{"id":63897,"date":"2021-09-06T11:34:28","date_gmt":"2021-09-06T06:04:28","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=63897"},"modified":"2021-09-17T20:03:49","modified_gmt":"2021-09-17T14:33:49","slug":"rd-sharma-solutions-class-9-maths-chapter-21-surface-area-and-volume-of-sphere","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-21-surface-area-and-volume-of-sphere\/","title":{"rendered":"RD Sharma Solutions Class 9 Maths Chapter 21 &#8211; Surface Area And Volume of Sphere (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-124665\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/RD-Sharma-Solutions-Class-9-Maths-Chapter-21-Surface-Area-And-Volume-of-Sphere.png\" alt=\"RD Sharma Solutions Class 9 Maths Chapter 21\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/RD-Sharma-Solutions-Class-9-Maths-Chapter-21-Surface-Area-And-Volume-of-Sphere.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/11\/RD-Sharma-Solutions-Class-9-Maths-Chapter-21-Surface-Area-And-Volume-of-Sphere-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 9 Maths chapter 21 &#8211; surface area and volume of sphere: <\/strong>The Chapter Sharma Solutions enable the students to understand all concepts &amp; topics available in this chapter in a better way. The solutions mentioned in <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> Chapter 21 are created by experienced and top faculty in a simple &amp; understandable manner for the students. to know more, read the whole blog.<\/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-69d7db478d8ed\" 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-69d7db478d8ed\"><\/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-21-surface-area-and-volume-of-sphere\/#download-rd-sharma-solutions-class-9-maths-chapter-21-pdf\" title=\"Download RD Sharma Solutions Class 9 Maths Chapter 21 PDF\">Download RD Sharma Solutions Class 9 Maths Chapter 21 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-21-surface-area-and-volume-of-sphere\/#exercise-wise-rd-sharma-solutions-class-9-maths-chapter-21-pdf\" title=\"Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 21 PDF\">Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 21 PDF<\/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-21-surface-area-and-volume-of-sphere\/#access-answers-of-rd-sharma-solutions-class-9-maths-chapter-21\" title=\"Access answers of RD Sharma Solutions Class 9 Maths Chapter 21\">Access answers of RD Sharma Solutions Class 9 Maths Chapter 21<\/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-21-surface-area-and-volume-of-sphere\/#rd-sharma-solutions-class-9-chapter-21-surface-areas-and-volume-of-a-sphere-ex-211\" title=\"RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1\">RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-21-surface-area-and-volume-of-sphere\/#rd-sharma-solutions-class-9-chapter-21-surface-areas-and-volume-of-a-sphere-mcqs\" title=\"RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere MCQS\">RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere MCQS<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-21-surface-area-and-volume-of-sphere\/#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-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-21-surface-area-and-volume-of-sphere\/#rd-sharma-class-9-chapter-21-exercise-21a\" title=\"RD Sharma class 9 chapter 21 exercise 21a:\">RD Sharma class 9 chapter 21 exercise 21a:<\/a><\/li><li class='ez-toc-page-1 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-21-surface-area-and-volume-of-sphere\/#rd-sharma-class-9-chapter-21-exercise-21b\" title=\"RD Sharma class 9 chapter 21 exercise 21b:\">RD Sharma class 9 chapter 21 exercise 21b:<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-21-surface-area-and-volume-of-sphere\/#important-topics-from-class-9-maths-chapter-21\" title=\"Important Topics From Class 9 Maths Chapter 21\">Important Topics From Class 9 Maths Chapter 21<\/a><\/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-21-surface-area-and-volume-of-sphere\/#faqs-on-rd-sharma-solutions-class-9-maths-chapter-21\" title=\"FAQs on RD Sharma Solutions Class 9 Maths Chapter 21\">FAQs on RD Sharma Solutions Class 9 Maths Chapter 21<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-21-surface-area-and-volume-of-sphere\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-21\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 21?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 21?<\/a><\/li><li class='ez-toc-page-1 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-21-surface-area-and-volume-of-sphere\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-21\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 21?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 21?<\/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-21-surface-area-and-volume-of-sphere\/#can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-21-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 9 Maths Chapter 21\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 9 Maths Chapter 21\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-21-pdf\"><\/span><strong>Download RD Sharma Solutions Class 9 Maths Chapter 21 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/21-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 21<\/a><\/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\/21-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-9-maths-chapter-21-pdf\"><\/span><strong>Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 21 PDF<\/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-21-class-9-maths-exercise-21-1-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma class 9 chapter 21 exercise 21a<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-21-class-9-maths-exercise-21-2-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma class 9 chapter 21 exercise 21b<\/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-21\"><\/span><strong>Access answers of RD Sharma Solutions Class 9 Maths Chapter 21<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-21-surface-areas-and-volume-of-a-sphere-ex-211\"><\/span>RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>Find the surface area of a sphere of radius.