{"id":26511,"date":"2021-06-26T10:28:02","date_gmt":"2021-06-26T04:58:02","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26511"},"modified":"2021-06-26T10:35:39","modified_gmt":"2021-06-26T05:05:39","slug":"ncert-solutions-for-class-9-maths-chapter-13-ex-13-8","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-8\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8"},"content":{"rendered":"

These 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>\n

NCERT 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\"NCERT
\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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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>\n

Question 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\"NCERT
\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\"NCERT
\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

\"NCERT<\/p>\n

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\"NCERT
\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>\n

Question 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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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>\n

Question 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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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\"NCERT
\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

\"NCERT<\/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\"NCERT
\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":{"spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 - NCERT Solutions<\/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:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-13-ex-13-8\/\" \/>\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 13.8 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"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. 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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.8 Question 1. 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