{"id":4586,"date":"2023-02-21T09:35:49","date_gmt":"2023-02-21T09:35:49","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4586"},"modified":"2023-02-21T09:38:43","modified_gmt":"2023-02-21T09:38:43","slug":"ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-6","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-6\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6"},"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 13.6 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-13-7\/\" 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-13-8\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.8<\/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; 13.6<\/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\" \/>, unless stated otherwise.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. The circumference of the base of cylindrical vessel is 132cm and its height is 25cm. <\/strong><strong>How many litres of water can it hold? (1000 cm<sup>3<\/sup>= 1 <em>l<\/em>)\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Let the radius of the base = <em>r<br \/>\n<\/em>Height of the cylinder, <em>h<\/em> = 25 cm<\/span><br \/>\n<span style=\"color: #000000;\">Circumference of the base = 132 cm<\/span><br \/>\n<span style=\"color: #000000;\">2\u03c0r = 132 cm<\/span><br \/>\n<span style=\"color: #000000;\">r = 132\/2\u03c0<\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{132}{2\\times&amp;space;\\frac{22}{7}}\" alt=\"\\frac{132}{2\\times \\frac{22}{7}}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{132\\times&amp;space;7}{2\\times&amp;space;22}\" alt=\"\\frac{132\\times 7}{2\\times 22}\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">= 21 cm<\/span><br \/>\n<span style=\"color: #000000;\">The\u00a0capacity of the cylindrical vessel\u00a0= \u03c0r<sup>2<\/sup>h<\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 21 cm \u00d7 21 cm \u00d7 25 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 34650 cm<sup>3<br \/>\n<\/sup>= 34650\/1000 (Since\u00a01000 cm<sup>3<\/sup> = 1 <em>l<\/em>)<\/span><br \/>\n<span style=\"color: #000000;\">= 34.65 <em>l<br \/>\n<\/em>Therefore, the vessel can hold 34.65 litres of water.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. The inner diameter of a cylindrical wooden pipe is 24cm and its outer diameter is 28 cm. The length of the pipe is 35cm. Find the mass of the pipe, if 1cm<sup>3<\/sup> of wood has a mass of 0.6g.\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Outer diameter of the pipe, <em>D <\/em>= 28 cm<br \/>\nOuter radius of the pipe, <em>R<\/em> = 28\/2 = 14 cm<br \/>\nInner diameter of the pipe, <em>d <\/em>= 24 cm<\/span><br \/>\n<span style=\"color: #000000;\">Inner\u00a0radius\u00a0of the pipe, <em>r<\/em> = 24\/2 = 12cm<\/span><br \/>\n<span style=\"color: #000000;\">Length of the pipe, <em>h<\/em> = 35 cm<br \/>\nVolume of the outer cylinder, V<sub>1<\/sub> = \u03c0R<sup>2<\/sup>h<\/span><br \/>\n<span style=\"color: #000000;\">V<sub>1<\/sub> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 14cm \u00d7 14cm \u00d7 35cm<\/span><br \/>\n<span style=\"color: #000000;\">= 21560 cm<sup>3<br \/>\n<\/sup>Volume of the inner cylinder, V<sub>2<\/sub>\u00a0= \u03c0r<sup>2<\/sup>h <\/span><br \/>\n<span style=\"color: #000000;\">V<sub>2<\/sub> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 12cm \u00d7 12cm \u00d7 35cm<\/span><br \/>\n<span style=\"color: #000000;\">= 15840 cm<sup>3<br \/>\n<\/sup>The volume of the wood used = Volume of the outer cylinder \u2013 Volume of the inner cylinder<\/span><br \/>\n<span style=\"color: #000000;\">= 21560 cm<sup>3<\/sup>\u00a0&#8211; 15840 cm<sup>3<br \/>\n<\/sup>= 5720 cm<sup>3<br \/>\n<\/sup>Mass of 1 cm<sup>3<\/sup>\u00a0wood is 0.6 g<\/span><br \/>\n<span style=\"color: #000000;\">Mass of 5720 cm<sup>3<\/sup> wood = 5720 \u00d7 0.6g <\/span><br \/>\n<span style=\"color: #000000;\">= 3432 g <\/span><br \/>\n<span style=\"color: #000000;\">= (3432\/1000) kg<\/span><br \/>\n<span style=\"color: #000000;\">= 3.432 kg <\/span><br \/>\n<span style=\"color: #000000;\">Thus, the mass of the wooden pipe is 3.432 kg<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. A soft drink is available in two packs \u2013<br \/>\n(i) a tin can with a rectangular base of length 5cm and width 4cm, having a height of 15 cm and<br \/>\n(ii) a plastic cylinder with circular base of diameter 7cm and height 10cm.<br \/>\nWhich container has greater capacity and by how much?\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong>The diagram of the plastic cylinder and rectangular base .<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4666\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.6-Ans3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"290\" height=\"217\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.6-Ans3.png 390w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.6-Ans3-300x224.png 300w\" sizes=\"auto, (max-width: 290px) 100vw, 290px\" \/><br \/>\nDimensions of the plastic cylinder with a circular base are: <\/span><br \/>\n<span style=\"color: #000000;\">The diameter of the cylindrical plastic can, <em>d <\/em>= 7 cm<\/span><br \/>\n<span style=\"color: #000000;\">The radius of the cylindrical plastic can, <em>r<\/em> = 7\/2 cm<\/span><br \/>\n<span style=\"color: #000000;\">Height of the cylindrical plastic can, <em>h<\/em> = 10cm<\/span><br \/>\n<span style=\"color: #000000;\">The volume of the cylindrical plastic can = \u03c0r<sup>2\u00a0<\/sup>h <\/span><br \/>\n<span style=\"color: #000000;\">= <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 10cm<\/span><br \/>\n<span style=\"color: #000000;\">= 385 cm<sup>3<br \/>\n<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Dimensions of tin can with a rectangular base are:<\/span><br \/>\n<span style=\"color: #000000;\">The Length of the cuboidal tin can, <em>l<\/em> = 5 cm<\/span><br \/>\n<span style=\"color: #000000;\">The breadth of the cuboidal tin can, <em>b<\/em> = 4 cm<\/span><br \/>\n<span style=\"color: #000000;\">Height of the cuboidal tin can, <em>h<\/em> = 15cm<\/span><br \/>\n<span style=\"color: #000000;\">The\u00a0volume\u00a0of the cuboidal\u00a0tin can = <em>l \u00d7 b\u00a0\u00d7 h<\/em><\/span><br \/>\n<span style=\"color: #000000;\">= 5 cm \u00d7 4 cm \u00d7 15 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 300 cm<sup>3<\/sup><\/span><\/p>\n<p><span style=\"color: #000000;\">Clearly, the plastic cylinder with a circular base has greater capacity than the tin container.<\/span><br \/>\n<span style=\"color: #000000;\">Difference = 385 cm<sup>3<\/sup>\u00a0&#8211; 300 cm<sup>3<\/sup>\u00a0= 85 cm<sup>3<br \/>\n<\/sup>The plastic cylindrical has\u00a0more capacity than the tin can by 85 cm<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. If the lateral surface of a cylinder is 94.2cm<sup>2<\/sup> and its height is 5cm, then find<br \/>\n<\/strong><strong>(i) radius of its base<br \/>\n(ii) its volume.