RS AGGARWAL CLASS 9 CHAPTER 15 VOLUMES AND SURFACE AREA OF SOLIDS EXERCISE 15B

  EXERCISE 15B

PAGE NO-573

Question 1:

The diameter of a cylinder is 28 cm and its height is 40 cm. Find the curved surface area, total surface area and the volume of the cylinder.

Answer 1:

Here, r = 28/2 = 14 cm; h = 40 cm

$$\begin{gathered} \mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder}=2\mathrm{\pi }rh \\ =2\times \frac{22}{7}\times 14\times 40 {\mathrm{cm}}^2 \\ =2\times 22\times 2\times 40 {\mathrm{cm}}^2 \\ =3520 {\mathrm{cm}}^2 \end{gathered}$$


$$\begin{gathered} \mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder}=2\mathrm{\pi }rh + 2\mathrm{\pi }{r}^2 \\ = \left(3520 +2\times \frac{22}{7}\times {14}^2\right) {\mathrm{cm}}^2 \\ =(3520+1232 )=4752 {\mathrm{cm}}^2 \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}=\mathrm{\pi }{r}^{\mathit{2}}h \\ =\frac{22}{7}\times {14}^2\times 40 {\mathrm{cm}}^3 \\ =22\times 14\times 2\times 40 {\mathrm{cm}}^3 \\ =24640 {\mathrm{cm}}^3 \end{gathered}$$

Question 2:

A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Answer 2:

Radius of the bowl, r = $\frac{7}{2} \mathrm{cm}$

Height of soup in the bowl, h = 4 cm

Volume of soup in one bowl = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}\times {\left(\frac{7}{2}\right)}^2\times 4=154 {\mathrm{cm}}^3$

∴ Amount of soup the hospital has to prepare daily to serve 250 patients

= Volume of soup in one bowl × 250

= 154 × 250

= 38500 cm³

Question 3:

The pillars of a temple are cylindrically shaped. Each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?

Answer 3:

Radius of each pillar, r = 20 cm = 0.2 m   (1 m = 100 cm)

Height of each pillar, h = 10 m

Volume of concrete mixture used in each pillar = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}\times {\left(0.2\right)}^2\times 10 {\mathrm{m}}^3$

∴ Amount of concrete mixture required to build 14 such pillars

= Volume of concrete mixture used in each pillar × 14

= $\frac{22}{7}\times {\left(0.2\right)}^2\times 10 \times 14$ 

= 17.6 m³

Question 4:

A soft drink is available in two packs: (i) a tin can with a rectangular base of length 5 cm, breadth 4 cm and height 15 cm, and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Answer 4:

(i) Length of tin can, l = 5 cm

Breadth of tin can, b = 4 cm

Height of tin can, h = 15 cm

∴ Volume of soft drink in tin can = l × b × h = 5 × 4 × 15 = 300 cm³

(ii) Radius of plastic cylinder, r = $\frac{7}{2}$ cm

Height of plastic cylinder, h = 10 cm

∴ Volume of soft drink in plastic cylinder = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}\times {\left(\frac{7}{2}\right)}^2\times 10$ = 385 cm³

So, the capacity of the plastic cylinder pack is greater than the capacity of the tin can pack.

Difference in the capacities of the two packs = 385 − 300 = 85 cm³

Thus, the capacity of the plastic cylinder pack is 85 cm³ more than the capacity of the tin can pack.

Question 5:

There are 20 cylindrical pillars in a building, each having a diameter of 50 cm and height 4 m. Find the cost of cleaning them at ₹ 14 per m².

Answer 5:

Radius of each pillar, r = $\frac{50}{2}$ = 25 cm = 0.25 m        (1 m = 100 cm)

Height of each pillar, h = 4 m

∴ Surface area of each pillar = $2\mathrm{\pi }rh=2\times \frac{22}{7}\times 0.25\times 4 {\mathrm{cm}}^2$

Surface area of 20 pillars = Surface area of each pillar × 20 = $2\times \frac{22}{7}\times 0.25\times 4 \times 20 {\mathrm{cm}}^2$

Rate of cleaning = ₹ 14 per m²

∴ Total cost of cleaning the 20 pillars

= Surface area of 20 pillars × Rate of cleaning 

= $2\times \frac{22}{7}\times 0.25\times 4 \times 20 \times 14$

= ₹ 1760

Question 6:

The curved surface area of a right circular cylinder is 4.4 m². If the radius of its base is 0.7 m, find its (i) height and (ii) volume.

