{"id":124931,"date":"2023-09-13T06:48:00","date_gmt":"2023-09-13T01:18:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=124931"},"modified":"2023-11-15T10:25:28","modified_gmt":"2023-11-15T04:55:28","slug":"rd-sharma-class-9-solutions-chapter-21-exercise-21-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-21-exercise-21-2\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125675\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-21-Exercise-21.2.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-21-Exercise-21.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-21-Exercise-21.2-768x432.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 21 Exercise 21.2&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 21 Exercise 21.2<\/strong>: <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-21-surface-area-and-volume-of-sphere\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 21 Surface Area And Volume of Sphere<\/a> Exercise 21.2. These solutions are created by our experts. They provided a very helpful answer to their Class 9 exam. The problem is solved in an organized way. These solutions are available for free PDF forms, which can be easily stored for a long time.<\/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-69e718f6d69a1\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-21-exercise-21-2\/#can-i-open-rd-sharma-class-9-solutions-chapter-21-exercise-212-pdf-on-my-smartphone\" title=\"Can I open RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2 PDF on my smartphone?\">Can I open RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2 PDF on my smartphone?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-21-exercise-212\"><\/span>Download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2&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 21 Exercise 21.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>\u00a0<\/p>\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\/21.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/21.2.pdf\">RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-21-exercise-212\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2&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 21 Exercise 21.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1: Find the volume of a sphere whose radius is:<\/strong><\/p>\n<p><strong>(i) 2 cm (ii) 3.5 cm (iii) 10.5 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The volume of a sphere = 4\/3\u03c0r<sup>3<\/sup>\u00a0Cubic Units<\/p>\n<p>Where, r = radius of a sphere<\/p>\n<p><strong>(i)<\/strong>\u00a0Radius = 2 cm<\/p>\n<p>Volume = 4\/3 \u00d7 22\/7 \u00d7 (2)<sup>3<\/sup><\/p>\n<p>= 33.52<\/p>\n<p>Volume = 33.52 cm<sup>3<\/sup><\/p>\n<p><strong>(ii)<\/strong>\u00a0Radius = 3.5cm<\/p>\n<p>Therefore volume = 4\/3\u00d722\/7\u00d7(3.5)<sup>3<\/sup><\/p>\n<p>= 179.666<\/p>\n<p>Volume = 179.666 cm<sup>3<\/sup><\/p>\n<p><strong>(iii)<\/strong>\u00a0Radius = 10.5 cm<\/p>\n<p>Volume = 4\/3\u00d722\/7\u00d7(10.5)<sup>3<\/sup><\/p>\n<p>= 4851<\/p>\n<p>Volume = 4851 cm<sup>3<\/sup><\/p>\n<p><strong>Question 2: Find the volume of a sphere whose diameter is:<\/strong><\/p>\n<p><strong>(i) 14 cm (ii) 3.5 dm (iii) 2.1 m<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The volume of a sphere = 4\/3\u03c0r<sup>3<\/sup>\u00a0Cubic Units<\/p>\n<p>Where, r = radius of a sphere<\/p>\n<p><strong>(i)<\/strong>\u00a0diameter =14 cm<\/p>\n<p>So, radius = diameter\/2 = 14\/2 = 7cm<\/p>\n<p>Volume = 4\/3\u00d722\/7\u00d7(7)<sup>3<\/sup><\/p>\n<p>= 1437.33<\/p>\n<p>Volume = 1437.33 cm<sup>3<\/sup><\/p>\n<p><strong>(ii)<\/strong>\u00a0diameter = 3.5 dm<\/p>\n<p>So, radius = diameter\/2 = 3.5\/2 = 1.75 dm<\/p>\n<p>Volume = 4\/3\u00d722\/7\u00d7(1.75)<sup>3<\/sup><\/p>\n<p>= 22.46<\/p>\n<p>Volume = 22.46 dm<sup>3<\/sup><\/p>\n<p><strong>(iii)<\/strong>\u00a0diameter = 2.1 m<\/p>\n<p>So, radius = diameter\/2 = 2.1\/2 = 1.05 m<\/p>\n<p>Volume = 4\/3\u00d722\/7\u00d7(1.05)<sup>3<\/sup><\/p>\n<p>= 4.851<\/p>\n<p>Volume = 4.851 m<sup>3<\/sup><\/p>\n<p><strong>Question 3: A hemispherical tank has an inner radius of 2.8 m. Find its capacity in litres.