{"id":63861,"date":"2023-09-13T05:40:00","date_gmt":"2023-09-13T00:10:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=63861"},"modified":"2023-11-28T11:08:53","modified_gmt":"2023-11-28T05:38:53","slug":"rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/","title":{"rendered":"RD Sharma Solutions Class 9 Maths Chapter 19 &#8211; Surface Area And Volume of A Right Circular Cylinder (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-124614\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-19-Surface-Area-And-Volume-of-A-Right-Circular-Cylinder.png\" alt=\"RD Sharma Solutions Class 9 Maths Chapter 19\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-19-Surface-Area-And-Volume-of-A-Right-Circular-Cylinder.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-19-Surface-Area-And-Volume-of-A-Right-Circular-Cylinder-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 9 Maths Chapter 19: <\/strong>Every student can secure excellent marks in the mathematics final exams by practising <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> Chapter 19. The top mathematics faculty has solved these solutions in order to assist the students in their exam preparation. These solutions enable students in their problem-solving skills &amp; examine their knowledge of the mathematics subject.<\/p>\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-69e0dba716b2c\" 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-69e0dba716b2c\"><\/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-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#download-rd-sharma-solutions-class-9-maths-chapter-19-pdf\" title=\"Download RD Sharma Solutions Class 9 Maths Chapter 19 PDF\">Download RD Sharma Solutions Class 9 Maths Chapter 19 PDF<\/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-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#exercise-wise-rd-sharma-solutions-class-9-maths-chapter-19\" title=\"Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 19\">Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 19<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#access-answers-of-rd-sharma-solutions-class-9-maths-chapter-19\" title=\"Access answers of RD Sharma Solutions Class 9 Maths Chapter 19\">Access answers of RD Sharma Solutions Class 9 Maths Chapter 19<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#rd-sharma-solutions-class-9-chapter-19-surface-areas-and-volume-of-a-circular-cylinder-ex-191\" title=\"RD Sharma Solutions Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1\">RD Sharma Solutions Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1<\/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-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#rd-sharma-class-9-solution-chapter-19-surface-areas-and-volume-of-a-circular-cylinder-ex-192\" title=\"RD Sharma Class 9 Solution Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2\">RD Sharma Class 9 Solution Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 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-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#rd-sharma-class-9-book-chapter-19-surface-areas-and-volume-of-a-circular-cylinder-vsaqs\" title=\"RD Sharma Class 9 Book Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS\">RD Sharma Class 9 Book Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#rd-sharma-solutions-class-9-chapter-19-surface-areas-and-volume-of-a-circular-cylinder-mcqs\" title=\"RD Sharma Solutions Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS\">RD Sharma Solutions Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#surface-area-and-volume-of-a-right-circular-cylinder-ex-191\" title=\"Surface Area and Volume of A Right Circular Cylinder Ex 19.1\">Surface Area and Volume of A Right Circular Cylinder Ex 19.1<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#detailed-exercise-wise-explanation-with-listing-of-important-topics\" title=\"Detailed Exercise-wise Explanation with Listing of Important Topics\">Detailed Exercise-wise Explanation with Listing of Important Topics<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#rd-sharma-class-9-chapter-19-exercise-19a\" title=\"RD Sharma class 9 chapter 19 exercise 19a:\">RD Sharma class 9 chapter 19 exercise 19a:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#rd-sharma-class-9-chapter-19-exercise-19b\" title=\"RD Sharma class 9 chapter 19 exercise 19b\">RD Sharma class 9 chapter 19 exercise 19b<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#important-topics-from-class-9-maths-chapter-19\" title=\"Important Topics from Class 9 Maths Chapter 19\">Important Topics from Class 9 Maths Chapter 19<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#faqs-on-rd-sharma-solutions-class-9-maths-chapter-19\" title=\"FAQs on RD Sharma Solutions Class 9 Maths Chapter 19\">FAQs on RD Sharma Solutions Class 9 Maths Chapter 19<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-19\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 19?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 19?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-19\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 19?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 19?