{"id":28244,"date":"2021-07-03T18:42:15","date_gmt":"2021-07-03T13:12:15","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=28244"},"modified":"2022-03-02T10:30:23","modified_gmt":"2022-03-02T05:00:23","slug":"ncert-solutions-for-class-8-maths-chapter-11-ex-11-3","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-8-maths-chapter-11-ex-11-3\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 11 Mensuration Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.3<\/h2>\n

Question 1.
\nThere are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
\n\"NCERT
\n\"NCERT
\nAnswer:
\n(a) Here, l = 60 cm, b = 40 cm; h = 50 cm Total surface area of the cuboid
\n= 2 (lb + bh + hl)
\n= 2 [(60 \u00d7 40) + (40 \u00d7 50) + (50 \u00d7 60)] cm2<\/sup>
\n= 2 [2400 + 2000 + 3000] cm2<\/sup>
\n= 2 (7400) cm2<\/sup> = 14800 cm2<\/sup><\/p>\n

(b) Here, l = 50 cm, b = 50 cm, h = 50 cm
\nTotal surface area of the cuboid
\n= 2 (lb + bh + lh)
\n= 2 [(50 \u00d7 50)+ (50 \u00d7 50)+ (50 \u00d7 50)] cm2<\/sup>
\n= 2 [2500 + 2500 + 2500] cm2<\/sup>
\n= 2 \u00d7 7500 cm2<\/sup> = 15000 cm2<\/sup>
\nSince, the total surface area of the second (b) is greater:
\n\u2234 Cuboid (a) will required lesser material.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nA suitcase with measures 80 cm \u00d7 48 cm \u00d7 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
\nAnswer:
\nHere, length = 80 cm, breadth = 48 cm, height = 24 cm
\nTotal surface are a of asuitcase = 2(lb+bh+lh)
\n= 2[80 \u00d7 48 + 48 \u00d7 24 + 24 \u00d7 80]
\n= 2 [3840 + 1152 + 1920]cm2<\/sup>
\n= 2[6912] cm2<\/sup> = 13824 cm2<\/sup>
\nRequired tarpaulin for 100 suitcases = 13824 x 100 cm2<\/sup>
\nWidth of the tarpaulin = 96 cm
\nLength, tarpaulin required = \\(\\frac{13824 \\times 100}{96}\\)cm
\n[1 m = 100 cm]
\n= \\(\\frac{13824 \\times 100}{96 \\times 100}\\)m = \\(\\frac{13824}{96}\\)m = 144 m
\nThe required length of tarpaulin for 100 suitcases = 144 m<\/p>\n

Question 3.
\nFind the side of a cube whose surface area is 600 cm2
\nAnswer:
\nLet the side of a cube be \u2018x\u2019 cm
\nTotal surface area of the cube = 600 cm2<\/sup>
\n6 \u00d7 x2<\/sup> = 600
\nx2<\/sup> = \\(\\frac{600}{6}\\)
\nx2<\/sup> = 100
\nx2<\/sup> = 102<\/sup>
\nx = 10
\nThe required side of the cube =10 cm.<\/p>\n