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The radius of hemispherical tank = 2.8 m<\/p>\n<p>Capacity of hemispherical tank = 2\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>=2\/3\u00d722\/7\u00d7(2.8)<sup>3\u00a0<\/sup>m<sup>3<\/sup><\/p>\n<p>= 45.997 m<sup>3<\/sup><\/p>\n<p>[Using 1m<sup>3\u00a0<\/sup>= 1000 liters]<\/p>\n<p>Therefore, capacity in litres = 45997 litres<\/p>\n<p><strong>Question 4: A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The inner radius of a hemispherical bowl = 5 cm<\/p>\n<p>Outer radius of a hemispherical bowl = 5 cm + 0.25 cm = 5.25 cm<\/p>\n<p>The volume of steel used = Outer volume \u2013 Inner volume<\/p>\n<p>= 2\/3\u00d7\u03c0\u00d7((5.25)<sup>3<\/sup>\u2212(5)<sup>3<\/sup>)<\/p>\n<p>= 2\/3\u00d722\/7\u00d7((5.25)<sup>3<\/sup>\u2212(5)<sup>3<\/sup>)<\/p>\n<p>= 41.282<\/p>\n<p>The volume of steel used is 41.282 cm<sup>3<\/sup><\/p>\n<p><strong>Question 5: How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Edge of a cube = 22 cm<\/p>\n<p>Diameter of bullet = 2 cm<\/p>\n<p>So, the radius of the bullet (r) = 1 cm<\/p>\n<p>Volume of the cube = (side)<sup>3<\/sup>\u00a0= (22)<sup>3<\/sup>\u00a0cm<sup>3\u00a0<\/sup>= 10648 cm<sup>3<\/sup><\/p>\n<p>And,<\/p>\n<p>The volume of each bullet which will be spherical in shape = 4\/3\u03c0r<sup>3<\/sup><\/p>\n<p>= 4\/3 \u00d7 22\/7 \u00d7 (1)<sup>3\u00a0<\/sup>cm<sup>3<\/sup><\/p>\n<p>= 4\/3 \u00d7 22\/7 cm<sup>3<\/sup><\/p>\n<p>= 88\/21 cm<sup>3<\/sup><\/p>\n<p>Number of bullets = (Volume of cube) \/ (Volume of bullet)<\/p>\n<p>= 10648\/88\/21<\/p>\n<p>= 2541<\/p>\n<p>Therefore, 2541 bullets can be made.<\/p>\n<p><strong>Question 6: A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The volume of laddoo having a radius of 5 cm (V1) = 4\/3\u00d722\/7\u00d7(5)<sup>3<\/sup><\/p>\n<p>= 11000\/21 cm<sup>3<\/sup><\/p>\n<p>Also, the volume of the laddoo having a radius of 2.5 cm (V2) = 4\/3\u03c0r<sup>3<\/sup><\/p>\n<p>= 4\/3\u00d722\/7\u00d7(2.5)<sup>3\u00a0<\/sup>cm<sup>3<\/sup><\/p>\n<p>= 1375\/21 cm<sup>3<\/sup><\/p>\n<p>Therefore,<\/p>\n<p>Number of laddoos of radius 2.5 cm that can be made = V1\/V2 = 11000\/1375 = 8<\/p>\n<p><strong>Question 7: A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 3\/2cm and 2 cm, find the diameter of the third ball.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The volume of the lead ball with the radius 3\/2 cm = 4\/3\u03c0r<sup>3<\/sup><\/p>\n<p>= 4\/3\u00d7\u03c0\u00d7(3\/2)<sup>3<\/sup><\/p>\n<p>Let, Diameter of the first ball (d1) = 3\/2cm<\/p>\n<p>The radius of the first ball (r1) = 3\/4 cm<\/p>\n<p>The diameter of the second ball (d2) = 2 cm<\/p>\n<p>Radius of second ball (r2) = 2\/2 cm = 1 cm<\/p>\n<p>Diameter of the third ball (d3) = d<\/p>\n<p>Radius of the third ball (r3) = d\/2 cm<\/p>\n<p>Now,<\/p>\n<p><img class=\"wp-image-1158298\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-21-ex-21-2.png\" sizes=\"(max-width: 583px) 100vw, 583px\" srcset=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-21-ex-21-2.png 583w, https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-21-ex-21-2-250x175.png 250w, https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-21-ex-21-2-120x84.png 120w\" alt=\"RD sharma class 9 maths chapter 21 ex 21.2\" width=\"583\" height=\"407\" \/><\/p>\n<p>So, the diameter of the third ball is 2.5 cm.<\/p>\n<p><strong>Question 8: A sphere of radius 5 cm is immersed in water filled in a cylinder, and the level of water rises 5\/3 cm. Find the radius of the cylinder.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Radius of sphere = 5 cm (Given)<\/p>\n<p>Let \u2018r\u2019 be the radius of the cylinder.