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-16\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/#can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-19-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 9 Maths Chapter 19\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 9 Maths Chapter 19\u00a0PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-9-maths-chapter-19-pdf\"><\/span><strong>Download RD Sharma Solutions Class 9 Maths Chapter 19 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/19-1.pdf\" data-wplink-edit=\"true\">RD Sharma Class 9 Solutions Chapter 19<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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\/19-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-9-maths-chapter-19\"><\/span><strong>Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 19<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-19-class-9-maths-exercise-19-1-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma class 9 chapter 19 exercise 19a<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-19-class-9-maths-exercise-19-2-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma class 9 chapter 19 exercise 19b<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-class-9-maths-chapter-19\"><\/span><strong>Access answers of RD Sharma Solutions Class 9 Maths Chapter 19<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-19-surface-areas-and-volume-of-a-circular-cylinder-ex-191\"><\/span>RD Sharma Solutions Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>The curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height. [NCERT]<br>Solution:<br>The curved surface area of a cylinder = 4.4 m<sup>2<\/sup><br>Radius (r) = 0.7 m<br><img src=\"https:\/\/farm2.staticflickr.com\/1937\/43828843210_f3b3003521_o.png\" alt=\"RD Sharma Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"330\" height=\"113\"><\/p>\n<p>Question 2.<br>In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. [NCERT]<br>Solution:<br>Diameter of the pipe = 5 cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1936\/43828843130_661bba5e50_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"337\" height=\"182\"><\/p>\n<p>Question 3.<br>A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of 12.50 per m<sup>2<\/sup>. [NCERT]<br>Solution:<br>The diameter of the cylindrical pillar = is 50 cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1969\/43828843040_750f678e0a_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"340\" height=\"331\"><\/p>\n<p>Question 4.<br>It 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? [NCERT]<br>Solution:<br>Height of cylinder (h) = 1 m = 100 cm<br>Diameter of box = 140 cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1961\/43828842930_eb852e1aff_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"291\" height=\"79\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1909\/43828842880_d8f298f4ba_o.png\" alt=\"Surface Areas and Volume of a Circular Cylinder Class 9 RD Sharma Solutions\" width=\"262\" height=\"134\"><\/p>\n<p>Question 5.<br>The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, the area of the base ring is 115.5 sq. cm and the height is 7 cm. Find the thickness of the cylinder.<br>Solution:<br>The total surface area of a hollow cylinder open from both sides = 4620 cm<sup>2<\/sup><br>Area of base of ring = 115.5 cm<sup>2<\/sup><br>Height (h) = 7 cm<br>Let outer radius (R) = R<br>and inner radius = r<br><img src=\"https:\/\/farm2.staticflickr.com\/1969\/45596600562_e6edaa8f28_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"365\" height=\"557\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1911\/43828842710_d1d421983a_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"306\" height=\"276\"><\/p>\n<p>Question 6.<br>Find the ratio between the total surface area of a cylinder to its curved surface area, given that its height and radius are 7.5 cm and 3.5 cm.<br>Solution:<br>The radius of the cylinder (r) = 3.5 cm<br>and height (h) = 7.5 cm<br>Total surface area = 2\u03c0r (h + r)<br>and curved surface area = 2\u03c0rh<br><img src=\"https:\/\/farm2.staticflickr.com\/1976\/45596600172_06d4054bb9_o.png\" alt=\"Class 9 Maths Chapter 19 Surface Areas and Volume of a Circular Cylinder RD Sharma Solutions\" width=\"316\" height=\"135\"><\/p>\n<p>Question 7.<br>A cylindrical vessel, without a lid, has to be tin-coated on both sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin-coating at the rate of \u20b93.50 per 1000 cm<sup>2<\/sup>.<br>Solution:<br>The radius of the base of a cylindrical vessel (r) = 70 cm<br>and height (h) = 1.4 m = 140 cm<br>Total surface area (excluding upper lid) on both sides = 2\u03c0rh x 2 + \u03c0r<sup>2<\/sup>&nbsp;x 2<br><img src=\"https:\/\/farm2.staticflickr.com\/1946\/43828842430_9ec68e44f5_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"303\" height=\"146\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1950\/45596599922_39146dbb25_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 - 7a\" width=\"355\" height=\"106\"><\/p>\n<p>Question 8.<br>The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find:<br>(i) inner curved surface area.<br>(ii) the cost of plastering this curved surface at the rate of \u20b940 per m<sup>2<\/sup>. [NCERT]<br>Solution:<br>Inner diameter of a well = 3.5 m<br><img src=\"https:\/\/farm2.staticflickr.com\/1975\/45596599832_effbd694ce_o.png\" alt=\"RD Sharma Class 9 Book Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"332\" height=\"248\"><\/p>\n<p>Question 9.<br>The students of a Vidyalaya were asked to participate s a competition for making and decorating pen holders in the shape of a cylinder with a base, using cardboard. Each pen holder 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? [NCERT]<br>Solution:<br>The radius of the cylindrical pen holder (r) = 3 cm<br>Height (h) = 10.5 cm<br>\u2234 The surface area of the pen holder<br><img src=\"https:\/\/farm2.staticflickr.com\/1902\/45596599642_7d6dd586fd_o.png\" alt=\"Surface Areas and Volume of a Circular Cylinder With Solutions PDF RD Sharma Class 9 Solutions\" width=\"260\" height=\"197\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1968\/43828842170_dbfa969d92_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"333\" height=\"108\"><\/p>\n<p>Question 10.