<br \/>(i) 10.5 cm<br \/>(ii) 5.6 cm<br \/>(iii) 14 cm<br \/>Solution:<br \/>In a sphere,<br \/>(i) Radius (r) = 10.5 cm<br \/>Surface area = 4\u03c0r<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1909\/45597193512_5a28b7b0cb_o.png\" alt=\"RD Sharma Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"284\" height=\"542\" \/><\/p>\n<p>Question 2.<br \/>Find the surface area of a sphere of diameter<br \/>(i) 14 cm<br \/>(ii) 21 cm<br \/>(iii) 3.5 cm<br \/>Solution:<br \/>(i) Diameter of a sphere = 14 cm<br \/>Radius (r) = \\(\\frac { 14 }{ 2 }\\) = 7 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1913\/31775266318_30bf7a2684_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"313\" height=\"493\" \/><\/p>\n<p>Question 3.<br \/>Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm. [Use \u03c0 = 3.14]<br \/>Solution:<br \/>(i) Radius of hemisphere = 10 cm<br \/>\u2234 Total surface area of hemisphere = 2\u03c0r<sup>2<\/sup><br \/>= 2 x 3.14 x 10 x 10 cm<sup>2<\/sup><br \/>= 628 cm<sup>2<\/sup><br \/>(ii) Total surface area of solid hemisphere<br \/>= 3\u03c0r<sup>2<\/sup>\u00a0= 3 x 3.14 x 10 x 10 cm<sup>2<\/sup><br \/>= 942 cm<sup>2<\/sup><\/p>\n<p>Question 4.<br \/>The surface area of a sphere in 5544 cm<sup>2<\/sup>, find the diameter.<br \/>Solution:<br \/>Let r be the radius of a sphere, then Surface area = 4\u03c0r<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1961\/45597192992_3e88360e6a_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"249\" height=\"143\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/31775266178_5082224a6f_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 21 Surface Areas and Volume of a Sphere\" width=\"331\" height=\"51\" \/><\/p>\n<p>Question 5.<br \/>A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of \u20b94 per 100 cm<sup>2<\/sup>. [NCERT]<br \/>Solution:<br \/>Inner diameter of a hemispherical bowl = 10.5 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1912\/45597192822_3e6cee459a_o.png\" alt=\"Surface Areas and Volume of a Sphere Class 9 RD Sharma Solutions\" width=\"322\" height=\"358\" \/><\/p>\n<p>Question 6.<br \/>The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of \u20b92 per sq. m.<br \/>Solution:<br \/>Radius of dome (hemispherical) = 63 dm<br \/>Area of curved surface<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1902\/31775266128_6f7e71678d_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 21 Surface Areas and Volume of a Sphere\" width=\"307\" height=\"187\" \/><\/p>\n<p>Question 7.<br \/>Assuming the earth to be a sphere of radius 6370 km, how many square kilometres is area of the land, if three-fourth of the earth\u2019s surface is covered by water?<br \/>Solution:<br \/>Radius of earth (sphere) = 6370 km<br \/>Water on the earth = \\(\\frac { 3 }{ 4 }\\) % total area<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1959\/45597192542_2d167d7590_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"290\" height=\"211\" \/><\/p>\n<p>Question 8.<br \/>A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the shape if the length of the shape be 7 cm.<br \/>Solution:<br \/>Total height of the so formed shape = 7 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1944\/44733444405_b557f75f49_o.png\" alt=\"Class 9 Maths Chapter 21 Surface Areas and Volume of a Sphere RD Sharma Solutions\" width=\"304\" height=\"451\" \/><\/p>\n<p>Question 9.<br \/>The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.<br \/>Solution:<br \/>Diameter of moon = \\(\\frac { 1 }{ 4 }\\) of diameter of earth<br \/>Let radius of earth = r km<br \/>Then radius of moon = \\(\\frac { 1 }{ 4 }\\) r km<br \/>Now surface area of earth = 4\u03c0r<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1963\/45597192272_281f30064e_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 21 Surface Areas and Volume of a Sphere\" width=\"330\" height=\"254\" \/><\/p>\n<p>Question 10.<br \/>A hemi-spherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is \u20b95 per 100 cm<sup>2<\/sup>. [NCERT]<br \/>Solution:<br \/>Circumference of the base of dome (r) = 17.6 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1940\/44733444275_d71934993b_o.png\" alt=\"RD Sharma Class 9 Book Chapter 21 Surface Areas and Volume of a Sphere\" width=\"312\" height=\"337\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1959\/31775265958_c97bd60aee_o.png\" alt=\"Surface Areas and Volume of a Sphere With Solutions PDF RD Sharma Class 9 Solutions\" width=\"284\" height=\"75\" \/><\/p>\n<p>Question 11.<br \/>A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at \u20b97 per 100 cm<sup>2<\/sup>.<br \/>Solution:<br \/>Diameter of toy = 16 cm<br \/>Radius (r) = \\(\\frac { 16 }{ 2 }\\) = 8 cm<br \/>Height of conical part (h) = 15 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1964\/45597191882_7319722cc7_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"318\" height=\"633\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1966\/31775265888_b0c09b88cf_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"322\" height=\"211\" \/><\/p>\n<p>Question 12.<br \/>A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of \u20b910 per m<sup>2<\/sup>.