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Q1. If the radius of a sphere is halved, the ratio of the surface area of original sphere to that of the second is

• 1) 2:1
• 2) 1:4
• 3) 4:1
• 4) 1:2

Solution

Let S be the surface area of the original sphere of radius r Then, S = 4ᴨr² When radius is halved, new radius is r/2. Let S₁ be the surface area of the new sphere. The ratio is 4:1.

Q2.

Solution

Q3. The radius and height of a cone are in the ratio 4 : 3. The area of its base is 154 cm². Find its curved surface area.

Solution

Let the radius and height of the cone be 4x and 3x respectively. Area of the base = r² l Curved surface area of the cone = rl

Q4. Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find : (i) Radius of the base (ii) Total surface area of the cone.

Solution

Curved surface area of cone = 308 cm² l = 14 cm (i) rl = 308 (ii) Total surface area of cone = r² + rl

Q5. The curved surface area of right circular cylinder of height 14 cm is 88 cm². The diameter of the base of the cylinder is:

• 1) 4 cm
• 2) 2 cm
• 3) 1 cm
• 4) 3 cm

Solution

Diameter = 2r = 21 cm = 2 cm

Q6. A cylindrical bucket is internally 14 cm in radius and 30 cm deep. It is full of sand and when emptied, the sand stands in the shape of a cone 21 cm in radius. Calculate the height of the heap.

Solution

Given: Radius of the cylindrical bucket (R) = 14 cm, depth (H) = 30 cm Radius of conical heap (r)  = 21 cm Let h be the height of the heap. Now, Volume of cylindrical bucket = volume of the conical heap

Q7.

Solution

Q8. Curved surface area of a cone of radius 4 cm is 20 cm². Find total surface area of the same cone in terms of .

Solution

CSA = TSA = r² + rl

Q9. The difference between the outer lateral surface area and inner lateral surface area of a cylindrical metallic pipe 28 cm long is 176 cm². Find the outer and inner radii of the pipe, if the pipe is made of 352 cm³ of metal.

Solution

Let R cm and r cm be the external and internal radii of the metallic pipe. h = length of the pipe = 28 cm Now, Outside surface area - inner surface area = 176 cm² It is given that volume of metal used = 352 cm³ Therefore,amount of material in the pipe= External volume - internal volume = 352 cm³ Solving (i) and (ii), R = 2.5 cm = Outer radius r = 1.5 cm = Inner radius.

Q10. The height and base diameter of a conical tent is 16 m and 24 m respectively. Find the cost of canvas required to make it at the rate of Rs. 210 per m²?

Solution

Height, h = 16 m Diameter = 24 m Slant height (l) = CSA of the tent = rl Cost of the canvas = Rs 210 = Rs. 158400

Q11. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find: (i) Its inner curved surface area (ii) The cost of plastering this curved surface at rate of Rs. 40 per m²

Solution

(i) Inner curved surface area = 2rh = 110 m² (ii) Cost of plastering = Rs. 40 110 = Rs. 4400

Q12. How many cubes of side 3 cm can be cut from a wooden solid cuboid with dimensions 12 cm 12 cm 9 cm?

Solution

Volume of a cuboid =  _{Volume of a cube = a³} No. of cubes =

Q13. The circumference of the base of a 10 m high conical tent is 44 m. Calculate the length of canvas used in making the tent if the canvas is 2 m wide.

Solution

Let h be the height of the conical tent and r be the radius of its base. Circumference of the base = 44 m 2r = 44 m Slant height, l= Area of canvas = C.S.A of cone L B = rl L = Thus, the length of the canvas used is 134.2 m.

Q14. The radius of a right circular cylinder with curved surface area of 440 cm² and an altitude of 5 cm is

• 1) 14 cm
• 2) 7 cm
• 3) 22 cm
• 4) 21 cm

Solution

Given: Curved surface area(S) = 440 cm², altitude (h) = 5 cm Let r be the radius of the cylinder.  

Q15. Volume of a right circular cone is 2200/7 cm² and its diameter is 10 cm. Find its curved surface area. .

Solution

d = 10 cm r = 5 cm Volume of cone = h = 12 cm

Q16. A sphere of radius 12 cm is melted and recasted into a cylinder of radius 6 cm. Find the height of the recasted cylinder.

