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

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.6 Textbook Exercise Questions and Answers.

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

Question 1.
The circumference of the base of a 7 cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm³ = 1l)
Answer:
For the given cylindrical vessel, height h = 25 cm and circumference of the base = 132 cm.
Circumference of base = 2πr
∴ 132 cm = 2 × \(\frac{22}{7}\) × r cm
∴ r = \(\frac{132 \times 7}{2 \times 22}\) cm
∴ r = 21 cm
Hence, for the cylindrical vessel, radius r = 21 cm
Capacity of cylindrical vessel
= Volume of cylinder
= πr²h
= \(\frac{22}{7}\) × 21 × 21 × 25 cm³
= 34650 cm³
= \(\frac{34650}{1000}\) liters
= 34.65 litres
Thus, the cylindrical vessel can hold 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.
Answer:
For the cylindrical wooden pipe,
outer radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{28}{2}\) cm = 14 cm,
inner radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{24}{2}\) cm = 12 cm and
height (length) h = 35 cm.
Volume of the cylindrical wooden pipe
= Volume of outer cylinder – Volume of inner cylinder
= πR²h – πr²h
= πh (R² – r²)
= πh (R + r) (R – r)
= \(\frac{22}{7}\) × 35 × (14 + 12) (14 – 12) cm³
= 110 × 26 × 2 cm³
= 5720 cm³
Now, mass of 1 cm³ of wood = 0.6 g
∴ Mass of 5720 cm³ of wood = 5720 × 0.6 g
= 3432 g
= 3.432 kg
Thus, the mass of the given pipe is 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) For the cuboidal container with rectangular base, length l = 5 cm; breadth b = 4 cm and height h = 15 cm.
Capacity of the cuboidal container
= Volume of cuboid
= l × b × h
= 5 × 4 × 15 cm³
= 300 cm³

(ii) For the cylindrical container,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) cm and
height h = 10 cm.
Capacity of the cylindrical container = Volume of cylinder
= πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) × 10 cm³
= 385 cm³
Hence, the capacity of the cylindrical container is more than the cuboidal container by 385 – 300 = 85 cm³.

Question 4.
If the lateral surface area of a cylinder is 94.2 cm2 and its height is 5 cm, then find
(i) radius of its base and
(ii) its volume. (Use π = 3.14)
Answer:
(i) For the given cylinder, height h = 5 cm and lateral (curved) surface area = 94.2 cm².
Curved surface area of a cylinder = 2 πrh
∴ 94.2 cm² = 2 × 3.14 × r × 5 cm²
∴ r = \(\frac{94.2}{2 \times 3.14 \times 5}\) cm
∴ r = 3 cm
Thus, the radius of the cylinder is 3 cm.

(ii) Volume of a cylinder
= πr²h
= 3.14 × 3 × 3 × 5 cm³
= 141.3 cm³
Thus, the volume of the cylinder is 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 and
(iii) capacity of the vessel.
Answer:
(i) Area of the region painted at the cost of ₹ 20 = 1 m²
∴ Area of the region painted at the cost of ₹ 2200 = \(\frac{2200}{20}\) m² = 110m²
Thus, the inner curved surface area of the vessel is 110 m².

(ii) For the cylindrical vessel, height (depth) h- 10 m and curved surface area = 110 m².
Curved surface area of cylindrical vessel = 2πrh
∴ 110 m² = 2 × \(\frac{22}{7}\) × r × 10 m²
∴ r = \(\frac{110 \times 7}{2 \times 22 \times 10}\) m
∴ r = \(\frac{7}{4}\) m
∴ r = 1.75 m
Thus, the radius of the cylindrical vessel is 1.75 m.

(iii) Capacity of the cylindrical vessel
= Volume of a cylinder
= πr²h
= \(\frac{22}{7}\) × 1.75 × 1.75 × 10 m³
= 96.25 m³
= 96.25 kilolitres
Thus, the capacity of the cylindrical vessel is 96.25 kilolitres.

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:
For the closed cylindrical vessel, height h = 1 m and capacity = 15.4 litres
∴ Volume of the vessel = 15.4 litres
= \(\frac{15.4}{1000}\) m³
= 0.0154 m³
Volume of cylindrical vessel = πr²h
∴ 0.0154 m³ = \(\frac{22}{7}\) × r² × 1 m³
∴ r² = \(\frac{154}{10000} \times \frac{7}{22}\) m²
∴ r² = \(\frac{49}{10000}\) m²
∴ r² = \(\frac{7}{100}\) m
∴ r = 0.07 m
Thus, the radius of the cylindrical vessel is 0. 07 m.
Area of the metal sheet required to make closed cylindrical vessel
= Total surface area of a cylinder
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 0.07 (0.07 + 1) m²
= 0.44 × 1.07 m²
= 0.4708 m²
Thus, 0.4708 m² of metal sheet would be needed to make the closed cylindrical vessel.

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:
For the solid cylinder of graphite,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{1}{2}\) mm = \(\frac{1}{20}\) cm and
height h = 14 cm.
Volume of cylinder of graphite
= πr²h
= \(\frac{22}{7}\) × \(\frac{1}{20}\) × \(\frac{1}{20}\) × 14 cm³ = 011 cm³
For the hollow cylinder of wood,
outer radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) mm = \(\frac{7}{20}\) cm,
inner radius r = \(\frac{1}{20}\) cm and height h = 14 cm.
Volume of hollow cylinder of wood
= πR²h – πr²h
= πh (R² – r²)
= πh (R + r) (R – r)
= \(\frac{22}{7}\) × 14 \(\left(\frac{7}{20}+\frac{1}{20}\right)\left(\frac{7}{20}-\frac{1}{20}\right)\) cm³
= 44 × \(\frac{8}{20}\) × \(\frac{8}{20}\) cm³
= \(\frac{528}{100}\) cm³
= 5.28 cm³
Thus, in the given pencil, the volume of wood is 5.28 cm³ and the volume of graphite is 0.11cm³.

Question 8.
A patient in a hospital is given soup daily in a cylindrical howl of diameter 7 cm. If the bowl is Oiled with soup to a height of 4 cm, how much soup the hospital has ‘ to prepare daily to serve 250 patients ?
Answer:
For the soup served in cylindrical bowl, radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) cm and height h = 4 cm.
Volume of soup served to one patient = Volume of a cylinder
= πr²h
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 4 cm³
= 154 cm³
Thus, the volume of soup served to 1 patient =154 cm³
∴ The volume of soup served to 250 patients
= 154 × 250 cm³
= 38500 cm³
= \(\frac{38500}{1000}\) litres
= 38.5 litres
Thus, the hospital has to prepare 38500 cm³, i.e., 38.5 litres of soup daily.