<br>The diameter of the roller 1.5 m long is 84 cm. If it takes 100 revolutions to level a play\u00acground, find the cost of leveling this ground at the rate of 50 paise per square meter.<br>Solution:<br>The diameter of a roller = 1.5 m<br>\u2234 Radius =&nbsp;<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"mfrac\"><span id=\"MathJax-Span-4\" class=\"mn\">1.5<\/span><span id=\"MathJax-Span-5\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 0.75 m = 75 cm<br>and length (h) = 84 cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1915\/44922112724_7bd22339e4_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"349\" height=\"428\"><\/p>\n<p>Question 11.<br>Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and the height is 4 m. What will be the cost of cleaning them at the rate of \u20b92.50 per square meter? [NCERT]<br>Solution:<br>Number of pillars = 20<br>The diameter of one pillar = 0.50 m<br><img src=\"https:\/\/farm2.staticflickr.com\/1961\/43828842030_d345c5b923_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"329\" height=\"50\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44922112554_5ac549f9e5_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"346\" height=\"294\"><\/p>\n<p>Question 12.<br>A solid cylinder has a total surface area of 462 cm<sup>2<\/sup>. Its curved surface area is one-third of its total surface area. Find the radius and height of the cylinder.<br>Solution:<br>The total surface of the solid cylinder = 462 cm<sup>2<\/sup><br>Curved surface area =&nbsp;<span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-6\" class=\"math\"><span id=\"MathJax-Span-7\" class=\"mrow\"><span id=\"MathJax-Span-8\" class=\"mfrac\"><span id=\"MathJax-Span-9\" class=\"mn\">1<\/span><span id=\"MathJax-Span-10\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span> of the total surface area<br><img src=\"https:\/\/farm2.staticflickr.com\/1955\/43828841880_0c16a3d234_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"351\" height=\"375\"><\/p>\n<p>Question 13.<br>The total surface area of a hollow metal cylinder, open at both ends of an external radius of 8 cm and height of 10 cm is 338 \u03c0 cm<sup>2<\/sup>. Taking r to be the inner radius, obtain an equation in r and use it to obtain the thickness of the metal in the cylinder.<br>Solution:<br>The total surface area of a hollow metal cylinder = 338\u03c0 cm<sup>2<\/sup><br>Let R be the outer radius, r be the inner radius and h be the height of the cylinder the cylinder<br>\u2234 2\u03c0Rh + 2\u03c0rh + 2\u03c0R<sup>2<\/sup>&nbsp;\u2013 2\u03c0r<sup>2<\/sup>&nbsp;= 338\u03c0<br>R = 8 cm, h = 10 cm<br>\u21d2 2\u03c0h (R + r) + 2\u03c0(R<sup>2<\/sup>&nbsp;\u2013 r<sup>2<\/sup>) = 338\u03c0<br>\u21d2 Dividing by 2\u03c0 , we get<br>\u21d2 h(R + r) + (R<sup>2<\/sup>&nbsp;\u2013 r<sup>2<\/sup>) = 169<br>\u21d2 10(8 + r) + (8 + r) (8 \u2013 r) = 169<br>\u21d2 80 + 10r + 64 \u2013 r<sup>2<\/sup>&nbsp;= 169<br>\u21d2 10r \u2013 r<sup>2<\/sup>&nbsp;+ 144 \u2013 169 = 0<br>\u21d2 r<sup>2<\/sup>&nbsp;\u2013 10r + 25 = 0<br>\u21d2 (r-5)<sup>2<\/sup>&nbsp;= 0<br>\u21d2 r = 5<br>\u2234 Thickness of the metal = R \u2013 r = 8 \u2013 5 = 3 cm<\/p>\n<p>Question 14.<br>Find the lateral curved surface area of a cylindrical petrol storage tank that is 4.2m in diameter and 4.5 m high. How much steel was actually used, if <span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-11\" class=\"math\"><span id=\"MathJax-Span-12\" class=\"mrow\"><span id=\"MathJax-Span-13\" class=\"mfrac\"><span id=\"MathJax-Span-14\" class=\"mn\">1<\/span><span id=\"MathJax-Span-15\" class=\"mn\">12<\/span><\/span><\/span><\/span><\/span>&nbsp;of steel actually used was wasted in making the closed tank? [NCERT]<br>Solution:<br>The diameter of a cylindrical tank = 4.2 m<br><img src=\"https:\/\/farm2.staticflickr.com\/1947\/44922112404_08e83a599d_o.png\" alt=\"RD Sharma Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"311\" height=\"340\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1912\/43828841720_b097e87a78_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"352\" height=\"165\"><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solution-chapter-19-surface-areas-and-volume-of-a-circular-cylinder-ex-192\"><\/span>RD Sharma Class 9 Solution Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>A soft drink is available in two packs \u2013 (i) a tin can with a rectangular base of a length of 5 cm and width of 4 cm, having a height of 15 cm and<br>(ii) a plastic cylinder with a circular base of a diameter of 7 cm and height of 10 cm. Which container has greater capacity and by how much? [NCERT]<br>Solution:<br>In the first case, in a rectangular container of soft drink, the length of the 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 the second case, in a cylindrical container, the diameter of the 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 the second container is greater and how much greater = 385 cm \u2013 380 cm<sup>2<\/sup>&nbsp;= 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 of 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>&nbsp;of wood has a mass of 0.6 gm. [NCERT]<br>Solution:<br>The inner diameter of a cylindrical wooden pipe = is 