Solution

Given: Radius of the base of the cylinder (r) = 6 cm Radius of the sphere (R) = 12 cm Let h = height of the cylinder Sphere is melted and recasted into a cylinder. Therefore, Volume of cylinder = Volume of sphere  

Q17. A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical ball is dropped into the tub raising the level of water by 6.75 cm. What is the radius of the ball?

Solution

Let r be the radius of the spherical ball. When spherical ball is dropped into the tub, the level of water raises by 6.75 cm. Volume of spherical ball = Volume of water displaced in the cylindrical tub r³ = _{r³ = 729 cm³} r = 9 cm

Q18. A hemispherical bowl of radius 6 cm is filled with water. If the water is transferred into cylindrical vessel of base radius 3 cm, find the height to which the water rises in the cylindrical vessel. .

Solution

Let h be the height upto which the water rises in the cylindrical vessel. The water in the hemispherical bowl is transferred into a cylindrical vessel. Volume of hemispherical bowl = Volume of cylindrical vessel _{Thus, the water in the cylindrical vessel will rise up to the height of 16 cm.}

Q19. A tent is in the form of a cylinder surmounted by a cone of radius 12 m. The height of the conical portion is 7 m, which is equal to the height of the cylindrical portion of the tent. Calculate the volume of the tent.

Solution

Radius of the cylinder = Radius of cone (r) = 12 m  Height of cylinder = Height of cone (h) = 7 m Volume of the tent = Volume of conical part + volume of cylindrical part

Q20. The total surface area of a solid right circular cylinder is 231 cm². If curved surface area is two-third of the total surface area, find the radius of the base.

Solution

CSA = 231 = 154 cm² We know: TSA = CSA + 2 area of base 2 area of base = (231 - 154) 2 area of base = 77 2 r r = 77 r = cm = 3.5 cm

Q21. Find the volume and surface area of a sphere of radius 2.1 cm. .

Solution

r = 2.1 cm Volume = = 38.808 cm³ SA = 4r² = = 55.44 cm².

Q22. Length, breadth and height of a cuboid shape box of medicines is 20 cm, 12 cm and 10 cm respectively. Find the total surface area of this box.

• 1) 2400 cm²
• 2) 1120 cm²
• 3) 560 cm²
• 4) 1200 cm²

Solution

l = 20 cm, b = 12 cm, h = 10 cm  Total surface area of a box = 2(lb + bh + hl) = 2(20 × 12 + 12 × 10 + 20 × 10) = 2(240 + 120 + 200) = 2 × 560 = 1120 cm² Total surface area of a box is 1120 cm².

Q23. Find the number of 4 cm cubes that can be cut out of a cuboid of dimensions 12 cm × 12 cm × 8 cm.

Solution

Given: In a cuboid, Length (l) = 12 cm, breadth (b) = 12 cm and height (h) = 8 cm Volume of cuboid of (V1) == 1152 cm³ Volume of cube of edge 4 cm (V) = (edge)³ = 64 cm³ Let N number of 4 cm cubes that can be cut out of the cuboid. Then, N × V = V₁

Q24. If the volume and surface area of sphere is numerically equal then its radius is

• 1) 3 Units
• 2) 2 Units
• 3) 4 Units
• 4) 85 Units

Solution

Surface area of sphere = Volume of sphere

Q25. Calculate the volume of an ice cream cone of radius 2.8 cm and height 4 cm surmounted by a hemispherical top.

Solution

Given: Radius of the cone = Radius of the sphere (r) = 2.8 cm, Height of the cone (h) = 4 cm Now, Total volume = volume of conical part + volume of hemispherical part

Q26. How many spherical lead shots each 4.2 cm diameter can be obtained from a solid cuboid of lead with dimensions 66 cm, 42 cm and 21 cm?

Solution

For lead shots_, d = 4.2 cm r = 2.1 cm Volume of one spherical lead shot = Volume of solid cuboid = 66 42 21 cm³ No. of shots = = 1500

Q27. The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimension 22.5 cm _{10 cm}7.5 cm can be painted out of this container?

