Bihar Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.6 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.6

Question 1.

The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold ? (1000 cm³ = 1l)

Solution:

Let r cm be the radius of the base and h cm be the height of the cylinder.

Circumference of the base = 132 cm

i.e., 34.65 litres of water.

Question 2.

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.

Solution:

We have h = Height of the cylindrical pipe = 35 cm

R = External radius = (\(\frac { 28 }{ 2 }\)) cm = 14 cm

r = Internal radius = (\(\frac { 24 }{ 2 }\)) cm = 12 cm

Volume of the wood used in making the pipe = Volume of the external cylinder – Volume of the internal cylinder

= πR²h – πr²h = π(R² – r²)h

= \(\frac { 22 }{ 7 }\) x (14² – 12²) x 35 cm³

= \(\frac { 22 }{ 7 }\) x 26 x 2 x 35 cm³

= 5720 cm³

Weight of 1 cm³ = 0.6 g

∴ Weight of 5720 cm³ = (5720 x 0.6) g

= \(\left(\frac{5720 \times 0.6}{1000}\right)\) kg

= 3.432 kg

Question 3.

A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 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?

Solution:

(i) Capacity of tin can = lbh cm³

= (5 x 4 x 15) cm³

= 300 cm³

(ii) Capacity of plastic cylinder = πr²h cm³

= (\(\frac { 22 }{ 7 }\) x \(\frac { 7 }{ 2 }\) x \(\frac { 7 }{ 2 }\) x 10) cm

= 385 cm³

Thus, the plastic cylinder has greater capacity by 85 cm³.

Question 4.

If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find (i) radius of its base (ii) its volume (Use π = 3.14)

Solution:

(i) Let r be the radius of the base and h be the height of the cylinder. Then,

Lateral surface = 94.2 cm²

⇒ 2 πrh = 94.2

⇒ 2 x 3.14 x r x 5 = 94.2

⇒ r = \(\frac{94.2}{2 \times 3.14 \times 5}\) = 3

Thus, the radius of its base = 3 cm.

(ii) Volume of the cylinder = πr²h

= (3.14 x 32 x 5) cm³

= 141.3 cm³.

Question 5.

It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of paintin g is at the rate of Rs 20 per m², find

(i) inner curved surface area of the vessel.

(ii) radius of the base.

(iii) capacity of the vessel.

Solution:

(i) Inner curved surface area of the vessel

(ii) Let r be the radius of the base and h be the height of the cylindrical vessel.

Thus, the radius of the base = 1.75 m.

(iii) Capacity of the vessel = πr²h

= (\(\frac { 22 }{ 7 }\) x \(\frac { 7 }{ 4 }\) x \(\frac { 7 }{ 4 }\) x 10) m³

= 96.25 m³

Question 6.

The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Solution:

Capacity of a closed cylindrical vessel = 15.4 litres

= (15.4 x \(\frac { 1 }{ 1000 }\)) m³ = 0.0154 m³

Let r be the radius of the base and h be the height of the vessel. Then,

Volume = πr²h = πr² x 1 = πr² [∵ h = 1 m]

πr² = 0.0154

⇒ \(\frac { 22 }{ 7 }\) x r² = 0.0154

⇒ r² = \(\frac{0.0154 \times 7}{22}\) = 0.0049

⇒ r = \(\sqrt{0.0049}\) = 0.07

Thus, the radius of the base of vessel = 0.07 m Metal sheet needed to make the vessel

= Total surface area of the vessel

= 2πrh + 2πr² = 2πr(h + r)

= 2 x \(\frac { 22 }{ 7 }\) x 0.07 x (1 + 0.07) m²

= 44 x 0.01 x 1.07 m²

= 0.4708 m².

Question 7.

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Solution:

Diameter of the graphite cylinder = 1 mm = \(\frac { 1 }{ 10 }\) cm

∴ Radius = \(\frac { 1 }{ 20 }\) cm

Length of the graphite = 14 cm

Volume of the graphite cylinder = (\(\frac { 22 }{ 7 }\) x \(\frac { 1 }{ 20 }\) x \(\frac { 1 }{ 20 }\) x 14) cm³

Diameter of the pencil = 7 mm = \(\frac { 7 }{ 10 }\) cm = 0.11 cm³

∴ Radius of the pencil = \(\frac { 7 }{ 20 }\) cm

and, length of the pencil = 14 cm

∴ Volume of the pencil = (\(\frac { 22 }{ 7 }\) x \(\frac { 7 }{ 20 }\) x \(\frac { 7 }{ 20 }\) x 14) cm³

= 5.39 cm³

Volume of wood = Volume of the pencil – Volume of the graphite

= (5.39 – 0.11) cm³

= 5.28 cm³.

Question 8.

A patient in a hospital is given soup daily in a cylindrical bowl ofL 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 ?

Solution:

Diameter of the cylindrical bowl = 7 cm

∴ Radius = \(\frac { 7 }{ 2 }\) cm

Height of serving bowl = 4 cm

∴ Soup saved in on serving = Volume of the bowl

= πr²h

= (\(\frac { 22 }{ 7 }\) x \(\frac { 7 }{ 2 }\) x \(\frac { 7 }{ 2 }\) x 4) cm³

= 1.54 cm³

Soup-served to 250 patients = (250 x 1.54) cm³

= 38500 cm³

i.e., 38.5l.