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>&nbsp;and 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>The 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 liters. How many square meters of the metal sheet would be needed to make it? [NCERT]<br>Solution:<br>The capacity of a closed cylindrical vessel = is 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 have to prepare daily to serve 250 patients? [NCERT]<br>Solution:<br>The diameter of the cylindrical bowl = 7 cm<br>\u2234 Radius (r) =&nbsp;<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 The 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 a 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:<br>The ratio in radii of two cylinders = 2:3<br>and ratio in their heights = 5:3<br>Let the radius of the first cylinder (r<sub>1<\/sub>) = 2x<br>and radius of the second cylinder (r<sub>2<\/sub>) = 3x<br>and height of first cylinders (h<sub>1<\/sub>) = 5y<br>and height of the second cylinder (h<sub>2<\/sub>) = 3y<br>(i) Now volume of the first cylinder = \u03c0r<sup>2<\/sup>h = \u03c0(2x)<sup>2<\/sup>&nbsp;x 5y = 20\u03c0x<sup>2<\/sup>2y<br>and volume of the second cylinder = \u03c0(3x)<sup>2<\/sup>&nbsp;x 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 The 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:<br>The 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> and its base has a diameter of 21 cm. Find the height and the volume of the cylinder. [Use \u03c0 = 22\/7]<br>Solution:<br>The curved surface area of a cylinder = 1320 cm<sup>2<\/sup><br>The 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:<br>The ratio between the 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 to form a cylinder. Find the volume of the cylinder so formed.<br>Solution:<br>Length of sheet = 44 cm<br>Breadth = 20 cm<br>By rolling along the length, the height of a 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:<br>The 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 is the 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:<br>Volumes of two cylinders are equal Ratio in their height h<sub>1<\/sub>:h<sub>2<\/sub>&nbsp;= 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:<br>Height of a right circular cylinder = 10.5 m<br>3 x sum of areas of two circular faces<br>= 2 x area of the 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 meters of the earth must be dug out to sink a well 21 m deep and 6 m in diameter? Find the cost of plastering the inner surface of the well at \u20b99.50 per m<sup>2<\/sup>.<br>Solution:<br>The diameter of a well = 6 m<br>\u2234 Radius (r) =&nbsp;<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>&nbsp;= 3 m<br>Depth (h) = 21 m<br>\u2234 The 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:<br>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 a 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:<br>The diameter of the cylindrical container = is 56 cm<br>\u2234 Radius (r) =&nbsp;<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>&nbsp;= 28 cm<br>Dimensions of a rectangular solid are 32 cm x 22 cm x 14 cm<br>\u2234 The volume of solid = lbs<br>= 32 x 22 x 14 = 9856 cm<sup>3<\/sup><br>\u2234 The 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:<br>Length of metallic tube = 25 cm<br>Inner diameter = 10.4 cm<br>\u2234 Radius (r) =&nbsp;<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>&nbsp;= 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>&nbsp;\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 an inner radius of 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>The 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 the 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:<br>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>&nbsp;in one minute, if the speed of water in the pipe is 30 cm\/sec?<br>Solution:<br>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 the mouth of pipe = 5 cm<sup>2<\/sup><br>\u2234 Volume = 1800 x 5 = 9000 cm<sup>3<\/sup><br>The 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 a diameter of 3 m at the rate of \u20b93.60 per cubic meter. Find also the cost of cementing its inner curved surface at \u20b92.50 per square meter.<br>Solution:<br>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:<br>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&nbsp;<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>&nbsp;of the total surface area. Find the volume of the cylinder.<br>Solution:<br>The surface area of the solid cylinder = 231 cm<sup>2<\/sup><br>and curved surface area =&nbsp;<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>&nbsp;of 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 a 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:<br>The diameter of a well = 14 m<br>\u2234 Radius (r) = y = 7 m<br>Depth (h) = 8 m<br>\u2234 The 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 the 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:<br>Length of cylindrical tube = 14 cm<br>Difference between 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 meters per second into a cylindrical tank. The radius of the base is 60 cm. Find the rise in the level of water in 30 minutes.<br>Solution:<br>The internal diameter of the pipe = 2 cm<br>\u2234 Radius (r) =&nbsp;<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>&nbsp;= 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 meters per second. In how much time the tank will be filled?