RS Aggarwal solution class 8 chapter 20 Volume and Surface Area of Solid Exercise 20B

Exercise 20B

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Q1 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 1:

Find the volume, curved surface area and total surface area of each of the cylinders whose dimensions are:
(i) radius of the base = 7 cm and height = 50 cm
(ii) radius of the base = 5.6 m and height = 1.25 m
(iii) radius of the base = 14 dm and height = 15 m

Answer 1:

Volume of a cylinder = $\pi r^{2 }h$
Lateral surface$=\mathrm{}2\mathrm{\pi }rh$
Total surface area $=\mathrm{}2\mathrm{\pi }r(h+r)$

(i) Base radius = 7 cm; height = 50 cm
Now, we have the following:
Volume$=\frac{22}{7}\times 7\times 7\times 50=7700 c{m}^3$
Lateral surface area$=\mathrm{}2\mathrm{\pi }rh$$=2\times \frac{22}{7}\times 7\times 50=2200 c{m}^2$
Total surface area $=\mathrm{}2\mathrm{\pi }r(h+r)$$=2\times \frac{22}{7}\times 7\left(50+7\right)=2508 c{m}^2$

(ii) Base radius = 5.6 m; height = 1.25 m
Now, we have the following:
Volume$=\frac{22}{7}\times 5.6\times 5.6\times 1.25=123.2 m^3$
Lateral surface area$=\mathrm{}2\mathrm{\pi }rh$$=2\times \frac{22}{7}\times 5.6\times 1.25=44 m^2$
Total surface area $=\mathrm{}2\mathrm{\pi }r(h+r)$$=2\times \frac{22}{7}\times 5.6\left(1.25+5.6\right)=241.12 m^2$

(iii) Base radius = 14 dm = 1.4 m, height = 15 m
Now, we have the following:
Volume$=\frac{22}{7}\times 1.4\times 1.4\times 15=92.4 m^3$
Lateral surface area$=\mathrm{}2\mathrm{\pi }rh$$=2\times \frac{22}{7}\times 1.4\times 15=132 m^2$
Total surface area $=\mathrm{}2\mathrm{\pi }r(h+r)$$=2\times \frac{22}{7}\times 1.4\left(15+1.4\right)=144.32 c{m}^2$

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Q2 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 2:

A milk tank is in the form of a cylinder whose radius is 1.5 m and height is 10.5 m. Find the quantity of milk in litres that can be stored in the tank.

Answer 2:

$$\begin{gathered} r = 1.5 \mathrm{m} \\ h = 10.5 \mathrm{m} \end{gathered}$$

$\mathrm{Capacity} \mathrm{of} \mathrm{the} \mathrm{tank}=\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{tank}$

$= \pi r^{2}h = \frac{22}{7}\times 1.5\times 1.5\times 10.5$

$=74.25 {\mathrm{m}}^3$

$\mathrm{<br>}$

$\mathrm{We} \mathrm{know} \mathrm{that} 1 {\mathrm{m}}^3=1000 \mathrm{L}$

$\therefore 74.25 {\mathrm{m}}^3=74250 \mathrm{L}$

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Q3 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 3:

A wooden cylindrical pole is 7 m high and its base radius is 10 cm. Find its weight if the wood weighs 225 kg per cubic metre.

Answer 3:

Height = 7 m
Radius = 10 cm = 0.1 m
Volume$={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 0.1\times 0.1\times 7=0.22 {\mathrm{m}}^3$
Weight of wood = 225 kg/m³
∴ Weight of the pole$=0.22\times 225=49.5 kg$

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Q4 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 4:

Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm?

Answer 4:

Diameter = 2r = 140 cm
i.e., radius, r = 70 cm = 0.7 m

Now, volume ​$={\mathrm{\pi r}}^2\mathrm{h}=1.54 {\mathrm{m}}^3$

$$\begin{gathered} \Rightarrow \frac{22}{7}\times 0.7\times 0.7\times h=1.54 \\ \therefore h=\frac{1.54\times 7}{0.7\times 0.7\times 22}=\frac{154\times 7}{154\times 7}=1 m \end{gathered}$$

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Q5 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 5:

The volume of a circular iron rod of length 1 m is 3850 cm³. Find its diameter.

