{"id":4874,"date":"2023-05-17T15:14:29","date_gmt":"2023-05-17T09:44:29","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=4874"},"modified":"2023-05-18T17:59:10","modified_gmt":"2023-05-18T12:29:10","slug":"ap-board-9th-class-maths-solutions-chapter-10-ex-10-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-board-9th-class-maths-solutions-chapter-10-ex-10-2\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.2"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 10 Surface Areas and Volumes Ex 10.2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.2<\/h2>\n

\"AP<\/p>\n

Question 1.
\nA closed cylindrical tank of height 1.4 m and radius of the base is 56 cm is made up of a thick metal sheet. How much metal sheet is required ?
\n(Express in square metres).
\nSolution:
\nRadius of the tank r’ = 56 cm
\n= \\(\\frac { 56 }{ 100 }\\) m = 0.56m
\nHeight of the tank h = 1.4 m
\nT.S.A. of a cylinder = 2\u03c0r (r + h)
\n\u2234 Area of the metal sheet required = 2\u03c0r (r + h)
\nA = 2 \u00d7 \\(\\frac { 22 }{ 7 }\\) \u00d7 0.56 \u00d7 (0.56 + 1.4)
\n= 2 \u00d7 22 \u00d7 0.08 \u00d7 1.96
\n= 6.8992 m2<\/sup>
\n= 6.90 m2<\/sup><\/p>\n

\"AP<\/p>\n

Question 2.
\nThe volume of a cylinder is 308 cm3<\/sup> . Its height is 8 cm. Find its lateral surface area and total surface area.
\nSolution:
\nVolume of the cylinder V = \u03c0r2<\/sup>h
\n= 308 cm3<\/sup>
\nHeight of the cylinder h = 8 cm
\n\u2234 308 = \\(\\frac { 22 }{ 7 }\\) . r2<\/sup> \u00d7 8
\nr2<\/sup> = 308 \u00d7 \\(\\frac { 7 }{ 22 }\\) x \\(\\frac { 1 }{ 8 }\\)
\nr2<\/sup> = 12.25
\n\u2234 r = \\(\\sqrt{12.25}\\) = 3.5cm
\nL.S.A. = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac { 22 }{ 7 }\\) \u00d7 3.5 \u00d7 8 = 176cm2<\/sup>
\nT.S.A. = 2\u03c0r (r + h)
\n2 \u00d7 \\(\\frac { 22 }{ 7 }\\) \u00d7 3.5 (3.5 + 8)
\n= 2 \u00d7 22 \u00d7 0.5 \u00d7 11.5 = 253 cm2<\/sup><\/p>\n

Question 3.
\nA metal cuboid of dimensions 22 cm \u00d7 15 cm \u00d7 7.5 cm was melted and cast into a cylinder of height 14 cm. What is its radius ?
\nSolution:
\nDimensions of the metal cuboid
\n= 22 cm \u00d7 15 cm \u00d7 7.5 cm
\nHeight of the cylinder, h = 14 cm
\nCuboid made as cylinder
\n\u2234 Volume of cuboid = Volume of cylinder
\nlbh = 2\u03c0r2<\/sup>h
\n\u21d2 22 \u00d7 15 \u00d7 7.5 = \\(\\frac { 22 }{ 7 }\\) \u00d7 r2<\/sup> \u00d7 14
\n\u21d2 r2<\/sup> = \\(\\frac{22 \\times 15 \\times 7.5 \\times 7}{14 \\times 22}\\)
\n\u21d2 r2<\/sup> = 7.5 \u00d7 7.5
\nr = 7.5 cm<\/p>\n

