Comprehensive CBSE Question Bank in Mathematics X (Term-II)

SURFACE AREAS AND VOLUMES 159 \\ (Edge)3 = 64 = 4 ¥ 4 ¥ 4 = 43 fi Edge = 4 cm. As the two cubes are joined end-to-end to form the resulting cuboid. fi Length of the cuboid, l = (2 ¥ 4) cm = 8 cm Breadth of the cuboid, b = 4 cm Height of the cuboid, h = 4 cm Fig. 6.6 \\ Surface area of the cuboid, S = 2(lb + bh + hl) = 2 ( 8 ¥ 4 + 4 ¥ 4 + 4 ¥ 8) cm2 = 160 cm2 Hence, the required surface area of the resulting cuboid is 160 cm2. EXAMPLE 2. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm. SOLUTION. When the largest right circular cone is cut out of a cube, then the base of the largest right circular cone is the circle inscribed in a face of the cube. In such a situation, height of the cone equals the edge of the cube. Radius of the base of cone, r= 9 cm 2 Height of cone, h = 9 cm \\ Volume of the largest cone, V= 1 pr2h = Ê1 ¥ 22 ¥ 9 ¥ 9 ¥ 9¯ˆ˜ cm3 3 ËÁ 3 7 2 2 = 2673 cm3 = 190.93 cm3 Fig. 6.7 14 Hence, the required volume of the largest cone cut out of the given cube is 190.93 cm3. EXAMPLE 3. Find the length of the side of the largest cube that can be inscribed in a sphere of a diameter d cm. SOLUTION. When the largest cube is inscribed in a sphere, then the vertices of the cube lie on the sphere. In such a situation, Diameter of the sphere = Diagonal of the largest cube. \\ Diameter of the sphere = d cm. Given Let x cm be the length of the side of the largest cube, then \\ Diagonal of the largest cube = 3 × side = 3 ¥ x fi d = 3x fi x= d 3 Hence, the required length of the side of the largest cube inscribed in the given sphere is d cm. Fig. 6.8 3

160 MATHEMATICS-X EXAMPLE 4. A rectangular sheet of paper 44 cm ¥ 18 cm is rolled along its length and a cylinder is formed. Find the volume of the cylinder. SOLUTION. When a rectangular paper is rolled along its length, then, Length of the paper sheet = Circumference of the base of the cylinder and breadth of the paper sheet = Height of the cylinder Then, 2pr = 44 44 44 ¥ 7 Fig. 6.9 2p 2 ¥ 22 fi r= = = 7 cm \\ Volume of the cylinder = pr2h = Ê 22 ¥ 7 ¥ 7 ¥ 18˜ˆ¯ cm3 = 154 ¥ 18 = 2772 cm3. ÁË 7 Hence, the required volume of the cylinder is 2772 cm3. EXAMPLE 5. The volumes of two spheres are in the ratio 64 : 27. Find their radii if the sum of their radii is 21 cm. SOLUTION. Let the radii of two spheres be r1 and r2 respectively, then. V1 = 4 pr13 = 64 = Ê 4ˆ3 fi r13 = Ê 4ˆ3 fi r1 = 4 . V2 3 27 ÁË 3˜¯ r23 ÁË 3 ˜¯ r2 3 4 3 pr23 \\ Ratio of the radii = 4 : 3 Let r1 = 4x and r2 = 3x fi 4x + 3x = 21   r1 + r2 = 21 (Given) fi 7x = 21 fi x= 21 = 3. 7 \\ Radius of 1st sphere r1 = 4x = 4 ¥ 3 = 12 cm and Radius of 2nd sphere r2 = 3x = 3 ¥ 3 = 9 cm. Hence, the required radii of two spheres are 12 cm and 9 cm respectively. EXAMPLE 6. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of p. [NCERT; CBSE 2015] SOLUTION. Here, a solid is the combination of Fig. 6.10 (i) a cone, and (ii) a hemisphere. Then, we have Radius of the base of the cone, r = 1 cm Radius of the hemispherical portion, r = 1 cm

