{"id":126066,"date":"2021-09-13T20:12:00","date_gmt":"2021-09-13T14:42:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126066"},"modified":"2021-09-13T20:13:28","modified_gmt":"2021-09-13T14:43:28","slug":"rd-sharma-class-9-solutions-chapter-19-exercise-19-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-19-exercise-19-2\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126113\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-19-Exercise-19.2.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-19-Exercise-19.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-19-Exercise-19.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\"><strong>RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2<\/strong>: You can download <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 19<\/a>\u00a0Exercise 19.2 from the link provided below in thisa article.\u00a0<\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\"><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Maths Solutions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Maths Solutions<\/span><\/a><\/li>\n<\/ul>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d907b31a9c1\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69d907b31a9c1\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-19-exercise-19-2\/#download-rd-sharma-class-9-solutions-chapter-19-exercise-192\" title=\"Download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2\">Download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-19-exercise-19-2\/#access-rd-sharma-class-9-solutions-chapter-19-exercise-192\" title=\"Access RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2\">Access RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-19-exercise-19-2\/#faq-rd-sharma-class-9-solutions-chapter-19-exercise-192\" title=\"FAQ: RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2\">FAQ: RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-19-exercise-19-2\/#can-i-download-rd-sharma-class-9-solutions-chapter-19-exercise-192-pdf-free\" title=\"Can I download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2 PDF free?\">Can I download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2 PDF free?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-19-exercise-19-2\/#what-are-the-benefits-of-studying-rd-sharma-class-9-solutions-chapter-19-exercise-192\" title=\"What are the benefits of studying RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2?\">What are the benefits of studying RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-19-exercise-19-2\/#is-rd-sharma-enough-for-class-12-maths\" title=\"Is RD Sharma enough for Class 12 Maths?\">Is RD Sharma enough for Class 12 Maths?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-19-exercise-192\"><\/span>Download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/19.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/19.2.pdf\">RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-19-exercise-192\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br \/>A soft drink is available in two packs \u2013 (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and<br \/>(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much? [NCERT]<br \/>Solution:<br \/>In first case, in a rectangular container of soft drink, the length of base = 5 cm<br \/>and Width = 4 cm<br \/>Height = 15 cm<br \/>\u2234 Volume of soft drink = lbh = 5 x 4 x 15 = 300 cm<sup>3<\/sup><br \/>and in second case, in a cylindrical container, diameter of base = 7 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1946\/43828841600_a0894cc6a8_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"352\" height=\"153\" \/><br \/>\u2234 The soft drink in second container is greater and how much greater = 385 cm \u2013 380 cm<sup>2<\/sup>\u00a0= 85 cm<sup>2<\/sup><\/p>\n<p>Question 2.<br \/>The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars? [NCERT]<br \/>Solution:<br \/>Radius of each pillar (r) = 20 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1922\/44922112064_3b251683ee_o.png\" alt=\"Surface Areas and Volume of a Circular Cylinder Class 9 RD Sharma Solutions\" width=\"173\" height=\"85\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1960\/43828841540_efbac2d76e_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"335\" height=\"326\" \/><\/p>\n<p>Question 3.