Answer 6:

Radius of the cylinder, r = 0.7 m

Curved surface area of cylinder = 4.4 m²

(i) Let the height of the cylinder be h m.

$$\begin{gathered} \therefore 2\mathrm{\pi }rh=4.4 \\ \Rightarrow 2\times \frac{22}{7}\times 0.7\times h=4.4 \\ \Rightarrow h=\frac{4.4\times 7}{2\times 22\times 0.7}=1 \mathrm{m} \end{gathered}$$

Thus, the height of the cylinder is 1 m.

(ii) Volume of the cylinder = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}\times {\left(0.7\right)}^2\times 1$ = 1.54 m³

Thus, the volume of the cylinder is 1.54 m³.

Question 7:

The lateral surface area of a cylinder is 94.2 cm² and its height is 5 cm. Find (i) the radius of its base and (ii) its volume. (Take π = 3.14.)

Answer 7:

Height of the cylinder, h = 5 cm

Lateral (or curved) surface area of cylinder = 94.2 cm²

(i) Let the radius of the cylinder be r cm.

$$\begin{gathered} \therefore 2\mathrm{\pi }rh=94.2 {\mathrm{cm}}^2 \\ \Rightarrow 2\times 3.14\times r\times 5=94.2 \\ \Rightarrow r=\frac{94.2}{2\times 3.14\times 5}=3 \mathrm{cm} \end{gathered}$$
Thus, the radius of the cylinder is 3 cm.

(ii) Volume of the cylinder = $\mathrm{\pi }{r}^{2}h=3.14\times {\left(3\right)}^2\times 5$ = 141.3 cm³

Thus, the volume of the cylinder is 141.3 cm³.

Question 8:

The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. Find the area of the metal sheet needed to make it.

Answer 8:

Height of the cylinder, h = 1 m

Capacity of the cylinder = 15.4 L = 15.4 × 0.001 m³ = 0.0154 m³

$$\begin{gathered} \therefore \mathrm{\pi }{r}^{2}h=0.0154 {\mathrm{m}}^3 \\ \Rightarrow \frac{22}{7}\times r^2\times 1=0.0154 \\ \Rightarrow r=\sqrt{\frac{0.0154\times 7}{22}}=0.07 \mathrm{m} \end{gathered}$$
∴ Area of the metal sheet needed to make the cylinder 

= Total surface area of the cylinder

$$\begin{gathered} =2\mathrm{\pi }rh\left(r+h\right) \\ =2\times \frac{22}{7}\times 0.07\times 1\times \left(0.07+1\right) \\ =2\times \frac{22}{7}\times 0.07\times 1\times 1.07 \\ =0.4708 {\mathrm{m}}^2 \end{gathered}$$
Thus, the area of the metal sheet needed to make the cylinder is 0.4708 m².

Question 9:

The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.

Answer 9:

Inner radius of the wooden pipe, r = $\frac{24}{2}$ = 12 cm

Outer radius of the wooden pipe, R = $\frac{28}{2}$ = 14 cm

Length of the wooden pipe, h = 35 cm

∴ Volume of wood in the pipe = $\mathrm{\pi }\left(R^2\mathit{-}{r}^{\mathit{2}}\right)h=\frac{22}{7}\times \left({14}^2-{12}^2\right)\times 35=5720 {\mathrm{cm}}^3$

It is given that 1 cm³ of wood has a mass of 0.6 g.

∴ Mass of the pipe = Volume of wood in the pipe × 0.6 = 5720 × 0.6 = 3432 g = $\frac{3432}{1000}$ = 3.432 kg

Thus, the mass of the pipe is 3.432 kg.

Question 10:

In a water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Answer 10:

Length of the cylindrical pipe, h = 28 m

Radius of the cylindrical pipe, r = $\frac{5}{2}$ = 2.5 cm = 0.025 m         (1 m = 100 cm)

∴ Total radiating surface in the system

= Curved surface area of the cylindrical pipe

$$\begin{gathered} =2\mathrm{\pi }rh \\ =2\times \frac{22}{7}\times 0.025\times 28 \\ =4.4 {\mathrm{m}}^2 \end{gathered}$$
Thus, the total radiating surface in the system is 4.4 m².

Question 11:

Find the weight of a solid cylinder of radius 10.5 cm and height 60 cm if the material of the cylinder weighs 5 g per cm².

Answer 11:

Here, r = 10.5 cm; h = 60 cm

$$\begin{gathered} \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}=\mathrm{\pi }{r}^{\mathit{2}}h\mathit{ } \\ =\frac{22}{7}\times (10.5{)}^2\times 60 {\mathrm{cm}}^3 \\ =22\times 10.5\times 1.5\times 60 {\mathrm{cm}}^3=20790 {\mathrm{cm}}^3 \end{gathered}$$

∴ Weight of cylinder = volume of cylinder $\times$weight of cylinder per gram
                              $$\begin{gathered} =20790\times 5 \mathrm{g} \\ =103950 \mathrm{g} \\ =103.95 \mathrm{kg} \end{gathered}$$ 

Question 12:

The curved surface area of a cylinder is 1210 cm² and its diameter is 20 cm. Find its height and volume.