SURFACE AREAS AND VOLUMES 161 The height of the cone, h = r = 1 cm \\ The volume of the solid = (Volume of the cone) + (Volume of the hemisphere) = 1 p r2h + 2 p (r)3 = p [(1)2 (1) + 2(1)3] 3 3 3 = p [ 1 + 2] = p (3) = p cm3 3 3 Hence, the required volume of the solid is p cm3. EXAMPLE 7. How many 3 metre cubes can be cut from a cuboid measuring 18 m ¥ 12 m ¥ 9 m? SOLUTION. Here, we have Edge of each cube = 3 m Volume of each cube, v = (Edge)3 = (3)3 = 27 m3 Volume of the cuboid, V = (18 ¥ 12 ¥ 9) m3 = 1944 m3 Let n be the number of cubes that can be cut off from the given cuboid. Then, Volume of the cuboid = Number of cubes ¥ Volume of each cube V= n¥v fi n= V = 1944 = 72. v 27 Hence, the required number of cubes that can be cut from the given cuboid is 72. EXAMPLE 8. A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm, find the rise in water level in the vessel. SOLUTION. Here, we have ...(1) Edge of the given cube = 9 cm. \\ Volume of the cube, V1 = (9)3 = 729 cm3 When a cube is immersed in the vessel, then there is a rise in the water level. Let x cm be the rise in the water level in the vessel. ...(2) Now, volume of the cuboid, V2 = (15 ¥ 12 ¥ x) cm3 Then, Volume of the cube = Volume of the water replaced by the cube. V1 = V2 fi 729 = 15 ¥ 12 ¥ x Using (1) and (2) fi x = 729 = 81 = 4.05 cm 15 ¥ 12 20 Hence, the required rise in the water level in the vessel is 4.05 cm. EXAMPLE 9. Three cubes of a metal whose edges are in the ratio 3 : 4 : 5 are melted and converted into a single cube whose diagonal is 12 3 cm. Find the edges of the three cubes. [NCERT (EP)] SOLUTION. Let the edges of three cubes (in cm) be 3x, 4x and 5x, respectively \\ Volume of the cubes after melting = (3x)3 + (4x)3 + (5x)3 = 216x3 cm3

162 MATHEMATICS-X Let a be the side of new cube so formed after melting. fi a = 6x ...(1) fi a3 = 216 x3 fi (a)3 = (6x)3 Diagonal of single cube = a2 + a2 + a2 = a 3 fi 12 3 = a 3   Diagonal of new cube = 12 3 (Given) fi a = 12 fi 6x = 12 Using (1) fi \\ x = 12 = 2 6 Edges of the 3 cubes are 6 cm, 8 cm and 10 cm respectively. TEST YOUR KNOWLEDGE 1. Two cubes each of volume 27 cm3 are joined end-to-end. Find the surface area of the resulting cuboid. [CBSE 2015] 2. A toy is in the form of a cone mounted on a hemisphere, of same radius 7 cm. If the total height of the toy is 31 cm, find its total surface area. [CBSE 2013] 3. A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere. [NCERT] 4. How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm ¥ 11 cm ¥ 12 cm? [NCERT (EP)] 5. How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter? 6. Two cones with same base-radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so formed. 7. The largest possible sphere is carved out of a wooden solid cube of side 7 cm. Find the volume wood left. [CBSE 2014] 8. 1 cm3. Find the The volume of a right circular cylinder with its height equal to the radius is 25 7 height of the cylinder. ÈÍÎUse p = 22 ˘ [CBSE 2020] 7 ˙˚ SHORT ANSWER TYPE-II QUESTIONS Illustrative Examples EXAMPLE 1. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. [NCERT, CBSE 2012] SOLUTION. Here, a given solid is the combination of (i) a cubical block, and (ii) a hemisphere Then, we have Radius of the base circle of the hemisphere, r = 7/2 cm Edge of the cubical block = 7 cm Fig. 6.11

SURFACE AREAS AND VOLUMES 163 fi The greatest diameter of the hemisphere that can be placed on the cubical block = side of the cube = 7 cm. TSA of the solid = (TSA of the cube) – (Base area of the hemisphere) + (CSA of the hemisphere) = [(6 ¥ edge2) – pr2 + 2pr2] cm2 = [{6 ¥ (7)2} + pr2] cm2 ÈÍ294 + 22 Ê 7 ˆ 2 ˘ cm2 = ÍÈÎ294 + 77 ˘ cm2 ÎÍ 7 ËÁ 2 ¯˜ ˙ 2 ˙˚ = ¥ ˚˙ = [294 + 38.5] cm2 = 332.5 cm2 Hence, the required TSA of the combined solid is 332.5 cm2. EXAMPLE 2. The decorative block shown in Fig. 6.12, is made of two solids, a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere, fixed on the top, has a diameter of 4.2 cm. Find the total surface area of the block. [NCERT, CBSE 2012, 19] SOLUTION. Here, a decorative block is the combination of: (i) a cube, and (ii) a hemisphere. Then, we have Edge of a cube = 5 cm Given Radius of the hemisphere, r = 4.2/2 cm Fig. 6.12 \\ TSA of cube = 6 ¥ (edge)2 = 6 ¥ 5 ¥ 5 cm2 = 150 cm2 Now, TSA of the block = (TSA of the cube) – (Base area of the hemisphere) + (CSA of the hemisphere) = 150 – pr2 + 2pr2 = (150 + pr2) cm2 = 150 cm2 + Ê 22 ¥ 4.2 ¥ 4.2 ˆ cm2 ÁË 7 2 2 ˜¯ = (150 + 13.86) cm2 = 163.86 cm2 Hence, the required total surface area of the block is 163.86 cm2. In Examples 1 and 2, the readers are advised to note carefully that the part of the cube where the hemisphere is attached has not been included while computing the TSA of the combined solid. EXAMPLE 3. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. [NCERT, CBSE 2012] SOLUTION. Here, a wooden block is the combination of (i) a cube, and (ii) a hemisphere.

