{"id":26145,"date":"2021-06-25T11:13:06","date_gmt":"2021-06-25T05:43:06","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=26145"},"modified":"2022-03-02T10:47:35","modified_gmt":"2022-03-02T05:17:35","slug":"ncert-solutions-for-class-9-maths-chapter-13-ex-13-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-2\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2<\/h2>\n

Question 1.
\nThe curved surface area of a right circular cylinder of height 14 cm is 88 cm2<\/sup>. Find the diameter of the base of the cylinder.
\nSolution:
\nWe know that
\n\"NCERT
\nCurved surface area of cylinder = 2\u03c0rh
\n\u21d2 88 = 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 14 \u00d7 r
\n\u21d2 88 = 88 \u00d7 r
\n\u21d2 r = 1 cm
\nDiameter of cylinder = 2r = 2 \u00d7 1 = 2 cm<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nIt is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet is required for the same?
\nSolution:
\nHeight of the cylinder = 1 m
\nand diameter of cylinder = 140 cm = 1.4 m
\n\"NCERT
\n\u2234 radius of cylinder = \\(\\frac{14}{2}\\) = 0.7 m
\n\u2234 Total surface are of cylinder = 2r(h + r)
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.7(1 + 0.7)
\n= 4.4 \u00d7 1.7
\n= 7.48 m2<\/sup>
\nTherefore, 7.48 m2<\/sup> of sheets are required to make the closed circular cylinder.<\/p>\n

Question 3.
\nA metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm (see fig. 13.11) find its.
\n(i) inner curved surface area.
\n(ii) Outer curved surface area.
\n(iii) Total surface area.
\n\"NCERT
\nSolution:
\n(i) We have given,
\nHeight of cylinder = 77 cm
\n\"NCERT
\nand inner diameter = 4 cm
\ninner radius = \\(\\frac{4}{2}\\) = 2 cm
\n\u2234 Inner curved surface area = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 2 \u00d7 77
\n= 968 cm2<\/sup>.
\n(ii) We have given
\nHeight of cylinder = 77 cm
\nand Outer diameter = 4.4 cm
\n\u2234 Outer curved surface area of cylinder = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 2.2 \u00d7 77
\n= 1064.8 cm2<\/sup>.
\n(iii) Total surface ara of pipe = Inner curved surface area + Outer curved surface area + Area of the two bases.
\n= 2\u03c0rh + 2\u03c0Rh + 2\u03c0(R2<\/sup> – r2<\/sup>)
\n= 2 \u00d7 \u03c0 \u00d7 2 \u00d7 77 + 2 \u00d7 \u03c0 \u00d7 2.2 \u00d7 77 + 2\u03c0((2.2)2<\/sup> – (2)2<\/sup>)
\n= 2 \u00d7 \u03c0 \u00d7 77(2 + 2.2) + 2 \u00d7 \u03c0 \u00d7 (4.84 – 4)
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 88 \u00d7 4.2 + 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.84
\n= 2032.8 + 5.28
\n= 2038.08 cm2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 4.
\nThe diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2<\/sup>.
\nSolution:
\nClearly, the roller is a right circular cylinder of height h = 120 cm = 1.2 m
\nand diameter of its base = 84 cm = 0.84 m
\n\u2234 radius of its base = \\(\\frac{84}{2}\\) = 0.42 m.
\n\u2234 Area covered by roller in one revolution = Curved surface area of the roller = 2\u03c0rh
\n\"NCERT
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.42 \u00d7 1.2
\n= 3.168 m2<\/sup>
\nSo, Area covered by roller in 500 revolutions = 3.168 \u00d7 500 = 1584 m2<\/sup>.
\nHence, Area of playground = 1584 m2<\/sup>.<\/p>\n

Question 5.
\nA cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface area of the pillar at the rate of Rs. 12.50 per m2<\/sup>.
\nSolution:
\nWe have given that,
\nthe height of pillar = 3.5m
\nand the diameter of pillar = 50 cm = 0.5 m
\n\u2234 Radious of the pillar = \\(\\frac{0.5}{2}\\) = 0.25 m
\n\u2234 Curved surface area of pillar = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.25 \u00d7 3.5
\n= 5.5 m2<\/sup>
\nNow, Rate of painting the pillar = Rs. 12.50 per m2<\/sup>.
\n\u2234 Total cost of painting = 5.5 \u00d7 12.50 = Rs. 68.75.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nThe curved surface area of a right circular cylinder is 4.4 m2<\/sup>. If the radius of the base of the cylinder is 0.7 m, find its height.
\nSolution:
\nWe have given that
\nThe curved surface area of cylinder = 4.4 m2<\/sup>
\nand radius of its base = 0.7 m
\nHeight =?
\nWe know that
\nCurved surface area of cylinder = 2\u03c0rh
\n\u21d2 4.4 = 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.7 \u00d7 h (\u2235 \u03c0 = \\(\\frac {22}{7}\\))
\n\u21d2 h = \\(\\frac{4.4 \\times 7}{2 \\times 22 \\times 0.7}\\) = 1 m
\n\u2234 Height of cylinder = 1 m.<\/p>\n