Answer 12:

Curved surface area = 1210 cm²
Suppose that the height of cylinder is h cm.
We have r = 10 cm
$$\begin{gathered} \mathrm{Now}, 1210 =2\mathrm{\pi }rh \\ \Rightarrow 1210 =2\times \frac{22}{7}\times 10\times h\mathit{ } \\ \Rightarrow h =\frac{1210\times 7}{2\times 22\times 10}=\frac{11\times 7}{2\times 2}=19.25 \mathrm{cm} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}=\mathrm{\pi }{r}^{\mathit{2}}h\mathit{ } \\ =\frac{22}{7}\times {10}^2\times 19.25 {\mathrm{cm}}^3 \\ = 2200\times 2.75 {\mathrm{cm}}^3 =6050 {\mathrm{cm}}^3 \end{gathered}$$

Question 13:

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

Answer 13:

Let r be the radius and h be the height of the cylinder.
Circumference of its base(circle) = 110 cm.
$\Rightarrow 2\mathrm{\pi }r=110 \Rightarrow r=\frac{110}{2\mathrm{\pi }}\Rightarrow r=\frac{110}{2\times {\displaystyle \frac{22}{7}}}\Rightarrow r=\frac{110\times 7}{2\times 22}\Rightarrow r=\frac{35}{2}\mathrm{cm}$
Curved surface area of a cylinder = 4400 cm².
$$\begin{gathered} \Rightarrow 2\mathrm{\pi }rh=4400\Rightarrow h=\frac{4400}{2\mathrm{\pi }r}\Rightarrow h=\frac{4400}{2\times {\displaystyle \frac{22}{7}}\times {\displaystyle \frac{35}{2}}} \\ \Rightarrow h=\frac{4400\times 7\times 2}{2\times 22\times 35}\Rightarrow h=40 \mathrm{cm} \end{gathered}$$
Also, Volume of the cylinder = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}\times {\left(\frac{35}{2}\right)}^2\times 40=\frac{22\times 35\times 35\times 40}{7\times 2\times 2}=38500 {\mathrm{cm}}^3.$

Question 14:

The radius of the base and the height of a cylinder are in the ratio 2 : 3. If its volume is 1617 cm³, find the total surface area of the cylinder.

Answer 14:

Suppose that the radius of the base and the height of the cylinder are 2x cm and 3x cm, respectively.
$$\begin{gathered} \mathrm{Then} 1617={\mathrm{\pi r}}^2\mathrm{h} =\frac{22}{7}\times (2\mathrm{x}{)}^2\times 3\mathrm{x} \\ = \frac{22}{7}\times 12{\mathrm{x}}^3 \\ \Rightarrow {\mathrm{x}}^3=\frac{1617\times 7}{22\times 12}=42.875 \\ \Rightarrow \mathrm{x}=\sqrt[3]{42.875} = 3.5 \mathrm{cm} \end{gathered}$$

Hence, radius = 7 cm; height of the cylinder = 10.5 cm

$$\begin{gathered} \therefore \mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder}=2\mathrm{\pi }rh+2\mathrm{\pi }{r}^{\mathit{2}} \\ =2\times \frac{22}{7}(7\times 10.5+7\times 7) {\mathrm{cm}}^2 \\ =\frac{44}{7}\times (73.5+49) {\mathrm{cm}}^2 =770 {\mathrm{cm}}^2 \end{gathered}$$

Question 15:

The total surface area of a cylinder is 462 cm². Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.

Answer 15:

Total surface area = 462 cm²

Given: Curved surface area =$\frac{1}{3}$$\times$total surface area = $\frac{1}{3}\times 462=154 {\mathrm{cm}}^2$

Now, total surface area − curved surface area = $2\mathrm{\pi rh} +2{\mathrm{\pi r}}^2 -2\mathrm{\pi rh}$
$$\begin{gathered} \Rightarrow \Rightarrow 462 -154 =2\mathrm{\pi }{r}^2 \\ \Rightarrow 308 =2\times \frac{22}{7}\times r^2 \\ \Rightarrow r^{\mathit{2}}=\frac{308\times 7}{44}=49 \\ \Rightarrow r = 7 \mathrm{cm} \end{gathered}$$

Now, curved surface area = 154 cm²

$$\begin{gathered} \Rightarrow 2\mathrm{\pi }rh = 154 \\ \Rightarrow 2\times \frac{22}{7}\times 7\times h=154 \\ \Rightarrow h =\frac{154}{44}=3.5 \mathrm{cm} \end{gathered}$$

 $$\begin{gathered} \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}=\mathrm{\pi }{r}^{\mathit{2}}h\mathit{ } \\ =\frac{22}{7}\times 7^2\times 3.5 \\ =539 {\mathrm{cm}}^3 \end{gathered}$$

Question 16:

The total surface area of a solid cylinder is 231 cm² and its curved surface area is $\frac{2}{3}$ of the total surface area. Find the volume of the cylinder.