<\/p>\n<p>We know, Volume of the sphere = 4\/3\u03c0r<sup>3<\/sup><\/p>\n<p>By putting values, we get<\/p>\n<p>= 4\/3\u00d7\u03c0\u00d7(5)<sup>3<\/sup><\/p>\n<p>The height (h) of water rises is 5\/3 cm (Given)<\/p>\n<p>The volume of water rises in the cylinder = \u03c0r<sup>2<\/sup>h<\/p>\n<p>Therefore, the volume of water rises in the cylinder = Volume of the sphere<\/p>\n<p>So, \u03c0r<sup>2<\/sup>h = 4\/3\u03c0r<sup>3<\/sup><\/p>\n<p>\u03c0 r<sup>2<\/sup>\u00a0\u00d7 5\/3 = 4\/3 \u00d7 \u03c0 \u00d7 (5)<sup>3<\/sup><\/p>\n<p>or r<sup>2\u00a0<\/sup>= 100<\/p>\n<p>or r = 10<\/p>\n<p>Therefore, the radius of the cylinder is 10 cm.<\/p>\n<p><strong>Question 9: If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let r be the radius of the first sphere, then 2r be the radius of the second sphere.<\/p>\n<p>Now,<\/p>\n<p><img class=\"wp-image-1158299\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/ratio-of-volume-of-the-first-sphere-to-the-second.png\" sizes=\"(max-width: 346px) 100vw, 346px\" srcset=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/ratio-of-volume-of-the-first-sphere-to-the-second.png 346w, https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/ratio-of-volume-of-the-first-sphere-to-the-second-250x55.png 250w, https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/ratio-of-volume-of-the-first-sphere-to-the-second-120x26.png 120w\" alt=\"Ratio of volume of the first sphere to the second sphere \" width=\"346\" height=\"76\" \/><\/p>\n<p>The ratio of the volume of the first sphere to the second sphere is 1:8.<\/p>\n<p><strong>Question 10: A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The volume of the cone = Volume of the hemisphere (Given)<\/p>\n<p>1\/3\u03c0r<sup>2<\/sup>h = 2\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>(Using respective formulas)<\/p>\n<p>r<sup>2<\/sup>h = 2r<sup>3<\/sup><\/p>\n<p>or h = 2r<\/p>\n<p>Since a cone and a hemisphere have equal bases it implies they have the same radius.<\/p>\n<p>h\/r = 2<\/p>\n<p>or h : r = 2 : 1<\/p>\n<p>Therefore, the ratio of their heights is 2:1<\/p>\n<p><strong>Question 11: A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm, respectively. Find the height to which the water will rise in the cylinder.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The volume of water in the hemispherical bowl = Volume of water in the cylinder \u2026 (Given)<\/p>\n<p>The inner radius of the bowl ( r<sub>1<\/sub>) = 3.5cm<\/p>\n<p>The inner radius of cylinder (r<sub>2<\/sub>) = 7cm<\/p>\n<p>The volume of water in the hemispherical bowl = Volume of water in the cylinder<\/p>\n<p>2\/3\u03c0r<sub>1<\/sub><sup>3<\/sup>\u00a0= \u03c0r<sub>2<\/sub><sup>2<\/sup>h<\/p>\n<p>[Using respective formulas]<\/p>\n<p>Where h is the height to which water rises in the cylinder.<\/p>\n<p>2\/3\u03c0(3.5)<sup>3<\/sup>\u00a0= \u03c0(7)<sup>2<\/sup>h<\/p>\n<p>or h = 7\/12<\/p>\n<p>Therefore, 7\/12 cm is the height to which water rises in the cylinder.<\/p>\n<p><strong>Question 12: A cylinder whose height is two-thirds of its diameter has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Radius of a sphere (R)= 4 cm (Given)<\/p>\n<p>Height of the cylinder = 2\/3 diameter (given)<\/p>\n<p>We know, Diameter = 2(Radius)<\/p>\n<p>Let h be the height and r be the base radius of a cylinder, then<\/p>\n<p>h = 2\/3\u00d7 (2r) = 4r\/3<\/p>\n<p>The volume of the cylinder = Volume of the sphere<\/p>\n<p>\u03c0r<sup>2<\/sup>h = 4\/3\u03c0R<sup>3<\/sup><\/p>\n<p>\u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 (4r\/3) = 4\/3 \u03c0 (4)<sup>3<\/sup><\/p>\n<p>(r)<sup>3<\/sup>\u00a0= (4)<sup>3<\/sup><\/p>\n<p>or r = 4<\/p>\n<p>Therefore, the radius of the base of the cylinder is 4 cm.<\/p>\n<p><strong>Question 13: A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are 6 cm and 4 cm. Find the height of water in the cylinder.