<br \/>Solution:<br \/>Diameter of the tank = 1.4 m<br \/>\u2234 Radius (r) = \\(\\frac { 1.4 }{ 2 }\\) m = 0.7 m<br \/>and height of cylindrical portion = 8m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1954\/44733443935_b6489262ea_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"307\" height=\"464\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/31775265828_7378731bca_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"341\" height=\"191\" \/><\/p>\n<p>Question 13.<br \/>The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm<sup>2<\/sup>\u00a0and black paint costs 5 paise per cm<sup>2<\/sup>. [NCERT]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1942\/31775265778_b0297abdb6_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"208\" height=\"153\" \/><br \/>Solution:<br \/>Diameter of each spheres = 21 cm<br \/>\u2234 Radius (R) = \\(\\frac { 21 }{ 2 }\\) cm<br \/>Radius of each cylinder (r) = 1.5 cm<br \/>and height (h) = 7 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/31775265718_5b3654ee3a_o.png\" alt=\"Class 9 Maths Chapter 21 Surface Areas and Volume of a Sphere RD Sharma Solutions Ex 21.1\" width=\"155\" height=\"188\" \/><br \/>Now surface area of one sphere = 4\u03c0R<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1938\/44733443645_1a4f335ddd_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 21 Surface Areas and Volume of a Sphere\" width=\"352\" height=\"686\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-21-surface-areas-and-volume-of-a-sphere-ex-212\"><\/span>RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>Find the volume of a sphere whose radius is<br \/>(i) 2 cm<br \/>(ii) 3.5 cm<br \/>(iii) 10.5 cm<br \/>Solution:<br \/>(i) Radius of sphere (r) = 2 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1907\/31775265608_e7c6517aa7_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 21 Surface Areas and Volume of a Sphere\" width=\"348\" height=\"114\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1974\/44922732364_1224e0ebdd_o.png\" alt=\"Surface Areas and Volume of a Sphere Class 9 RD Sharma Solutions\" width=\"359\" height=\"377\" \/><\/p>\n<p>Question 2.<br \/>Find the volume of a sphere whose diameter is,<br \/>(i) 14 cm<br \/>(ii) 3.5 dm<br \/>(iii) 2.1 m<br \/>Solution:<br \/>(i) Diameter of a sphere = 14 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1968\/31775265548_e58d60f3cf_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 21 Surface Areas and Volume of a Sphere\" width=\"361\" height=\"453\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1933\/44922731994_4050cd80d7_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"320\" height=\"164\" \/><\/p>\n<p>Question 3.<br \/>A hemspherical tank has inner radius of 2.8 m. Find its capacity in litres.<br \/>Solution:<br \/>Radius of hemispherical tank (r) = 2.8 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1959\/31775265508_f99bd3c834_o.png\" alt=\"Class 9 Maths Chapter 21 Surface Areas and Volume of a Sphere RD Sharma Solutions\" width=\"346\" height=\"329\" \/><\/p>\n<p>Question 4.<br \/>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.<br \/>Solution:<br \/>Thickness of steel = 0.25 cm = \\(\\frac { 1 }{ 4 }\\)cm<br \/>Inside radius of the hemispherical bowl (r) = 5 cm<br \/>\u2234 Outer radius (R) = 5 + 0.25 = 5.25 cm<br \/>\u2234 Volume of the steel used = \\(\\frac { 1 }{ 4 }\\)\u03c0(R<sup>3<\/sup>\u00a0\u2013 r<sup>3<\/sup>)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1909\/31775265438_4fccf824f8_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 21 Surface Areas and Volume of a Sphere\" width=\"319\" height=\"121\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1946\/31775265398_9f137eeecc_o.png\" alt=\"RD Sharma Class 9 Book Chapter 21 Surface Areas and Volume of a Sphere\" width=\"281\" height=\"50\" \/><\/p>\n<p>Question 5.<br \/>How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?<br \/>Solution:<br \/>Edge of cube (r) = 22 cm<br \/>\u2234 Volume = a<sup>3<\/sup>\u00a0= (22)<sup>3<\/sup>\u00a0cm<sup>3<\/sup><br \/>= 22 x 22 x 22 = 10648 cm<sup>3<\/sup><br \/>Diameter of a bullet = 2 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1972\/44733442795_b7c5111100_o.png\" alt=\"Surface Areas and Volume of a Sphere With Solutions PDF RD Sharma Class 9 Solutions\" width=\"339\" height=\"299\" \/><\/p>\n<p>Question 6.<br \/>A shopkeeper has one laddoo of radius 5 cm. With the same material how many laddoos of radius 2.5 cm can be made?<br \/>Solution:<br \/>Radius of bigger laddoo (R) = 5 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1957\/44922731214_cd7c8186bd_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"330\" height=\"344\" \/><br \/>Question 7.<br \/>A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. It the diameters of two balls be \\(\\frac { 3 }{ 2 }\\) cm and 2 cm, find the diameter of the third ball.<br \/>Solution:<br \/>Diameter of a spherical ball of lead = 3 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1980\/44733442655_79a13c92be_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"336\" height=\"603\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1917\/44922730924_38cf03af09_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"301\" height=\"203\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1907\/30706715547_92be2abe63_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"335\" height=\"414\" \/><\/p>\n<p>Question 8.<br \/>A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises \\(\\frac { 5 }{ 3 }\\) cm. Find the radius of the cylinder.