Answer 16:

Curved surface area = $\frac{2}{3}$$\times$total surface area = $\frac{2}{3}\times 231 = 2\times 77 =154 {\mathrm{cm}}^2$

Now, total surface area − curved surface area = $2\mathrm{\pi }rh +2\mathrm{\pi }{r}^2-2\mathrm{\pi }rh$

$$\begin{gathered} \mathrm{Then} 231-154=2\mathrm{\pi }{r}^2 \\ \Rightarrow 2\times \frac{22}{7}\times r^2=77 \\ \Rightarrow r^2=\frac{77\times 7}{44}=12.25 \\ \Rightarrow r=3.5 \mathrm{cm} \\ \mathrm{Also}, \mathrm{curved} \mathrm{surface} \mathrm{area}=154 {\mathrm{cm}}^2 \\ \Rightarrow 2\mathrm{\pi }rh = 154 \\ \Rightarrow 2\times \frac{22}{7}\times 3.5\times h\mathit{ }=154 \\ \Rightarrow h=\frac{154\times 7}{44\times 3.5}=7 \mathrm{cm} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}=\mathrm{\pi }{r}^{\mathit{2}}h\mathit{ } \\ =\frac{22}{7}\times (3.5{)}^2\times 7=269.5 {\mathrm{cm}}^3 \end{gathered}$$

PAGE NO-574

Question 17:

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder if its total surface area is 616 cm².

Answer 17:

Suppose that the curved surface area and the total surface area of the right circular cylinder are x cm² and 2x cm².
Then we have:
2x = 616
x = 308 sq cm
Hence, the curved surface area of the cylinder is 308 sq cm.
Let  r cm and h cm be the radius and height of the cylinder, respectively.
$$\begin{gathered} \mathrm{Then} 2\mathrm{\pi }rh +2\mathrm{\pi }{r}^2 =616 {\mathrm{cm}}^2 \mathrm{and} 2\mathrm{\pi }rh = 308 {\mathrm{cm}}^2 \\ \therefore 2\mathrm{\pi }rh +2\mathrm{\pi }{r}^2 - 2\mathrm{\pi }rh =616-308 \\ \Rightarrow 2\mathrm{\pi }{r}^2=308 \\ \Rightarrow 2\times \frac{22}{7}\times r^2=308 \\ \Rightarrow r^2=\frac{308\times 7}{44}=49 \\ \Rightarrow r= 7 \mathrm{cm} \\ \mathrm{Now}, 2\mathrm{\pi }rh = 308 {\mathrm{cm}}^2 \\ \Rightarrow 2\times \frac{22}{7}\times 7\times h\mathit{ }=308 \\ \Rightarrow h\mathit{ }=\frac{308}{44}=7 \mathrm{cm} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}=\mathrm{\pi }{r}^{\mathit{2}}h \mathrm{cubic} \mathrm{cm} \\ =\frac{22}{7}\times 7^2\times 7 \\ =1078 {\mathrm{cm}}^3 \end{gathered}$$

Question 18:

A cylindrical bucket, 28 cm in diameter and 72 cm and 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 in the tank.

Answer 18:

 Given: Diameter of the cylindrical bucket = 28 cm
i.e., radius = 14 cm
Height of the cylindrical bucket, h₁ = 72 cm
Length of the rectangular tank, l = 66 cm
Breadth of the rectangular tank, b = 28 cm
Let the height of the rectangular tank be h cm.
The water from the cylindrical bucket is emptied into the rectangular tank.
i.e., volume of the bucket = volume of the tank
$$\begin{gathered} \Rightarrow \mathrm{\pi }{r}^{\mathit{2}}{h}_{\mathit{1}} =l\mathit{\times }b\mathit{\times }h \\ \Rightarrow \frac{22}{7}\times {14}^2\times 72=66\times 28\times h \\ \Rightarrow h =\frac{22\times 14\times 2\times 72}{66\times 28}=24 \mathrm{cm} \end{gathered}$$

∴ Height of the rectangular tank = 24 cm

Question 19:

The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre?

