{"id":4588,"date":"2023-02-21T09:35:54","date_gmt":"2023-02-21T09:35:54","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4588"},"modified":"2023-02-21T09:38:19","modified_gmt":"2023-02-21T09:38:19","slug":"ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-8","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-8\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 13 (Surface Areas and Volumes)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 13 Surface Areas and Volumes <\/strong>Exercise 13.8 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 13: Surface Areas and Volumes<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.4<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.5<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-6\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.6<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-7\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.7<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-9\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.9<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 13.8<\/span><\/h2>\n<p style=\"text-align: center;\"><span style=\"color: #000000;\"><strong>(Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>, <span class=\"fontstyle0\">unless stated otherwise<\/span>)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Find the volume of a sphere whose radius is<br \/>\n<\/strong><strong>(i) 7 cm<br \/>\n(ii) 0.63 m <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong><strong>(i)<\/strong> Radius of the sphere, <em>r<\/em> = 7 cm<br \/>\nVolume of sphere = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>3<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 7<sup>3\u00a0<\/sup>cm<sup>3<\/sup><br \/>\n= 4312\/3 cm<sup>3<\/sup><br \/>\n= 1437.33 cm<sup>3<\/sup><br \/>\nHence, the volume of the sphere is 1437.33 cm<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Radius of the sphere, r = 0.63 m<br \/>\nvolume of sphere = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>3<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 0.63<sup>3<br \/>\n<\/sup>= 1.0478 m<sup>3\u00a0<\/sup><br \/>\nHence, the volume of the sphere is 1.05 m<sup>3\u00a0<\/sup>(approx).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find the amount of water displaced by a solid spherical ball of diameter<br \/>\n<\/strong><strong>(i) 28 cm<br \/>\n(ii) 0.21 m<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong><strong>(i)<\/strong> Diameter = 28 cm<br \/>\nRadius, r = 28\/2 cm = 14cm<br \/>\nVolume of the solid spherical ball = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/> \u03c0r<sup>3<br \/>\n<\/sup>Volume of the ball = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 14<sup>3<br \/>\n<\/sup>= 34496\/3<br \/>\n= 11498.66\u00a0cm<sup>3<\/sup><br \/>\nHence, the volume of the ball is 11498.66\u00a0cm<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Diameter = 0.21 m<br \/>\nRadius of the ball =0.21\/2 m= 0.105 m<br \/>\nVolume of the ball = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>3<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 0.105<sup>3<\/sup>\u00a0m<sup>3<br \/>\n<\/sup>= 0.004851 m<sup>3<\/sup><br \/>\nHence, the volume of the ball = 0.004851 m<sup>3<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3.The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm<sup>3<\/sup>?\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Diameter of a metallic ball = 4.2 cm<br \/>\nRadius(r) of the metallic ball, r = 4.2\/2 cm = 2.1 cm<br \/>\nVolume formula = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/> \u03c0r<sup>3<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 2.1 cm<sup>3<br \/>\n<\/sup>= 38.808 cm\u00b3<br \/>\nVolume of the metallic ball = 38.808 cm<sup>3<\/sup>.<br \/>\nNow, Mass = Volume \u00d7 Density<\/span><br \/>\n<span style=\"color: #000000;\">Mass of the metallic ball = 38.808 cm<sup>3<\/sup> \u00d7 8.9g \/ cm\u00b3 <\/span><br \/>\n<span style=\"color: #000000;\">= 345.3912 g<\/span><br \/>\n<span style=\"color: #000000;\">= 345.39 g (approx.)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the mass of the ball is 345.39 g.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. 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?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Let the diameter of the earth = <em>d<\/em>.<br \/>\nDiameter of the moon = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{4}\" alt=\"\\frac{1}{4}\" align=\"absmiddle\" \/> \u00d7 diameter of the earth<\/span><br \/>\n<span style=\"color: #000000;\">The radius of the moon = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{4}\" alt=\"\\frac{1}{4}\" align=\"absmiddle\" \/> \u00d7 radius of the earth\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Since, diameter = 2 \u00d7 Radius]<\/span><br \/>\n<span style=\"color: #000000;\">r = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{4}\" alt=\"\\frac{1}{4}\" align=\"absmiddle\" \/> \u00d7 R = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{R}{4}\" alt=\"\\frac{R}{4}\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">The volume of the earth = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0 R<sup>3<\/sup><\/span><\/p>\n<p>The volume of the moon = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0 r<sup>3<br \/>\n<\/sup>=<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{R}{4}&amp;space;\\right&amp;space;)^3\" alt=\"\\left ( \\frac{R}{4} \\right )^3\" align=\"absmiddle\" \/>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[ replacing r with <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{R}{4}\" alt=\"\\frac{R}{4}\" align=\"absmiddle\" \/> ]<\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{64}\" alt=\"\\frac{1}{64}\" align=\"absmiddle\" \/> R<sup>3<br \/>\n<\/sup>The volume of the moon = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{64}\" alt=\"\\frac{1}{64}\" align=\"absmiddle\" \/> \u00d7 Volume of the earth<\/span><br \/>\n<span style=\"color: #000000;\">Hence the volume of the moon is <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{64}\" alt=\"\\frac{1}{64}\" align=\"absmiddle\" \/> times the volume of the earth.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume \u03c0 = 22\/7)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Diameter of the hemispherical ball, d = 10.5 cm<\/span><br \/>\n<span style=\"color: #000000;\">Radius of the hemispherical ball, r = 10.5\/2 cm = 5.25 cm<\/span><br \/>\n<span style=\"color: #000000;\">Volume of the hemispherical ball = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}\" alt=\"\\frac{2}{3}\" align=\"absmiddle\" \/> \u03c0r<sup>3<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}\" alt=\"\\frac{2}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 5.25 cm \u00d7 5.25 cm \u00d7 5.25 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 303.1875 cm<sup>3<br \/>\n<\/sup>= 303.1875 \/1000 ( 1000cm<sup>3<\/sup> = 1 L)<\/span><br \/>\n<span style=\"color: #000000;\">= 0.3031875 litres\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">= 0.303 litres (approx.)<\/span><br \/>\n<span style=\"color: #000000;\">The hemispherical bowl can hold 0.303 litres of milk.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. A hemi spherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Inner Radius of the tank, (r) = 1 m<br \/>\nThickness of iron = 1cm = 1\/100 m = 0.01 m<\/span><br \/>\n<span style=\"color: #000000;\">Outer radius of the tank, R = 1 m + 0.01m = 1.01 m<\/span><br \/>\n<span style=\"color: #000000;\">Volume of the iron used to make the tank = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}\" alt=\"\\frac{2}{3}\" align=\"absmiddle\" \/> \u03c0R<sup>3<\/sup> &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}\" alt=\"\\frac{2}{3}\" align=\"absmiddle\" \/> \u03c0r<sup>3<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}\" alt=\"\\frac{2}{3}\" align=\"absmiddle\" \/> \u03c0 (R<sup>3<\/sup>\u00a0&#8211; r<sup>3<\/sup>)<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}\" alt=\"\\frac{2}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 [(1.01m)<sup>3<\/sup>\u00a0&#8211; (1m)<sup>3<\/sup>]<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}\" alt=\"\\frac{2}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 [1.030301 m<sup>3<\/sup>\u00a0&#8211; 1 m<sup>3<\/sup>]<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}\" alt=\"\\frac{2}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 0.030301 m<sup>3<br \/>\n<\/sup>= 0.06348 m<sup>3<\/sup>\u00a0(approx.)<\/span><br \/>\n<span style=\"color: #000000;\">So, the volume of the iron used in the hemispherical tank is 0.06348 m<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. Find the volume of a sphere whose surface area is 154 cm<sup>2<\/sup>.\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Let the radius of a sphere = r<br \/>\nSurface area of the sphere = 4\u03c0r<sup>2<\/sup>\u00a0= 154cm\u00b2<br \/>\n4\u03c0r<sup>2\u00a0<\/sup>= 154 cm<sup>2<\/sup><br \/>\nr<sup>2<\/sup> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{154\\times&amp;space;7}{4\\times&amp;space;22}\" alt=\"\\frac{154\\times 7}{4\\times 22}\" align=\"absmiddle\" \/>\u00a0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{2}\" alt=\"\\frac{7}{2}\" align=\"absmiddle\" \/><\/span><\/p>\n<p>Radius is <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{2}\" alt=\"\\frac{7}{2}\" align=\"absmiddle\" \/> cm<br \/>\nTherefore, volume of the sphere = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>3<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{2}\" alt=\"\\frac{7}{2}\" align=\"absmiddle\" \/> cm \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{2}\" alt=\"\\frac{7}{2}\" align=\"absmiddle\" \/> cm \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{2}\" alt=\"\\frac{7}{2}\" align=\"absmiddle\" \/> cm<\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{539}{3}\" alt=\"\\frac{539}{3}\" align=\"absmiddle\" \/> cm<sup>3<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?179\\frac{2}{3}\" alt=\"179\\frac{2}{3}\" align=\"absmiddle\" \/> cm<sup>3<\/sup>.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, volume of the sphere is <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?179\\frac{2}{3}\" alt=\"179\\frac{2}{3}\" align=\"absmiddle\" \/> cm<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. A dome of a building is in the form of a hemi sphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing is Rs20 per square meter, find the<br \/>\n<\/strong><strong>(i) inside surface area of the dome<br \/>\n(ii) volume of the air inside the dome<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Cost of white-washing the dome from inside = Rs 4989.60<br \/>\nCost of white-washing 1 m<sup>2<\/sup>\u00a0area = Rs 20<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) <\/strong>Inside surface\u00a0area\u00a0of the dome = Total cost for whitewashing the dome\u00a0inside \/\u00a0Rate of whitewashing<\/span><br \/>\n<span style=\"color: #000000;\">= 4989.60\/20 <\/span><br \/>\n<span style=\"color: #000000;\">= 249.48 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Let the inner radius of the hemispherical dome = r.