164 MATHEMATICS-X Then, we have 11 22 l Diameter of the circular base of the hemisphere l 1 = edge of the cube = l units 2 fi Radius of the circular base of the hemisphere, r= l units. 2 \\ TSA of the remaining cubical wooden block l (after hemispherical depression) Fig. 6.13 = (TSA of the cube) – (Base area of the hemisphere) + (CSA of hemisphere) = 6l2 – pr2 + 2pr2 = 6l2 + pr2 ÈÍ6l2 Ê l ˆ 2 ˘ ÍÈ6l2 l2 ˘ 1 ÎÈ24l2 pl2 ˘˚ l2 ÎÍ ÁË 2 ˜¯ ˙ ÎÍ 4 4 = + p ¥ ˙˚ = + p ¥ 4 ˙ = + = ÎÈ24 + p˚˘ sq. units ˙˚ Hence, the required TSA of the cubical block after the removal of the hemispherical depression is l2 ÈÎ24 + p˘˚ sq. units. 4 EXAMPLE 4. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 6.14. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article. [NCERT, CBSE 2012] Fig. 6.14 Fig. 6.15 SOLUTION. Here, a wooden article is the combination of (i) a cylinder, and (ii) two equal hemispheres. Then, we have Radius of the cylinder, r = 3.5 = 7 cm 2 Height of the cylinder, h = 10 cm \\ TSA of the wooden article = (CSA of the cylinder) + (CSA of 2 equal hemispheres) = 2prh + 2(2pr2) = 2pr(h + 2r)

SURFACE AREAS AND VOLUMES 165 = 2¥ 22 ¥ 7 ÈÎÍ10 + 2 Ê 7 ˆ ˘ = 2¥ 22 ¥ 7 (17) = 22 ¥ 17 = 374 cm2 7 2 ËÁ 2 ¯˜ ˙˚ 7 2 Hence, the required total surface area of the wooden article is 374 cm2. In the above Example 4, we have added the two curved surfaces of two hemispheres since they are created in addition to the curved surface of the cylinder. EXAMPLE 5. Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see Fig. 6.16). The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. [NCERT] SOLUTION. Here, a bird-bath is the combination of (i) a cylinder, and (ii) a hemisphere. Then, we have Height of the cylinder, h = 1.45 m = 145 cm   1 m = 100 cm Common radius of the cylinder and hemisphere, r = 30 cm Fig. 6.16 \\ TSA of the bird-bath = (CSA of the cylinder) + (CSA of the hemisphere) = 2prh + 2pr2 = 2pr(h + r) = 2 ¥ 22 ¥ 30 (145 + 30) cm2 7 =2¥ 22 ¥ 30 (175) cm2 = 33000 cm2 7 Hence, the required surface area of the bird-bath is 33000 cm2. EXAMPLE 6. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner area of the vessel. [NCERT, CBSE 2012] SOLUTION. Here, a vessel is the combination of (i) a hollow hemisphere, and (ii) a hollow cylinder. Then, we have Total, height of the vessel, H = 13 cm Diameter of the hemisphere = 14 cm \\ Radius of the hemisphere, r= 14 = 7 cm 2 Height of the cylindrical-portion, Fig. 6.17 h = (Total height of the vessel) – (Radius of the hemisphere) = (H – r) = (13 – 7) cm = 6 cm

166 MATHEMATICS-X Total inner surface-area of the vessel = (Inner CSA of the hemisphere) + (Inner CSA of the cylinder) = 2pr2 + 2prh = 2pr(r + h) =2¥ 22 ¥ 7 (7 + 6) =2¥ 22 ¥ 7(13) 7 7 = 2 ¥ 22 ¥ 13 = 572 cm2 Hence, the required inner area of the vessel is 572 cm2. EXAMPLE 7. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends see Fig 6.18. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. [NCERT, CBSE 2012] SOLUTION. Here, a medicine capsule is the combination of (i) a cylinder, and (ii) two hemispheres. Then, we have The total length of the capsule, L = 14 mm Diameter of the capsule = 5 mm \\ Radius of the hemisphere, r = 5 mm 2 Fig. 6.18 Radius of the cylinder, r = 5 mm 2 Length (or height) of the cylinder, h = (L – 2r) = ÎÈÍ14 - 2 Ê 5ˆ ˘ mm ÁË 2 ¯˜ ˚˙ = (14 – 5) mm = 9 mm TSA of the capsule = (CSA of the cylinder) + 2(Surface area of the hemisphere) = 2prh + 2 ¥ 2pr2 = 2pr [h + 2r] = 2p 5 ¥ È 9 + 2 Ê 5ˆ È =2¥ 22 ¥ 5 È 9 + 2 Ê 5ˆ È 2 Í ÁË 2˜¯ Í 7 2 Í ËÁ 2¯˜ Í Î Î Î Î =2¥ 22 ¥ 5 ¥ [14] = 110 ¥ 14 7 2 7 = 220 sq. mm Hence, the required surface area of the capsule is 220 sq. mm. EXAMPLE 8. A golf ball has diameter equal to 4.1 cm. Its surface has Fig. 6.19 150 dimples each of radius 2 mm. Calculate total surface area which is exposed to the surroundings assuming that the dimples are hemispherical. SOLUTION. Here, a golf ball is the combination of (i) a sphere and (ii) a hemispheres.