Answer 19:

Height of the barrel = h = 7 cm
Radius of the barrel = r = 2.5 mm = 0.25 cm
Volume of the barrel = $\mathrm{\pi }{r}^{\mathit{2}}h$
                               $$\begin{gathered} =\frac{22}{7}\times 0.25\times 0.25\times 7 \\ =22\times 0.25\times 0.25 \\ =1.375 {\mathrm{cm}}^3 \end{gathered}$$

i.e., 1.375 cm³ of ink is used for writing 330 words
Now, number of words that could be written with one-fifth of a litre, i.e., $\frac{1}{5}\times 1000$ cm³ = $330\times \frac{1}{1.375}\times \frac{1}{5}\times 1000=48000$  
∴ 48000 words would use up a bottle of ink containing one-fifith of a litre.  

Question 20:

1 cm³ of gold is drawn into a wire 0.1 mm in diameter. Find the length of the wire.

Answer 20:

Let r cm be the radius of the wire and h cm be the height of the wire.
Volume of the gold = 1 cm³
1 cm³ of gold is drawn into a wire of diameter 0.1 mm.
Here, r =  $0.1 \mathrm{mm}=\frac{0.1}{20}\mathrm{cm}=\frac{1}{200}\mathrm{cm}$
$$\begin{gathered} \therefore \mathrm{\pi }{r}^{\mathit{2}}h=1 \\ \Rightarrow \frac{22}{7}\times \frac{1}{200}\times \frac{1}{200}\times h =1 \\ \Rightarrow h=\frac{40000\times 7}{22}=12727.27 \mathrm{cm} \end{gathered}$$

Hence, length of the wire = 127.27 m

Question 21:

If 1 cm³ of case iron weighs 21 g, find the weight of a cast iron pipe of length 1 m with a bore of 3 cm in which the thickness of the metal is 1 cm.

Answer 21:

Internal radius of the pipe = 1.5 cm
External radius of the pipe = (1.5 + 1) cm = 2.5 cm
Height of the pipe = 1 m = 100 cm
Volume of the cast iron = total volume of the pipe − internal volume of the pipe
                                   $$\begin{gathered} =\mathrm{\pi }\times (2.5{)}^2\times 100-\mathrm{\pi }\times {\left(1.5\right)}^2\times 100 \\ =\frac{22}{7}\times 100\times \left[6.25-2.25\right] \\ =\frac{2200}{7}\times 4=\frac{8800}{7}{\mathrm{cm}}^3 \end{gathered}$$
1 cm³ of cast iron weighs 21 g.

∴ Weight of $\frac{8800}{7}{\mathrm{cm}}^3$cast iron = $\frac{8800}{7}\times \frac{21}{1000} \mathrm{kg}=88\times 0.3=26.4 \mathrm{kg}$

Question 22:

A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.

Answer 22:

Inner radius of the cylindrical tube, r = 5.2 cm
Height of the cylindrical tube, h = 25 cm
Outer radius of the cylindrical tube, R = (5.2 + 0.8) cm = 6 cm

$$\begin{gathered} \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{metal}=\mathrm{external} \mathrm{volume} - \mathrm{internal} \mathrm{volume} \\ = \mathrm{\pi }{R}^{\mathit{2}}h -\mathrm{\pi }{r}^{\mathit{2}}h \mathit{ }\mathit{ } (\mathrm{where}\mathit{ }R\mathit{ }\mathrm{and}\mathit{ }r \mathrm{are} \mathrm{the} \mathrm{outer} \mathrm{and} \mathrm{inner} \mathrm{radii}, \mathrm{respectively}) \\ =\frac{22}{7}\times 25\times \left(6^2-5.2^2\right) \\ =\frac{22}{7}\times 25\times \left(36-27.04\right) \\ =\frac{22}{7}\times 25\times 8.96=704 {\mathrm{cm}}^3 \end{gathered}$$

Question 23:

It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

Answer 23:

Height of the cylindrical tank, h = 1 m

Radius of the cylindrical tank, r = $\frac{140}{2}$ = 70 cm = 0.7 m       (1 m = 100 cm)

∴ Area of the metal sheet required to make the cylindrical tank

= Total surface area of the cylindrical tank

$$\begin{gathered} =2\mathrm{\pi }rh\left(r+h\right) \\ =2\times \frac{22}{7}\times 0.7\times 1\times \left(0.7+1\right) \\ =2\times \frac{22}{7}\times 0.7\times 1\times 1.7 \\ =7.48 {\mathrm{m}}^2 \end{gathered}$$
Thus, the area of metal sheet required to make the cylindrial tank is 7.48 m².

Question 24:

A juiceseller has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses of radius 3 cm up to a height of 8 cm, and sold for ₹ 15 each. How much money does he received by selling the juice completely?