<br \/>\nInner\u00a0surface area\u00a0of the hemispherical dome = 2\u03c0r<sup>2<br \/>\n<\/sup>2\u03c0r<sup>2\u00a0<\/sup>=\u00a0249.48 m<sup>2<br \/>\n<\/sup>\u21d2 r<sup>2<\/sup>\u00a0= 249.48\/2\u03c0 m<sup>2<br \/>\n<\/sup>\u21d2 r<sup>2<\/sup> = 249.48\u00a0\u00f7 (2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/>)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 r<sup>2<\/sup>\u00a0= 39.69<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 r = \u221a39.69<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 r = 6.3 m<\/span><br \/>\n<span style=\"color: #000000;\">The\u00a0volume\u00a0of the air inside the dome will be the same as the volume of the hemisphere.<\/span><br \/>\n<span style=\"color: #000000;\">Now the\u00a0volume of the air inside the dome = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}\" alt=\"\\frac{2}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>3<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}\" alt=\"\\frac{2}{3}\" align=\"absmiddle\" \/>\u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 6.3 m \u00d7 6.3 m \u00d7 6.3 m<\/span><br \/>\n<span style=\"color: #000000;\">= 523.9 m<sup>3<\/sup> (approx.) <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the inner surface area of the dome is 249.48 m<sup>2<\/sup>\u00a0and he volume of the air inside the dome is 523.9 m<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S\u2019. Find the<br \/>\n<\/strong><strong>(i) radius r\u2019 of the new sphere,<br \/>\n<\/strong><strong>(ii) ratio of Sand S\u2019.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>The surface area of a sphere\u00a0= 4\u03c0r<sup>2<br \/>\n<\/sup>The\u00a0volume of a sphere = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>3<br \/>\n<\/sup>Therefore, The volume of 27 solid spheres with radius r = 27 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/> \u03c0r<sup>3<\/sup>\u00a0= 36\u03c0r<sup>3<\/sup>\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (1)<\/span><br \/>\n<span style=\"color: #000000;\">Also, the volume of the new sphere with radius r&#8217; = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0r&#8217;<sup>3<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (2)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> Volume of the new sphere = Volume of 27 solid spheres<\/span><br \/>\n<span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0r&#8217;<sup>3<\/sup>\u00a0= 36\u03c0r<sup>3<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[From equation (1) and (2)]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 r&#8217;<sup>3<\/sup> = 36\u03c0r\u00b3 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3}{4}\" alt=\"\\frac{3}{4}\" align=\"absmiddle\" \/>\u03c0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 r&#8217;<sup>3<\/sup>\u00a0= 27r\u00b3<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 r&#8217; = \u221b27r\u00b3<\/span><br \/>\n<span style=\"color: #000000;\">r&#8217; = 3r <\/span><br \/>\n<span style=\"color: #000000;\">Radius of the new sphere, r&#8217; = 3r<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii)\u00a0<\/strong>Ratio of S and S\u2032 <\/span><br \/>\n<span style=\"color: #000000;\">Now, surface area of each iron sphere, S = 4\u03c0r<sup>2<br \/>\n<\/sup>Surface area of the new sphere, S&#8217; = 4\u03c0r&#8217;<sup>2<\/sup>\u00a0= 4\u03c0(3r)<sup>2<\/sup>\u00a0= 36\u03c0r<sup>2<br \/>\n<\/sup>The ratio of the S and S\u2019 = 4\u03c0r<sup>2<\/sup>\/36\u03c0r<sup>2<\/sup>\u00a0= 1\/9<\/span><br \/>\n<span style=\"color: #000000;\">Hence, the radius of\u00a0new sphere is 3r and the\u00a0ratio of S and S\u2019 is 1 : 9.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm<sup>3<\/sup>) is needed to fill this capsule?\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Diameter of capsule = 3.5 mm<br \/>\nRadius of capsule, say r = diameter\/ 2 = (3.5\/2) mm = 1.75 mm<br \/>\nVolume of spherical capsule = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>3<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}\" alt=\"\\frac{4}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 (1.75)<sup>3<\/sup><br \/>\n= 22.458<br \/>\n= 22.46mm<sup>3<\/sup>\u00a0(approx.)<br \/>\nThe volume of the spherical capsule is 22.46 mm<sup>3<\/sup>.<\/span><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 13 (Surface Areas and Volumes)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 13 Surface Areas and Volumes Exercise 13.8 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 13: Surface Areas and Volumes NCERT Solution Class 9 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[725,702,186,726,701,5],"class_list":["post-4588","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-13-surface-areas-and-volumes-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-13-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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