<br \/>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 cm<sup>3<\/sup>\u00a0of wood has a mass of 0.6 gm. [NCERT]<br \/>Solution:<br \/>Inner diameter of a cylindrical wooden pipe = 24 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1961\/44922111944_477301acd0_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"308\" height=\"342\" \/><\/p>\n<p>Question 4.<br \/>If the lateral surface of a cylinder is 94.2 cm<sup>2<\/sup>\u00a0and its height is 5 cm, find:<br \/>(i) radius of its base<br \/>(ii) volume of the cylinder [Use \u03c0 = 3.14] [NCERT]<br \/>Solution:<br \/>Lateral surface area of a cylinder = 94.2 cm2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1971\/43828841270_0ebc79945e_o.png\" alt=\"Class 9 Maths Chapter 19 Surface Areas and Volume of a Circular Cylinder RD Sharma Solutions\" width=\"348\" height=\"199\" \/><\/p>\n<p>Question 5.<br \/>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? [NCERT]<br \/>Solution:<br \/>The capacity of a closed cylindrical vessel = 15.4 l<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1901\/43828841210_b3fc7c643d_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"338\" height=\"323\" \/><\/p>\n<p>Question 6.<br \/>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? [NCERT]<br \/>Solution:<br \/>Diameter of the cylindrical bowl = 7 cm<br \/>\u2234 Radius (r) =\u00a0<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-16\" class=\"math\"><span id=\"MathJax-Span-17\" class=\"mrow\"><span id=\"MathJax-Span-18\" class=\"mfrac\"><span id=\"MathJax-Span-19\" class=\"mn\">7<\/span><span id=\"MathJax-Span-20\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>cm<br \/>Level of soup in it = 4 cm<br \/>\u2234 Volume of soup in one bowl for one patient<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1960\/43828841030_669a4bf823_o.png\" alt=\"RD Sharma Class 9 Book Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"345\" height=\"50\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1971\/44922111074_7a92bf97b5_o.png\" alt=\"Surface Areas and Volume of a Circular Cylinder With Solutions PDF RD Sharma Class 9 Solutions\" width=\"341\" height=\"104\" \/><\/p>\n<p>Question 7.<br \/>A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.<br \/>Solution:<br \/>Width of hollow cylinder (A) = 63 cm<br \/>Girth = 440 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1947\/43828840910_a622a8cb2c_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"336\" height=\"228\" \/><\/p>\n<p>Question 8.<br \/>The cost of painting the total outside surface of a closed cylindrical oil tank at 50 paise per square decimetre is \u20b9 198. The height of the tank is 6 times the radius of the base of the tank. Find the volume corrected to 2 decimal places.<br \/>Solution:<br \/>Rate of painting = 50 paise per dm<sup>2<\/sup><br \/>Total cost = \u20b9198<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1965\/44922110924_856a74c058_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"300\" height=\"342\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1907\/43828840830_92cef43a80_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"259\" height=\"208\" \/><\/p>\n<p>Question 9.<br \/>The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5:3. Calculate the ratio of their volumes and the ratio of their curved surfaces.<br \/>Solution:<\/p>\n<p>Ratio in radii of two cylinders = 2:3<br \/>and ratio in their heights = 5:3<br \/>Let radius of the first cylinder (r<sub>1<\/sub>) = 2x<br \/>and radius of second cylinder (r<sub>2<\/sub>) = 3x<br \/>and height of first cylinders (h<sub>1<\/sub>) = 5y<br \/>and height of second cylinder (h<sub>2<\/sub>) = 3y<\/p>\n<p>(i) Now volume of the first cylinder = \u03c0r<sup>2<\/sup>h = \u03c0(2x)<sup>2<\/sup>\u00a0x 5y = 20\u03c0x<sup>2<\/sup>2y<br \/>and volume of tlie second cylinder = \u03c0(3x)<sup>2<\/sup>\u00a0x 3y = \u03c0 x 9\u00d72 x 3y = 27\u03c0x<sup>2<\/sup>y<br \/>Now ratio in their volume<br \/>= 20\u03c0x<sup>2<\/sup>y : 21\u03c0x<sup>2<\/sup>y = 20 : 27<br \/>(ii) Curved surface area of first cylinder = 2\u03c0rh = 2\u03c0 x 2x x 5y =20\u03c0xy<br \/>and curved surface area of second cylinder = 2\u03c0 x 3x x = 1 8\u03c0xy<br \/>\u2234 Ratio in their curved surface area<br \/>= 20\u03c0xy : 18\u03c0xy = 10 : 9<\/p>\n<p>Question 10.<br \/>The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder, if its total surface area is 616 cm<sup>2<\/sup>.<br \/>Solution:<\/p>\n<p>Ratio in curved surface area and total surface area of a cylinder =1:2<br \/>Total surface area = 616 cm<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1911\/44922110794_0a1a92fa7d_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"319\" height=\"496\" \/><\/p>\n<p>Question 11.