Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6 Textbook Exercise Questions and Answers.
Rajasthan Board RBSE Solutions for Class 9 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 9. Students can also read RBSE Class 9 Maths Important Questions for exam preparation. Students can also go through RBSE Class 9 Maths Notes to understand and remember the concepts easily. Practicing the class 9 math chapter 13 hindi medium textbook questions will help students analyse their level of preparation.
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 × 227 × r = 132
or r = (132×72×22)cm = 21cm
Volume of the cylinder = πr2h = (227 × 21 × 21 × 25) cm3 = 34650 cm3
∴ Vessel can hold = (346501000) 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 = (282) cm = 14 cm
r = Internal radius = (242) = 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
= 227 × (142 - 122) × 35 cm3
= 227 × 26 × 2 × 35 cm3 = 5720 cm3
Weight of" 1 cm3 = 0.6 g
∴ Weight of 5720 cm3 = (5720 × 0.6) g = (5720×0.61000) 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 = (227×72×72 × 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 = 94.22×3.14×5 = 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
= Total cost of painting Rate of painting =(220020)m2 = 110m2.
(ii) Let r be the radius of the base and h be the height of the cylindrical vessel.
∴ 2πrh = 110
or 2 × 227× r × 10 = 110
r = 110×72×22×10m = 74m = 1.75m
Thus, the radius of the base is 1.75 m.
(iii) Capacity of the vessel = πr2h = (227×74×74 × 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
⇒ 227 × r2 = 0.0154
⇒ r2 = 0.0154×722 = 0.0049
⇒ r = √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 × 227 × 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 = 110 cm
∴ Radius = 120 cm
Length of the graphite = 14 cm
Volume of the graphite cylinder = (227×120×120 × 14) cm3 = 0.11 cm3
Diameter of the pencil = 7 mm = 710 cm
∴ Radius of the pencil = 720 cm
and, Length of the pencil = 14 cm
∴ Volume of the pencil = (227×720×720 × 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 = 72 cm
Height of the serving bowl = 4 cm
∴ Soup served to 1 patient = Volume of the bowl
= πr2h = (227×72×72 × 4) cm3 = 154 cm3
Soup served to 250 patients = (250 × 154) cm3
= 38500 cm3, i.e. 38.5 litres.