SURFACE AREAS AND VOLUMES 167 Then, we have Diameter of the spherical golf-ball = 4.1 cm Radius of the spherical golf-ball, R= 4.1 cm 2 Surface area of the golf-ball = 4pR2 = 4p ¥ Ê 4.1ˆ 2 = 16.81 p cm2 ÁË 2 ˜¯ \\ Radius of each dimple, r = 2 mm = 2 cm 10  The dimples are hemispherical. fi In the case of each dimple, surface area of pr2 is removed from the surface of the ball and the surface area of the hemisphere (i.e., 2pr2) is exposed to the surroundings. fi TSA exposed to the surroundings = (Surface area of the ball) - (Surface area of all the dimples) + (Surface area of 150 hemispheres) = 16.81p – 150pr2 + (150 ¥ 2pr2) = 16.81p + 150pr2 Ê 2 ˆ 2 cm2 cm2 ÁË 10 ¯˜ = [16.81p + 150p ] = (16.81p + 6p) = 22.81p cm2 Hence, the required TSA of the golf-ball exposed to the surroundings is 22.81p cm2. EXAMPLE 9. A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the cone and of the remaining solid left out after the cone carved out. [NCERT (EP)] SOLUTION. The cone of maximum size that is carved out from a cube of edge 14 cm will be of base-radius 7 cm and the height 14 cm. Surface area of the cone = prl + pr2 = pr [l + r] = 22 ¥ 7 ¥ È 72 + 142 + 7 ˘ cm 2 = 22 ¥ È 245 + 7 ˘ cm2 7 ÎÍ ˚˙ Î ˚ = 22 ¥ È 49 ¥ 5 + 7 ˘ cm2 = 22 ¥ 7 È 5 + 1˘˚ cm2 Î ˚ Î = 154 È 5 + 1˚˘ cm2 ...(1) Î ...(2) Surface area of the cube = 6 ¥ (14)2 cm2 = 6 ¥ 196 cm2 = 1176 cm2 \\ Surface area of the remaining solid left out after the cone is carved out = [1176 – 154 – 154 5 ] cm2 Using (1) and (2) = (1022 – 154 5 ) cm2.

168 MATHEMATICS-X EXAMPLE 10. In Fig. 6.20, the shape of a solid copper piece (made of two pieces with dimensions) is shown. The face ABCDEFA is the uniform crosssection. Assume that the angles at A, B, C, D, E and F are right angles. Calculate the volume of the piece. SOLUTION. Here, the given body is the combination of (i) a horizontal piece of cuboid and (ii) a vertical piece of a cuboid. Then, for the horizontal piece, we have Length l = 8 cm, Breadth b = 22 cm Height h = 3 cm \\ Volume of the horizontal piece, ...(1) Fig. 6.20 V1 = (l ¥ b ¥ h) cm3 = (8 ¥ 22 ¥ 3) cm3 = 528 cm3 For the vertical piece, we have Length L = 22 cm, Breadth B = 2 cm Height H =(5 + 3) cm = 8 cm \\ Volume of the vertical piece, ...(2) V2 = (L ¥ B ¥ H) cm3 = (22 ¥ 2 ¥ 8) cm3 = 352 cm3 Total volume of the whole piece = (V1 + V2) = (528 + 352) cm3 Using (1) and (2) = 880 cm3 Hence, the required total volume of the whole piece is 880 cm3. EXAMPLE 11. A pen-stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The diameter of each of the depressions is 1 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 6.21). [CBSE 2012] 10 1 cm 15 1.4 cm 3.5 Fig. 6.21

174 MATHEMATICS-X EXAMPLE 2. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of the hemisphere and conical parts are the same as that of the cylindrical part. Find the surface area of the toy, if the total of the height of the toy is 30 cm. SOLUTION. Here, a toy is the combination of (i) a right circular cylinder (ii) a hemisphere, and (iii) a cone. Then, we have Radius of the hemisphere, r = 5 cm Radius of the base of the cylinder, r = 5 cm Height of the cylindrical part, H = 13 cm Radius of the base of the cone, r = 5 cm Height of the hemisphere = Radius of the hemisphere = 5 cm Total height of the toy, H = 30 cm Let h cm be the height of the cone. Then, H = (5 + 13 + h) cm 30 = (18 + h) h = 30 – 18 = 12 cm \\ Height of the cone, h = 12 cm Fig. 6.26 \\ Slant height of the cone, l = (r)2 + (h)2 = (5)2 + (12)2 cm = 169 cm = 13 cm Total surface area of the toy, TSA = (CSA of the hemisphere) + (CSA of the cylinder) + (CSA of the cone) = (2pr2 + 2prH + prl) unit2 = pr(2r + 2H + l) unit2 = 22 ¥ 5 [(2 ¥ 5) + (2 ¥ 13) + 13] = 22 ¥ 5 [10 + 26 + 13] 7 7 = 22 ¥ 5 [49] = 22 ¥ 245 = 770 cm2 7 7 Hence, the required total surface area of the toy is 770 cm2. EXAMPLE 3. A wooden toy rocket is in the shape of Fig. 6.27 a cone mounted on a cylinder, as shown in Fig. 6.27. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylin- drical portion yellow, find the area of the rocket painted with each of these colours. (Take p = 3.14) [NCERT] SOLUTION. Here, a wooden toy rocket is the combination of