Answer 5:

Volume$={\mathrm{\pi r}}^2\mathrm{h}=3850 {\mathrm{cm}}^3$
Height = 1 m =100 cm

Now, radius, $r=\sqrt{\frac{3850}{\mathrm{\pi }\times \mathrm{h}}}=\sqrt{\frac{3850\times 7}{22\times 100}}=1.75\times 7=3.5 cm$
∴ Diameter =2(radius) $=2\times 3.5=7 cm$

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Q6 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 6:

A closed cylindrical tank of diameter 14 m and height 5 m is made from a sheet of metal. How much sheet of metal will be required?

Answer 6:

Diameter = 14 m
Radius $=\frac{14}{2}=7 m$
Height = 5 m

∴ Area of the metal sheet required = total surface area

 $$\begin{gathered} =2\mathrm{\pi r}(\mathrm{h}+\mathrm{r}) \\ =2\times \frac{22}{7}\times 7(5+7) m^2 \\ =44\times 12 m^2 \\ =528 {\mathrm{m}}^2 \end{gathered}$$

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Q7 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 7:

The circumference of the base of a cylinder is 88 cm and its height is 60 cm. Find the volume of the cylinder and its curved surface area.

Answer 7:

Circumference of the base = 88 cm
Height = 60 cm

Area of the curved surface $=circumference\times height=88\times 60=5280 c{m}^2$
Circumference $=2\mathrm{\pi r}=88 cm$
Then radius$=r=\frac{88}{2\mathrm{\pi }}=\frac{88\times 7}{2\times 22}=14 cm$
∴ Volume$={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 14\times 14\times 60=36960 {\mathrm{cm}}^3$

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Q8 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 8:

The lateral surface area of a cylinder of length 14 m is 220 m². Find the volume of the cylinder.

Answer 8:

Length = height = 14 m
Lateral surface area$=2\mathrm{\pi rh}=220 {\mathrm{m}}^2$
Radius $=r=\frac{220}{2\mathrm{\pi h}}=\frac{220\times 7}{2\times 22\times 14}=\frac{10}{4}=2.5 m$
∴ Volume$={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 2.5\times 2.5\times 14=275 {\mathrm{m}}^3$  

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Q9 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 9:

The volume of a cylinder of height 8 cm is 1232 cm³. Find its curved surface area and the total surface area.

Answer 9:

Height = 8 cm
Volume$={\mathrm{\pi r}}^2\mathrm{h}=1232 {\mathrm{cm}}^3$
Now, radius$=r=\sqrt{\frac{1232}{\mathrm{\pi h}}}=\sqrt{\frac{1232\times 7}{22\times 8}}=\sqrt{49}=7cm$
Also, curved surface area $=2\mathrm{\pi rh}=2\times \frac{22}{7}\times 7\times 8=352 {\mathrm{cm}}^2$

∴ Total surface area $=2\mathrm{\pi r}(\mathrm{h}+\mathrm{r})=\left(2\times \frac{22}{7}\times 7\times 8\right)+\left(2\times \frac{22}{7}\times {\left(7\right)}^2\right)$

$=352+308=660 {\mathrm{cm}}^2$

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Q10 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 10:

The radius and height of a cylinder are in the ratio 7 : 2. If the volume of the cylinder is 8316 cm³, find the total surface area of the cylinder.

Answer 10:

We have: $\frac{radius}{height}=\frac{7}{2}$
i.e., $r=\frac{7}{2}h$
Now, volume $={\mathrm{\pi r}}^2\mathrm{h}=\mathrm{\pi }{\left(\frac{7}{2}\mathrm{h}\right)}^2\mathrm{h}=8316 {\mathrm{cm}}^3$
$\Rightarrow \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times h^3=8316$

$\Rightarrow h^3=\frac{8316\times 2}{11\times 7}=108\times 2=216$

$\Rightarrow h=\sqrt[3]{216}=6 cm$

Then $r=\frac{7}{2}h=\frac{7}{2}\times 6=21 cm$
∴ Total surface area $=2\mathrm{\pi r}(h+r)=2\times \frac{22}{7}\times 21\times (6+21)=3564 {\mathrm{cm}}^2$

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Q11 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 11:

The curved surface area of a cylinder is 4400 cm² and the circumference of its base is 110 cm. Find the volume of the cylinder.

