{"id":28970,"date":"2021-07-14T18:34:55","date_gmt":"2021-07-14T13:04:55","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=28970"},"modified":"2022-03-02T10:30:06","modified_gmt":"2022-03-02T05:00:06","slug":"ncert-solutions-for-class-10-maths-chapter-13-ex-13-3","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-13-ex-13-3\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.3 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 13.3<\/h2>\n

\"NCERT<\/p>\n

Unless stated otherwise, take \u03c0 = \\(\\frac { 22 }{ 7 }\\)<\/p>\n

Question 1.
\nA metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
\nSolution:
\nGiven: radius of metallic sphere = 4.2 cm
\n\"NCERT
\n\u2234 Volume = \\(\\frac { 4 }{ 3 }\\)\u03c0(4.2)\u00b3 …. (i)
\n\u2235 Sphere is melted and recast into a cylinder of radius 6 cm and height h.
\n\u2234 Volume of the cylinder =\u03c0r\u00b2h = \u03c0(6)\u00b2 x h … (ii)
\nAccording to question,
\nVolume of the cylinder = Volume of the sphere
\n310.464 cm\u00b3 = \\(\\frac{22 \\times 36}{7}\\) h
\n310.464 cm\u00b3 = 113.142 cm\u00b3h
\nh = \\(\\frac{310.464}{113.142}\\) cm\u00b3
\nh = 2.74 cm.
\nHeight of cylinder = 2.74 cm.<\/p>\n

Question 2.
\nMetallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
\nSolution:
\nRadius of sphere (r1<\/sub>) = 6 cm.
\n\"NCERT
\nSo the volume = \\(\\frac { 4 }{ 3 }\\)\u03c0 x (r1<\/sub>)\u00b3
\n= \\(\\frac { 4 }{ 3 }\\)\u03c0 (6)\u00b3 cm.
\nRadius of sphere (r2<\/sub>) = 8 cm.
\n\"NCERT
\nSo the volume = \\(\\frac { 4 }{ 3 }\\)\u03c0 x (r2<\/sub>)\u00b2
\n= \\(\\frac { 4 }{ 3 }\\)\u03c0 (6)\u00b2 cm.
\nRadius of sphere (r3 <\/sub>) = 10 cm.
\n\"NCERT
\nSo the volume = \\(\\frac { 4 }{ 3 }\\)\u03c0 x (r3<\/sub>)\u00b2
\n= \\(\\frac { 4 }{ 3 }\\)\u03c0 (10)\u00b2 cm.
\nNow, spheres are melted and form a single
\nSphere of radii R. Volume of single sphere = \\(\\frac { 4 }{ 3 }\\)\u03c0R\u00b3
\n\"NCERT
\nRadius of single sphere = 12 cm.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nA 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
\nSolution:
\nGiven: diameter of the well = 7 m Radius = \\(\\frac { 7 }{ 2 }\\)m
\nand depth of the well = 20 m
\nVolume of the earth taken out from the well = \u03c0r\u00b2
\n\"NCERT
\n= \\(\\frac { 22 }{ 7 }\\) x \\(\\frac { 7 }{ 2 }\\) x \\(\\frac { 7 }{ 2 }\\) x 20
\n= 770 cm\u00b3.
\nTo form a platform of 22m , 14 m ,and h height.
\n\u2234 Volume = l x b x h
\nVolume = 22 x 14 x h
\nVolume of earth = Volume of platform
\n770m\u00b3 = 22 x 14 x h
\nSo, h = \\(\\frac{770 m^{3}}{308 m^{3}}\\)
\nh= 2.5 m
\nHeight of the platform = 2.5 m<\/p>\n