<br>Solution:<br>The diameter of the cylindrical tank = 1.4 m<br>\u2234 Radius (r) =&nbsp;<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>&nbsp;= 0.7 m<br>and height (h) = 2.1 m<br>\u2234 The 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 the 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:<br>The sum of the 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>The 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 a 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:<br>Diameter of the well = 10 m 10<br>\u2234 Radius (r) =&nbsp;<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>&nbsp;= 5 m<br>Depth (h) = 8.4 m<br>\u2234 The 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<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-book-chapter-19-surface-areas-and-volume-of-a-circular-cylinder-vsaqs\"><\/span>RD Sharma Class 9 Book Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>Write the number of surfaces of a right circular cylinder.<br>Solution:<br>Three, two circular and one curved.<\/p>\n<p>Question 2.<br>Write the ratio of the total surface area to the curved surface area of a cylinder of radius r and height h.<br>Solution:<br>\u2235 Radius = r<br>and height = h<br>\u2234 Curved surface area = 2\u03c0rh<br>and total surface area = 2\u03c0r(h + r)<br>\u2234 Ratio = 2\u03c0r(h + r) : 2\u03c0rh<br>= h + r: h<\/p>\n<p>Question 3.<br>The ratio between the radius of the base and the height of a cylinder is 2 : 3. If its volume is 1617 cm3, find the total surface area of the cylinder.<br>Solution:<br>The ratio in radius and height of the cylinder = 2 : 3<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\/1971\/43828843780_7f803682c1_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS\" width=\"295\" height=\"228\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1911\/43828843490_1c89320b79_o.png\" alt=\"Class 9 Maths Chapter 19 Surface Areas and Volume of a Circular Cylinder RD Sharma Solutions VSAQS\" width=\"334\" height=\"306\"><\/p>\n<p>Question 4.<br>If the radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3, then find the ratio of their volumes.<br>Solution:<br>The ratio of radii of two cylinders = 2:3<br>Let the radius of the first cylinder (r<em><sub>1<\/sub><\/em>) = 2x<br>and second cylinder (r<sub>2<\/sub>) = 3x<br>and ratio in their heights = 5:3<br>Let the height of the first cylinder (h<sub>1<\/sub>) = 5y<br>and height of second (h<sub>2<\/sub>) = 3y<br>\u2234 The volume of the first cylinder =\u03c0r<sup>2<\/sup>h<br>= \u03c0 x (2x)<sup>2<\/sup>&nbsp;x 5y = 20\u03c0x<sup>2<\/sup>y<br>and volume of second cylinder = \u03c0(3x)2 x 3y = 27\u03c0x<sup>2<\/sup>y<br>Now ratio between them,<br>= 20\u03c0x<sup>2<\/sup>y: 21\u03c0x<sup>2<\/sup>y<br>= 20: 27<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-19-surface-areas-and-volume-of-a-circular-cylinder-mcqs\"><\/span>RD Sharma Solutions Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Mark the correct alternative in each of the following:<br>Question 1.<br>In a cylinder, if the radius is doubled and the height is halved, the curved surface area will be<br>(a) halved<br>(b) doubled<br>(c) same<br>(d) four times<br>Solution:<br>Let the radius of the first cylinder (r<sub>1<\/sub>) = r<br>and height (h<sub>1<\/sub>) = h<br>Surface area = 2\u03c0rh<br>If the radius is doubled and the height is halved<br><img src=\"https:\/\/farm2.staticflickr.com\/1979\/31774683818_19bc973ae3_o.png\" alt=\"Surface Areas and Volume of a Circular Cylinder With Solutions PDF RD Sharma Class 9 Solutions\" width=\"324\" height=\"157\"><br>\u2234 Their surface area remains the same (c)<\/p>\n<p>Question 2.<br>Two cylindrical jars have their diameters in the ratio of 3:1, but their height is 1:3. Then the ratio of their volumes is<br>(a) 1: 4<br>(b) 1: 3<br>(c) 3: 1<br>(d) 2: 5<br>Solution:<br>Sol. The ratio in the diameters of two cylinders = 3:1<br>and ratio in their heights = 1:3<br>Let the radius of the first cylinder (r<sub>1<\/sub>) = 3x<br>and radius of a second (r<sub>2<\/sub>) = x<br>and height of the first (h<sub>1<\/sub>) = y<br>and height of the second (h<sub>2<\/sub>) = 3y<br>Now the volume of the first cylinder = \u03c0r<sup>2<\/sup>h<br>= \u03c0(3x)<sup>2<\/sup>&nbsp;x y = 9\u03c0x<sup>2<\/sup>y<br>and of second cylinder = \u03c0(x<sup>2<\/sup>) (3y)<br>\u2234 Ratio between then = 9\u03c0x<sup>2<\/sup>y : 3\u03c0x<sup>2<\/sup>y<br>= 3 : 1 (c)<\/p>\n<p>Question 3.<br>The number of surfaces in the right cylinder is<br>(a) 1<br>(b) 2<br>(c) 3<br>(d) 4<br>Solution:<br>The number of surfaces of a right cylinder is three. (c)<\/p>\n<p>Question 4.<br>A vertical cross-section of a right circular cylinder is always a<br>(a) square<br>(b) rectangle<br>(c) rhombus<br>(d) trapezium<br>Solution:<br>The vertical cross-section of a right circular cylinder is always a rectangle. (b)<\/p>\n<p>Question 5.<br>If r is the radius and h is the height of the cylinder the volume will be<br><img src=\"https:\/\/farm2.staticflickr.com\/1907\/43828846390_62eab0519c_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"253\" height=\"89\"><br>Solution:<br>Volume of a cylinder = \u03c0r<sup>2<\/sup>h (b)<\/p>\n<p>Question 6.<br>The number of surfaces of a hollow cylindrical object is<br>(a) 1<br>(b) 2<br>(c) 3<br>(d) 4<br>Solution:<br>The number of surfaces of a hollow cylindrical object is 4. (d)<\/p>\n<p>Question 7.<br>If the radius of a cylinder is doubled and the height remains the same, the volume will be<br>(a) doubled<br>(b) halved<br>(c) same<br>(d) four times<br>Solution:<br>If r is the radius and h is the height, then volume = \u03c0r2h<br>If a radius is doubled and the height remains the same,<br>the volume will be<br>= \u03c0(2r)<sup>2<\/sup>h = \u03c0 x 4r<sup>2<\/sup>h<br>= 4\u03c0r<sup>2<\/sup>h = 4 x Volume<br>The volume is four times (d)<\/p>\n<p>Question 8.