check the answer vbcbcfb6Explanation:. Find the volume of a sphere whose radius is(i) 7 CM (ii) 0.63 m(Assume π =22/7)Solution:(i) Radius of sphere, r = 7 cmUsing, Volume of sphere = (4/3) πr3= (4/3)×(22/7)×73= 4312/3Hence, volume of the sphere is 4312/3 cm3(ii) Radius of sphere, r = 0.63 mUsing, volume of sphere = (4/3) πr3= (4/3)×(22/7)×0.633= 1.0478Hence, volume of the sphere is 1.05 m3 (approx).2. Find the amount of water displaced by a solid spherical ball of diameter(i) 28 cm (ii) 0.21 m(Assume π =22/7)Solution:(i) Diameter = 28 cmRadius, r = 28/2 cm = 14cmVolume of the solid spherical ball = (4/3) πr3Volume of the ball = (4/3)×(22/7)×143 = 34496/3Hence, volume of the ball is 34496/3 cm3(ii) Diameter = 0.21 mRadius of the ball =0.21/2 m= 0.105 mVolume of the ball = (4/3 )πr3Volume of the ball = (4/3)× (22/7)×0.1053 m3Hence, volume of the ball = 0.004851 m33.The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3? (Assume π=22/7)Solution:Given,Diameter of a metallic ball = 4.2 cmRadius(r) of the metallic ball, r = 4.2/2 cm = 2.1 cmVolume formula = 4/3 πr3Volume of the metallic ball = (4/3)×(22/7)×2.1 cm3Volume of the metallic ball = 38.808 cm3Now, using relationship between, density, mass and volume,Density = Mass/VolumeMass = Density × volume= (8.9×38.808) g= 345.3912 gMass of the ball is 345.39 g (approx).4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?Solution:Let the diameter of earth be “d”. Therefore, the radius of earth will be will be d/2Diameter of moon will be d/4 and the radius of moon will be d/8Find the volume of the moon :Volume of the moon = (4/3) πr3 = (4/3) π (d/8)3 = 4/3π(d3/512)Find the volume of the earth :Volume of the earth = (4/3) πr3= (4/3) π (d/2)3 = 4/3π(d3/8)Fraction of the volume of the earth is the volume of the moonNcert solutions class 9 chapter 13-18Answer: Volume of moon is of the 1/64 volume of earth.5. How many litres of MILK can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22/7)Solution:Diameter of hemispherical bowl = 10.5 cmRadius of hemispherical bowl, r = 10.5/2 cm = 5.25 cmFormula for volume of the hemispherical bowl = (2/3) πr3Volume of the hemispherical bowl = (2/3)×(22/7)×5.253 = 303.1875Volume of the hemispherical bowl is 303.1875 cm3Capacity of the bowl = (303.1875)/1000 L = 0.303 litres(approx.)Therefore, hemispherical bowl can hold 0.303 litres of milk.6. A hemi spherical tank is made up of an iron sheet 1CM thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume π = 22/7)Solution:Inner Radius of the tank, (r ) = 1mOuter Radius (R ) = 1.01mVolume of the iron used in the tank = (2/3) π(R3– r3)Put values,Volume of the iron used in the hemispherical tank = (2/3)×(22/7)×(1.013– 13) = 0.06348So, volume of the iron used in the hemispherical tank is 0.06348 m3.7. Find the volume of a sphere whose surface area is 154 cm2. (Assume π = 22/7)Solution:Let r be the radius of a sphere.Surface area of sphere = 4πr24πr2 = 154 cm2 (given)r2 = (154×7)/(4 ×22)r = 7/2Radius is 7/2 cmNow,Volume of the sphere = (4/3) πr3Ncert solutions class 9 chapter 13-198. A dome of a building is in the form of a hemi sphere. From inside, it was white-washed at the COST of Rs. 4989.60. If the cost of white-washing isRs20 per square meter, find the(i) inside surface area of the dome (ii) volume of the air inside the dome(Assume π = 22/7)Solution:(i) Cost of white-washing the dome from inside = Rs 4989.60Cost of white-washing 1m2 area = Rs 20CSA of the inner side of dome = 498.96/2 m2 = 249.48 m2(ii) Let the inner radius of the hemispherical dome be r.CSA of inner side of dome = 249.48 m2 (from (i))Formula to find CSA of a hemi sphere = 2πr22πr2 = 249.482×(22/7)×r2 = 249.48r2 = (249.48×7)/(2×22)r2 = 39.69r = 6.3So, radius is 6.3 mVolume of air inside the dome = Volume of hemispherical domeUsing formula, volume of the hemisphere = 2/3 πr3= (2/3)×(22/7)×6.3×6.3×6.3= 523.908= 523.9(approx.)Answer: Volume of air inside the dome is 523.9 m3.9. Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the(i) radius r’ of the new sphere,(ii) ratio of Sand S’.Solution:Volume of the solid sphere = (4/3)πr3Volume of twenty seven solid sphere = 27×(4/3)πr3 = 36 π r3(i) New solid iron sphere radius = r’Volume of this new sphere = (4/3)π(r’)3(4/3)π(r’)3 = 36 π r3(r’)3 = 27r3r’= 3rRadius of new sphere will be 3r (thrice the radius of original sphere)(ii) Surface area of iron sphere of radius r, S =4πr2Surface area of iron sphere of radius r’= 4π (r’)2NowS/S’ = (4πr2)/( 4π (r’)2)S/S’ = r2/(3r’)2 = 1/9The ratio of S and S’ is 1: 9.