{"id":2566,"date":"2020-09-09T11:58:53","date_gmt":"2020-09-09T06:28:53","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=2566"},"modified":"2021-07-02T15:03:35","modified_gmt":"2021-07-02T09:33:35","slug":"kseeb-solutions-class-9-maths-chapter-13-ex-13-6","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-13-ex-13-6\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6"},"content":{"rendered":"<h2>Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6<\/h2>\n<p>(Assume \u03c0 = \\(\\frac{22}{7}\\) unless stated otherwise)<\/p>\n<p>Question 1.<br \/>\nThe circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold ? (1000 cm<sup>3 <\/sup>= 1 l)<br \/>\nSolution:<br \/>\nThe circumference of the base of cylindrical vessel, C = 132 cm.<br \/>\nheight, h = 25 cm<br \/>\nC = 2\u03c0r = 132<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2573\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 1\" width=\"248\" height=\"89\" \/><br \/>\nr = 21 cm<br \/>\n\u2234 Volume of Cylinder, V = \u03c0r<sup>2<\/sup>h<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2574\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-1.1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 1.1\" width=\"177\" height=\"107\" \/><br \/>\n= 34650 cm<sup>3<\/sup><br \/>\n1000 cubic cm. = 1 litre<br \/>\n\u2234 34650 cm<sup>3<\/sup> = &#8230; ? &#8230;<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2575\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-1.2.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 1.2\" width=\"240\" height=\"48\" \/><\/p>\n<p>Question 2.<br \/>\nThe inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.<br \/>\nSolution:<br \/>\nLet the inner diameter of a cylindrical vessel be d<sub>1<\/sub> and outer diameter be d<sub>2<\/sub>.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2577\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-2.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 2\" width=\"272\" height=\"73\" \/><br \/>\nheight of pipe, h = 35 cm.<br \/>\n\u2234 volume of pipe, V = \\(\\pi r_{2}^{2} h-\\pi r_{1}^{2} h\\)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2578\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-2.1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 2.1\" width=\"189\" height=\"164\" \/><br \/>\n= 110 \u00d7 52<br \/>\n\u2234 V = 5720 cm<sup>3<\/sup><br \/>\nMass of pipe of 1 ccm is 0.6 gm.<br \/>\n\u2234 Mass of 5720 cm<sup>3<\/sup>. &#8230; ? &#8230;<br \/>\n= 5720 \u00d7 0.6<br \/>\n= 3432 Kg.<\/p>\n<p>Question 3.<br \/>\nA soft drink is available in two packs &#8211;<br \/>\n(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and<br \/>\n(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.<br \/>\nWhich container has greater capacity and by how much ?<br \/>\nSolution:<br \/>\nLength of rectangular tin, l = 5 cm,<br \/>\nbreadth, b = 4 cm,<br \/>\nheight, h = 15 cm.<br \/>\n\u2234 Volume of rectangular tin, V = l \u00d7 b \u00d7 h<br \/>\nV = 5 \u00d7 4 \u00d7 15<br \/>\n= 300 cm<sup>3<\/sup><br \/>\nDiameter of plastic cylinder, d = 7 cm,<br \/>\n\u2234 r = \\(\\frac{7}{2}\\)<br \/>\nVolume of cylinder, v = \\(\\pi \\mathrm{r}^{2} \\mathrm{h}\\)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2579\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-3.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 3\" width=\"137\" height=\"52\" \/><br \/>\n= 11 \u00d7 7 \u00d7 5<br \/>\n= 385 cm<sup>3<\/sup>.<br \/>\n\u2234 Plastic cylinder\u2019s volume is great.<br \/>\n= Volume of cylindrical tin &#8211; Volume of rectangular tin.<br \/>\n= 385 &#8211; 300<br \/>\n= 85 cm<sup>3<\/sup>.<br \/>\n\u2234 Volume of plastic tin is greater than rectangular tin by 85 cm<sup>3<\/sup>.<\/p>\n<p>Question 4.<br \/>\nIf the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find<br \/>\n(i) radius of its base,<br \/>\n(ii) its volume. (Use \u03c0 = 3.14)<br \/>\nSolution:<br \/>\nCurved surface area of cylinder = 94.2 cm<sup>2<\/sup><br \/>\nheight, h = 5 cm.<br \/>\nr = ?<br \/>\nV = ?<br \/>\n(i) Curved Surface area of cylinder, L.S.A.<br \/>\nA= 2\u03c0rh<br \/>\n94.2 = 2 \u00d7 3.14 \u00d7 5 \u00d7 r<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2582\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-4.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 4\" width=\"180\" height=\"44\" \/><br \/>\n= 3.01 cm.<br \/>\n= 3 cm.<br \/>\n(ii) Volume of cylinder, V = \u03c0r<sup>2<\/sup>h<br \/>\n= 3.14 \u00d7 (3)2 \u00d7 5<br \/>\n= 3.14 \u00d7 9 \u00d7 5<br \/>\n= 141.3 cm<sup>3<\/sup>.<\/p>\n<p>Question 5.<br \/>\nIt costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs. 20 per m<sup>2<\/sup>, find<br \/>\n(i) inner curved surface area of the vessel.<br \/>\n(ii) radius of the base,<br \/>\n(iii) capacity of the vessel.