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Radius of a bowl (R)= 6 cm (Given)<\/p>\n<p>Radius of a cylinder (r) = 4 cm (given)<\/p>\n<p>Let h be the height of a cylinder.<\/p>\n<p>Now,<\/p>\n<p>The volume of water in the hemispherical bowl = Volume of the cylinder<\/p>\n<p>2\/3 \u03c0 R<sup>3<\/sup>\u00a0= \u03c0r<sup>2<\/sup>\u00a0h<\/p>\n<p>2\/3 \u03c0 (6)<sup>3<\/sup>\u00a0= \u03c0(4)<sup>2<\/sup>\u00a0h<\/p>\n<p>or h = 9<\/p>\n<p>Therefore, the height of the water in the cylinder is 9 cm.<\/p>\n<p><strong>Question 14: A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub, and thus the level of water is raised by 9 cm. What is the radius of the ball?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let r be the radius of the iron ball.<\/p>\n<p>Radius of the cylinder (R) = 16 cm (Given)<\/p>\n<p>A spherical iron ball is dropped into the cylinder, and thus the level of water is raised by 9 cm. So, height (h) = 9 cm<\/p>\n<p>From statement,<\/p>\n<p>The volume of the iron ball = Volume of water raised in the hub<\/p>\n<p>4\/3\u03c0r<sup>3<\/sup>\u00a0= \u03c0R<sup>2<\/sup>h<\/p>\n<p>4\/3 r<sup>3<\/sup>\u00a0= (16)<sup>2\u00a0<\/sup>\u00d7 9<\/p>\n<p>or r<sup>3<\/sup>\u00a0= 1728<\/p>\n<p>or r = 12<\/p>\n<p>Therefore, the radius of the ball = 12 cm.<\/p>\n<p>We have included all the information regarding <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2<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 queries 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-21-exercise-212\"><\/span>FAQ: RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2<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-1630935478605\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-the-rd-sharma-class-9-solutions-chapter-21-exercise-212-pdf-for-free\"><\/span>Can I download the RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2 PDF for 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 21 Exercise 21.2 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630935527581\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-from-rd-sharma-solutions-class-12\"><\/span>What are the benefits of studying from RD Sharma Solutions Class 12?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing these solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630935528630\" 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 id=\"faq-question-1630935532068\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-open-rd-sharma-class-9-solutions-chapter-21-exercise-212-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2 PDF on my smartphone?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can open RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2 PDF on any device.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2: RD Sharma Solutions Class 9 Maths Chapter 21 Surface Area And Volume of Sphere Exercise 21.2. These solutions are created by our experts. They provided a very helpful answer to their Class 9 exam. The problem is solved in an organized way. These solutions are available &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-21-exercise-21-2\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":125675,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76791,76796],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124931"}],"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=124931"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124931\/revisions"}],"predecessor-version":[{"id":507283,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/124931\/revisions\/507283"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125675"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=124931"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=124931"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=124931"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}