<br \/>Solution:<br \/>Radius of sphere (r1) = 5 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1955\/44733442375_3b6a30c463_o.png\" alt=\"RD Sharma Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"287\" height=\"364\" \/><br \/>Level of water rises in the cylinder after immersing the sphere in it<br \/>\u2234 Height of water level = \\(\\frac { 5 }{ 3 }\\) cm<br \/>Let r be radius of the cylinder, then Volume of water = Volume of the sphere<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1914\/44922730404_1e6d170351_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"301\" height=\"216\" \/><\/p>\n<p>Question 9.<br \/>If the radius of a sphere is doubled, what is the ratio of the volumes of the first sphere to that of the second sphere?<br \/>Solution:<br \/>Let r2 be the radius of the given sphere<br \/>then volume = \\(\\frac { 4 }{ 3 }\\) \u03c0r<sup>3<\/sup><br \/>By doubling the radius the radius of the new sphere = 2r<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1910\/30706715067_d4b7bc3470_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"218\" height=\"287\" \/><\/p>\n<p>Question 10.<br \/>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.<br \/>Solution:<br \/>Radius of hemispherical bowl (r) = 3.5 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1904\/44733442015_1a755131eb_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"342\" height=\"403\" \/><\/p>\n<p>Question 11.<br \/>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.<br \/>Solution:<br \/>Radius of a sphere (r) = 4 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1952\/30706714817_ea566c8af2_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 21 Surface Areas and Volume of a Sphere\" width=\"339\" height=\"406\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1960\/31775264978_5dcc2112bc_o.png\" alt=\"Surface Areas and Volume of a Sphere Class 9 RD Sharma Solutions\" width=\"305\" height=\"165\" \/><\/p>\n<p>Question 12.<br \/>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 respectively 6 cm and 4 cm. Find the height of water in the cylinder.<br \/>Solution:<br \/>Radius of hemispherical bowl (r) = 6 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1972\/44922729914_98c6223447_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 21 Surface Areas and Volume of a Sphere\" width=\"345\" height=\"307\" \/><\/p>\n<p>Question 13.<br \/>The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.<br \/>Solution:<br \/>Diameter of a copper sphere = 18 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1978\/30706714427_e0f6e39870_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"345\" height=\"200\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1905\/44922729534_5e358e8ddb_o.png\" alt=\"Class 9 Maths Chapter 21 Surface Areas and Volume of a Sphere RD Sharma Solutions\" width=\"306\" height=\"269\" \/><\/p>\n<p>Question 14.<br \/>The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.<br \/>Solution:<br \/>Diameter of a sphere = 6 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/30706714187_03195d1d94_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 21 Surface Areas and Volume of a Sphere\" width=\"326\" height=\"539\" \/><\/p>\n<p>Question 15.<br \/>The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted and recast into a solid cylinder of height 2 \\(\\frac { 2 }{ 3 }\\) cm. Find the diameter of the cylinder.<br \/>Solution:<br \/>Internal radius of the hollow spherical shell (r) = 3 cm<br \/>and external radius (R) = 5 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1974\/31775264458_91098be341_o.png\" alt=\"RD Sharma Class 9 Book Chapter 21 Surface Areas and Volume of a Sphere\" width=\"344\" height=\"412\" \/><\/p>\n<p>Question 16.<br \/>A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.<br \/>Solution:<br \/>Radius of hemisphere (r) = 7 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1944\/30706713857_bc8fcb6540_o.png\" alt=\"Surface Areas and Volume of a Sphere With Solutions PDF RD Sharma Class 9 Solutions\" width=\"282\" height=\"306\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1914\/31775264288_7dc82b4c6f_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"285\" height=\"84\" \/><\/p>\n<p>Question 17.<br \/>A hollow sphere of internal and external radius 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.<br \/>Solution:<br \/>Internal radius of a hollow sphere (r) = 2 cm<br \/>and external radius (R) = 4 cm<br \/>\u2234 Volume of the metal used<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1950\/44922729014_d59dc95af3_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"298\" height=\"482\" \/><\/p>\n<p>Question 18.<br \/>A metallic sphere of radius 10.5 cm is melted and thus recast into small cones each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.<br \/>Solution:<br \/>Radius of a metallic sphere (R) = 10.5 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1956\/31775264118_357c3bdce7_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"165\" height=\"103\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1972\/30706713437_63c7e45fb1_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"356\" height=\"393\" \/><\/p>\n<p>Question 19.<br \/>A cone and a hemisphere have equal bases and equal volumes. Find the ratio Of their heights.<br \/>Solution:<br \/>Let r be the radius and h be the height of the cone, hemisphere<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1953\/31775263888_388962c4cd_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"319\" height=\"324\" \/><\/p>\n<p>Question 20.<br \/>The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.<br \/>Solution:<br \/>By carving a largest sphere out of the cube, the diameter of the sphere = 10.5<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/30706713167_00c2ec397e_o.png\" alt=\"RD Sharma Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"321\" height=\"416\" \/><\/p>\n<p>Question 21.