<br \/>\nSolution:<br \/>\n(i) Cost of painting for cylindrical vessel = Rs. 2200<br \/>\nCost of painting is at the rate of Rs. 20 per m<sup>2<\/sup>.<br \/>\n\u2234 Curved Surface area of vessel = \\(\\frac{2200}{20}\\)<br \/>\n= 110 m<sup>2<\/sup><\/p>\n<p>(ii) Depth of vessel, h = 10 m<br \/>\nr = ?<br \/>\nCurved surface area of vessel, V = 2\u03c0rh<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2583\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-5.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 5\" width=\"216\" height=\"130\" \/><br \/>\n(iii) Capacity of the Vessel = Volume of Vessel<br \/>\nV = \u03c0r<sup>2<\/sup>h<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2584\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-5.1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 5.1\" width=\"149\" height=\"141\" \/><br \/>\nV = 96.25 m<sup>3<\/sup>.<\/p>\n<p>Question 6.<br \/>\nThe capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it ?<br \/>\nSolution:<br \/>\nVolume of a cylindrical vessel = 15.4 litres<br \/>\n1000 c.cm. = 1 litre.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2585\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-6.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 6\" width=\"171\" height=\"48\" \/><br \/>\nheight, h = 1 m,<br \/>\nradius, r = ?<br \/>\nCurved surface area =?<br \/>\nVolume of cylinder, V = \u03c0r<sup>2<\/sup>h<br \/>\n0. 0154 = \\(\\frac{22}{7}\\) \u00d7 r<sup>2<\/sup> \u00d7 1<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2587\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-6.1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 6.1\" width=\"185\" height=\"135\" \/><br \/>\nr = 0.7 m<br \/>\nNow, curved surface area of cylinder vessel,<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2589\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-6.2.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 6.2\" width=\"221\" height=\"112\" \/><br \/>\n= 0.44 \u00d7 1.07<br \/>\nA = 0.4708 m<sup>2<\/sup><\/p>\n<p>Question 7.<br \/>\nA lead pencil consists of a cylinder of wood with a 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.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2590\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-7.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 7\" width=\"131\" height=\"175\" \/><br \/>\nSolution:<br \/>\nDiameter of pencil, d = 7 mm = 0.7 m.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2591\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-7.1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 7.1\" width=\"238\" height=\"48\" \/><br \/>\nlength of pencil, h = 14 cm<br \/>\ndiameter of pencil, d = 1 mm = 0.1 cm<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2599\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-7.2.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 7.2\" width=\"211\" height=\"47\" \/><br \/>\nVolume of wood in pencil, V<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-medium wp-image-2604\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-7.3-279x300.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 7.3\" width=\"279\" height=\"300\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-7.3-279x300.png 279w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-7.3.png 296w\" sizes=\"(max-width: 279px) 100vw, 279px\" \/><\/p>\n<p>Question 8.<br \/>\nA 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?<br \/>\nSolution:<br \/>\nDiameter of pencil, d = 7 cm<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2606\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.6-Q-8.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 8\" width=\"163\" height=\"44\" \/><br \/>\nheight of soup, h = 4 cm<br \/>\nVolume of soup given for 1 patient is 154 cm<sup>3<\/sup><br \/>\nVolume of soup given for 250 patients .. ? &#8230;.<br \/>\n= 154 \u00d7 250 = 38500 cm<sup>3<\/sup><br \/>\nFor 1000 cm<sup>3<\/sup>, 1 litre<br \/>\nFor 38500 cm<sup>3<\/sup> &#8230;.. ? &#8230;<br \/>\n= \\(\\frac{38500}{1000}\\)<br \/>\n= 38.5 litres soup.<\/p>\n<h5><a href=\"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths\/\">KSEEB Solutions for Class 9 Maths<\/a><\/h5>\n","protected":false},"excerpt":{"rendered":"<p>Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 (Assume \u03c0 = unless stated otherwise) Question 1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold ? (1000 cm3 = 1 l) Solution: The &hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2566"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2566"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2566\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2566"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2566"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2566"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}