<br \/>The curved surface area of a cylinder is 1320 cm<sup>2<\/sup>\u00a0and its base had diameter 21 cm. Find the height and the volume of the cylinder. [Use \u03c0 = 22\/7]<br \/>Solution:<\/p>\n<p>Curved surface area of a cylinder = 1320 cm<sup>2<\/sup><br \/>Diameter of the base = 21 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/43828840570_23d5e6b7d9_o.png\" alt=\"RD Sharma Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"343\" height=\"298\" \/><\/p>\n<p>Question 12.<br \/>The ratio between the radius of the base and the height of a cylinder is 2 : 3. Find the total surface area of the cylinder, if its volume is 1617 cm<sup>3<\/sup>.<br \/>Solution:<\/p>\n<p>Ratio between radius and height of a cylinder = 2:3<br \/>Volume =1617 cm<sup>3<\/sup><br \/>Let radius (r) = 2x<br \/>Then height (h) = 3x<br \/>\u2234 Volume = \u03c0r<sup>2<\/sup>h<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1939\/44922110604_526d3ea713_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"304\" height=\"574\" \/><\/p>\n<p>Question 13.<br \/>A rectangular sheet of paper, 44 cm x 20 cm, is rolled along its length of form a cylinder. Find the volume of the cylinder so formed.<br \/>Solution:<\/p>\n<p>Length of sheet = 44 cm<br \/>Breadth = 20 cm<br \/>By rolling along length, the height of cylinder (h) = 20cm<br \/>and circumference of the base = 44cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1957\/43828840380_290f8130a9_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"332\" height=\"206\" \/><\/p>\n<p>Question 14.<br \/>The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the diameter and the height of the pillar.<br \/>Solution:<\/p>\n<p>Curved surface area of a pillar = 264 m<sup>2<\/sup><br \/>and volume = 924 m<sup>3<\/sup><br \/>Let r be the radius and It be height, then 2\u03c0rh = 264<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1905\/43828840310_2c68cf5b52_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"351\" height=\"339\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1943\/44922110134_65c843801c_o.png\" alt=\"Surface Areas and Volume of a Circular Cylinder Class 9 RD Sharma Solutions\" width=\"281\" height=\"107\" \/><\/p>\n<p>Question 15.<br \/>Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Find the ratio of their radii.<br \/>Solution:<\/p>\n<p>Volumes of two cylinders are equal Ratio in their height h<sub>1<\/sub>\u00a0:h<sub>2<\/sub>\u00a0= 1: 2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1961\/43828840210_6dcb10270e_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"350\" height=\"401\" \/><\/p>\n<p>Question 16.<br \/>The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the\u201d area of the curved surface. Find the volume of the cylinder.<br \/>Solution:<\/p>\n<p>Height of a right circular cylinder = 10.5 m<br \/>3 x sum of areas of two circular faces<br \/>= 2 x area of curved surface<br \/>Let r be that radius,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1934\/44922109934_ec7fb4e6ef_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"220\" height=\"124\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1970\/44922109724_94cb01ce82_o.png\" alt=\"Class 9 Maths Chapter 19 Surface Areas and Volume of a Circular Cylinder RD Sharma Solutions\" width=\"319\" height=\"75\" \/><\/p>\n<p>Question 17.<br \/>How many cubic metres of earth must be dugout to sink a well 21 m deep and 6 m diameter? Find the cost of plastering the inner surface of the well at \u20b99.50 per m<sup>2<\/sup>.<br \/>Solution:<\/p>\n<p>Diameter of a well = 6 m<br \/>\u2234 Radius (r) =\u00a0<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-21\" class=\"math\"><span id=\"MathJax-Span-22\" class=\"mrow\"><span id=\"MathJax-Span-23\" class=\"mfrac\"><span id=\"MathJax-Span-24\" class=\"mn\">6<\/span><span id=\"MathJax-Span-25\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 3 m<br \/>Depth (h) = 21 m<br \/>\u2234 Volume of earth dugout = \u03c0r<sup>2<\/sup>h<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1978\/44922109634_b49bd59d96_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"341\" height=\"200\" \/><\/p>\n<p>Question 18.<br \/>The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the trunk is 3 m. Find the volume of the timber that can be obtained from the trunk.