Answer 11:

Curved surface area $=2\mathrm{\pi }rh=4400 c{m}^2$
Circumference $=2\mathrm{\pi r}=110 \mathrm{cm}$
Now, height$=h=\frac{curved surface area}{circumference}=\frac{4400}{110}=40 cm$

Also, radius, $r=\frac{4400}{2\mathrm{\pi h}}=\frac{4400\times 7}{2\times 22\times 40}=\frac{35}{2}$

∴ Volume$={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times \frac{35}{2}\times \frac{35}{2}\times 40$

$=22\times 5\times 35\times 10=38500 {\mathrm{cm}}^3$

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Q12 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 12:

A particular brand of talcum powder is available in two packs, a plastic can with a square base of side 5 cm and of height 14 cm, or one with a circular base of radius 3.5 cm and of height 12 cm. Which of them has greater capacity and by how much?

Answer 12:

For the cubic pack:
Length of the side, a = 5 cm
Height = 14 cm
Volume$=a^{2}h=5\times 5\times 14=350 c{m}^3$

For the cylindrical pack:
Base radius $=r=3.5 cm$
Height = 12 cm
Volume$={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 3.5\times 3.5\times 12=462 c{\mathrm{m}}^3$

We can see that the pack with a circular base has a greater capacity than the pack with a square base.
Also, difference in volume$=462-350=112 c{m}^3$

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Q13 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 13:

Find the cost of painting 15 cylindrical pillars of a building at Rs 2.50 per square metre if the diameter and height of each pillar are 48 cm and 7 metres respectively.

Answer 13:

Diameter = 48 cm
Radius = 24 cm = 0.24 m
Height = 7 m

Now, we have:
Lateral surface area of one pillar$=\mathrm{\pi dh}=\frac{22}{7}\times 0.48\times 7=10.56 {\mathrm{m}}^2$
Surface area to be painted = total surface area of 15 pillars $=10.56\times 15=158.4 m^2$
∴ Total cost$=\mathrm{Rs} (158.4\times 2.5)=\mathrm{Rs} 396$

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Q14 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 14:

A rectangular vessel 22 cm by 16 cm by 14 cm is full of water. If the total water is poured into an empty cylindrical vessel of radius 8 cm, find the height of water in the cylindrical vessel.

Answer 14:

Volume of the rectangular vessel $=22\times 16\times 14=4928 c{m}^3$
Radius of the cylindrical vessel = 8 cm
Volume$={\mathrm{\pi r}}^2\mathrm{h}$

As the water is poured from the rectangular vessel to the cylindrical vessel, we have:
Volume of the rectangular vessel  = volume of the cylindrical vessel

∴ Height of the water in the cylindrical vessel$=\frac{volume}{{\mathrm{\pi r}}^2}=\frac{4928\times 7}{22\times 8\times 8}=\frac{28\times 7}{8}=\frac{49}{2}=24.5 cm$

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Q15 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 15:

A piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 11 cm. It is drawn out into a wire of diameter 1 mm. What will be the length of the wire so obtained?

Answer 15:

Diameter of the given wire = 1 cm
Radius = 0.5 cm
Length = 11 cm
Now, volume$={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 0.5\times 0.5\times 11=8.643 c{m}^3$
The volumes of the two cylinders would be the same.
Now, diameter of the new wire = 1 mm = 0.1 cm
Radius = 0.05 cm
∴ New length $=\frac{\mathrm{volume}}{{\mathrm{\pi r}}^2}=\frac{8.643\times 7}{22\times 0.05\times 0.05}=1100.02 cm$ ≅ 11 m

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Q16 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 16:

A solid cube of metal each of whose sides measures 2.2 cm is melted to form a cylindrical wire of radius 1 mm.  Find the length of the wire so obtained.

Answer 16:

Length of the edge, a = 2.2 cm
Volume of the cube $=a^3={\left(2.2\right)}^3=10.648 c{m}^3$
Volume of the wire$={\mathrm{\pi r}}^2\mathrm{h}$
Radius = 1 mm = 0.1 cm
As volume of cube = volume of wire, we have:

$h=\frac{volume}{{\mathrm{\pi r}}^2}=\frac{10.648\times 7}{22\times 0.1\times 0.1}=338.8 cm$

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Q17 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 17:

How many cubic metres of earth must be dug out to sink a well which is 20 m deep and has a diameter of 7 metres? If the earth so dug out is spread over a rectangular plot 28 m by 11 m, what is the height of the platform so formed?