Question 4.
\nRukshar painted the outside of the cabinet of measure 1m \u00d7 2m \u00d7 1.5m. How much surface area did she cover if she painted all except the bottom of the cabinet.
\n\"NCERT
\nAnswer:
\nTotal surface area of the cabinet = 2 \u00d7 [lb + bh + lh] sq m
\nArea not to be painted = Bottom of the cabinet = lb
\nArea to be painted = T.S.A – lb
\n= 2 (lb + bh + lh) – lb
\n= 2[1 \u00d7 2 + 2 \u00d7 1.5 + 1.5 \u00d7 1]-(1 \u00d7 2)m2<\/sup>
\n= 2(2 + 3 + 1.5) -2 m2<\/sup>
\n= 2 \u00d7 6.5 – 2 m2<\/sup>
\n= 13 – 2 m2<\/sup>
\n= 11 m2<\/sup>
\nArea to be painted = 11 m2<\/sup>.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nDaniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint, 100 m2<\/sup> of area is painted. How many cans of paint will she need to paint the room?
\nAnswer:
\nHere l = 15 m, b = 10 m and h = 7 m.
\nArea to the painted = Area of four walls + Area of ceiling.
\n= 2 (bh + hl) + lb
\n2 [10 \u00d7 7 +7 \u00d7 15] + 15 \u00d7 10 m2<\/sup>
\n= 2 [70 + 105] + 150 m2<\/sup>
\n= 2(175) + 150 m2<\/sup>
\n= 350 + 150 m2<\/sup> = 500 m2<\/sup>
\n= Number of cans needed = \\(\\frac{\\text { Area to be painted }}{\\text { Area painted by } 1 \\text { can }}\\) = \\(\\frac{500}{100}\\) = 5
\nNumber of cans needed = 5.<\/p>\n

Question 6.
\nDescribe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?
\n\"NCERT
\nAnswer:
\nSimilarity: Both have the same height
\nDifference: Cylinder has curved and circular surfaces.
\nbut in cube, all faces are identical squares
\nLateral surface area of the cylinder
\n= 2\u03c0rh = 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{7}{2}\\) \u00d7 7 cm2<\/sup>
\n= 7 \u00d7 22 cm2<\/sup> =154 cm2<\/sup>
\nLateral surface area of the cube = 4l2 sq.m
\n= 4 \u00d7 7 \u00d7 7 cm2<\/sup> = 196 cm2<\/sup><\/p>\n

Question 7.
\nA closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
\nAnswer:
\nHere r = 7m, h = 3m
\nTotal surface area of the cylinder = 2 \u03c0 r (r + h)sq. units
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 7(7 + 3) = 44 \u00d7 10 m2<\/sup> = 440 m2<\/sup>
\nMetal required for the tank = 440 m2<\/sup>.<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nThe lateral surface area of a hollow cylinder is 4224 cm2<\/sup>. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
\nAnswer:
\nLateral surface area of the cylinder = 4224 cm2<\/sup>
\nLet the length of the rectangular sheet be l cm
\nWidth of the sheet = 33 cm.
\nArea of the rectangular sheet = L.S.A of the cylinder
\nl \u00d7 33 = 4224
\nl = \\(\\frac{4224}{33}\\) cm = 128 cm
\nPerimeter of the rectangular sheet = 2 (l + b)
\n= 2 (128 + 33) cm = 2 \u00d7 161 cm = 322 cm<\/p>\n

Question 9.
\nA road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1m.
\n\"NCERT
\nAnswer:
\nRadius of the roller = \\(\\frac{84}{2}\\) = 42cm
\nLength of the roller = 1 m = 100 cm
\nLateral surface area = 2 \u03c0 rh sq unit
\n[road roller is a cylinder]
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 42 \u00d7 100 cm2<\/sup>
\n= 44 \u00d7 6 \u00d7 100 cm2<\/sup> = 26400 cm2<\/sup>
\nArea levelled by 750 revolutions
\n= 750 \u00d7 26400 cm2<\/sup>
\n=\\(\\frac{750 \\times 26400}{100 \\times 100}\\)m2<\/sup> = \\(\\frac{75 \\times 264}{10}\\) = 1980 m2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 10.
\nA company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label.
\n\"NCERT
\nAnswer:
\nRadius of the label = \\(\\frac{14}{2}\\) = 7 cm
\nHeight of the label = [20 – (2 + 2)] cm
\n= (20 – 4) cm = 16 cm
\nArea of the label = Lateral surface area of the cylinder.
\n= 2\u03c0rh = 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 16
\n= 44 \u00d7 16 cm2<\/sup> = 704 cm2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.3 Question 1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3<\/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":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 - 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-8-maths-chapter-11-ex-11-3\/\" \/>\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 8 Maths Chapter 11 Mensuration Ex 11.3 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts. 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