<br>If the height of a cylinder is doubled and the radius remains the same, then the volume will be<br>(a) doubled<br>(b) halved<br>(c) same<br>(d) four times<br>Solution:<br>If r is the radius and h is the height, then the volume of a cylinder = \u03c0r<sup>2<\/sup>h<br>If the height is doubled and the radius remains the same, then volume = \u03c0r<sup>2<\/sup>(2h) = 2\u03c0r<sup>2<\/sup>h<br>\u2234 Its doubled (a)<\/p>\n<p>Question 9.<br>In a cylinder, if the radius is halved and the height is doubled, the volume will be<br>(a) same<br>(b) doubled<br>(c) halved<br>(d) four times<br>Solution:<br>Let r be the radius and h be the height, then Volume = \u03c0r<sup>2<\/sup>h<br>If the radius is halved and the height is doubled<br><img src=\"https:\/\/farm2.staticflickr.com\/1935\/44922117164_f5c4fd02d9_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"347\" height=\"142\"><\/p>\n<p>Question 10.<br>If the diameter of the base of a closed right circular cylinder is equal to its height h, then its whole surface area is<br><img src=\"https:\/\/farm2.staticflickr.com\/1931\/43828846020_03e9cee139_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"282\" height=\"106\"><br>Solution:<br>Let the diameter of the base of a cylinder (r) = h<br>Then its height (h) = h<br><img src=\"https:\/\/farm2.staticflickr.com\/1938\/44922116834_c33f2f8b95_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"327\" height=\"190\"><\/p>\n<p>Question 11.<br>A right circular cylindrical tunnel of diameter 2 m and length 40 m is to be constructed from a sheet of iron. The area of the iron sheet required in m<sup>2<\/sup>&nbsp;is<br>(a) 40\u03c0<br>(b) 80\u03c0<br>(c) 160\u03c0<br>(d) 200\u03c0<br>Solution:<br>The diameter of a cylindrical tunnel = 2 m<br>\u2234 Radius (r) =&nbsp;<span id=\"MathJax-Element-13-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-61\" class=\"math\"><span id=\"MathJax-Span-62\" class=\"mrow\"><span id=\"MathJax-Span-63\" class=\"mfrac\"><span id=\"MathJax-Span-64\" class=\"mn\">2<\/span><span id=\"MathJax-Span-65\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 1m<br>and length (h) = 40 m<br>Curved surfae area = 2\u03c0rh = 2 x \u03c0 x 1 x 40 = 80\u03c0 (b)<\/p>\n<p>Question 12.<br>Two circular cylinders of equal volume have their heights in the ratio 1: 2. Ratio of their radii is<br><img src=\"https:\/\/farm2.staticflickr.com\/1934\/43828845860_defa65a8b4_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"275\" height=\"55\"><br>Solution:<br>Let r<sub>1<\/sub>&nbsp;and h<sub>1<\/sub>&nbsp;be the radius and height of the<br>first cylinder, then<br>Volume = \u03c0r<sub>1<\/sub><sup>2<\/sup>h<sub>1<\/sub><br>Similarly, r<sub>1<\/sub>&nbsp;and h<sub>2<\/sub>&nbsp;are the radius and height of the second cylinder<br>\u2234 Volume = \u03c0r<sup>2<\/sup>h<sub>2<\/sub><br>But their volumes are equal,<br><img src=\"https:\/\/farm2.staticflickr.com\/1913\/43828845840_4a74c885dd_o.png\" alt=\"RD Sharma Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS\" width=\"355\" height=\"399\"><\/p>\n<p>Question 13.<br>The radius of a wire is decreased to one-third. If the volume remains the same, the length will become<br>(a) 3 times<br>(b) 6 times<br>(c) 9 times<br>(d) 27 times<br>Solution:<br>In the first case, r and h1, be the radius and height of the cylindrical wire<br>\u2234 Volume = \u03c0r<sup>2<\/sup>h<sub>1<\/sub>&nbsp;\u2026(i)<br>In the second case, the radius is decreased to one-third<br><img src=\"https:\/\/farm2.staticflickr.com\/1968\/31774682238_9e6b052c5c_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"243\" height=\"310\"><br>\u2234 In the second case height is 9 times (c)<\/p>\n<p>Question 14.<br>If the height of a cylinder is doubled, by what number must the radius of the base be multiplied so that the resulting cylinder has the same volume as the original cylinder?<br><img src=\"https:\/\/farm2.staticflickr.com\/1946\/43828845360_7976572e36_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"245\" height=\"105\"><br>Solution:<br>Let r be the radius and h be the height then volume = \u03c0r<sup>2<\/sup>h<br>If the height is doubled and volume is the same and let x be the radius then \u03c0r<sup>2<\/sup>h = \u03c0(x)<sup>2<\/sup>&nbsp;x 2h<br><img src=\"https:\/\/farm2.staticflickr.com\/1911\/31774681888_c2e8b7097f_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"344\" height=\"121\"><\/p>\n<p>Question 15.<br>The volume of a cylinder of radius r is 1\/4 of the volume of a rectangular box with a square base of side length x. If the cylinder and the box have equal heights, what is r in terms of x?<br><img src=\"https:\/\/farm2.staticflickr.com\/1936\/31774681428_a8d86736e4_o.png\" alt=\"Surface Areas and Volume of a Circular Cylinder Class 9 RD Sharma Solutions\" width=\"273\" height=\"117\"><br>Solution:<br>Let r be the radius and h be the height, then volume = \u03c0r<sup>2<\/sup>h<br>This volume is&nbsp;<span id=\"MathJax-Element-14-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-66\" class=\"math\"><span id=\"MathJax-Span-67\" class=\"mrow\"><span id=\"MathJax-Span-68\" class=\"mfrac\"><span id=\"MathJax-Span-69\" class=\"mn\">1<\/span><span id=\"MathJax-Span-70\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>&nbsp;of the volume of a rectangular box<br>\u2234 Volume of box = 4\u03c0r<sup>2<\/sup>h<br>Let the side of the base of the box = x and height h,<br>then volume = x<sup>2<\/sup>h<br>\u2234 4\u03c0r<sup>2<\/sup>h = x<sup>2<\/sup>h<br><img src=\"https:\/\/farm2.staticflickr.com\/1933\/31774681218_9b9d537847_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"232\" height=\"58\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1912\/31774681228_e38c0a0268_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"342\" height=\"61\"><\/p>\n<p>Question 16.