Solution

Surface area of one brick = 2(l b + b h + h l) = 2(22.510 + 107.5 + 7.522.5) cm² = 2[225 + 75 + 168.75] cm² = 2[300 + 168.75] cm² = 937.5 cm² Number of bricks that can be painted = = 100

Q28. The height of a right circular cylinder with lateral surface area 792 sq cm is 21 cm. The diameter of the base is:

• 1) 6 cm
• 2) 3 cm
• 3) 12 cm
• 4) 24 cm

Solution

Lateral surface area of a right circular cylinder =

Q29. The volume of a cylinder of altitude 15 cm is 1200 cm³. The area of its base is

• 1) 30 cm²
• 2) 60 cm²
• 3) 40 cm²
• 4) 80 cm²

Solution

Volume of the cylinder (V) = 1200 cm³ Altitude (h) = 15 cm                         Thus, area of the base of the cylinder is 80 cm²

Q30. Find the curved surface area of a cylindrical pillar which is 1.2 m high and has the diameter of the base as 28 cm. [Use ]

Solution

r = cm _{h = 1.2 m = 120 cm} CSA of the cylinder = 2rh = 10560 cm²

Q31. If the lateral surface area of a cube is 1600 cm², then its edge is:

• 1) 18 cm
• 2) 25 cm
• 3) 15 cm
• 4) 20 cm

Solution

Lateral surface area of a cube = 4b² 4b² = 1600 cm² b² = 400 cm² b = 20 cm

Q32. A rectangular piece of paper 33 cm long and 16 cm wide is rolled along its breadth to get a cylinder of height 16 cm. Find the volume of the cylinder.

Solution

On rolling a rectangular paper along its breadth, length becomes the circumference of the cylinder and breadth becomes its height. Given: Circumference of the base of the cylinder (C) = 33 cm, height (h) = 16 cm Let V denote the volume of the cylindrical vessel.               Thus, the volume of the cylinder so formed is 1386 cm³.

Q33. The volume of a sphere is 905 cm³. Find its radius.

Solution

Volume of the sphere = r³ = 216 r = r = 6 cm

Q34. A man built a cubical water tank with lid for his house, with each outer edge 1.5 m long. He gets the outer surface of tank excluding the base covered with square tiles of side 25 cm. Find how much he would spend for the tiles, if cost of tiles is Rs. 480 per dozen.

Solution

Surface Area of tank = 5a²      (Area of 5 faces only) Area of one square tile = _lb = _{25 25 cm²} Cost of 180 tiles =  = Rs. 7200

Q35. An isosceles triangle with a base of 0.4 m and an altitude of 0.3 m is revolved about a line through the vertex and perpendicular to the base. What is the volume of the solid generated by the revolving triangle?

Solution

The solid generated by revolving the triangle is a cone with height (h) = 0.3 m and radius (r) = 0.4 m    

Q36. A river 3 m deep and 40 m wide is flowing at the rate of 2 km/hour. How much water will fall into the sea in one minute?

Solution

Depth = 3 m Width = 40 m Volume of water that flows into the sea in 1 minute = 4000  1000 litres = 4000 kilo litres

Q37. The area of coloured paper required to completely cover a solid cubical box of edge 'x' units is:

• 1) 4x²
• 2) 5x²
• 3) 6x²
• 4) 8x²

Solution

6x² as area of cube = 6a² where a is the edge of the cube.

Q38. A wooden box of dimensions 1.5 m × 1.25 m × 0.65 m and open at the top is to be made. Determine the cost of wood required for it, if 1 m² of wood costs Rs 20.

Solution

Given: length (l)=1.5 m, breadth (b)=1.25 m and height (h)=0.65 m Area of four walls and base Cost of 1 m² wood = Rs 20 Therefore, cost of 5.45 m² of wood =Rs {5.45×20}                                                             =Rs 109

Q39. A cubical water tank with lid has each edge 2 m long. If the cost of the tiles is Rs 240 per dozen. Calculate the amount spent to cover the outer surface of the tank excluding the base, with square tiles of side 50 cm.

Solution

Given: Edge of the tank (a) =2 m Area of four walls and lid=                                                                                   Area of each tile of edge 50 cm =2500 cm² Let “N” numbers of tiles are required to cover the required area. N×2500=200000 N=80 Cost of 12 tiles= Rs 240 Cost of 80 tiles= Rs 1600

Q40. The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10 m, how much concrete mixture would be required to build 14 such pillars?