<br \/>A cube, of side 4 cm, contains a sphere touching its sides. Find the volume of the gap in between.<br \/>Solution:<br \/>Side of cube = 4 cm<br \/>\u2234 Volume = (side)<sup>3<\/sup>\u00a0= 4x4x4 = 64 cm<sup>3<\/sup><br \/>Diameter of the largest sphere touching its sides = 4 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1969\/45647700111_376ee7d1e3_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"335\" height=\"316\" \/><\/p>\n<p>Question 22.<br \/>A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m, then find the volume of the iron used to make the tank. (NCERT)<br \/>Solution:<br \/>Thickness of hemispherical tank = 1 cm<br \/>Inner radius (r) = 1 m = 100 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1934\/43829454700_bbaae3bf42_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"354\" height=\"540\" \/><\/p>\n<p>Question 23.<br \/>A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule? (NCERT)<br \/>Solution:<br \/>Diameter of a medicine spherical capsule = 3.5 mm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1954\/30706712627_23913f0dda_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 21 Surface Areas and Volume of a Sphere\" width=\"284\" height=\"84\" \/><\/p>\n<p>Question 24.<br \/>The diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon? (NCERT)<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1966\/43829454280_6107f36ebd_o.png\" alt=\"Surface Areas and Volume of a Sphere Class 9 RD Sharma Solutions\" width=\"357\" height=\"492\" \/><\/p>\n<p>Question 25.<br \/>A cone and a hemisphere have equal bases and equal volumes. Find the ratio in their heights.<br \/>Solution:<br \/>Let r be the radius of cone and hemisphere and let h be the height of the cone then<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1967\/45647699661_f204ee5db7_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 21 Surface Areas and Volume of a Sphere\" width=\"292\" height=\"187\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1941\/43829453930_791f19e7a1_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"303\" height=\"81\" \/><\/p>\n<p>Question 26.<br \/>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 level of water is raised by 9 cm. What is the radius of the ball?<br \/>Solution:<br \/>Radius of cylinderical tub (r) = 16 cm<br \/>Height of water in it (h) = 30 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1909\/45647699571_a789987d8d_o.png\" alt=\"Class 9 Maths Chapter 21 Surface Areas and Volume of a Sphere RD Sharma Solutions\" width=\"324\" height=\"441\" \/><\/p>\n<p>Question 27.<br \/>A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use \u03c0 = 22\/7).<br \/>Solution:<br \/>Radius of cylinder (r) = 12 cm<br \/>Depth of water in it (h) = 20 cm<br \/>By dropping a ball, the water level rose by 6.75 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1950\/43829453660_c2f2323608_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 21 Surface Areas and Volume of a Sphere\" width=\"281\" height=\"98\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1944\/45647699431_a630bd7126_o.png\" alt=\"RD Sharma Class 9 Book Chapter 21 Surface Areas and Volume of a Sphere\" width=\"336\" height=\"190\" \/><\/p>\n<p>Question 28.<br \/>A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimetres?<br \/>Solution:<br \/>Radius of cylinderical jar (r) = 6 cm<br \/>Level of oil in it (h) = 2 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1964\/44922726374_caaea7d966_o.png\" alt=\"Surface Areas and Volume of a Sphere With Solutions PDF RD Sharma Class 9 Solutions\" width=\"341\" height=\"273\" \/><br \/>Question 29.<br \/>A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm eacfy are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar?<br \/>Solution:<br \/>Diameter of measuring jar = 10 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/45647699251_4f567e60d9_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"301\" height=\"182\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1917\/44922725924_6a2b9680f9_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"324\" height=\"105\" \/><br \/>Now after swing the ball in the water of jar Let volume of water raised, by h cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1953\/43829453040_afda045b68_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"314\" height=\"355\" \/><\/p>\n<p>Question 30.<br \/>A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2:3.<br \/>Solution:<br \/>\u2235 Bases and heights of a cones hemisphere and a cylinder are equal<br \/>Let r be the radius and h be their heights<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1947\/44922725664_e35e48eacf_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"347\" height=\"274\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/43829452720_64868039bc_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"335\" height=\"156\" \/><\/p>\n<p>Question 31.<br \/>A cylinderical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?<br \/>Solution:<br \/>Radius of the cylinderical tub (r) = 12 cm<br \/>Depth of water in it (h) = 20 cm<br \/>By dropping a spherical ball in it, the water raised by 6.75 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1914\/44922725324_0def7b8fc9_o.png\" alt=\"RD Sharma Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"293\" height=\"277\" \/><\/p>\n<p>Question 32.<br \/>A sphere, a cylinder and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes.<br \/>Solution:<br \/>Diameter of a sphere, cylinder and a cone are equal<br \/>Let each as diameter = 2r<br \/>Then radius of each = r<br \/>Height of cylinder = diameter = 2r<br \/>and height of cone = 2r<br \/>Now volume of sphere = \\(\\frac { 4 }{ 3 }\\)\u03c0r<sup>3<\/sup><br \/>Volume of cylinder = \u03c0r<sup>2<\/sup>h<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1954\/43829452490_4647458e39_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"352\" height=\"308\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"surface-areas-and-volume-of-a-sphere-class-9-rd-sharma-solutions-vsaqs\"><\/span>Surface Areas and Volume of a Sphere Class 9 RD Sharma Solutions VSAQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>Find the surface area of a sphere of radius 14 cm.<br \/>Solution:<br \/>Radius of a sphere (r) = 14 cm<br \/>\u2234 Surface area = 4\u03c0r<sup>2<\/sup>\u00a0= 4 x \\(\\frac { 22 }{ 7 }\\) x 14 x 14 cm<sup>2<\/sup><br \/>= 2464 cm<sup>3<\/sup><\/p>\n<p>Question 2.<br \/>Find the total surface afea of a hemisphere of radius 10 cm.<br \/>Solution:<br \/>Radius of hemisphere (r) = 10 cm<br \/>\u2234 Total surface area = 3\u03c0r<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1974\/31775266878_8139d90f87_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"246\" height=\"102\" \/><\/p>\n<p>Question 3.<br \/>Find the radius of a sphere whose surface area is 154 cm<sup>2<\/sup>.<br \/>Solution:<br \/>Surface area of a sphere = 154 cm<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1955\/44733445365_e9a6a3c432_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 21 Surface Areas and Volume of a Sphere\" width=\"345\" height=\"129\" \/><\/p>\n<p>Question 4.<br \/>The hollow sphere, in which the circus motor cyclist performs his stunts, has a diameter of 7 m. Find the area available to the motor cyclist for riding.<br \/>Solution:<br \/>Diameter of hollow sphere = 7 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/31775266828_c3880a15fb_o.png\" alt=\"Surface Areas and Volume of a Sphere Class 9 RD Sharma Solutions\" width=\"257\" height=\"186\" \/><\/p>\n<p>Question 5.<br \/>Find the volume of a sphere whose surface area is 154 cm<sup>2<\/sup>.<br \/>Solution:<br \/>Surface area of a sphere = 154 cm<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1915\/45597195472_40a1d7b022_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 21 Surface Areas and Volume of a Sphere\" width=\"329\" height=\"326\" \/><\/p>\n<p>Question 6.<br \/>How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?<br \/>Solution:<br \/>Edge of a solid cube = 44 cm<br \/>\u2234 Volume = a<sup>2<\/sup>\u00a0= (44)<sup>2<\/sup>\u00a0cm<sup>2<\/sup><br \/>= 44 \u00d7 44 \u00d7 44 cm<sup>3<\/sup><br \/>Diameter of a spherical bullet = 4 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1956\/31775266748_1d767956d2_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"302\" height=\"113\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1980\/31775266688_94113e7ae7_o.png\" alt=\"Class 9 Maths Chapter 21 Surface Areas and Volume of a Sphere RD Sharma Solutions\" width=\"313\" height=\"266\" \/><\/p>\n<p>Question 7.<br \/>If a sphere of radius 2r has the same volume as that of a cone with circular base of radius r, then find the height of the cone.<br \/>Solution:<br \/>Radius of a sphere (R) = 2r<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1971\/45597194752_de13f2e3b4_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 21 Surface Areas and Volume of a Sphere\" width=\"327\" height=\"424\" \/><\/p>\n<p>Question 8.<br \/>If a hollow sphere of intefnal and external diameters 4 cm and 8 cm respectively melted into a cone of base diameter 8 cm, then find the height of the cone.<br \/>Solution:<br \/>Internal diameter of a hollow sphere = 4cm<br \/>\u2234 Internal radius = \\(\\frac { 4 }{ 2 }\\) = 2 cm<br \/>Similarly the outer radius (R) = \\(\\frac { 8 }{ 2 }\\) = 4 cm<br \/>\u2234 Volume of melted used in hollow sphere<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/31775266618_cf7ba4fb6d_o.png\" alt=\"RD Sharma Class 9 Book Chapter 21 Surface Areas and Volume of a Sphere\" width=\"362\" height=\"501\" \/><\/p>\n<p>Question 9.<br \/>The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height of the cone.<br \/>Solution:<br \/>Radius of a sphere (r) = 5 cm<br \/>\u2234 Surface area = 4\u03c0r<sup>2<\/sup><br \/>= 4\u03c0 x 5 x 5 = 100\u03c0 cm<sup>2<\/sup><br \/>Radius of cone (r<sub>1<\/sub>) = 4 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1943\/45597194352_2343edd2c3_o.png\" alt=\"Surface Areas and Volume of a Sphere With Solutions PDF RD Sharma Class 9 Solutions\" width=\"245\" height=\"193\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1912\/31775266538_45ec0b4b19_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"260\" height=\"119\" \/><\/p>\n<p>Question 10.<br \/>If a sphere is inscribed in a cube, find the ratio of the volume of cube to the volume of the sphere.<br \/>Solution:<br \/>Let edge of a cube = a<br \/>Then its volume = a<sup>3<\/sup><br \/>\u2235 A sphere is inscribed in the cube<br \/>\u2234 Diameter of sphere = a<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1963\/45597193962_6ce196eb09_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"282\" height=\"320\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-21-surface-areas-and-volume-of-a-sphere-mcqs\"><\/span>RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere 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 sphere, the number of faces is<br \/>(a) 1<br \/>(b) 2<br \/>(c) 3<br \/>(d) 4<br \/>Solution:<br \/>Number of faces of a sphere is 1 (a)<\/p>\n<p>Question 2.<br \/>The total surface area of a hemisphere of radius r is<br \/>(a) \u03c0r<sup>2<\/sup><br \/>(b) 2\u03c0r<sup>2<\/sup><br \/>(c) 3\u03c0r<sup>2<\/sup><br \/>(d) 4\u03c0r<sup>2<\/sup><br \/>Solution:<br \/>Total surface area of a hemisphere is 37\u03c0r<sup>2<\/sup>\u00a0(c)<\/p>\n<p>Question 3.