Answer 17:

Diameter = 7 m
Radius = 3.5 m
​Depth = 20 m

​Volume of the earth dug out $={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 3.5\times 3.5\times 20=770 {\mathrm{m}}^3$
Volume of the earth piled upon the given plot$=28\times 11\times h=770 m^3$

$\therefore h=\frac{770}{28\times 11}=\frac{70}{28}=2.5 m$

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Q18 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 18:

A well of inner diameter 14 m is dug to a depth of 12 m. Earth taken out of it has been evenly spread all around it to a width of 7 m to form an embankment. Find the height of the embankment so formed.

Answer 18:

Inner diameter = 14 m
i.e., radius = 7 m
Depth = 12 m
​Volume of the earth dug out$={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 7\times 7\times 12=1848 {\mathrm{m}}^3$

Width of embankment = 7 m
Now, total radius $=7+7=14 m$

$\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{embankment}$

$=\mathrm{total} \mathrm{volume} - \mathrm{inner} \mathrm{volume}$

$$\begin{gathered} ={{\mathrm{\pi r}}_{\mathrm{o}}}^2\mathrm{h}-{{\mathrm{\pi r}}_{\mathrm{i}}}^2\mathrm{h}=\mathrm{\pi h}\left({{\mathrm{r}}_{\mathrm{o}}}^2-{{\mathrm{r}}_{\mathrm{i}}}^2\right) \\ =\frac{22}{7}\mathrm{h}\left({14}^2-7^2\right) \end{gathered}$$

$$\begin{gathered} =\frac{22}{7}\mathrm{h}\left(196-49\right) \\ =\frac{22}{7}\mathrm{h}\times 147 \end{gathered}$$

$$\begin{gathered} =21\times 22\mathrm{h} \\ =462\times \mathrm{h} {\mathrm{m}}^3 \end{gathered}$$

Since volume of embankment = volume of earth dug out

,we have:
$1848=462 h$
$\Rightarrow h=\frac{1848}{462}=4 m$
∴ Height of the embankment = 4 m

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Q19 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 19:

A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of the road roller is 84 cm and its length is 1 m.

Answer 19:

Diameter = 84 cm
i.e., radius = 42 cm
Length = 1 m = 100 cm
Now, lateral surface area $=2\mathrm{\pi rh}$

$=2\times \frac{22}{7}\times 42\times 100=26400 {\mathrm{cm}}^2$

∴ Area of the road $=\mathrm{lateral} \mathrm{surface} \mathrm{area} \times \mathrm{no}. \mathrm{of} \mathrm{rotations}$

$=26400\times 750=19800000 {\mathrm{cm}}^2=1980 {\mathrm{m}}^2$

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Q20 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 20:

A cylinder is open at both ends and is made of 1.5-cm-thick metal. Its external diameter is 12 cm and height is 84 cm. What is the volume of metal used in making the cylinder? Also, find the weight of the cylinder if 1 cm³ of the metal weighs 7.5 g.

Answer 20:

Thickness of the cylinder = 1.5 cm
External diameter = 12 cm
i.e., radius = 6 cm
also, internal radius = 4.5 cm
Height = 84 cm

Now, we have the following:
Total volume$={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 6\times 6\times 84=9504 {\mathrm{cm}}^3$
Inner volume $={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 4.5\times 4.5\times 84=5346 {\mathrm{cm}}^3$

Now, volume of the metal = total volume − inner volume  

 $=9504-5346=4158 c{m}^3$

∴ Weight of iron $=4158\times 7.5= 31185 \mathrm{g}$

$= 31.185 \mathrm{kg}$   [Given: $1 c{m}^3=7.5g$]

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Q21 | Ex-20B | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 21:

The length of a metallic tube is 1 metre, its thickness is 1 cm and its inner diameter is 12 cm. Find the weight of the tube if the density of the metal is 7.7 grams per cubic centimetre.

Answer 21:

Length = 1 m = 100 cm
Inner diameter = 12 cm
Radius = 6 cm
Now, inner volume$={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 6\times 6\times 100=11314.286 {\mathrm{cm}}^3$
Thickness = 1 cm
Total radius = 7 cm

Now, we have the following:
Total volume$={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 7\times 7\times 100=15400 {\mathrm{cm}}^3$

Volume of the tube $=\mathrm{total} \mathrm{volume}-\mathrm{inner} \mathrm{volume}$

$= 15400-11314.286=4085.714 c{m}^3$

Density of the tube = 7.7 g/cm³

∴ Weight of the tube $=\mathrm{volume}\times \mathrm{density}=4085.714\times 7.7$

$=31459.9978 \mathrm{g}=31.459 \mathrm{kg}$