{"id":26334,"date":"2021-06-25T16:21:46","date_gmt":"2021-06-25T10:51:46","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26334"},"modified":"2021-06-25T16:27:29","modified_gmt":"2021-06-25T10:57:29","slug":"ncert-solutions-for-class-9-maths-chapter-13-ex-13-6","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-6\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6"},"content":{"rendered":"

These 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>\n

NCERT 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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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>\n

Question 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\"NCERT
\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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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>\n

Question 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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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>\n

Question 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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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":{"spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-6\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"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. 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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6 Question 1. 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