<br>The height ft of a cylinder equals the circumference of the cylinder. In terms of ft, what is the volume of the cylinder?<br><img src=\"https:\/\/farm2.staticflickr.com\/1905\/43828844850_a73f0052c1_o.png\" alt=\"Class 9 Maths Chapter 19 Surface Areas and Volume of a Circular Cylinder RD Sharma Solutions\" width=\"276\" height=\"118\"><br>Solution:<br>In a cylinder,<br>h = circumference of the cylinder<br><img src=\"https:\/\/farm2.staticflickr.com\/1938\/31774680878_7028f9e747_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"351\" height=\"190\"><\/p>\n<p>Question 17.<br>A cylinder with radius r and height ft is closed on the top and bottom. Which of the following expressions represents the total surface area of this cylinder?<br>(a) 2\u03c0r(r + h)<br>(b) \u03c0r(r + 2h)<br>(c) \u03c0r(2r + h)<br>(d) 2\u03c0r<sup>2<\/sup>&nbsp;+ h<br>Solution:<br>r is the radius of the base and it is the height of a closed cylinder<br>The total surface area = 2\u03c0r(r + h ) (a)<\/p>\n<p>Question 18.<br>The height of sand in a cylindrical shape can drop 3 inches when 1 cubic foot of sand is poured out. What is the diameter, in inches, of the cylinder?<br><img src=\"https:\/\/farm2.staticflickr.com\/1922\/43828844590_e79962fb51_o.png\" alt=\"RD Sharma Class 9 Book Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"273\" height=\"122\"><br>Solution:<br>Let h be the height and d be the diameter of a cylinder, then<br><img src=\"https:\/\/farm2.staticflickr.com\/1952\/45596601962_9ff1a4ff9a_o.png\" alt=\"Surface Areas and Volume of a Circular Cylinder With Solutions PDF RD Sharma Class 9 Solutions\" width=\"364\" height=\"493\"><\/p>\n<p>Question 19.<br>Two steel sheets each of length a<sub>1<\/sub>&nbsp;and breadth a<sub>2<\/sub>&nbsp;are used to prepare the surfaces of two right circular cylinders \u2013 one having volume v<sub>1&nbsp;<\/sub>and height a<sub>2<\/sub> and the other having volume v<sub>2<\/sub>&nbsp;and height a<sub>1<\/sub>. Then,<br><img src=\"https:\/\/farm2.staticflickr.com\/1946\/43828844300_ef29a44abf_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"308\" height=\"78\"><br>Solution:<br>Length of each sheet = a<sub>1<\/sub><br>and breadth = a<sub>2<\/sub><br>Volume of cylinder = \u03c0r<sup>2<\/sup>h<br>In the first case,<br>v<sub>1<\/sub>&nbsp;is volume and a<sub>2<\/sub>&nbsp;is the height<br><img src=\"https:\/\/farm2.staticflickr.com\/1964\/43828844190_0a274d4e3c_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"349\" height=\"168\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1979\/43828844050_4593e672e3_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"339\" height=\"381\"><\/p>\n<p>Question 20.<br>The altitude of a circular cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface of the cylinder increases, is<br><img src=\"https:\/\/farm2.staticflickr.com\/1926\/43828843970_34e26d1749_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 19 Surface Areas and Volume of a Circular Cylinder\" width=\"240\" height=\"101\"><br>Solution:<br>In the first case,<br>Let r be the radius and h be the height of the cylinder. Then,<br>\u2234 Lateral surface area = 2\u03c0rh<br>In the second case,<br><img src=\"https:\/\/farm2.staticflickr.com\/1905\/43828843910_8300ed8f29_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS\" width=\"341\" height=\"318\"><\/p>\n<p>&nbsp;<\/p>\n<h3><span class=\"ez-toc-section\" id=\"surface-area-and-volume-of-a-right-circular-cylinder-ex-191\"><\/span>Surface Area and Volume of A Right Circular Cylinder Ex 19.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><img class=\"alignnone size-full wp-image-6748\" src=\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2014\/12\/RD-Sharma-Class-9-solutions-Chapter-19-Surface-Area-and-volume-of-A-Right-Circular-cylinder-Ex-19..1-1.png\" sizes=\"(max-width: 468px) 100vw, 468px\" srcset=\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2014\/12\/RD-Sharma-Class-9-solutions-Chapter-19-Surface-Area-and-volume-of-A-Right-Circular-cylinder-Ex-19..1-1.png 468w, https:\/\/www.learncbse.in\/wp-content\/uploads\/2014\/12\/RD-Sharma-Class-9-solutions-Chapter-19-Surface-Area-and-volume-of-A-Right-Circular-cylinder-Ex-19..1-1-197x300.png 197w\" alt=\"RD Sharma Class 9 solutions Chapter 19 Surface Area and volume of A Right Circular cylinder Ex 19.1 1\" width=\"468\" height=\"712\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8584\/15901106999_5de2d80447_o.png\" alt=\"RD Sharma Class 9 solutions Chapter 19 Surface Area and volume of A Right Circular cylinder Ex 19.1 2\" width=\"485\" height=\"672\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7535\/16085282611_b910668520_o.png\" alt=\"RD Sharma Class 9 solutions Chapter 19 Surface Area and volume of A Right Circular cylinder Ex 19.1 3\" width=\"438\" height=\"623\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8617\/15901106349_0e220928eb_o.png\" alt=\"RD Sharma Class 9 solutions Chapter 19 Surface Area and volume of A Right Circular cylinder Ex 19.1 4\" width=\"462\" height=\"645\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7491\/15901106179_5e4b81d43a_o.png\" alt=\"RD Sharma Class 9 solutions Chapter 19 Surface Area and volume of A Right Circular cylinder Ex 19.1 