Answer 24:

Radius of the vassel, R = 15 cm

Height of orange juice in the vassel, H = 32 cm

∴ Volume of orange juice in the vassel = $\mathrm{\pi }{R}^{2}H=\mathrm{\pi }\times {\left(15\right)}^2\times 32 {\mathrm{cm}}^3$

Radius of the glass, r = 3 cm

Height of orange juice in the glass, h = 8 cm

∴ Volume of orange juice in each glass = $\mathrm{\pi }{r}^{2}h=\mathrm{\pi }\times {\left(3\right)}^2\times 8 {\mathrm{cm}}^3$

Number of glasses of orange juice sold by the juiceseller 

$$\begin{gathered} =\frac{\mathrm{Volume} \mathrm{of} \mathrm{orange} \mathrm{juice} \mathrm{in} \mathrm{the} \mathrm{vassel}}{\mathrm{Volume} \mathrm{of} \mathrm{orange} \mathrm{juice} \mathrm{in} \mathrm{each} \mathrm{glass}} \\ =\frac{\mathrm{\pi }\times {\left(15\right)}^2\times 32}{\mathrm{\pi }\times {\left(3\right)}^2\times 8} \\ =100 \end{gathered}$$
Rate of each glass of orange juice = ₹ 15

∴ Total money received by the juiceseller

= Number of glasses of orange juice sold by the juiceseller × Rate of each glass of orange juice

= 100 × 15

= ₹ 1,500

Thus, the total money received by the juiceseller by selling the juice completely is ₹ 1,500.

Question 25:

A well with inside diameter 10 m is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.

Answer 25:

Inner radius of the well, r = $\frac{10}{2}$ = 5 m

Depth of the well, h = 8.4 m

Suppose the outer radius of the embankment is R m.

Width of the embankment = 7.5 m

∴ R − r = 7.5 m

⇒ R = 7.5 + 5 = 12.5 m

Let the height of the embankment be H m.

Now,

Volume of earth used to form the embankment = Volume of earth dugged out of the well

$$\begin{gathered} \therefore \mathrm{\pi }\left(R^2-r^2\right)H=\mathrm{\pi }{r}^{2}h \\ \Rightarrow H=\frac{5^2\times 8.4}{{\left(12.5\right)}^2-{\left(7.5\right)}^2} \\ \Rightarrow H=\frac{210}{100} \\ \Rightarrow H=2.1 \mathrm{m} \end{gathered}$$
Thus, the height of the embankment is 2.1 m.

Question 26:

How many litres of water flows out of a pipe having an area of cross section of 5 cm² in 1 minute, if the speed of water in the pipe is 30 cm/sec?

Answer 26:

Speed of the water = 30 cm/s

Area of the cross section = 5 cm²

Volume of the water flowing out of the pipe in one second = Area of the cross section × 30 cm = 5 × 30 = 150 cm³

Now, 1 minute = 60 seconds

∴ Volume of the water flowing out of the pipe in 60 seconds

= Volume of the water flowing out of the pipe in one second × 60

= 150 × 60

= 9000 cm³

= $\frac{9000}{1000}$              (1 L = 1000 cm³)

= 9 L

Thus, 9 L of water flows out of the given pipe in 1 minute.

Question 27:

A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 m per second. In how much time will the tank be filled?

Answer 27:

Radius of the water tank, R = $\frac{1.4}{2}$ = 0.7 m

Height of the water tank, H = 2.1 m

∴ Capacity of the water tank = $\mathrm{\pi }{R}^{2}H=\mathrm{\pi }{\left(0.7\right)}^2\times 2.1$ m³

Speed of the water flow = 2 m/s

Radius of the pipe, r = $\frac{3.5}{2}$ = 1.75 cm = 0.0175 m 

Area of the cross section of the pipe = $\mathrm{\pi }{r}^2=\mathrm{\pi }{\left(0.0175\right)}^2$ m²

Volume of the water flowing out of the pipe in one second = Area of the cross section of the pipe × 2 m = $\mathrm{\pi }{\left(0.0175\right)}^2\times 2$ m³

Let the time taken to fill the tank be t seconds.

∴ Volume of the water flowing out of the pipe in t seconds

= Volume of the water flowing out of the pipe in one second × t

= $\mathrm{\pi }{\left(0.0175\right)}^2\times 2\times t$ m³

Now,

Volume of the water flowing out of the pipe in t seconds = Capacity of the water tank

$$\begin{gathered} \therefore \mathrm{\pi }{\left(0.0175\right)}^2\times 2\times t=\mathrm{\pi }{\left(0.7\right)}^2\times 2.1 \\ \Rightarrow t=\frac{{\left(0.7\right)}^2\times 2.1}{{\left(0.0175\right)}^2\times 2} \\ \Rightarrow t=1680 \mathrm{s} \end{gathered}$$
$\Rightarrow t=\frac{1680}{60}$

⇒ t = 28 minutes

Thus, the tank will be filled in 28 minutes.