SURFACE AREAS AND VOLUMES 175 (i) a conical portion, and (ii) a cylindrical portion. Then, we have Radius of the conical portion, R = 2.5 cm Height of the conical portion, h = 6 cm Slant height of the conical portion, l is given by l = (R)2 + (h)2 = (2.5)2 + (6)2 cm = 6.5 cm Total height of the entire rocket, H = 26 cm Height of the cylindrical portion h¢ = (H – h) = (26 – 6) = 20 cm. Radius of the cylindrical portion, r = 1.5 cm Here the conical portion has its circular base resting on the base of the cylinder, but the base of the cone is larger than the base of the cylinder. So, a part of the base of the cone (a ring) is to be painted. Now, the area to be painted orange = (CSA of the cone) + (Base area of the cone) – (Base area of the cylinder) = [p Rl + pR2 – p r2] = [p (Rl + R2 – r2)] = p [2.5 ¥ 6.5 + (2.5)2 – (1.5)2] cm2 = p [20.25] cm2 = 3.14 ¥ 20.25 cm2 = 63.585 cm2 Now, the area to be painted yellow = (CSA of the cylinder) + (Area of one base of the cylinder) = 2prh¢ + p (r)2 = pr(2h¢ + r) = 3.14 ¥ 1.5 [2 ¥ 20 + 1.5] cm2 = 4.71 ¥ 41.5 cm2 = 195.465 cm2 Hence, the area of the rocket required to be painted with orange colour and yellow colour are 63.585 cm2 and 195.465 cm2 respectively. 8 cm EXAMPLE 4. A solid iron pole consists of a cylinder of height 220 cm high 60 cm Upper and the base of diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that mass of 1 cm3 of iron has 8 g mass. (Use p = 3.14) [NCERT, CBSE 2012, 19] SOLUTION. Here, a solid iron pole is the combination of two 220 cm Lower cylinders of different heights and radii 24 cm (i) a lower cylinder, and (ii) an upper cylinder. Fig. 6.28 Then, for the lower cylinder, we have Diameter of the base, d = 24 cm

176 MATHEMATICS-X Radius of the base, R= 24 cm = 12 cm 2 Height of the cylinder, H = 220 cm For the upper cylinder, we have Radius of the base, r = 8 cm Height of the cylinder, h = 60 cm \\ Volume of the solid iron pole, V = (Volume of the lower cylinder) + (Volume of the upper cylinder) = p R2H + p r2h = p [R2H + r2h] = p [(12)2 ¥ 220 + (8)2 ¥ 60] = 3.14[144 ¥ (220) + 64 ¥ (60)] = 3.14[31680 + 3840] = 3.14 ¥ 35520 = 111532.8 cm3 Mass of 1 cm3 of iron = 8 g \\ Mass of 111532.8 cm3 of iron = 8 ¥ 111532.8 = 892262.4 g = 892.26 kg Hence, the required mass of an iron pole is 892.26 kg. EXAMPLE 5. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm, is hollowed out. Find the volume of the remaining solid correct to two places of decimal. Also, find the total surface area of the remaining solid. (Take p = 3.1416) SOLUTION. Here, the solid is the combination of (i) a circular cylinder, and (ii) a cone Then, we have Height of the cylinder, h = 8 m = Height of the cone Radius of the base of the cylinder, r = 6 cm = Radius of the cone Slant height of the cone, l = r2 + h2 = (6)2 + (8)2 Fig. 6.29 = 36 + 64 = 100 = 10 cm (i) Volume of the remaining solid = (Volume of the cylinder) – (Volume of the cone) = (p r2h - 1 p r2h) unit3 = ( 2 p r2h) cm3 3 3 = ( 2 p (6)2 ¥ 8) cm3 = ( 2 p (36 ¥ 8)) cm3 3 3 = ( 2 ¥ 3.1416 ¥ 36 ¥ 8) cm3 Putting p = 3.1416 3 Correct up to two places of decimal = 603.1872 cm3 = 603.19 cm3