Question 7.
\nThe inner diameter of the circular well is 3.5 m. It is a 10 m deep find.
\n(i) It’s inner curved surface area.
\n(ii) The cost of plastering this curved surface at the rate of Rs. 40 per m2<\/sup>.
\nSolution:
\n(i) We have given that
\nThe inner diameter of circular well = 3.5 m
\n\"NCERT
\nInner radius circular well = \\(\\frac{3.5}{2}\\) m
\nand height of circular well = 10 m.
\nNow, we know that,
\nCurved surface area of well = 2\u03c0rh
\n= \\(2 \\times \\frac{22}{7} \\times \\frac{3.5}{2} \\times 10\\)
\n= 110 m2<\/sup>
\nSo, inner curved surface area of well = 110 m2<\/sup>
\n(ii) Rate of plastering = Rs. 40 per m2<\/sup>
\nTotal cost of plastering its curved surface area = 110 \u00d7 40 = Rs. 4400<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nIn a hot water heating system, there is a cylindrical pipe of length 28m and diameter 5 cm. Find the total radiating surface in the system.
\nSolution:
\nWe have given that
\nLength of the cylindrical pipe (h) = 28m
\nand diameter of its base = 5 cm = 0.05 m
\nRadius of its base = \\(\\frac{.05}{2}\\) m
\n\"NCERT
\n\u2234 Curved surface area of cylindrical pipe = 2\u03c0rh
\n= \\(2 \\times \\frac{22}{7} \\times \\frac{.05}{2} \\times 28\\)
\n= 4.4 m2<\/sup><\/p>\n

Question 9.
\nFind
\n(i) The lateral or curved surface area of a cylindrical petrol Storage tank that is 4.2 m in diameter and 1.5 m high.
\n(ii) How much steel was actually used if \\(\\frac{1}{12}\\) of the steel actually used was wasted in making the closed tank.
\nSolution:
\n(i) We have given that,
\nHeight of cylindrical tank = 4.5 m
\n\"NCERT
\nDiameter of cylindrical tank = 4.2 m
\nRadius of cylindrical tank = 2.1 m
\nSo, Curved surface area of tank = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 4.5 \u00d7 21
\n= 59.4 m2<\/sup>
\n(ii) Let the actual area of steel used be x m2<\/sup>.
\nSince \\(\\frac{1}{12}\\) of the actual steel used was wasted = \\(\\frac{1}{12}\\) of x = \\(\\frac{x}{12}\\)
\n\u2234 Area of steel which has gone into the tank = x – \\(\\frac{x}{12}\\) = \\(\\frac{11x}{12}\\)
\nTotal surface area of tank = 2\u03c0r(r + h)
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 2.1(2.1 + 4.5)
\n= 87.12 m2<\/sup>
\nThe actual area of steel used be
\n\\(\\frac{11x}{12}\\) = 87.12
\n\u21d2 x = \\(\\frac{87.12 \\times 12}{11}\\)
\n\u21d2 x = 95.04 m2<\/sup>
\nHence, total steel was used in tank = 95.04 m2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 10.
\nIn Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and a height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and button of the frame. Find how much cloth is required for covering the lampshade.
\n\"NCERT
\nSolution:
\nWe have given that
\nDiameter of lampshade = 20 cm
\nRadius of lampshade = 10 cm
\nand height of lampshade = 30 + 2.5 + 2.45 = 35 cm
\nTherefore, Curved surface area of cylindrical lampshade = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 10 \u00d7 35
\n= 2200 cm2<\/sup>
\nTherefore, the cloth is required for covering the lampshade is 2200 cm2<\/sup>.<\/p>\n

\"NCERT<\/p>\n

Question 11.
\nThe students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
\nSolution:
\nTotal surface area of one penholder is 2\u03c0rh + \u03c0r2<\/sup>
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 3 \u00d7 10.5 + \\(\\frac{22}{7}\\) \u00d7 3 \u00d7 3
\n= 198 + 28.28
\n= 226.28 cm2<\/sup>
\n\"NCERT
\n\u2234 Total cardboard is used to making one penholder is 226.28 cm2<\/sup>
\nTherefore, total cardboard is used to making 35 penholders is = 226.28 \u00d7 35 = 7919.8 cm2<\/sup>.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2 Question 1. The curved surface area of a right circular cylinder of height 14 cm …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts. 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