Class IX Math NCERT Solution for Surface Areas and Volumes 13.4

EXERCISE 13.4 (Page 225)
Q1.   Find the surface area of a sphere of radius:
           (i) 10.5 cm                                 (ii) 5.6 cm
Sol: (i) Here r = 10.5 cm
                 ∴ Surface area of the sphere = 4πr2
           
           
Q2.   Find the surface area of a sphere of diameter:
           (i) 14 cm                                 (ii) 21 cm
Sol: (i) Here, Diameter = 14 cm
           
Q3.   Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Sol: Here, radius (r) = 10 cm
           
Q4.   The radius of a 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.
Sol: Case I: When radius (r1) = 7 cm
           
           
Q5.   A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm2.
Sol: Inner diameter of the hemisphere = 10.5 m
           
           ∵ Curved surface area of a hemisphere = 2πr2
           ∴ Inner curved surface area of hemispherical bowl
           
Q6.   Find the radius of a sphere whose surface area is 154 cm2.
Sol: Let the radius of the sphere be 'r' cm.
           ∴ Surface area = 4πr2
           ⇒ 4πr2 = 154
           
           Thus, the required radius of the sphere is 3.5 cm
Q7.   The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.
Sol: Let the radius of the earth = r
           
           ∵ Surface area of a sphere = 4πr2
           Since, the earth as well as the moon are considered to be spheres.
           ∴ Surface area of the earth = 4πr2
           
           or [Surface area of the moon] : [Surface area of the earth] = 1 : 16
           Thus, the required ratio = 1 : 16
Q8.   A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Sol: Inner radius (r) = 5cm
           Thickness = 0.25 cm
           ∴ Outer radius (R) = [5.00 + 0.25] cm = 5.25 cm
           ∴ Outer surface area of the bowl = 2πR2
                 
Q9.   A right circular cylinder just encloses a sphere of radius r (see the figure) find
           (i) surface area of the sphere,
           (ii) curved surface area of the cylinder,
           (iii) ratio of the areas obtained in (i) and (ii).
Sol: (i) For the sphere:
                 Radius = r
                 ∴ Surface area of the sphere = 4 πr2
           (ii) For the right circular cylinder:
                 ∵ Radius of the cylinder = Radius of the sphere
                 ∴ Radius of the cylinder = r
                 Height of the cylinder = Diameter of the sphere
                 ⇒ Height of the cylinder (h) = 2r
                 Since, curved surface area of a cylinder = 2πrh = 2πr(2r) = 4πr2
           

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