<br \/>\n[Use \u03c0 = 3.14]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong><strong>(i)<\/strong> Let the radius of the cylinder = <em>r<br \/>\n<\/em>Height of the cylinder, <em>h<\/em> = 5 cm<\/span><br \/>\n<span style=\"color: #000000;\">Lateral surface area = 92.4 cm<sup>2<br \/>\n<\/sup>2\u03c0<em>rh<\/em> = 92.4<\/span><br \/>\n<span style=\"color: #000000;\"><em>r<\/em>\u00a0= 92.4 \/ 2<em>h<\/em>\u03c0\u00a0<\/span><br \/>\n<span style=\"color: #000000;\"><em>r<\/em> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{92.4}{2\\times&amp;space;5\\times&amp;space;3.14}\" alt=\"\\frac{92.4}{2\\times 5\\times 3.14}\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">r = 3 cm (approx.)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii)\u00a0<\/strong>Volume of cylinder\u00a0= \u03c0<em>r<\/em><sup>2<\/sup><em>h<br \/>\n<\/em>= 3.14 \u00d7 3 cm \u00d7 3 cm \u00d7 5 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 141.3 cm<sup>3<br \/>\n<\/sup>Thus, the radius of the base is 3 cm and the volume is 141.3 cm<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10m deep. If the cost of painting is at the rate of Rs 20 per m<sup>2<\/sup>, find<br \/>\n<\/strong><strong>(i) inner curved surface area of the vessel<br \/>\n<\/strong><strong>(ii) radius of the base<br \/>\n<\/strong><strong>(iii) capacity of the vessel<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Total cost to paint inner CSA = \u20b9 2200<br \/>\nRate of painting = \u20b9 20 per m<sup>2<br \/>\n<\/sup>Height of the vessel, <em>h<\/em> = 10 m<br \/>\n<strong>(i)\u00a0<\/strong>Inner CSA of the\u00a0cylindrical vessel = 2200\/20 = 110 m<sup>2<\/sup><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) radius<br \/>\n<\/strong>Inner CSA of the vessel = 110 m<sup>2<br \/>\n<\/sup>2\u03c0<em>rh<\/em> = 110 m<sup>2<br \/>\n<\/sup><em>r<\/em> = 110 \/ 2\u03c0<em>h<\/em><\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{110}{2\\times&amp;space;10\\times&amp;space;\\frac{22}{7}}\" alt=\"\\frac{110}{2\\times 10\\times \\frac{22}{7}}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{110\\times&amp;space;7}{2\\times&amp;space;10\\times&amp;space;22}\" alt=\"\\frac{110\\times 7}{2\\times 10\\times 22}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000;\">= 1.75 m<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iii) Capacity of the vessel<br \/>\n<\/strong>Volume of the vessel = \u03c0<em>r<\/em><sup>2<\/sup><em>h<br \/>\n<\/em>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 1.75 m \u00d7 1.75 m \u00d7 10 m<\/span><br \/>\n<span style=\"color: #000000;\">= 96.25 m<sup>3<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. The capacity of a closed cylindrical vessel of height 1m is 15.4 liters. How many square meters of metal sheet would be needed to make it?\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Let the radius of the vessel = <em>r<\/em><br \/>\nHeight of the cylindrical vessel, h = 1 m<br \/>\nCapacity of the vessel = 15.4 litres<br \/>\n= 15.4 \/ 1000 m<sup>3<\/sup> (Since,\u00a01000 l = 1 m<sup>3<\/sup>)<\/span><br \/>\n<span style=\"color: #000000;\">= 0.0154 m<sup>3<br \/>\n<\/sup>Volume of the vessel = 0.0154 m<sup>3<br \/>\n<\/sup>\u03c0r<sup>2<\/sup>h = 0.0154 m<sup>3<br \/>\n<\/sup>r<sup>2<\/sup> = 0.0154 \/ \u03c0h <\/span><br \/>\n<span style=\"color: #000000;\">r<sup>2<\/sup> = <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{0.0154}{\\frac{22}{7}\\times&amp;space;1}\" alt=\"\\frac{0.0154}{\\frac{22}{7}\\times 