SURFACE AREAS AND VOLUMES 177 (ii) TSA of the remaining solid = (CSA of cylinder) + (Area of the base of the cylinder) + (CSA of cone) = (2 p rh + p r2 + p rl) unit2 = p r(2h + r + l) unit2 = 6p [(2(8) + 6 + 10) cm2 = 6 ¥ 3.1416 ¥ [16 + 6 + 10] cm2 Taking p = 3.1416 = 6 ¥ 3.1416 ¥ 32 cm2 = 603.1872 cm2 = 603.19 cm2 Correct up to two places of decimal Hence, the required volume and TSA of the remaining solid are 603.19 cm3 and 603.19 cm2 respectively. EXAMPLE 6. A building is in the form of a cylinder surmounted by a hemispherical dome and contains 41 19 m3 of air. If the internal diameter of the building is equal to its total height above 21 the floor. Find the height of the building. SOLUTION. Here, a building is the combination of (i) a cylinder, and (ii) a hemisphere. Let r m be the internal radius of the building. fi Let h m be the height of the cylindrical portion. Then, we have Total height of the building, H = (h + r) m ...(1) According to the problem, fi H = 2r m ...(2) Fig. 6.30   Total height of the building above the floor = Internal diameter of the building From (1) and (2), we get 2r = h + r fi h = 2r – r fi h=r ...(3) Volume of the building, V = (Volume of the cylindrical part) + (Volume of the hemispherical part) V= È p r 2h + 2 pr3 ˘ units3 = È p r 2 (r) + 2 pr 3 ˘ units3 From (3), h = r ÎÍ 3 ˚˙ ÍÎ 3 ˚˙ = Ê pr 3 + 2 p r 3 ˆ units3 fi V= 5 pr3 units3 ËÁ 3 ¯˜ 3 Also, volume of the building, V= 41 19 m3 Given 21 fi 5 pr3 = 41 19 fi 5 pr3 = 880 m3 3 21 3 21

178 MATHEMATICS-X fi r3 = 880 ¥ 3 ¥ 7 =8 fi (r)3 = (2)3 21 5 22 fi r=2   (a)3 = (b)3 fi a = b From (2), H = 2r m fi H = 2 ¥ 2 = 4 m Hence, the required height of the building is 4 m. EXAMPLE 7. A cylindrical vessel of diameter 14 cm and height 42 cm 16 cm is fixed symmetrically inside another cylindrical vessel of diameter 16 cm and height 42 cm. The total space between the two vessels is filled h = 42 cm with cork dust for heating insulations purpose. How many cubic centi- metres of cork dust is required for the purpose? 14 cm [CBSE 2015, 17] SOLUTION. Here, we have External diameter = 16 cm fi External radius, R = 8 cm Internal diameter = 14 cm fi Internal radius, r = 7 cm Fig. 6.31 Height of vessel, h = 42 cm Volume of cork dust = (Volume of external cylinder) – (Volume of internal cylinder) = pR2h – pr2h = ph(R2 – r2) = 22 ¥ 42 ¥ (82 – 72) = 22 ¥ 6 ¥ 15 7 = 1980 cm3. Hence, the required cork dust is 1980 cm3. EXAMPLE 8. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. [NCERT, CBSE 2012, 20] SOLUTION. Here, we have Radius of sphere, r = 4.2 cm fi Volume of the sphere, V1 = 4 p r3 = 4 ¥ p ¥ (4.2)3 ...(1) Also, we have 3 3 Radius of cross-section of the cylinder, r = 6 cm Let the height of the cylinder be h cm ...(2) fi Volume of the cylinder V2 = pr2h = p ¥ (6)2 ¥ h cm3 = 36p h cm3  A sphere is converted (recast) into a cylinder. fi Volume of the cylinder = Volume of the sphere fi V2 = V1 fi 36 ¥ p ¥ h = 4 ¥ p ¥ (4.2)3 Using (1) and (2) 3

SURFACE AREAS AND VOLUMES 189 According to the problem, we have Total volume of ice-cream in the container = (Number of cones) ¥ (Volume of an ice-cream cone) fi V=n¥v Using (1) and (2) fi 540p = 10 ¥ 2p x3   a3 = b3 fi a = b fi 540p = 20p x3 fi x3 = 27 = (3)3 fi x = 3 fi Diameter of the ice-cream cone, d = 2x = 2 ¥ 3 = 6 cm The required diameter of the ice-cream cone is 6 cm. EXAMPLE 21. Sushant has a vessel, of the form of an inverted cone, open at the top, of height 11 cm and radius of top as 2.5 cm and a full of water. Metallic balls, each of diameter 0.5 cm are put in the vessel due to which 2 th of the water in the vessel flows out. Find how many balls 5 were put in the vessel? [CBSE 2014] SOLUTION. Here, we have r 2.5 cm Height of conical vessel, h = 11 cm h 11 cm Top-radius of conical vessel R = 2.5 cm \\ Volume of water in conical vessel = Volume of cone = 1 pR2h 3 = 1 ¥ p ¥ 2.5 ¥ 2.5 ¥ 11 cm3 3 Volume of water that flows out, V1 = 2 ¥ Volume of water in vessel Fig. 6.42 5 = 2 ¥ 1 ¥ p ¥ (2.5)2 ¥ 11 cm3 ...(1) 5 3 Diameter of one metallic spherical ball, d = 0.5 cm fi Radius of one spherical ball, r= 0.5 cm 2 Let the number of metallic spherical balls be n. Then, volume of one metallic spherical ball, 4 3 4 Ê 0.5 ˆ 3 cm3 3 3 ÁË 2 ˜¯ v= pr = ¥ p ¥ ...(2) Volume of n metallic spherical balls, 4 Ê 0.5 ˆ 3 3 ÁË 2 ˜¯ V2 = nv = n ¥ ¥ p ¥ cm3 ...(3)