<br \/>The ratio of the total surface area of a sphere and a hemisphere of same radius is<br \/>(a) 2 : 1<br \/>(b) 3 : 2<br \/>(c) 4 : 1<br \/>(d) 4 : 3<br \/>Solution:<br \/>Total surface area of a sphere = 4\u03c0r<sup>2<\/sup><br \/>and total surface area of a hemisphere = 3m<sup>2<\/sup><br \/>\u2234 Ratio 4\u03c0r<sup>2<\/sup>: 3\u03c0r<sup>2<\/sup><br \/>= 4 : 3 (d)<\/p>\n<p>Question 4.<br \/>A sphere and a cube are of the same height. The ratio of their volumes is<br \/>(a) 3 :4<br \/>(b) 21 : 11<br \/>(c) 4 : 3<br \/>(d) 11 : 21<br \/>Solution:<br \/>Let r be the height of a sphere and cube<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1972\/45597199312_c97e168878_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere MCQS\" width=\"346\" height=\"353\" \/><\/p>\n<p>Question 5.<br \/>The largest sphere is cut off from a cube of side 6 cm. The volume of the sphere will be<br \/>(a) 27\u03c0 cm<sup>3<\/sup><br \/>(b) 36\u03c0 cm<sup>3<\/sup><br \/>(c) 108\u03c0 cm<sup>3<\/sup><br \/>(d) 12\u03c0 cm<sup>3<\/sup><br \/>Solution:<br \/>Side of cube = 6 cm<br \/>\u2234 Diameter of sphere cut off from it = 6 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1973\/44733446535_378c57151c_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 21 Surface Areas and Volume of a Sphere\" width=\"347\" height=\"169\" \/><\/p>\n<p>Question 6.<br \/>A cylmderical rod whose height is 8 times of its radius is melted and recast into spherical balls of same radius. The number of balls will be<br \/>(a) 4<br \/>(b) 3<br \/>(c) 6<br \/>(d) 8<br \/>Solution:<br \/>Let r be the radius of a cylindrical rod = r<br \/>Then its height (h) = 8r<br \/>Volume = \u03c0r<sup>2<\/sup>h = \u03c0r<sup>2<\/sup>\u00a0x 8r = 8\u03c0r<sup>3<\/sup><br \/>Radius of spherical ball = r<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1976\/45597198782_8753eca8a6_o.png\" alt=\"Surface Areas and Volume of a Sphere Class 9 RD Sharma Solutions\" width=\"345\" height=\"195\" \/><\/p>\n<p>Question 7.<br \/>If the ratio of volumes of two spheres is 1 : 8, then the ratio of their surface areas is<br \/>(a) 1 : 2<br \/>(b) 1 : 4<br \/>(c) 1 : 8<br \/>(d) 1 : 16<br \/>Solution:<br \/>Let r<sub>1<\/sub>\u00a0and r<sub>2<\/sub>\u00a0be the radius of two spheres<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1967\/44733446465_af27476ccd_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 21 Surface Areas and Volume of a Sphere\" width=\"331\" height=\"414\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/45597198342_e40a38e0ce_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"351\" height=\"86\" \/><\/p>\n<p>Question 8.<br \/>If the surface area of a sphere is 144\u03c0 m<sup>2<\/sup>\u00a0then its volume (in. m3) is<br \/>(a) 288\u03c0<br \/>(b) 316\u03c0<br \/>(c) 300\u03c0<br \/>(d) 188\u03c0<br \/>Solution:<br \/>Surface area of a sphere = 144\u03c0 m<sup>2<\/sup><br \/>Let r be the radius, then<br \/>4\u03c0r<sup>2<\/sup>\u00a0= 144\u03c0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1973\/44733446355_0cdc7de1a0_o.png\" alt=\"Class 9 Maths Chapter 21 Surface Areas and Volume of a Sphere RD Sharma Solutions\" width=\"344\" height=\"221\" \/><\/p>\n<p>Question 9.<br \/>If a solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then the surface area of each ball (in sq. cm) is<br \/>(a) 100\u03c0<br \/>(b) 75\u03c0<br \/>(c) 60\u03c0<br \/>(d) 50\u03c0<br \/>Solution:<br \/>Radius of a sphere (r) = 10 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1946\/31775267448_6ef81b2487_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 21 Surface Areas and Volume of a Sphere\" width=\"325\" height=\"315\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1940\/44733446165_7f57123e95_o.png\" alt=\"RD Sharma Class 9 Book Chapter 21 Surface Areas and Volume of a Sphere\" width=\"331\" height=\"178\" \/><\/p>\n<p>Question 10.<br \/>If a sphere is inscribed in a cube, then the ratio of the volume of the sphere to the volume of the cube is<br \/>(a) \u03c0 : 2<br \/>(b) \u03c0 : 3<br \/>(c) \u03c0 : 4<br \/>(d) \u03c0 : 6<br \/>Solution:<br \/>Let side of a cube = a<br \/>Then volume of cube = a<sup>3<\/sup><br \/>The diameter of inscribed sphere = a<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1967\/44733446005_4b5dcb13fe_o.png\" alt=\"Surface Areas and Volume of a Sphere With Solutions PDF RD Sharma Class 9 Solutions\" width=\"348\" height=\"305\" \/><\/p>\n<p>Question 11.<br \/>If a solid sphere of radius r is melted and cast into the shape of a solid cone of height r, then the radius of the base of the cone is<br \/>(a) 2r<br \/>(b) 3r<br \/>(c) r<br \/>(d) 4r<br \/>Solution:<br \/>Radius of a sphere = r<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1903\/31775267198_cdb2cf537c_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"293\" height=\"133\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1923\/44733445915_068c123c92_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"348\" height=\"255\" \/><\/p>\n<p>Question 12.<br \/>A sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder. If the radius of the sphere is r, then the volume of the cylinder is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/31775267138_f3756220d7_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"265\" height=\"80\" \/><br \/>Solution:<br \/>Radius of sphere = r<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1966\/31775267098_fd64e24c3b_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"345\" height=\"357\" \/><\/p>\n<p>Question 13.