1}\" width=\"50\" height=\"44\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{0.0154&amp;space;\\times&amp;space;7}{22\\times&amp;space;1}\" alt=\"\\frac{0.0154 \\times 7}{22\\times 1}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000;\">r<sup>2<\/sup> = 0.0049 m<sup>2<br \/>\n<\/sup>r = 0.07 m <\/span><br \/>\n<span style=\"color: #000000;\">TSA of the cylinder = 2\u03c0r(r + h)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 0.07m \u00d7 (0.07m + 1m)<\/span><br \/>\n<span style=\"color: #000000;\">= 0.44 m \u00d7 1.07 m<\/span><br \/>\n<span style=\"color: #000000;\">= 0.4708 m<sup>2<br \/>\n<\/sup>Therefore, 0.4708 m<sup>2<\/sup>\u00a0of the metal sheet would be required to make the cylindrical vessel.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. A lead pencil consists of a cylinder of wood with solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong><strong>For graphite:<br \/>\n<\/strong>Diameter of the graphite, d = 1 mm<\/span><br \/>\n<span style=\"color: #000000;\">Radius (r) = d\/2 = 1 mm \/2 = 0.5 mm = 0.5\/10 cm = 0.05 cm<\/span><br \/>\n<span style=\"color: #000000;\">h = 14 cm<\/span><br \/>\n<span style=\"color: #000000;\">Volume of the graphite =\u00a0\u03c0r<sup>2\u00a0<\/sup>h <\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 0.05 cm \u00d7 0.05 cm \u00d7 14 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 0.11 cm\u00b3<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>For pencil:<br \/>\n<\/strong>Diameter of the pencil, D = 7 mm<\/span><br \/>\n<span style=\"color: #000000;\">Radius (R) = 7 \/\u00a02 mm = 3.5\/10 cm = 0.35 cm<\/span><br \/>\n<span style=\"color: #000000;\">h = 14 cm<\/span><br \/>\n<span style=\"color: #000000;\">Volume of the pencil = \u03c0 R<sup>2<\/sup>\u00a0h<\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 0.35 cm \u00d7 0.35 cm \u00d7 14 cm <\/span><br \/>\n<span style=\"color: #000000;\">= 5.39 cm<sup>3<br \/>\n<\/sup>Volume of wood = Volume of the pencil \u2013 Volume of the graphite <\/span><br \/>\n<span style=\"color: #000000;\">= 5.39 cm<sup>3<\/sup>\u00a0&#8211; 0.11cm<sup>3<br \/>\n<\/sup>= 5.28 cm<sup>3<br \/>\n<\/sup>The volume of the wood is 5.28 cm<sup>3<\/sup>\u00a0and the volume of graphite is 0.11 cm<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Diameter of the bowl, d = 7 cm<\/span><br \/>\n<span style=\"color: #000000;\">Radius\u00a0of the bowl, r = 7\/2 cm = 3.5cm<\/span><br \/>\n<span style=\"color: #000000;\">Height of the bowl, h = 4 cm<\/span><br \/>\n<span style=\"color: #000000;\">The volume of soup in each bowl = \u03c0r<sup>2<\/sup>h <\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 3.5 cm \u00d7 3.5 cm \u00d7 4 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 154 cm<sup>3<br \/>\n<\/sup>The volume of soup for 1 patient = 154 cm<sup>3<br \/>\n<\/sup>Thus, the volume of soup for 250 patients <\/span><br \/>\n<span style=\"color: #000000;\">= 250 \u00d7 154 cm<sup>3<br \/>\n<\/sup>= 38500 cm<sup>3<br \/>\n<\/sup>= 38500 \/ 1000\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(Since, 1000 cm<sup>3<\/sup> = 1 <em>l<\/em>) <\/span><br \/>\n<span style=\"color: #000000;\">= 38.5 l<\/span><br \/>\n<span style=\"color: #000000;\">The hospital has to prepare 38.5 litres of soup daily to serve 250 patients.<\/span><\/p>\n<p style=\"text-align: justify;\">\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 13.6 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-4586","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 v27.3) - 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