190 MATHEMATICS-X Using (1) and (3) According to the problem, V2 = V1 fi n¥ 4 ¥ p ¥ Ê 0.5 ˆ 3 = 2 ¥ 1 ¥ p ¥ (2.5)2 ¥ 11 3 ÁË 2 ˜¯ 5 3 fi n= 8 ¥ 2 ¥ 2.5 ¥ 2.5 ¥ 11 4 ¥ 0.5 ¥ 0.5 ¥ 0.5 = 4 ¥ 5 ¥ 5 ¥ 22 = 2200 Hence, the required number of balls put in the vessel is 2200. TEST YOUR KNOWLEDGE 1. A wooden lattu is shaped like a cone surrouned by a hemisphere. The total height of the lattu is 5 cm and the diameter of the lattu is 3.5 cm. A boy wants to paint it. Find the area he will have to paint. [CBSE 2016] Fig. 6.43 OR Rasheed got a toy, a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour with his crayons. The toy is shaped like a cone sur- mounted by a hemisphere (See Fig. 6.43). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour.[NCERT] 2. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 100 cm and the diameter of the hemispherical ends is 28 cm, find the cost of polishing the surface of the solid at the rate of 5 paise per cm2. [CBSE 2012] 3. A military tent of height 8.25 m is in the form of a right circular cyclinder of base diameter 30 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the length of the canvas used in making the tent, if the breadth of the convas is 1.5 m. [CBSE 2012] 4. From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. [CBSE 2014] 5. A solid is in the shape of a hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7 cm and height of the cone is 3.5 cm, find the volume of the solid. [CBSE 2020] 6. A toy is in the form of a cone mounted on a cylinder of same radius 7 cm. If the

STATISTICS 223 ( )(ii) To Find the Mean X Let the assumed mean, A = 2750 and the step-deviation, h = 500 (= 1500 – 1000) Putting the given data in tabular form as follows: Class intervals Mid - values di Number of 500 (Expenditure in `) xi = li + ui di = xi - A di¢ = families ( fi ) fi di¢ 2 1000 -1500 -1500 24 -72 1500 - 2000 1250 -1000 -3 40 -80 2000 - 2500 -500 -2 33 -33 1750 -1 2500 - 3000 0 28 0 2250 500 0 +30 3000 - 3500 1000 30 +44 3500 - 4000 2750 = A 1500 +1 22 +48 4000 - 4500 2000 +2 16 +28 4500 - 5000 3250 +3 7 S fi di¢ = - 35 3750 +4 S fi = 200 4250 4750 Using the Step-Deviation formula, we have X =A + S fi di¢ ¥ h = 2750 + -35 ¥ 500 S fi 200 =2750 + -175 = 2750 – 87.50 = 2662.50 2 Thus, mean = 2662.50 Hence, the required mean monthly expenditure is ` 2662.50. EXAMPLE 7. On the sports day of a school, 300 students participated. Their ages are given in the following distribution: Ages (in years) 5–7 7–9 9 – 11 11 – 13 13 – 15 15 – 17 17 – 19 No. of students 67 33 41 95 36 13 15 Find the mean and mode of the data. [CBSE 2015] SOLUTION. (i) To Find the Mean (X) : Let the assumed mean, A = 12 and the step-deviation, h = 2

224 MATHEMATICS-X Mid-values di 2 C-I xi = li + ui di = xi – A di = fi fi di 2 5–7 –6 67 – 201 7–9 6 –4 –3 33 – 66 9 – 11 8 –2 –2 41 – 41 11 – 13 10 0 –1 95 0 13 – 15 12 = A 2 0 36 36 15 – 17 14 4 1 13 26 17 – 19 16 6 2 15 45 18 3 Total fi = 300 fi di = – 201 Using the Step-Deviation Formula, we have X =A+ Â fi di ¢ ¥h = 12 + È - 201˘ ¥ 2 = 10.66 Â fi ÍÎ 300 ˚˙ Hence, the required mean is 10.66. (ii) To Find the Mode (Mo): Modal frequency, fm = Maximum frequency in the data = 95 fi Modal class = 11–13 Frequency of the pre-modal class, f1 = 41 Frequency of the post-modal class, f2 = 36 Width of the modal class, h=2 Using the Mode-formula, we get Mo = l + È fm - f1 ˘ ¥ h Í fm - f1 - ˙ Î 2 f2 ˚ = 11 + È 95 - 41 ˘ ¥ 2 Í ˙ Î 2(95) - 41 - 36 ˚ = 11 + 54 ¥2 = 11 + 0.96 = 11.96 113 Hence, the required mode is 11.96 EXAMPLE 8. Determine missing frequencies F1, F2 from the following data, when mode is 67 and total of frequencies is 47. Classes 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 Frequency 5 F1 15 F2 7 SOLUTION. Here, we have: Mode of the data, Mo = 67