Solution

Radius of each pillar = 20 cm = Volume of each pillar = r²h Volume of 14 pillars = 14 Volume of one pillar So, 14 pillars would need 17.6 m³ of concrete mixture.

Q41. The length of the longest rod that can be put in a hall of dimension 23 m 10 m 10 m is:

• 1) 27 m
• 2) 15 m
• 3) 20 m
• 4) 23 m

Solution

The longest rod will have length as the diagonal of the cuboidal room. Diagonal = D = Thus, the length of the longest rod is 27 m.

Q42. If the volume of a cube is then total surface area of the cube is

• 1)
• 2)
• 3) 18 a²
• 4) 6 a²

Solution

_{Volume of cube = (side)³ = Therefore, side = Surface area = 6 ×(side)²}

Q43. How many litres of water flow out through a pipe having an area of cross-section of 5 cm² in one minute, if the speed of water in pipe is 30 cm/sec?

Solution

Area of cross section of pipe = 5 cm² Speed of water flowing out of the pipe = 30 cm/sec Volume of water that flows out in 1 sec.= 5 30 cm³ = 150 cm³ volume of water that flows out in 1 minute = 150 60 = 9000 cm³ = 9 litres

Q44. A cylindrical bucket of diameter 28 cm and height 12 cm is full of water. The water is emptied into a rectangular tub of length 66 cm and breadth 28 cm. Find the height to which water rises in the tub.  

Solution

Given: Diameter of cylindrical vessel = 28 cm ⇒ radius of cylindrical vessel (R) = 14 cm Height of cylindrical vessel (H) = 12 cm Let V denote the volume of the cylindrical vessel.                          Length of the rectangular tub (l) = 66 cm Breadth of the rectangular tub (b) = 28 cm Let h be the height to which the water rises and V₁ denote the volume of the rectangular tub.                          Now, volume of water in the rectangular tub is equal to the volume of water in the cylindrical tub. Therefore,

Q45. The total surface are of a solid cylinder is 231 cm² and its curved surface area is of the total surface area. Find the volume of the cylinder.

Solution

TSA of cylinder = 231 cm² 2r (h + r) = 231 ….(i) CSA of cylinder = = 154 cm² 2rh = 154 …..(ii) Substituting (ii) in (i) 2rh + 2r² = 231 154 + 2r² = 2312 r² = 231 - 154 = 77 = 269.5 cm³

Q46. An iron pipe 20 cm long has exterior diameter 25 cm. If the thickness of the pipe is 1 cm, find the whole surface area of the pipe.

Solution

Total surface area of pipe = 2Rh + 2rh + 2(R² - r²) = 2h(R + r) + 2[(R + r) (R - r)] = 2(R + r) (h + R - r)

Q47. Metal spheres, each of radius 2 cm are packed into a rectangular box of dimensions 16 cm 8 cm 8 cm. When 16 spheres are packed into the box, it is filled with preservative liquid. Find the volume of this liquid to the nearest integer. [Use= 3.14]

Solution

Volume of cubical box = l b h = 16 8 8 cm³ = 1024 cm³ Volume of sphere = _{= 33.49 cm³} Volume of 16 spheres = Volume of liquid = Volume of box Volume of spheres _{= (1024 535.84) cm³ = 488.16 cm³}

Q48. Two cubes of side 6 cm each are joined end to end. Find the surface area of the resultant cuboid.

Solution

Length of cuboid formed = (6 + 6) cm= 12 cm Base = 6 cm, Height = 6 cm Surface Area of Cuboid = 2 (lb + bh + hl) Surface Area of Cuboid = 2 [(12 6) + (6 6) + (6 12)]= 360 cm²

Q49. The curved surface area of a cylinder is 2200 sq cm and circumference of its base is 220 cm. Then the height of the cylinder is:

• 1) 22 cm
• 2) 5 cm
• 3) 2.2 cm
• 4) 10 cm

Solution

Q50. The radius and height of cylinder are in the ratio 5 : 7 and its volume is 550 cm³. Find its diameter. .