<br \/>Solution:<\/p>\n<p>Circumference of a cylindrical trunk of a tree = 176 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1908\/43828839920_07fbac074f_o.png\" alt=\"RD Sharma Class 9 Book Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"350\" height=\"288\" \/><\/p>\n<p>Question 19.<br \/>A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the rise in the level of the water when the solid is completely submerged.<br \/>Solution:<\/p>\n<p>Diameter of cylindrical container = 56 cm<br \/>\u2234 Radius (r) =\u00a0<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-26\" class=\"math\"><span id=\"MathJax-Span-27\" class=\"mrow\"><span id=\"MathJax-Span-28\" class=\"mfrac\"><span id=\"MathJax-Span-29\" class=\"mn\">56<\/span><span id=\"MathJax-Span-30\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 28 cm<br \/>Dimensions of a rectangular solid are = 32 cm x 22 cm x 14 cm<br \/>\u2234 Volume of solid = lbh<br \/>= 32 x 22 x 14 = 9856 cm<sup>3<\/sup><br \/>\u2234 Volume of water in the container = 9856 cm<sup>3<\/sup><br \/>Let h be the level of water, then<br \/>\u03c0r2h = 9856<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1942\/44922109464_381bd76276_o.png\" alt=\"Surface Areas and Volume of a Circular Cylinder With Solutions PDF RD Sharma Class 9 Solutions\" width=\"251\" height=\"132\" \/><\/p>\n<p>Question 20.<br \/>A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.<br \/>Solution:<\/p>\n<p>Length of metallic tube = 25 cm<br \/>Inner diameter = 10.4 cm<br \/>\u2234 Radius (r) =\u00a0<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-31\" class=\"math\"><span id=\"MathJax-Span-32\" class=\"mrow\"><span id=\"MathJax-Span-33\" class=\"mfrac\"><span id=\"MathJax-Span-34\" class=\"mn\">10.4<\/span><span id=\"MathJax-Span-35\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 5.2 cm<br \/>Thickness of metal = 8 mm<br \/>\u2234 Outer radius (R) = 5.2 + 0.8 = 6.0 cm<br \/>Volume of metal used = \u03c0(R<sup>2<\/sup>\u00a0\u2013 r<sup>2<\/sup>) x h<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1914\/43828839770_4336e49d34_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"345\" height=\"160\" \/><\/p>\n<p>Question 21.<br \/>From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour.<br \/>Solution:<br \/>Inner radius of a tap = 0.75 cm<br \/>Speed of flow of water in it = 7 m\/s<br \/>Time = 1 hour<br \/>\u2234 Length of flow of water (h)<br \/>= 7 x 60 x 60 m = 25200 m<br \/>\u2234 Volume of water = \u03c0r<sup>2<\/sup>h<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1951\/43828839700_39cc873759_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"285\" height=\"207\" \/><\/p>\n<p>Question 22.<br \/>A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways i.e., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.<br \/>Solution:<\/p>\n<p>Size of rectangular sheet = 30 cm x 18 cm<br \/>\u2234 Length of sheet = 30 cm<br \/>and breadth = 18 cm<br \/>By folding length wise,<br \/>Height = 18 cm<br \/>and circumference = 30 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1906\/44922109294_cf0f32ee76_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"330\" height=\"267\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1978\/43828839610_937fee17a0_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"241\" height=\"391\" \/><\/p>\n<p>Question 23.<br \/>How many litres of water flow out of a pipe having an area of cross-section of 5 cm<sup>2<\/sup>\u00a0in one minute, if the speed of water in the pipe is 30 cm\/sec?<br \/>Solution:<\/p>\n<p>Area of the cross-section of the pipe = 5 cm<sup>2<\/sup><br \/>Speed of water flow = 30 cm\/sec<br \/>Period = 1 minute<br \/>\u2234 Flow of water in 1 minute = 30 x 60 cm = 1800 cm<br \/>Area of mouth of pipe = 5 cm<sup>2<\/sup><br \/>\u2234 Volume = 1800 x 5 = 9000 cm<sup>3<\/sup><br \/>Volume of water in litres = 9000 ml<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1921\/43828839580_92c4764f84_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"329\" height=\"81\" \/><\/p>\n<p>Question 24.<br \/>Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of \u20b93.60 per cubic metre. Find also the cost of cementing its inner curved surface at \u20b92.50 per square metre.<br \/>Solution:<\/p>\n<p>Depth of well (h) = 280 m<br \/>Diameter = 3 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1910\/30706118327_361b800f70_o.png\" alt=\"RD Sharma Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"174\" height=\"51\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1949\/44922108994_74e196cd31_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"341\" height=\"306\" \/><\/p>\n<p>Question 25.