NCERT Solutions for Class 9 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.8 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.8<\/h2>\n Question 1. \nFind the volume of a sphere whose radius is \n(i) 7 cm \n(ii) 0.63 m \nSolution: \n(i) We have radius of sphere = 7 cm \n \n\u2234 Volume of sphere = \\(\\frac{4}{3} \\pi r^{3}\\) \n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 7 \\times 7 \\times 7\\) \n= 1437 \\(\\frac {1}{3}\\) cm3<\/sup><\/p>\n(ii) We have \nRadius of sphere = 0.63 m \n \n\u2234 Volume of sphere = \\(\\frac{4}{3} \\pi r^{3}\\) \n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 0.63 \\times 0.63 \\times 0.63\\) \n= 1.0478 m3<\/sup> \n= 1.05 m3<\/sup> (approx)<\/p>\n <\/p>\n
Question 2. \nFind the amount of water displaced by a solid spherical ball of diameter. \n(i) 28 cm \n(ii) 0.21 m \nSolution: \n(i) We have given that \n \nDiameter of spherical ball = 28 cm \n\u2234 Radius of spherical ball = \\(\\frac {28}{2}\\) = 14 cm \nVolume of spherical ball = \\(\\frac{4}{3} \\pi r^{3}\\) \n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 14 \\times 14 \\times 14\\) \n= 11498\\(\\frac {2}{3}\\) cm3<\/sup> \nTherefore, amount of water displaced by solid sphere is 11498\\(\\frac {2}{3}\\) cm3<\/sup>.<\/p>\n(ii) Diameter of spherical ball = 0.21 m \n \n\u2234 Radius of spherical a ball = \\(\\frac {0.21}{2}\\) = 0.105 m \nSo, volume of spherical ball = \\(\\frac{4}{3} \\pi r^{3}\\) \n= \\(\\frac{4}{3} \\times \\frac{22}{7}\\) \u00d7 0.105 \u00d7 0.105 \u00d7 0.105 \n= 0.004851 m3<\/sup> \nTherefore, the amount of water displaced by the solid sphere is 0.004851 m3<\/sup><\/p>\n <\/p>\n
Question 3. \nThe diameter of a metallic ball is 4.2. What is the mass of the ball, if the density of the metal is 8.9 g per cm3<\/sup>? \nSolution: \n \nWe have given that diameter of metallic ball = 4.2 cm \nRadius of metallic ball = \\(\\frac {4.2}{2}\\) = 2.1 cm \nvolume of metallic ball = \\(\\frac{4}{3} \\pi r^{3}\\) \n= \\(\\frac{4}{3} \\times \\frac{22}{7}\\) \u00d7 2.1 \u00d7 2.1 \u00d7 2.1 \n= 38.808 cm3<\/sup> \nThe density of the metal is 8.9 g per cm3<\/sup>. \nMass of the metallic ball = 38.808 \u00d7 8.9 = 345.39 g (approx)<\/p>\nQuestion 4. \nThe diameter of the moon is approximately one-fourth the diameter of the earth. What fraction of the volume of the earth is the volume of the moon. \nSolution: \nLet the diameter of earth = x \n \n\u2234 Radius of earth = \\(\\frac{x}{2}\\) \nVolume of earth = \\(\\frac{4}{3} \\pi r^{3}\\) \n= \\(\\frac{4}{3} \\pi \\times \\frac{x}{2} \\times \\frac{x}{2} \\times \\frac{x}{2}\\) \n= \\(\\frac{\\pi x^{3}}{6}\\) \nAgain, the diameter of the moon is one-fourth of the diameter of the earth. \n \nDiameter of moon = \\(\\frac{x}{4}\\) \nRadius of moon = \\(\\frac{x}{8}\\) \n\u2234 Volume of moon = \\(\\frac{4}{3} \\pi r^{3}\\) \n= \\(\\frac{4}{3} \\pi \\times \\frac{x}{8} \\times \\frac{x}{8} \\times \\frac{x}{8}\\) \n= \\(\\frac{x^{3} \\pi}{384}\\) \nTherefore, fraction of the volume of the moon and the volume of earth is = \\(\\frac{\\frac{\\pi x^{3}}{384}}{\\frac{\\pi x^{3}}{6}}=\\frac{6}{384}=\\frac{1}{64}\\)<\/p>\n
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Question 5. \nHow many liters of milk can a hemispherical bowl of diameter 10.5 cm hold. \nSolution: \nWe have a diameter of hemispherical bowl = 10.5 cm \n \n\u2234 Radius of hemispherical = \\(\\frac{10.5}{2}\\) = 5.25 cm \n\u2234 Volume of hemisphere = \\(\\frac{2}{3} \\pi r^{3}\\) \n= \\(\\frac{2}{3} \\times \\frac{22}{7}\\) \u00d7 5.25 \u00d7 5.25 \u00d7 5.25 \n= 303.187 cm3<\/sup> \nWe know that 1000 cm3<\/sup> = 1 litre \n\u2234 303.18 cm3<\/sup> = 0.303 litre (approx) \nTherefore, a hemispherical bowl of diameter 10.2 cm holds 0.303 liters of milk.<\/p>\nQuestion 6. \nA hemispherical tank is made up of an iron sheet 1 cm thick if the inner radius is 1 m, then find the volume of the iron used to make the tank. \nSolution: \nWe have given that \nThe inner radius of hemispherical tank = 1 m \nand thickness of iron sheet = 1 cm = 0.01 m \nThe outer radius of hemispherical tank = 1.01 m \n \nSo, volume of iron used to make the tank = Volume of out spherical tank – Volume of inner spherical tank \n= \\(\\left.\\frac{2}{3} \\pi(1.01)^{3}\\right)-\\frac{2}{3} \\pi(1)^{3}\\) \n= \\(\\frac{2}{3}\\) \u00d7 \u03c0 \u00d7 1.030301 – \\(\\frac{2}{3}\\) \u00d7 \u03c0 \u00d7 1 \n= \\(\\frac{2}{3}\\) \u00d7 \u03c0 \u00d7 (1.030301 – 1) \n= \\(\\frac{2}{3} \\times \\frac{22}{7} \\times 0.030301\\) \n= 0.06348 m3<\/sup> (approx) \nTherefore, the volume of the iron used to make the tank is 0.06348 m3<\/sup> (approx)<\/p>\n <\/p>\n