Question 4.
\nA well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
\nSolution:
\nDiameter of the well = 3 m
\nInner radius of well = \\(\\frac { 3 }{ 2 }\\)m
\n\"NCERT
\nVolume of the earth dug out = (\u03c0r\u00b2h) m\u00b3
\nWidth of circular ring = 4 m
\nOuter radius of well = \\(\\frac { 3 }{ 2 }\\) + 4 = \\(\\frac { 11 }{ 2 }\\) m
\nVolume of the earth = \\(\\frac { 22 }{ 1 }\\) x \\(\\frac { 3 }{ 2 }\\) x \\(\\frac { 3 }{ 2 }\\) x 14 m
\n= 99m\u00b3
\nArea of shaded region = \u03c0(R\u00b2 – r\u00b2)
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nA container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
\nSolution:
\nWe have
\nRadius of the cylinder = 6 cm
\nHeight of the cylinder = 15 cm
\n\u2234 Volume of the cylinder = \u03c0r\u00b2h
\n= \u03c0 x 6\u00b2 x 15 cm.
\n= 540\u03c0 cm\u00b3
\nRadius of the ice cream cone = 3 cm
\nHeight of the ice cream cone = 12 cm
\n\"NCERT<\/p>\n

Question 6.
\nHow many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?
\nSolution:
\nGiven: diameter of each coin = 1.75 cm \u21d2 radius = \\(\\frac { 1.75 }{ 2 }\\)cm
\nand thickness of each coin = 2 mm
\nLet the number of coins be n. So volume of coins of
\n= \\(\\frac { 22 }{ 7 }\\) x (0.875)2 x (0.2) cm x n
\n= \\(\\left(\\frac{4.4}{7} \\times 0.765\\right) n\\)
\n= \\(\\frac{(3.3687) n}{7}\\)
\n= (0.48125) n
\nCoins melted to form a cuboid of dimensions 5.5 cm, 10 cm, 3.5 cm,
\n= 5.5 x 10 cm x 3.5 cm
\n= 192.5 cm\u00b3.
\n\u2234 0.48125 n = 192.5
\n\u2234 n = \\(\\frac{192.5}{0.48125}\\) = 400
\nNo. of coins = 400<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nA cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
\nSolution:
\nGiven: radius of the cylindrical bucket = 18 cm
\nand height = 32 cm
\n\"NCERT
\nHeight of conical heap = 24 cm. Let the radius of conical heap be r1<\/sup> and slant height, l then the volume of conical heap
\n\"NCERT
\nVolume of cylinder = Volume of conical heap
\n3258.14 cm\u00b3 = 25.1428 r\u00b2 cm\u00b2.
\n\"NCERT
\nSlant height of conical heap = 12\\(\\sqrt{13}\\) cm.<\/p>\n

Question 8.
\nWater in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km\/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
\nSolution:
\nGiven: width of canal = 6m, depth = 1.5 m
\nRate of flowing water – 10 km\/h
\nVolume of the water flowing in 30 minutes = \\(\\frac { 6\u00d71.5\u00d730\u00d710 }{ 60 }\\)km\u00b3
\n= \\(\\frac { 6\u00d71.5\u00d710\u00d71000\u00d730 }{ 10\u00d760 }\\)km\u00b3 = 45000 m\u00b3
\nWe require water for standing up to height = 8 cm = \\(\\frac { 8 }{ 100 }\\) m
\nLet the required area he A
\n\u2234 Volume of water required = A(\\(\\frac { 8 }{ 100 }\\))m\u00b3
\nAccording to question. 45000 = \\(\\frac { A\u00d78 }{ 100 }\\)
\n\u21d2 \\(\\frac { 45000\u00d7100 }{ 8 }\\) = A \u21d2 A = 562500 m\u00b2
\nArea will it irrigate in 30 minutes = 562500 m\u00b2 or 56.25 hectares.<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nA farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his Held, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km\/h, in how much time will the tank be filled?
\nSolution:
\nSuppose the tank is filled in x hours. Since water is flowing at the rate of 3 km\/hr. Therefore, length of the water of the water column in x hours – 3x km = 3000x meters. Clearly, the water column forms a cylinder of radius r = \\(\\frac { 20 }{ 2 }\\) cm = 10 cm = \\(\\frac { 1 }{ 10 }\\) m and h = height (length) = 300x meters.
\n\u2234 Volume of the water that flows in the tank in x hours
\n\"NCERT
\nSince the tank is filled in x hours
\n\u2234 Volume of the water that flows in the tank in x hours = Volume of the tank
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3 Unless stated otherwise, take \u03c0 = Question 1. A metallic sphere of radius 4.2 cm …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3<\/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 13.3 - 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-13-3\/\" \/>\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 13.3 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Questions and Answers are prepared by our highly skilled subject experts. 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