STATISTICS 225 Sum (total) of frequencies, S fi = 47 Modal frequency, fm = Maximum frequency in the data = 15 Modal class = 60-70   Mode = 67, which lies in the class interval 60-70 Lower limit of the modal class, l = 60 Frequency of the preceding class/pre-modal class, f1 = F1 Frequency of the succeeding class/post-modal class, f2 = F2 Width of the modal class, h = 10 (= 70 - 60) S fi = 27 + F1 + F2 fi 47 = 27 + F1 + F2 F1 + F2 = 20 fi F1 + F2 = 47 – 27 fi ...(1) Using the mode formula, we get Mo = l + È fm - f1 ˘ ¥h Í fm - f1 - ˙ Î 2 f2 ˚ fi 67 = 60 + È 2 15 - F1 F2 ˘ ¥ 10   Mo = 67 (given) Í - F1 - ˙ ÎÍ (15) ˙˚ = 60 + È 15 - F1 ˘ ¥ 10 Í - (F1 + ˙ Î 30 F2 ) ˚ = 60 + (15 - F1) ¥ 10  Using (1) 30 - 20 = 60 + (15 - F1) ¥ 10 10 fi 67 = 60 + 15 - F1 ...(2) fi 67 – 60 = 15 - F1 fi 7 = 15 - F1 fi 15 – 7 = F1 fi F1 = 8 Putting the value of F1 from (2) in (1), we get 8 + F2 = 20 fi F2 = 20 - 8 = 12. Hence, the required missing frequencies are F1 = 8 and F2 = 12. EXAMPLE 9. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. Monthly consumption 65 – 85 85 – 105 105 – 125 125 – 145 145 – 165 165 – 185 185 – 205 (in units) 45 13 20 14 84 Number of consumers [NCERT]

226 MATHEMATICS-X SOLUTION. Number of (c. f .) Class mark li + ui fi xi Monthly consumers ( fi ) 4 xi = 2 300 consumption (in units) 4 9 (= 4 + 5) 75 475 5 22 (= 9 + 13) 1495 65 - 85 13 42 (= 22 + 20) 95 2700 85 - 105 20 56 (= 42 + 14) 2170 105 - 125 14 64 (= 56 + 8) 115 1400 125 - 145 8 68 (= 64 + 4) 780 145 - 165 4 135 S fi xi = 9320 165 - 185 N = S fi = 68 185 - 205 155 175 195 (i) To find the median: Making (less than type) cumulative frequency distribution table as follows: Here, we have Median number: N = 68 = 34 2 2 c.f. just > the median number (34) = 42 fi Median class = The class interval corresponding to c.f. of 42 fi Median class = 125 – 145 Lower limit of the median class, l = 125 Frequency of the median class, fm = 20 Width of the median class, h = 20 (= 145 – 125) Cumulative frequency of the pre-median class, C = 22 Using median formula, we get Me = l+ h ÊN - C˜ˆ¯ = 125 + 20 (34 – 22) fm ÁË 2 20 = 125 + 12 = 137 Median = 137 units Hence, the required median is 137. (ii) To find the mean: Using the direct method formula for mean, we have X = S xi fi = 9320 = 137.06 S fi 68 Mean = 137.06 units Hence, the required mean is 137.06.

STATISTICS 227 (iii) To find the mode: Here, we have modal frequency, fm = maximum frequency in the data = 20 Modal class = 125 – 145 Lower limit of the modal class, l = 125 Frequency of the pre-modal class, f1 = 13 Frequency of the post-modal class, f2 = 14 Width of the modal class interval, h = 20 (= 145 – 125) Using the mode formula, we get Mo = l + È fm - f1 ˘ ¥ h = 125 + È 20 - 13 ˘ ¥ 20 Í fm - f1 - ˙ Í ˙ Î 2 f2 ˚ ÍÎ 2 (20) - 13 - 14 ˚˙ = 125 + È 7 ˘ ¥ 20 = 125 + È140 ˘ Í - ˙ ÍÎ 13 ˙˚ Î 40 27 ˚ = 125 + 10.77 = 135.77 fi Mode = 135.77 units Hence, the required mode is 135.77. On comparing the computed values of the median, mean and mode of the given data as above, we find that the three measures are approximately the same in this case. EXAMPLE 10. If the median for the following frequency distribution is 28.5, find the values of x and y. Class interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Total Frequency 5 x 20 15 y 5 60 [CBSE 2012, 13] SOLUTION. First, we make the cumulative frequency table as follows: Variable Frequency c. f . (Less than type) 0 - 10 5 5 10 - 20 x 5+x 20 - 30 20 25 + x 30 - 40 15 40 + x 40 - 50 y 40 + x + y 50 - 60 5 45 + x + y Total N = 45 + x + y Since, N = 60. Also, N = 45 + x + y ... (1) fi 45 + x + y = 60 fi x + y = 60 – 45 = 15

ISBN: 978-93-93268-83-9 789393 268839 T10-0960-299-COMP.CBSE QB MATHS T-II X