{"id":26221,"date":"2021-06-25T12:47:49","date_gmt":"2021-06-25T07:17:49","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=26221"},"modified":"2022-03-02T10:47:34","modified_gmt":"2022-03-02T05:17:34","slug":"ncert-solutions-for-class-9-maths-chapter-13-ex-13-4","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-4\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4"},"content":{"rendered":"

These NCERT Solutions for Class 9 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 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4<\/h2>\n

Question 1.
\nFind the surface area of a sphere of radius.
\n(i) 10.5 cm
\n(ii) 5.6 cm
\n(iii) 14 cm
\nSolution:
\n(i) We have given that
\n\"NCERT
\nRadius of sphere = 10.5 cm
\nSurface area of sphere = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac {22}{7}\\) \u00d7 5.6 \u00d7 5.6
\n= 394.24 cm2<\/sup><\/p>\n

(ii) We have given that
\n\"NCERT
\nRadius of sphere = 5.6 cm
\nSurface area of sphere = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac {22}{7}\\) \u00d7 5.6 \u00d7 5.6
\n= 394.24 cm2<\/sup><\/p>\n

(iii) We have given that
\n\"NCERT
\nRadius of sphere = 14 cm
\nSurface area of sphere = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac {22}{7}\\) \u00d7 14 \u00d7 14
\n= 2464 cm2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 2.
\nFind the surface area of a sphere of diameter.
\n(i) 14 cm
\n(ii) 21 cm
\n(iii) 3.5 cm
\nSolution:
\n(i) We have given that
\n\"NCERT
\nDiameter of sphere = 14 cm
\nRadius of sphere = 7 cm
\nSurface area of sphere = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac {22}{7}\\) \u00d7 7 \u00d7 7
\n= 616 cm2<\/sup><\/p>\n

(ii) We have given that
\n\"NCERT
\nDiameter of sphere = 21 cm
\nRadius of sphere = 10.5 cm
\nTherefore,
\nSurface area of sphere = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac {22}{7}\\) \u00d7 10.5 \u00d7 10.5
\n= 1386 cm2<\/sup><\/p>\n

(iii) We have given that
\n\"NCERT
\nDiameter of sphere = 3.5 m
\nRadius os sphere = 1.75 m
\nTherefore,
\nSurface area of sphere = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac {22}{7}\\) \u00d7 1.75 \u00d7 1.75
\n= 38.5 m2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 3.
\nFind the total surface area of a hemisphere of radius 10 cm.
\nSolution:
\nWe have given that
\nRadius of hemisphere = 10 cm
\n\"NCERT
\nTherefore, Total surface area of hemisphere = 3\u03c0r2<\/sup>
\n= 3 \u00d7 3.14 \u00d7 10 \u00d7 10 (\u2234 \u03c0 = 3.14)
\n= 942 cm2<\/sup><\/p>\n

Question 4.
\nThe radius of the spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
\nSolution:
\nIn first case,
\n\"NCERT
\nRadius of spherical balloon = 7 cm
\nSurface area of spherical balloon = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac {22}{7}\\) \u00d7 7 \u00d7 7
\n= 616 cm2<\/sup>
\nIn second case,
\n\"NCERT
\nradius of spherical balloon = 14 cm
\nsurface area of spherical balloon = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac {22}{7}\\) \u00d7 14 \u00d7 14
\n= 2464 cm2<\/sup>
\nTherefore the ratio of the surface area of balloons in two cases is 616 : 2464 = 1 : 4<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nA hemispherical bowl made of brass has an inner diameter of 10.2. Find the cost of tin plating it on the inside at the rate of Rs. 16 per 100 cm2<\/sup>.
\nSolution:
\nWe have given that
\nDiameter of hemispherical bowl = 10.5 cm
\nThe radius of hemispherical bowl = 5.25 cm
\n\"NCERT
\nCurved surface area of hemispherical bowl = 2\u03c0r2<\/sup>
\n= 2 \u00d7 \\(\\frac {22}{7}\\) \u00d7 5.25 \u00d7 5.25
\n= 173.25 cm2<\/sup>
\nRate of tin plating = Rs. 16 per 100 cm2<\/sup>
\nTotal cost of tin plating the hemispherical bowl is = 173.25 \u00d7 \\(\\frac{16}{100}\\) = Rs. 27.72<\/p>\n

Question 6.
\nFind the radius of a sphere whose surface area is 154 cm2<\/sup>.
\nSolution:
\nWe have given that
\n\"NCERT
\nSurface area of sphere = 154 cm2<\/sup>
\nBut, Surface area of sphere = 2\u03c0r2<\/sup>
\n\u21d2 154 = 4 \u00d7 \\(\\frac {22}{7}\\) \u00d7 r2<\/sup>
\n\u21d2 r2<\/sup> = \\(\\frac{154 \\times 7}{22 \\times 4}\\) = 12.25
\n\u21d2 r = \u221a12.25 = 3.5 cm
\nTherefore, the radius of the sphere whose surface area is 154 cm2<\/sup> is 3.5 cm.<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nThe diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.
\nSolution:
\nLet the diameter of earth = r cm
\nDiameter of moon = \\(\\frac{r}{4}\\) cm
\nAgain, radius earth = \\(\\frac{1}{2}\\) cm
\nand radius of moon = \\(\\frac{r}{8}\\) cm
\nSurface area of earth = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \u03c0 \u00d7 \\(\\left(\\frac{r}{2}\\right)^{2}\\)
\n= \u03c0r2<\/sup>
\nand surface area of moon = 4\u03c0 \\(\\left(\\frac{r}{8}\\right)^{2}\\) = \\(\\frac{\\pi r^{2}}{16}\\)
\nRatio of their surface areas = \\(\\frac{\\pi r^{2}}{\\frac{16}{\\pi r^{2}}}\\) = \\(\\frac{1}{16}\\) = 1 : 16
\nRatio of moon and earth surface area = 1 : 16<\/p>\n

Question 8.
\nA hemispherical bowl is made of steel, 0.25 cm thick. The inner, radius of the bowl is 5 cm. Find the outer curves surface area of the bowl.
\nSolution:
\n\"NCERT
\nWe have given tha
\nInner radius of hemispherical bowl = 5 cm
\nand thickness of steel = 0.25 cm.
\nOuter curved surface area of hemispherical bowl = 5.25 cm
\nTherefore, Outer curved surface area of hemisperical bowl = 2\u03c0r2<\/sup>
\n= 2 \u00d7 \\(\\frac {22}{7}\\) \u00d7 5.25 \u00d7 5.25
\n= 173.25 cm2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 9.
\nA right circular cylinder just encloses a sphere of radius r (see fig. 13.22) Find.
\n(i) surface area of the sphere.
\n(ii) Curved surface area of the cylinder
\n(iii) ratio of areas obtained in (i) and (ii)
\n\"NCERT
\nSolution:
\n(i) We have given that
\n\"NCERT
\nRadius of sphere = r
\nSurface area of sphere = 4\u03c0r2<\/sup>
\n(ii) Surface area of cylinder = 4\u03c0rh
\n= 2 \u00d7 \u03c0 \u00d7 r \u00d7 2r (\u2235 h = 2r)
\n= 4\u03c0r2<\/sup>
\n(iii) Ratio of the areas obtaining in (i) and (ii) is
\n\\(\\frac{4 \\pi r^{2}}{4 \\pi r^{2}}\\) = 1 : 1<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 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 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4 Question 1. Find the surface area of a sphere of radius. (i) 10.5 cm (ii) …<\/p>\n

NCERT Solutions for Class 9 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":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 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-9-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 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 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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