<br \/>The ratio between the volume of a sphere and volume of a circumscribing right circular cylinder is<br \/>(a) 2 : 1<br \/>(b) 1 : 1<br \/>(c) 2 : 3<br \/>(d) 1 : 2<br \/>Solution:<br \/>Let r be the radius of the sphere, then 4<br \/>Volume = \\(\\frac { 4 }{ 3 }\\)\u03c0r<sup>3<\/sup><br \/>Diameter of circumscribed cylinder = 2r<br \/>\u2234 Radius = r<br \/>and height (h) = 2r<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1953\/44733445725_caa2faaee3_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"347\" height=\"194\" \/><\/p>\n<p>Question 14.<br \/>A cone and a hemisphere have equal bases and equal volumes the ratio of their heights is<br \/>(a) 1 : 2<br \/>(b) 2 : 1<br \/>(c) 4 : 1<br \/>(d) \\(\\sqrt { 2 } \\) : 1<br \/>Solution:<br \/>Let radius of hemisphere and a cone be r<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1906\/31775267058_6dfbf3c5e4_o.png\" alt=\"RD Sharma Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"343\" height=\"352\" \/><\/p>\n<p>Question 15.<br \/>A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is<br \/>(a) 1 : 2 : 3<br \/>(b) 2 : 1 : 3<br \/>(c) 2 : 3 : 1<br \/>(d) 3 : 2 : 1<br \/>Solution:<br \/>\u2235 Bases of a cone, hemisphere and a cylinder are same<br \/>Let radius of each = r<br \/>and height of each = r<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1903\/44733445595_ced3c5854e_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere\" width=\"255\" height=\"162\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1966\/31775266958_76d1d1d863_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 21 Surface Areas and Volume of a Sphere\" width=\"340\" height=\"192\" \/><\/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-21-exercise-21a\"><\/span>RD Sharma class 9 chapter 21 exercise 21a:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>This exercise is based on the topics related to sphere, hemisphere, surface area of a sphere, &amp; surface area of a hemisphere. RD Sharma class 9 chapter 21 exercise 21a assist the students in constructive preparation &amp; learning.<\/p>\n<p>The students should practice all the questions &amp; match their solution with the answers given in the RD Sharma textbook. The students can grasp all topics entirely by practicing these solutions regularly.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-21-exercise-21b\"><\/span>RD Sharma class 9 chapter 21 exercise 21b:<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>This exercise includes topics related to the volume of a sphere, hemisphere, &amp; spherical shell. These questions as well as answers are solved by the top experts of Maths in a simplified manner. The solutions are designed in such a way that students can easily understand all concepts of this exercise.\u00a0<\/p>\n<h2><span class=\"ez-toc-section\" id=\"important-topics-from-class-9-maths-chapter-21\"><\/span><strong>Important Topics From Class 9 Maths Chapter 21<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>RD Sharma Solutions Class 9 Maths Chapter 21 cover some important concepts that are listed below:<\/p>\n<ul>\n<li style=\"font-weight: 400;\">Introduction of sphere<\/li>\n<li style=\"font-weight: 400;\">The segment of a sphere by a plane<\/li>\n<li style=\"font-weight: 400;\">Surface Area of a sphere, hemisphere, &amp; spherical shell<\/li>\n<li style=\"font-weight: 400;\">Volume of a sphere, hemisphere, &amp; spherical shell<\/li>\n<\/ul>\n<p>This is the complete blog of RD Sharma Solutions Class 9 Chapter 21. To know more about the <a href=\"http:\/\/cbse.nic.in\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> 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-21\"><\/span><strong>FAQs on RD Sharma Solutions Class 9 Maths Chapter 21<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630908750090\" 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-21\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 21?<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>\n<\/div>\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630908773372\" 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-21\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 21?<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-1630908802098\" 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-21-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 9 Maths Chapter 21\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 21 &#8211; surface area and volume of sphere: The Chapter Sharma Solutions enable the students to understand all concepts &amp; topics available in this chapter in a better way. The solutions mentioned in RD Sharma Solutions Class 9 Maths Chapter 21 are created by experienced and top faculty &#8230; <a title=\"RD Sharma Solutions Class 9 Maths Chapter 21 &#8211; Surface Area And Volume of Sphere (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-21-surface-area-and-volume-of-sphere\/\" aria-label=\"More on RD Sharma Solutions Class 9 Maths Chapter 21 &#8211; Surface Area And Volume of Sphere (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":124665,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,73411,73410],"tags":[3081,3037,3086],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63897"}],"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=63897"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63897\/revisions"}],"predecessor-version":[{"id":128916,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63897\/revisions\/128916"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/124665"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=63897"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=63897"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=63897"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}