NCERT Solutions for Class 9 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.6 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6<\/h2>\n Question 1. \nThe circumference of the base of a cylinderical vessel is 132 cm and its height is 25 cm. How many liters of water can it hold? (1000 cm3<\/sup> = 1L) \nSolution: \nWe have given that \n \nCircumference of base of cylinder = 132 cm \n2\u03c0r = 132 \n\u21d2 r = \\(\\frac{132 \\times 7}{22 \\times 2}\\) = 21 cm \nand height of cylinderical vessel = 25 cm \nVolume of cylindrical vessel = \u03c0r2<\/sup>h \n= \\(\\frac {22}{7}\\) \u00d7 21 \u00d7 21 \u00d7 25 \n= 34650 cm3<\/sup> \nNow, we know that 1000 cm3<\/sup> = 1L \n34650 cm3<\/sup> = 34.65 L \nTherefore, the cylindrical vessel can hold 34.65 liters of water.<\/p>\n <\/p>\n
Question 2. \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<\/sup> is wood has a mass of 0.6 g. \nSolution: \n \nVolume of required wood = volume of outer cylinder – volume of inner cylinder. \n= \u03c0(14)2<\/sup> \u00d7 35 – \u03c0(12)2<\/sup> \u00d7 35 \n= \\(\\frac {22}{7}\\) \u00d7 35((14)2<\/sup> – (12)2<\/sup>) \n= \\(\\frac {22}{7}\\) \u00d7 35 \u00d7 (196 – 144) \n= \\(\\frac {22}{7}\\) \u00d7 35 \u00d7 52 \n= 5720 cm3<\/sup> \nNow, we have given that, mass of 1 cm3<\/sup> of wood = 0.6 \nTherefore, mass of 5720 cm3<\/sup> of wood = 0.6 \u00d7 5720 g. \n= 3432g \n= 3.432 kg.<\/p>\nQuestion 3. \nA soft drink is available in two packs \u2014 (i) a tin can with a rectangular base of length 5 cm and wide 4 cm having a height of 15 cm and (ii) a plastic cylinder with a circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much. \nSolution: \nIn the first case, \nThe tin can with a rectangular base of length 5 cm and wide 4 cm having a height of 15 cm. \n\u2234 The volume of this can = l \u00d7 b \u00d7 h \n= 15 \u00d7 4 \u00d7 5 \n= 300 cm3<\/sup> …….(i) \n \nIn the second case, \nThe circular plastic cylinder has a base diameter is 7 cm and the height of the cylinder is 10 cm. \n \n\u2234 Volume of plastic cylinder = \u03c0r2<\/sup>h \n= \\(\\frac {22}{7}\\) \u00d7 3.5 \u00d7 3.5 \u00d7 10 \n= 385 cm3<\/sup> ……(ii) \nFrom (i) and (ii), it is clear that \nThe volume of the circular cylinder has a greater capacity of 85 cm3<\/sup>.<\/p>\n <\/p>\n
Question 4. \nIf the lateral surface of a cylinder is 94.2 cm3<\/sup> and its height is 5 cm, then find \n(i) radius of its base \n(ii) volume of the cylinder. \nSolution: \n(i) We have given that lateral surface of cylinder = 94.2 cm3<\/sup> and height is 5 cm. \n \nNow, we know that, lateral surface area of cylinder = 2\u03c0rh \n94.2 = 2 \u00d7 3.14 \u00d7 r \u00d7 5 \nr = \\(\\frac{94.2}{2 \\times 3.14 \\times 5}\\) = 3 cm \n(ii) We know that \nVolume of cylinder = 2\u03c0rh \n= 3.14 \u00d7 3 \u00d7 3 \u00d7 5 \n= 141.3 cm3<\/sup>.