Solution

Let the radius be 5x and height be 7x respectively. Volume of cylinder = 550 cm³ r²h = 550 radius = 5x = 5 1 = 5 cm Diameter = 2r = 2 5 = 10 cm

Q51.

Solution

Q52. Three cubes of metal whose edges are 2 cm, 3 cm and 4 cm are melted and a new cube is made. Find the volume of the new cube.

Solution

Volume of a cube of edge = Volume of a cube of edge 2 cm (V₁) = 8 cm³ Volume of a cube of edge 3 cm (V₂) = 27 cm³ Volume of a cube of edge 4 cm (V₃) = 64 cm³ Three cubes of metal whose edges are 2 cm, 3 cm and 4 cm are melted and a new cube is made. Let V be the volume of the new cube. Therefore, V = V₁ + V₂ + V₃                  V = 99 cm³

Q53. If the radius of the base and altitude of a right circular cylinder and a right circular cone are equal. What is the relation between the volumes of a cylinder to that of a cone?

Solution

Radius of the cylinder = Radius of cone =r  Height of cylinder = Height of cone=h Let V denote the volume of the cylinder and V₁ denote the volume of the cone  and ⇒ ⇒ Thus, Volume of cylinder is 3 times the volume of cone.

Q54. The height of the cone is 15 cm. If its volume is 1570 cm³, find the diameter of the base. (use = 3.14)

Solution

Let the radius of the base of the cone be r cm. Height = 15 cm Volume of cone = 1570 cm³ Diameter = 2 10 cm = 20 cm.

Q55. How many litres of milk can a hemispherical bowl of radius 10.5 cm hold?

Solution

_{Radius of hemisphere = 10.5 cm} = 2425.5 cm³ = 2.4255 litres

Q56. A solid rectangular block of iron of dimensions 7 cm x 4 cm x 2.75 cm is recast into a cylinder of radius 7 cm. Find the height of the recast cylinder.

Solution

Dimensions of the rectangular block are Length (l) = 7 cm, breadth (b) = 4 cm and height (h) = 2.75 cm Volume of the rectangular block (V) = lbh = 77 cm³ Radius of recast cylinder (R) = 7 cm Let H be the height of the recast cylinder and V₁ denotes its volume. Rectangular block is recasted into a cylinder. Therefore volume of the cylinder is equal to the volume of the rectangular block. Thus, the height of the recast cylinder is 0.5 cm.

Q57. Find the cost of white washing the four walls of a room with dimensions 5 m 4m 3m at the rate of Rs. 12/m².

Solution

Area of four walls= 2(l + b)h                    (ceiling and floor is not to be included) = 2 3 (5 + 4) = 6 9 = 54 m² Cost @ Rs. 12/m² = Rs. 54 12 = Rs 648

Q58. From a right circular cylinder with height 8 cm and radius 6 cm, a right circular cone of the same height and base is removed. Find the total surface area of the remaining solid.

Solution

Cylinder - h = 8 cm,r=6 cm length of cone = T.S.A of remaining solid = curved surface area of cylinder + curved surface area of cone + area of 1(upper )base

Q59. The radius of a spherical balloon increases from 2.8 cm to 3.5 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases

Solution

Case I: when radius (r) = 2.8 cm Case II When radius (R) = 3.5 cm   Therefore, S_1: S₂ = 16:25

Q60. The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the cylinder is 1628 m², find the volume of the cylinder. .

Solution

Let r and h respectively be the radius and the height of the cylinder. We have: r + h = 37 m TSA = 1628 m² 2r (h + r) = 1628 2r(37) = 1628 Volume of the cylinder = r²h

Q61. The circumference of the base of the cone of slant height 35 cm is 88 cm. Find its curved surface area.

Solution

Slant height (l) = 35 cm. Let r  = radius of the cone Circumference of the base = 88 cm   The curved surface area is 1540 cm²

Q62. The curved surface area of a cylinder is 176 cm² and area of its base is 38.5 cm². Find the volume of the cylinder. (Take )

Solution

_{Area of the base =}r² = 38.5 cm² Volume = r²h = 38.5 = 308 cm³ Therefore, the volume of the cylinder is 308 cm³.

