{"id":28905,"date":"2021-07-13T17:50:50","date_gmt":"2021-07-13T12:20:50","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=28905"},"modified":"2022-03-02T10:30:07","modified_gmt":"2022-03-02T05:00:07","slug":"ncert-solutions-for-class-10-maths-chapter-13-ex-13-4","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-13-ex-13-4\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nA drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
\nSolution:
\nDiameters of two circular ends are 4 cm and 2 cm
\n\u2234 Radius are 2 cm and 1 cm.
\nVolume of the furstum of the cone
\n\"NCERT<\/p>\n

Question 2.
\nThe slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
\nSolution:
\nLet r1<\/sub> and r1<\/sub> be the radii of the circular bases of the frustum, l be the slant height and h be its height.
\nWe have, l = 4 cm, 2\u03c0r1<\/sub> = 18 and 2\u03c0r2<\/sub> = 6
\n\u21d2 l = 4 cm, r1<\/sub> = \u2014 and r2<\/sub> = \u2014
\nCurved surface area = l
\n= \u03c0\\(\\left(\\frac{9}{\\pi}+\\frac{3}{\\pi}\\right)\\) x 4 cm\u00b2
\n= \u03c0\\(\\left(\\frac{9+3}{\\pi}\\right)\\) x 4 cm\u00b2
\n= (12 x 4) cm\u00b2
\n= 48 cm\u00b2.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nA fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
\n\"NCERT
\nSolution:
\nRadius of open side (r1<\/sub>) = 10 cm
\nRadius of upper base (r2<\/sub>) = 4 cm
\nSlant height (l) = 15 cm
\nArea of material used for making = Curved surface area of frustum of cone + area of closed side
\n\"NCERT
\nArea of material used for cap = 710\\(\\frac { 2 }{ 3 }\\) cm\u00b2.<\/p>\n

Question 4.
\nA container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of \u20b9 20 per litre. Also find the cost of metal sheet used to make the container, if it costs \u20b9 8 per 100 cm\u00b2. (Take \u03c0 = 3.14)
\nSolution:
\nRadius of the lower end (r1<\/sub>) = 8 cm
\nRadius of the upper end (r2<\/sub>) = 20 cm
\nHeight of the frustum (h) = 16 cm
\n\"NCERT
\nCost of milk at the rate of \u20b9 20 per litre = \u20b9(20 x 10.45) = \u20b9 209
\nNow, Total surface area of the frustum = \u03c0(r1<\/sub>) + r2<\/sub>) l + \u03c0r2<\/sub>\u00b2 [\u2235 Top is open]
\n= [3.14 (20 + 8) x 20 + 3.14 x 8\u00b2] cm\u00b2
\n= 3.14 x (560 + 64) cm\u00b2
\n= 3.14 x 624 cm\u00b2
\n= 1959.36 cm\u00b2
\nCost of metal used = \u20b9\\(\\left(\\frac{1959.36 \\times 8}{100}\\right)\\) = \u20b9 156.75<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nA metallic right circular cone 20 cm high and whose vertical angle is 60\u00b0 is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \\(\\frac { 1 }{ 16 }\\) cm, find the length of the wire.
\nSolution:
\nHeight of right circular cone = 20 cm
\nIn \u2206ABP,
\n\"NCERT
\nDiameter of wire = \\(\\frac { 1 }{ 16 }\\) cm
\nRadius of wire = \\(\\frac { 1 }{ 32 }\\) cm
\nVolume of wire = \u03c0r\u00b2h (h = 1)
\n= \u03c0(\\(\\frac { 1 }{ 32 }\\))\u00b2 x l … (2)
\nFrom (1) and (2)
\n\u03c0(\\(\\frac { 1 }{ 32 }\\))\u00b2 x l = \\(\\frac { 7000\u03c0 }{ 9 }\\)
\nl = \\(\\frac{7000 \\times 32 \\times 32}{9}\\)
\n= 79644.44 cm = 7964.4 m
\n\u2234 Length of the wire = 7964.4 m<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4 Question 1. A drinking glass is in the shape of a frustum of a cone …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4<\/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":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-13-ex-13-4\/\" \/>\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 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts. 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