5\" width=\"463\" height=\"649\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7566\/16086480092_4763c720e3_o.png\" alt=\"RD Sharma Class 9 solutions Chapter 19 Surface Area and volume of A Right Circular cylinder Ex 19.1 6\" width=\"456\" height=\"642\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8625\/16087202675_7b1025b013_o.png\" alt=\"RD Sharma Class 9 solutions Chapter 19 Surface Area and volume of A Right Circular cylinder Ex 19.1 7\" width=\"471\" height=\"704\"><br><img src=\"https:\/\/farm9.staticflickr.com\/8648\/15901105449_f7b7869514_o.png\" alt=\"RD Sharma Class 9 solutions Chapter 19 Surface Area and volume of A Right Circular cylinder Ex 19.1 8\" width=\"437\" height=\"586\"><br><img src=\"https:\/\/farm8.staticflickr.com\/7512\/15901440377_28fdce06b6_o.png\" alt=\"RD Sharma Class 9 solutions Chapter 19 Surface Area and volume of A Right Circular cylinder Ex 19.1 9\" width=\"458\" height=\"434\"><\/p>\n<h2><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-explanation-with-listing-of-important-topics\"><\/span><strong>Detailed Exercise-wise Explanation with Listing of Important Topics<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-19-exercise-19a\"><\/span>RD Sharma class 9 chapter 19 exercise 19a:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>This exercise includes topics related to the surface area of a cylinder. Solving problems on surface areas of a circular cylinder with the help of RD Sharma solutions assists the students to have a better understanding of each concept.<\/p>\n<p>These solutions are framed by subject experts to assist the students in increasing their scoring potential in the final exam. The students can develop problem-solving skills to answer any type of question with ease.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-19-exercise-19b\"><\/span>RD Sharma class 9 chapter 19 exercise 19b<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>RD Sharma&#8217;s solutions for class 9 Maths are considered the perfect tools to prepare for the exam. Each question is solved in such a way that assists the students in understanding the topics easily.<\/p>\n<p>Every student must consider these while practicing to prepare for the exam more effectively. Each student must practice these exercises regularly as this assists them to excel in the exams.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"important-topics-from-class-9-maths-chapter-19\"><\/span>Important Topics from Class 9 Maths Chapter 19<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>RD Sharma Solutions Class 9 Maths Chapter 19 covers some important concepts that are listed below:<\/p>\n<ul>\n<li style=\"font-weight: 400;\">Introduction of Right Circular Cylinder&nbsp;<\/li>\n<li style=\"font-weight: 400;\">Important terms definition of Base, Radius, Axis, Height, and Lateral Surface.<\/li>\n<li style=\"font-weight: 400;\">Surface Area of a Cylinder<\/li>\n<li style=\"font-weight: 400;\">The volume of a Cylinder<\/li>\n<\/ul>\n<p>The students must practice RD Sharma Solutions <a href=\"http:\/\/cbse.nic.in\" target=\"_blank\" rel=\"noopener noreferrer\">CBSE<\/a> Class 9 Maths Chapter 19 regularly to assist them in enhancing their confidence level to secure excellent ranks in the final exams. To know more about the Class 9 Maths exam, ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-9-maths-chapter-19\"><\/span><strong>FAQs on RD Sharma Solutions Class 9 Maths Chapter 19<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>&nbsp;<\/p>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630758152096\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-19\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 19?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630758176979\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-19\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 19?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630758292061\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-19-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 9 Maths Chapter 19\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 9 Maths Chapter 19: Every student can secure excellent marks in the mathematics final exams by practising RD Sharma Solutions Class 9 Maths Chapter 19. The top mathematics faculty has solved these solutions in order to assist the students in their exam preparation. These solutions enable students in their problem-solving skills &#8230; <a title=\"RD Sharma Solutions Class 9 Maths Chapter 19 &#8211; Surface Area And Volume of A Right Circular Cylinder (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-19-surface-area-and-volume-of-a-right-circular-cylinder\/\" aria-label=\"More on RD Sharma Solutions Class 9 Maths Chapter 19 &#8211; Surface Area And Volume of A Right Circular Cylinder (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":124614,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3081,3037,3086],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63861"}],"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\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=63861"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63861\/revisions"}],"predecessor-version":[{"id":510055,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63861\/revisions\/510055"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/124614"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=63861"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=63861"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=63861"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}