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

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

## Gujarat Board Textbook Solutions Class 9 Maths 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 the base radius of the cylindrical vessel be r cm. Then, the circumference of the base of the cylindrical vessel = 2πr cm.
According to the question,
2πr = 132
2 x $$\frac {22}{7}$$ x r = 132
r = $$\frac {132 x 7}{2 x 22}$$
⇒  r = 21 cm; h = 25 cm
∴ Capacity of the cylindrical vessel = πr2h
= $$\frac {22}{7}$$ (21)2 (25) cm3 = 34650 cm3
= $$\frac {34650}{1000}$$ L = 34.65 L
Hence, the cylindrical vessel can hold 34.65 L 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:
∴ Inner diameter = 24 cm
∴ Inner radius (r) = $$\frac {24}{2}$$ cm = 12 cm
∴ Outer diameter = 28 cm
∴ Outer radius (R) = $$\frac {28}{2}$$ cm = 14 cm
Length of the pipe (h) = 35 cm
∴ Inner volume = πr2h
= $$\frac {22}{7}$$ x (12)2 x 35
= 15840 cm3
and outer volume = πR2h
= $$\frac {22}{7}$$ x (14)2 x 35
= 21560 cm3
∴ Volume of the wood used
= Outer volume – Inner volume = 21560 cm3 – 15840 cm3 = 5720 cm3
∴ Mass of the pipe
= 5720 x 0.6 g = 3432 g = 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 a circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
Solution:
(i) For tin can
1 = 5 cm
b = 4 cm
h = 15 cm
Capacity = l x b x h
= 5 x 4 x 15 cm3 = 300 cm3

(ii) For plastic cylinder
Diameter = 7 cm
∴ Radius (r) = $$\frac {7}{2}$$ cm
Height (h) = 10 cm
∴ Capacity = πr2h
= $$\frac {22}{7}$$ x ($$\frac {7}{2}$$)2x 10 = 385 cm2
Clearly the second container i.e., a plastic cylinder has greater capacity than the first container i.e., a tin can by 385 – 300 = 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 the radius of the base of the cylinder be r cm.
h = 5 cm
Lateral surface = 94.2 cm2
⇒ 2πrh = 94.2
⇒ 2 x 3.14 x r x 5 = 94.2
r = $$\frac{94.2}{2 \times 3.14 \times 5}$$
r = $$\frac{94.2}{31.4}$$ = r = 3cm
Hence, the radius of the base is 3 cm.

(ii) r = 3 cm
h = 5 cm
∴ Volume of the cylinder = πr2h
= 3.14 x (3)2 x 5 = 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,
(iii) capacity of the vessel.
Solution:
(i) Inner curved surface area of the vessel
= $$\frac{2200}{20}$$ = 110 m2

(ii) Let the radius of the base be r m.
h = 10 m
Inner curved surface area = 110 m2
⇒ 2πrh = 110
⇒  2 x $$\frac{22}{7}$$ x r x 10 = 110
r = $$\frac{110 \times 7}{2 \times 22 \times 10}$$
r = $$\frac{7}{4}$$
r = 1.75 m
Hence, the radius of the base is 1.75 m. (iii) r = 1.75 m
h = 10 m
∴ Capacity of the vessel = πr2h
= $$\frac{22}{7}$$ x (1.75)2 x 10
= 96.25 m3
Hence, the capacity of the vessel is 96.25 m3 (or 96.25 kL).

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of the metal sheet would be needed to make it?
Solution:
h = 1 m
Capacity = 15.4 litres
$$\frac{15.4}{1000}$$ m3 = 0.0154 m3
Let the radius of the base be r m.
Capacity = 0.0154 m3
⇒ πr2h = 0.0154
$$\frac{22}{7}$$ x r2 x 1 = 0.0154
⇒ r2 = $$\frac{0.0154 \times 7}{22}$$
⇒ r2 = 0.0049
⇒ r = $$\sqrt{0.0049}$$
⇒ r = 0.07 m ∴ Total surface area = 2πrh + 2πr2
= 2 x $$\frac{22}{7}$$ x 0.07 x 1 + 2 x $$\frac{22}{7}$$ x (0.07)2
= 0.44 + 0.0308 = 0.4708 m2
Hence, 0.4708 m2 of the metal sheet should be needed.

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:
For solid cylinder of graphite
Diameter = 1 mm
∴ Radius (r) = $$\frac{1}{2}$$ mm
Length of the pencil (h) = 14 cm = 140 mm
∴ Volume of the graphite = πr2 h
= $$\frac{22}{7}$$ x ($$\frac{1}{2}$$)2 x 140 = 110 mm3
= $$\frac{110}{10 \times 10 \times 10}$$cm3 = 0.11 cm3

For cylinder of wood
Diameter = 7 mm
∴ Radius (R) = $$\frac{7}{2}$$ mm
Length of the pencil (h) = 14 cm = 140 mm
∴ Volume of the wood
= π(R2 – r2)h
= $$\frac{22}{7}$$[($$\frac{7}{2}$$)2 – ($$\frac{7}{2}$$)2] 140
= 5280 mm3 Question 8.
A patient in a hospital is given soup daily in a cylindrical bowl of a 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 = 7 cm
∴ Radius (r) = $$\frac{7}{2}$$ cm
Height (h) = 4 cm
∴ Volume of soup in the cylindrical bowl = πr2h
= $$\frac{22}{7}$$ x ($$\frac{7}{2}$$)2 x 4 cm3 = 154 cm3
∴ Volume of soup to be prepared daily to serve 250 patients
= 154 x 250 cm3 = 38500 cm3 (or 38.5 L)
Hence, the hospital has to prepare 38500 cm3 (or 38.5 L) of soup daily to serve 250 patients.