Question 7. \nFind the volume of a sphere whose surface area is 154 cm2<\/sup>. \nSolution: \nWe have given that \nThe surface area of a sphere is 154 cm2<\/sup> \nbut, the surface area of sphere = 4\u03c0r2<\/sup> \n \nVolume of sphere = \\(\\frac{4}{3} \\pi r^{3}\\) \n= \\(\\frac{4}{3} \\times \\frac{22}{7}\\) \u00d7 3.5 \u00d7 3.5 \u00d7 3.5 \n= 179\\(\\frac{2}{3}\\) cm3<\/sup><\/p>\nQuestion 8. \nA dome of a building is in the form of a hemisphere. From inside it was while washed at the cost of Rs. 498.96. If the cost of whitewashing is Rs. 2.00 per square meter find the: \n(i) Inside the surface area of the dome. \n(ii) Volume of the air inside the dome. \nSolution: \n(i) We have given that Rs. 2 is the rate of whitewashing of 1 m2<\/sup> area \nRs. 1 is the rate of white-washing \\(\\frac{1}{2}\\) m2<\/sup> area \n \nRs. 498.96 is the rate of white-washing = \\(\\frac{1}{2}\\) \u00d7 498.96 = 249.48 m2<\/sup> area. \nTherefore, according to question, inside surface area of dome = 249.48 m2<\/sup><\/p>\n(ii) We have, \nsurface area of hemispherical dome = 249.48 m2<\/sup> \nbut surface area of hemispherical dome = 2\u03c0r2<\/sup> \n\u21d2 2\u03c0r2<\/sup> = 249.48 \n\u21d2 r2<\/sup> = \\(\\frac{249.48 \\times 7}{2 \\times 22}\\) \n\u21d2 r2<\/sup> = 39.69 \n\u21d2 r = \u221a39.69 = 6.3 m \nTherefore, \nVolume of hemispherical dome = \\(\\frac{2}{3} \\pi r^{3}\\) \n= \\(\\frac{2}{3} \\times \\frac{22}{7}\\) \u00d7 6.3 \u00d7 6.3 \u00d7 6.3 \n= 523.908 m3<\/sup> \nTherefore, the volume of the air inside the dome is 523.9 m3<\/sup> (approx).<\/p>\n <\/p>\n
Question 9. \nTwenty-seven solid iron spheres, each of radius r and surface area s are melted to form a sphere with surface area s’. Find the: \n(i) radius r’ of the new sphere \n(ii) ratio of s and s’. \nSolution: \nWe have given that \n \nThe radius of sphere = r \nSurface area(s) = 4\u03c0r2<\/sup> \nand volume of sphere = \\(\\frac{4}{3} \\pi r^{3}\\) \nVolume of 27 such speres = \\(\\frac{4}{3} \\times \\pi r^{3} \\times 27\\) = 36\u03c0r3<\/sup> \n \nAccording to question Volume of 27 small sphere = volume of big sphere \nor, 36\u03c0r3<\/sup> = \\(\\frac{4}{3} \\pi\\left(r^{\\prime}\\right)^{3}\\) (\u2235 radius = r’) \n\u21d2 (r’)3<\/sup> = \\(\\frac{36 \\pi r^{3} \\times 3}{4 \\times \\pi}\\) \n(\u2234 volume of big sphere = volume of 27 small sphere) \n\u21d2 (r’)3<\/sup> = 27r3<\/sup> \n\u21d2 r’ = 3r \nTherefore, radius r’ of new sphere is equal to 3r.<\/p>\n(ii) Surface area of new-sphere (s’) =4\u03c0(r’)2<\/sup> \nand surface area of small sphere = 4\u03c0r2<\/sup> \n\u2234 Ratio between them = \\(\\left(\\frac{4 \\pi r^{2}}{4 \\pi\\left(r^{\\prime}\\right)^{2}}\\right)\\) \n= \\(\\left(\\frac{4 \\pi r^{2}}{4 \\pi \\times 3 r \\times 3 r}\\right)\\) (\u2234 r’ = 3r) \n= \\(\\frac{1}{9}\\) \n= 1 : 9<\/p>\n <\/p>\n
Question 10. \nA capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3<\/sup>) is needed to fill this capsule. \nSolution: \nWe have given \n \nDiameter of spherical capsule = 3.5 mm \n\u2234 Radius of spherical capsule = \\(\\frac{3.5}{2}\\) = 1.75 mm \n\u2234 Volume of spherical capsule = \\(\\frac{4}{3} \\pi r^{3}\\) \n= \\(\\frac{4}{3} \\times \\frac{22}{7}\\) \u00d7 1.75 \u00d7 1.75 \u00d7 1.75 \n= 22.4583 mm3<\/sup> (approx) \nTherefore, medicine is needed to fill the capsule is 22.4583 mm3<\/sup> (approx)<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.8 Question 1. Find the volume of a sphere whose radius is (i) 7 cm (ii) …<\/p>\n
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8<\/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 13.8 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n