RBSE Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.6

Question 1.
The circumference of the hase of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm3 = 1 litre)
Answer:
Let r be the radius of the base and h be the height of the cylinder.
Circumference of the base = 132 cm
So, 2πr = 132 ⇒ 2 × \(\frac{22}{7}\) × r = 132
or r = \(\left(\frac{132 \times 7}{2 \times 22}\right)\)cm = 21cm
Volume of the cylinder = πr²h = (\(\frac{22}{7}\) × 21 × 21 × 25) cm³ = 34650 cm³
∴ Vessel can hold = \(\left(\frac{34650}{1000}\right)\) litres, i.e. 34.65 litresof water.

Questions 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.
Answer:
We have : h = Length of the cylindrical pipe = 35 cm
R = External radius = \(\left(\frac{28}{2}\right)\) cm = 14 cm
r = Internal radius = \(\left(\frac{24}{2}\right)\) = 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
= \(\frac{22}{7}\) × (14² - 12²) × 35 cm³
= \(\frac{22}{7}\) × 26 × 2 × 35 cm³ = 5720 cm³
Weight of" 1 cm³ = 0.6 g
∴ Weight of 5720 cm³ = (5720 × 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?
Answer:
(i) Capacity of tin can = lbh = (5 × 4 × 15) cm³ = 300 cm³
(ii) Capacity of plastic cylinder = πr²h = (\(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) × 10) cm³ = 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)
Answer:
(i) Let r be the radius of the base and h be the height of the cylinder. Then, Lateral surface area = 94.2 cm²
So, 2πrh = 94.2
or 2 × 3.14 × r × 5 = 94.2
⇒ r = \(\frac{94.2}{2 \times 3.14 \times 5}\) = 3
Thus, the radius of the base = 3 cm.

(ii) Volume of the cylinder = πr²h = (3.14 × 32 × 5) cm³ = 141.3 cm³.

Question 5.
It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m2, find :
(i) inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.
Answer:
(i) Inner curved surface area of the vessel
= \(\frac{\text { Total cost of painting }}{\text { Rate of painting }}=\left(\frac{2200}{20}\right)\)m² = 110m².

(ii) Let r be the radius of the base and h be the height of the cylindrical vessel.
∴ 2πrh = 110
or 2 × \(\frac{22}{7} \)× r × 10 = 110
r = \(\frac{110 \times 7}{2 \times 22 \times 10}\)m = \(\frac{7}{4}\)m = 1.75m
Thus, the radius of the base is 1.75 m.

(iii) Capacity of the vessel = πr²h = (\(\frac{22}{7} \times \frac{7}{4} \times \frac{7}{4}\) × 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?
Answer:
Capacity of a closed cylindrical vessel = 15.4 litres

Let r be the radius of the base and h be the height of the vessel.
Then, Volume = πr²h = πr² × 1 = πr² [∵ h = 1 m]
∴ πr² = 0.0154
⇒ \(\frac{22}{7}\) × 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 is 0.07 m.
Metal sheet needed to make the vessel = Total surface area of the vessel
= 2πrh + 2πr² = 2πr(h + r)
= 2 × \(\frac{22}{7}\) × 0.07 × (1 + 0.07) m³
= 44 × 0.01 × 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.
Answer:
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 = (\(\frac{22}{7} \times \frac{1}{20} \times \frac{1}{20}\) × 14) cm³ = 0.11 cm³
Diameter of the pencil = 7 mm = \(\frac{7}{10}\) cm
∴ Radius of the pencil = \(\frac{7}{20}\) cm
and, Length of the pencil = 14 cm
∴ Volume of the pencil = (\(\frac{22}{7} \times \frac{7}{20} \times \frac{7}{20}\) × 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 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:
Diameter of the cylindrical bowl = 7 cm
∴ Radius = \(\frac{7}{2}\) cm
Height of the serving bowl = 4 cm

∴ Soup served to 1 patient = Volume of the bowl
= πr²h = (\(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) × 4) cm³ = 154 cm³

Soup served to 250 patients = (250 × 154) cm³
= 38500 cm³, i.e. 38.5 litres.