\"AP<\/p>\n

Question 4.
\nAn overhead water tanker is in the shape of a cylinder has capacity of 616 litres. The diameter of the tank is 5.6 m. Find the height of the tank.
\ncagp)
\nSolution:
\nVolume of the cylinder, V = \u03c0r2<\/sup>h = 616
\nDiameter of the tank = 5.6 m
\nThus its radius, r = \\(\\frac{d}{2}=\\frac{5.6}{2}\\) = 2.8 m
\nHeight = h (say)
\n\u2234 \u03c0r2<\/sup> h = 616
\n\\(\\frac{22}{7}\\) \u00d7 2.8 \u00d7 2.8 \u00d7 h = 616
\nh = \\(\\frac{616 \\times 7}{22 \\times 2.8 \\times 2.8}\\) = 25
\n\u2234 Height = 25 m<\/p>\n

Question 5.
\nA metal pipe is 77 cm long. The inner diametre of a cross section is 4 cm; the outer diameter being 4.4 cm (see figure). Find its
\n\"AP
\ni) Inner curved surface area
\nii) Outer curved surface area
\niii) Total surface area
\ni) Inner curved surface area
\nSolution:
\nHeight of the pipe = 77 cm
\nInner diameter = 4 cm
\nInner radius = \\(\\frac{d}{2}=\\frac{4}{2}\\) = 2 cm
\n\u2234 Inner C.S.A. = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 2 \u00d7 77
\n= 88 \u00d7 11 = 968cm2<\/sup><\/p>\n

\"AP<\/p>\n

ii) Outer curved surface area
\nSolution:
\nOuter diameter = 4.4 cm
\n\u2234 Outer radius, r = \\(\\frac{d}{2}=\\frac{4.4}{2}\\) = 2.2 cm
\nHeight of the pipe, h = 77 cm
\n\u2234 Outer C.S.A. = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 2.2 \u00d7 77
\n= 96.8 \u00d7 11
\n= 1064.8 cm2<\/sup><\/p>\n

iii) Total surface area Sol. Total surface area .
\n= Inner C.S.A + Outer C.S.A
\n= 968 + 1064.8
\n= 2032.8 cm2<\/sup><\/p>\n

Question 6.
\nA cylindrical pillar has a diameter of 56 cm and is of 35 m high. There are 16 pillars around the building. Find the cost of painting the curved surface area of all the pillars at the rate of \u20b9 5.50 per 1 m2<\/sup>.
\nSolution:
\nDiametre of the cylindrical pillar = 56 cm
\nThus its radius, r = \\(\\frac{d}{2}\\)
\n= \\(\\frac{56}{2}\\) = 28cm = \\(\\frac{28}{100}\\)m = 0.28m
\nHeight of the pillar, h = 35 m
\nTotal number of pillars =16
\nCost of painting = \u20b9 5.50 per sq. m.
\nC.S.A. of each pillar = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.28 \u00d7 35
\n= 2 \u00d7 22 \u00d7 0.04 \u00d7 35 = 61.6 m2<\/sup>
\n\u2234 C.S.A. of 16 pillars = 16 \u00d7 61.6 = 985.6 m2<\/sup>
\nCost of painting 16 pillars at the rate of \u20b9 5.5 per sq.m. = 985.6 \u00d7 5.5
\n= \u20b9 5420.8<\/p>\n

\"AP<\/p>\n

Question 7.
\nThe diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to roll once over the play ground to level. Find the area of the play ground in m2<\/sup>
\nSoi. Diameter of the roller = 84 cm
\nThus radius = \\(\\frac{84}{2}\\) = 42 cm
\n= \\(\\frac{42}{100}\\)m = 0.42m
\nLength of the roller =120 cm
\n= \\(\\frac{120}{100}\\) = 1.2m
\nIt takes 500 complete revolutions to roll over the play ground.
\nThus 500 \u00d7 L.S.A. of the roller
\n= Area of the play ground
\n\u2234 Area of the play ground = 500 \u00d7 2\u03c0rh
\n= 500 \u00d7 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.42 \u00d7 1.2 = 1584 m2<\/sup><\/p>\n