<\/p>\nQuestion 5. \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 m2<\/sup>, Find. \n(i) Inner curved surface area of the vessel. \n(ii) radius of the base. \n(iii) capacity of the vessel. \nSolution: \n(i) We have given that \n \nRate of painting = Rs. 20 per m2<\/sup> \nand total cost of curved surface area of cylindrical vessel = Rs. 2200 \nIt means that inner curved surface area of cylindrical vessel = \\(\\frac{2200}{20}\\) = 110 m2<\/sup><\/p>\n(ii) we know that curved surface ara of cylinder = 2\u03c0rh \n\u21d2 110 = 2 \u00d7 \\(\\frac {22}{7}\\) \u00d7 r \u00d7 10 \n\u21d2 r = \\(\\frac{110 \\times 7}{2 \\times 22 \\times 10}\\) = 1.75 m<\/p>\n
(iii) We know that volume of cylinder = \u03c0r2<\/sup>h \n= \\(\\frac {22}{7}\\) \u00d7 1.75 \u00d7 1.75 \u00d7 10 \n= 96.25 m3<\/sup> \nTherefore, Capacity of the vessel = 96.25 m3<\/sup> = 96.25 kl \n(1 m3<\/sup> = 1000 l =1 kl)<\/p>\n <\/p>\n
Question 6. \nThe capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of the metal sheet would be needed to make it? \nSolution: \nWe have given that \nCapacity of closed cylindrical vessel = 15.41 = \\(\\frac{15.4}{1000}\\) m3<\/sup> \n \nand height of the cylindrical vessel = 1 m. \nWe know that, \nvolume of cylinder = \u03c0r2<\/sup>h \n\u21d2 \\(\\frac{15.4}{1000}\\) = \\(\\frac {22}{7}\\) \u00d7 r2<\/sup> \u00d7 1 \n\u21d2 r2<\/sup> = 0.0049 \n\u21d2 r = 0.07m \nNow, Total surface area of cylindrical vessel = 2\u03c0r(h + r) \n= 2 \u00d7 \\(\\frac {22}{7}\\) \u00d7 0.07 (1 + 0.07) \n= 0.4708 m2<\/sup> \nTherefore, a metal sheet would be needed to make the required cylinder is 0.4708 m2<\/sup>.<\/p>\nQuestion 7. \nA lead pencil consists of a cylinder of wood will 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. \nSolution: \nWe have given that \nDiameter of pencil = 7 mm \nRadius of pencil = 3.5 mm \nand height of pencil = 14 cm = 140 mm \n \nVolume of pencil = \u03c0r2<\/sup>h \n= \\(\\frac {22}{7}\\) \u00d7 3.5 \u00d7 3.5 \u00d7 140 \n= 5390 mm3<\/sup> \nAgian, we have \nDiamter of lead = 1 mm \nRadius of lead = 0.5 mm \nand height of lead = 14 cm = 140 mm \nVolume of lead = \u03c0r2<\/sup>h \n= \\(\\frac {22}{7}\\) \u00d7 0.5 \u00d7 0.5 \u00d7 140 \n= 110 mm3<\/sup> \nVolume of graphite = 110 mm3<\/sup> = 0.11 cm3<\/sup> \nand volume of wood = 5390 – 110 \n= 5280 mm3<\/sup> \n= 5.28 cm3<\/sup><\/p>\n <\/p>\n
Question 8. \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? \nSolution: \n \nWe have given that \nDiameter of pencil = 7 mm \nRadius of cylindrical bowl = 3.5 cm \nand height of soup to serve = 4 cm. \nVolume of soup to serve one patient = \u03c0r2<\/sup>h \n= \\(\\frac {22}{7}\\) \u00d7 3.5 \u00d7 3.5 \u00d7 4 \n= 154 cm3<\/sup> \nTherefore, volume of soup to serve 250 patients are = 250 \u00d7 154 \n= 38500 cm3<\/sup>. \n= 38.5 litre. \nHence, Hospital has to prepare daily 38.5 litres of soup to serve 250 patients.<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6 Question 1. The circumference of the base of a cylinderical vessel is 132 cm and …<\/p>\n
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n