Question 28:

A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions (32 cm × 22 cm × 14 cm). Find the rise in the level of water when the solid is completely submerged.

Answer 28:

Volume of the rectangular solid of iron = 32 cm × 22 cm × 14 cm 

Radius of the container, r = $\frac{56}{2}$ = 28 cm

Let the rise in the level of water in the container when rectangular solid of iron is submerged in it be h cm.

∴ Volume of the water displaced in the container = $\mathrm{\pi }{r}^{2}h=\mathrm{\pi }\times {\left(28\right)}^2\times h$

When the rectangular solid of iron is submerged in the container, then the volume of water displaced in the container is equal to the volume of the rectangular solid of iron.

$$\begin{gathered} \therefore \frac{22}{7}\times {\left(28\right)}^2\times h=32\times 22\times 14 \\ \Rightarrow h=\frac{32\times 22\times 14\times 7}{22\times {\left(28\right)}^2} \\ \Rightarrow h=4 \mathrm{cm} \end{gathered}$$
Thus, the rise in the level of water in the container is 4 cm.

PAGE NO-575

Question 29:

Find the cost of sinking a tube-well 280 m deep, having a diameter 3 m at the rate of ₹ 15 per cubic metre. Find also the cost of cementing its inner curved surface at ₹ 10 per square metre.

Answer 29:

Height of the tube-well, h = 280 m

Radius of the tube-well, r = $\frac{3}{2}$ m

∴ Volume of the tube-well = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}\times {\left(\frac{3}{2}\right)}^2\times 280$ = 1980 m³

Rate of sinking the tube-well = ₹ 15 per cubic metre

∴ Cost of sinking the tube-well = Volume of the tube-well × Rate of sinking the tube-well = 1980 × 15 = ₹ 29,700

Thus, the cost of sinking the tube-well is ₹ 29,700.

Inner curved surface of the tube-well = $2\mathrm{\pi }rh=2\times \frac{22}{7}\times \frac{3}{2}\times 280$ = 2640 m²

Rate of cementing = ₹ 10 per square metre

∴ Cost of cementing the inner curved surface of the tube-well 

= Inner curved surface of the tube-well × Rate of cementing 

= 2640 × 10 

= ₹ 26,400

Thus, the cost of cementing the inner curved surface of the tube-well is ₹ 26,400.

Question 30:

Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic centimetre of copper weighs 8.4 g.

Answer 30:

Mass of copper wire = 13.2 kg = 13.2 × 1000 = 13200 g        (1 kg = 1000 g)

Volume of 8.4 g of copper wire = 1 cm³

∴ Volume of 13200 g (or 13.2 kg) of copper wire = $\frac{13200}{8.4}$ cm³

Let the length of the copper wire be l cm.

Radius of the copper wire, r = $\frac{4}{2}$ = 2 mm = 0.2 cm            (1 cm = 10 mm)

∴ Volume of the copper wire = $\mathrm{\pi }{r}^{2}l=\frac{22}{7}\times {\left(0.2\right)}^2\times l$ cm<sup>3

$$\begin{gathered} \Rightarrow \frac{22}{7}\times {\left(0.2\right)}^2\times l=\frac{13200}{8.4} \\ \Rightarrow l=\frac{13200\times 7}{22\times {\left(0.2\right)}^2\times 8.4} \\ \Rightarrow l=12500 \mathrm{cm} \end{gathered}$$</sup>

$$\begin{gathered} \Rightarrow l=\frac{12500}{100} \left(1 \mathrm{m} = 100 \mathrm{cm}\right) \\ \Rightarrow l=125 \mathrm{m} \end{gathered}$$
Thus, the length of the copper wire is 125 m.

Question 31:

It costs ₹ 3300 to paint the inner curved surface of a cylindrical vessel 10 m deep at the rate of Rs 30 per m². Find the
(i) inner curved surface area of the vessel,
(ii) inner radius of the base, and
(iii) capacity of the vessel.

Answer 31:

Total cost of paining the inner curved surface of the cylinderical vassel = ₹ 3,300

Rate of painting = ₹ 30 per m²

(i) Inner curved surface area of the vassel

$$\begin{gathered} =\frac{\mathrm{Total} \mathrm{cost} \mathrm{of} \mathrm{painting} \mathrm{the} \mathrm{inner} \mathrm{curved} \mathrm{surface} \mathrm{of} \mathrm{the} \mathrm{cylindrical} \mathrm{vassel}}{\mathrm{Rate} \mathrm{of} \mathrm{painting}} \\ =\frac{3300}{30} \\ =110 {\mathrm{m}}^2 \end{gathered}$$
Thus, the inner curved surface area of the vassel is 110 m².