Question 8.
\nThe inner diameter of a circular well is 3.5 m. It is 10 m deep. Find (i) its inner curved surface area (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2<\/sup>.
\nSolution:
\nInner diameter of the circular well, d = 3.5 m
\nThus its radius, r = \\(\\frac{d}{2}=\\frac{3.5}{2}\\) = 1. 75 m
\nDepth of the well (height) = 10 m
\ni) Inner C.S.A. = 2\u03c0rh
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 1.75 \u00d7 10
\n= 110 m2<\/sup>
\nii) Cost of plastering at the rate of
\n\u20b9 40 \/ m2<\/sup> = 110 \u00d7 40 = \u20b9 4400<\/p>\n

\"AP<\/p>\n

Question 9.
\nFind (i) the total surface area of a closed cylindrical petrol storage tank whose diameter 4.2 m and height 4.5 m.
\nSolution:
\nDiameter of the cylindrical tank \u2018d\u2019 = 4.2m
\nThus its radius, r = \\(\\frac{\\mathrm{d}}{2}=\\frac{4.2}{2}\\) = 2.1 m
\nHeight of the tank, h = 4.5 m
\nT.S.A. of the tank = 2\u03c0r (r + h)
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 2.1 (2.1 + 4.5)
\n= 2 \u00d7 22 \u00d7 0.3 \u00d7 6.6 = 87.12m2<\/sup><\/p>\n

ii) How much steel sheet was actually used, if \\(\\frac{1}{12}\\) of the steel was wasted in making the tank ?
\nSolution:
\n\\(\\frac{1}{12}\\) of the sheet was wasted.
\n=> 1 – \\(\\frac{1}{12}\\) = \\(\\frac{11}{12}\\) of the sheet was used
\nin making the tank.
\nLet the metal sheet originally brought was = x m2<\/sup>
\n\\(\\frac{11}{12}\\) x = 87.12m2<\/sup>
\n\u2234 x = 87.12 x \\(\\frac{12}{11}\\) = 95.04m2<\/sup><\/p>\n

\"AP<\/p>\n

Question 10.
\nA one side open cylindrical drum has inner radius 28 cm and height 2.1 m. How much water you can store in the drum? Express in litres.
\n(1 litre = 1000 c.c)
\nSolution:
\nInner radius of the cylindrical drum ‘r’ = 28 cm
\n. Its height, h = 2.1 m = 2.1 \u00d7 100 = 210 cm
\nVolume of the drum = \u03c0r2<\/sup>h
\n= \\(\\frac{22}{7}\\) \u00d7 28 \u00d7 28 \u00d7 210
\n= 22 \u00d7 4 \u00d7 28 \u00d7 210
\n= 517440 cc
\n= \\(\\frac{517440}{1000}\\)
\n= 517.44 lit.<\/p>\n

\"AP<\/p>\n

Question 11.
\nThe curved surface area of the cylinder is 1760 cm2<\/sup> and its volume is 12320 cm3<\/sup>. Find its height.
\nSolution:
\nC.S.A of the cylinder = 2\u03c0rh = 1760 cm2<\/sup>
\nVolume of the cylinder = \u03c0r2<\/sup>h
\n= 12320 cm3<\/sup>
\nHeight = h (say)
\n\\(\\frac{\\text { Volume }}{\\text { C.S.A. }}=\\frac{\\pi r^{2} h}{2 \\pi r h}=\\frac{12320}{1760}\\)
\n\u21d2 \\(\\frac{r}{2}\\) = 7
\n\u2234 r = 7 \u00d7 2 = 14cm
\nNow 2\u03c0rh = 1760cm2<\/sup>
\n2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 14h = 1760
\nh = \\(\\frac{1760 \\times 7}{2 \\times 22 \\times 14}\\) = 20cm
\n\u2234Height of the cylinder = 20cm<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.2 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.2 Question 1. A closed cylindrical tank of height 1.4 m and radius of the base is 56 cm is made … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4874"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=4874"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4874\/revisions"}],"predecessor-version":[{"id":40559,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4874\/revisions\/40559"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=4874"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=4874"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=4874"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}