<br \/>Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.<br \/>Solution:<\/p>\n<p>Weights of copper wire = 13.2 kg<br \/>Diamter = 4 mm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1911\/30706118087_17a140d0e6_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"321\" height=\"427\" \/><\/p>\n<p>Question 26.<br \/>A solid cylinder has a total surface area of 231 cm<sup>2<\/sup>. Its curved surface area is\u00a0<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-36\" class=\"math\"><span id=\"MathJax-Span-37\" class=\"mrow\"><span id=\"MathJax-Span-38\" class=\"mfrac\"><span id=\"MathJax-Span-39\" class=\"mn\">2<\/span><span id=\"MathJax-Span-40\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0of the total surface area. Find the volume of the cylinder.<br \/>Solution:<\/p>\n<p>Surface area of solid cylinder = 231 cm<sup>2<\/sup><br \/>and curved surface area =\u00a0<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-41\" class=\"math\"><span id=\"MathJax-Span-42\" class=\"mrow\"><span id=\"MathJax-Span-43\" class=\"mfrac\"><span id=\"MathJax-Span-44\" class=\"mn\">2<\/span><span id=\"MathJax-Span-45\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0of 231 cm<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1959\/44922108724_607fc2048a_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"337\" height=\"538\" \/><\/p>\n<p>Question 27.<br \/>A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.<br \/>Solution:<\/p>\n<p>Diameter of a well = 14 m<br \/>\u2234 Radius (r) = y = 7 m<br \/>Depth (h) = 8 m<br \/>\u2234 Volume of the earth dugout = \u03c0r<sup>2<\/sup>h<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1947\/30706117417_50b7907dbd_o.png\" alt=\"Surface Areas and Volume of a Circular Cylinder Class 9 RD Sharma Solutions\" width=\"301\" height=\"102\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1922\/44922108324_b3ed2c110c_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"342\" height=\"530\" \/><\/p>\n<p>Question 28.<br \/>The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq. cm. If the volume of the tube is 176 cubic cm, find the inner and outer radii of the tube.<br \/>Solution:<\/p>\n<p>Length of cylindrical tube = 14 cm<br \/>Difference betveen the outer surface and inner surface = 88 cm<sup>2<\/sup><br \/>and volume of the tube = 176 cm<sup>3<\/sup><br \/>Let R and r be the outer and inner radius of the tube<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1928\/43828838990_50a658f953_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"379\" height=\"160\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1910\/44922107734_f4e219466e_o.png\" alt=\"Class 9 Maths Chapter 19 Surface Areas and Volume of a Circular Cylinder RD Sharma Solutions\" width=\"346\" height=\"431\" \/><\/p>\n<p>Question 29.<br \/>Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank. The radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?<br \/>Solution:<\/p>\n<p>Internal diameter of the pipe = 2 cm<br \/>\u2234 Radius (r) =\u00a0<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-46\" class=\"math\"><span id=\"MathJax-Span-47\" class=\"mrow\"><span id=\"MathJax-Span-48\" class=\"mfrac\"><span id=\"MathJax-Span-49\" class=\"mn\">2<\/span><span id=\"MathJax-Span-50\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 1 cm<br \/>Speed of water flow = 6m per second Water in 30 minutes (h) = 6 x 60 x 30 m = 10800 m<br \/>Volume of water = \u03c0r<sup>2<\/sup>h<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1934\/43828838650_70279402dd_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"331\" height=\"225\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/44922107334_3d258b1898_o.png\" alt=\"RD Sharma Class 9 Book Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"328\" height=\"162\" \/><\/p>\n<p>Question 30.<br \/>A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled?<br \/>Solution:<\/p>\n<p>Diameter of cylindrical tank = 1.4 m<br \/>\u2234 Radius (r) =\u00a0<span id=\"MathJax-Element-11-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-51\" class=\"math\"><span id=\"MathJax-Span-52\" class=\"mrow\"><span id=\"MathJax-Span-53\" class=\"mfrac\"><span id=\"MathJax-Span-54\" class=\"mn\">1.4<\/span><span id=\"MathJax-Span-55\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 0.7 m<br \/>and height (h) = 2.1 m<br \/>\u2234 Volume of water in the tank = \u03c0r<sup>2<\/sup>h<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1938\/44922107154_2fb678c0bd_o.png\" alt=\"Surface Areas and Volume of a Circular Cylinder With Solutions PDF RD Sharma Class 9 Solutions\" width=\"334\" height=\"434\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1947\/43828838360_4dd990502c_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"318\" height=\"136\" \/><\/p>\n<p>Question 31.<br \/>The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the solid cylinder is 1628 m<sup>2<\/sup>. Find the volume of the cylinder.<br \/>Solution:<\/p>\n<p>Sum of radius and height of a cylinder = 37 m<br \/>Let r be the radius and h be the height, then r + h = 37m \u2026(i)<br \/>Total surface area of a solid cylinder = 1628m<sup>3<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1939\/44922106894_01d8c42455_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"301\" height=\"268\" \/><\/p>\n<p>Question 32.<br \/>A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.<br \/>Solution:<\/p>\n<p>Diameter of the well = 10 m 10<br \/>\u2234 Radius (r) =\u00a0<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-56\" class=\"math\"><span id=\"MathJax-Span-57\" class=\"mrow\"><span id=\"MathJax-Span-58\" class=\"mfrac\"><span id=\"MathJax-Span-59\" class=\"mn\">10<\/span><span id=\"MathJax-Span-60\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 5 m<br \/>Depth (h) = 8.4 m<br \/>\u2234 Volume of earth dugout = \u03c0r<sup>2<\/sup>h<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1909\/43828838150_b8a0a05e28_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"301\" height=\"224\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1935\/44922106474_fea2a4f3b4_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"338\" height=\"200\" \/><\/p>\n<p>We have included all the information regarding <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2<\/span><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">. If you have any query feel free to ask in the comment section.\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-class-9-solutions-chapter-19-exercise-192\"><\/span>FAQ: <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631191218414\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-rd-sharma-class-9-solutions-chapter-19-exercise-192-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2 PDF free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631191279321\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-rd-sharma-class-9-solutions-chapter-19-exercise-192\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2, students can earn higher academic grades. Our experts solve RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2 solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631191280318\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-enough-for-class-12-maths\"><\/span>Is RD Sharma enough for Class 12 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>RD Sharma is a good book that gives you thousands of questions to practice.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2: You can download RD Sharma Class 9 Solutions Chapter 19\u00a0Exercise 19.2 from the link provided below in thisa article.\u00a0 RD Sharma Class 9 Maths Solutions Download RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2 \u00a0 RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2 Access &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-19-exercise-19-2\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 19 Exercise 19.2 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":126113,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76837],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126066"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/244"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126066"}],"version-history":[{"count":3,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126066\/revisions"}],"predecessor-version":[{"id":127216,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126066\/revisions\/127216"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126113"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126066"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126066"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126066"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}