(ii) Depth of the vassel, h = 10 m

Let the inner radius of the base be r m.

∴ Inner curved surface area of the vasssel = $2\mathrm{\pi }rh=2\times \frac{22}{7}\times r\times 10$

$$\begin{gathered} \Rightarrow 2\times \frac{22}{7}\times r\times 10=110 \\ \Rightarrow r=\frac{110\times 7}{2\times 22\times 10} \\ \Rightarrow r=1.75 \mathrm{m} \end{gathered}$$
Thus, the inner radius of the base is 1.75 m.

(iii) Capacity of the vassel = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}\times {\left(1.75\right)}^2\times 10=$96.25 m³

Thus, the capacity of the vassel is 96.25 m³.

Question 32:

The difference between inside and outside surfaces of a cylindrical tube 14 cm long, is 88 cm². If the volume of the tube is 176 cm³, find the inner and outer radii of the tube.

Answer 32:

Let the inner and outer radii of the tube be r cm and R cm, respectively.

Length of the cylindrical tube, h = 14 cm

Outer curved surface of the cylinder − Inner curved surface of the cylinder = 88 cm²        (Given)

$$\begin{gathered} \therefore 2\mathrm{\pi }Rh-2\mathrm{\pi }rh=88 \\ \Rightarrow 2\times \frac{22}{7}\times 14\times \left(R\mathit{-}r\right)=88 \\ \Rightarrow R\mathit{-}r=\frac{88\times 7}{2\times 22\times 14} \\ \Rightarrow R-r=1 \mathrm{cm} .....\left(1\right) \end{gathered}$$
Volume of the tube = 176 cm³                (Given)

$$\begin{gathered} \therefore \mathrm{\pi }\left(R^2\mathit{-}{r}^{\mathit{2}}\right)h=176 \\ \Rightarrow \frac{22}{7}\times \left(R\mathit{-}r\right)\times \left(R+r\right)\times 14=176 \\ \Rightarrow R+r=\frac{176\times 7}{22\times 1\times 14} \left[\mathrm{Using} \left(1\right)\right] \\ \Rightarrow R+r=4 \mathrm{cm} .....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get

2R = 5

⇒ R = 2.5 cm

Putting R = 2.5 cm in (2), we get

2.5 + r = 4

⇒ r = 4 − 2.5 = 1.5 cm

Thus, the inner and outer radii of the tube are 1.5 cm and 2.5 cm, respectively.

Question 33:

A rectangular sheet of paper 30 cm × 18 cm can be transformed into the curved surface of a right circular cylinder in two ways namely, either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders, thus formed.

Answer 33:

The dimensions of the rectangular sheet of paper are 30 cm × 18 cm.

Let V₁ and V₂ be the volumes of the cylinders formed by rolling the rectangular sheet of paper along its length (i.e. 30 cm) and breadth (i.e. 18 cm), respectively.

Suppose r₁ and h₁ be the radius and height of the cylinder formed by rolling the rectangular sheet of paper along its length, respectively.

$\therefore 2\mathrm{\pi }{r}_1=30\Rightarrow r_1=\frac{30}{2\mathrm{\pi }} \mathrm{cm}$

h₁ = 18 cm

$\therefore V_1=\mathrm{\pi }{r}_1^2{h}_1=\mathrm{\pi }{\left(\frac{30}{2\mathrm{\pi }}\right)}^2\times 18 {\mathrm{cm}}^3$

Also, suppose r₂ and h₂ be the radius and height of the cylinder formed by rolling the rectangular sheet of paper along its breadth, respectively. 

$\therefore 2\mathrm{\pi }{r}_2=18\Rightarrow r_2=\frac{18}{2\mathrm{\pi }} \mathrm{cm}$

h₂ = 30 cm

$\therefore V_2=\mathrm{\pi }{r}_2^2{h}_2=\mathrm{\pi }{\left(\frac{18}{2\mathrm{\pi }}\right)}^2\times 30 {\mathrm{cm}}^3$

Now,

$\frac{V_1}{V_2}=\frac{\mathrm{\pi }{\left({\displaystyle \frac{30}{2\mathrm{\pi }}}\right)}^2\times 18}{\mathrm{\pi }{\left({\displaystyle \frac{18}{2\mathrm{\pi }}}\right)}^2\times 30}=\frac{5}{3}$

⇒ V₁ : V₂ = 5 : 3

Thus, the ratio of the volumes of the two cylinders thus formed is 5 : 3.