tank is half the capacity of the sump.

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