Q63. It costs Rs 15228 to paint the base of a cuboidal tank of height 10 m. If the cost of painting is Rs 9 per m², find the volume of the tank.

Solution

Given, height (h) =10 m Area of base of the tank painted for Rs. 9 = 1 m² Area of base of the tank painted for Rs. 15228 = 1692 m² Thus, Volume of the cuboidal tank = Area of base × h =1692×10 m³ = 16920 m³

Q64. The circumference of the base of a cone is 22 cm. If its height is 8 cm, find its volume.

Solution

Circumference of the base = 22 = 2r, where r is radius of cone h = 8 cm Volume of cone = ₌

Q65. The surface area of a sphere of radius 5 cm is five times the area of curved surface of a cone of radius 4 cm. Find the height and volume of a cone. .

Solution

Surface area of sphere = 5 (CSA of cone) 4 l 4 25 = 5 4 l l = 5 cm We know, h² = l² - r² = 5² - 4² = 9 h = 3 cm Volume of cone =

Q66. A metallic sheet is of the rectangular shape with dimensions 48 cm36 cm. From each one of its corner a square of 8 cm is cut off. An open box is made of the remaining sheet. Find the volume of the box.

Solution

Length of the box = 48 - 8 - 8 = 32 cm Breadth of the box = 36 - 8 - 8 = 20 cm Height = 8 cm Volume of the box formed = l b h = 32 20 8 = 5120 cm³

Q67. A cuboid of dimensions 50 cm × 40 cm × 12 cm is melted and transformed into three cubes of same size. Find the edge of the new cube.

Solution

Given: In a cuboid, Length (l) = 50 cm, breadth (b) = 40 cm, and let height (h) =12 cm Volume of the cuboid(V₁) = It is melted and transformed into three cubes of same size. Let ‘a’ be the edge of the formed cube. Volume of cube of edge ‘a’(V) = a³ 3 × V = V1  

Q68. The radius of a spherical balloon increases from 7 cm to 14 cm as air is pumped into it. Find the ratio of surface areas of the balloon in two cases.

Solution

Original surface areas S₁ = 4 New surface area S₂ =

Q69. A tent is in the form of a cylinder surmounted by a hemisphere. The height and radius of the cylindrical portion is 7.2 cm and 4.2 cm respectively. Calculate its volume.

Solution

Given: Radius of the cylindrical part = Radius of the hemispherical part (r) = 4.2 cm, Height of the cylindrical portion (h) = 7.2 cm Now, Total volume = volume of cylindrical part + volume of hemispherical part  

Q70. A joker's cap is in the form of a right circular cone of base with radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Solution

Radius = 7 cm, height = 24 cm Slant height l = Area of the sheet required for the cap = rl Sheet required for 10 such caps = 550 10 = 5500 cm².

Q71. A hollow cylindrical metal pipe is 80 cm long. Its internal and external radii are 8 cm and 10 cm respectively. It is melted and cast into a solid cylinder 20 cm long. Calculate the radius of the solid cylinder.

Solution

Given: Internal radius (r) =8 cm External radius (R) = 10 cm Length of the cylindrical tube (h) = 80 cm Let V denote the volume of metal in the tube.                                     Thus, volume of metal in the pipe is It is melted and recasted into a cylinder. Length of the cylinder (H) = 20 cm. Let R₁ be the radius of the recasted cylinder and V₁ denote its volume. Then, Thus, radius of the solid cylinder is 12 cm.

Q72. The hollow sphere, in which the circus motorcyclist performs his stunts, has a radius of 2.8 m. find the area available to the motorcyclist for riding.

Solution

Given:         Radius (r) = 2.8 m Let S be the area available to the motorcyclist in which he performs his stunts = Surface area of the sphere.                                                                          

Q73. The paint in a certain container is sufficient to paint an area equal to 74295 cm². How many bricks of dimensions 19.5 cm × 11 cm × 6.5 cm can be painted out of this container?

Solution

Given: length (l)=19.5 cm, breadth(b)=11 cm and height(h)=6.5 cm Surface area of the brick= 2{(19.5 × 11) + (11 × 6.5) + (19.5 × 6.5)} cm²                                                     = 825.50 cm² _{Let “N” number of bricks be painted by the given amount of paint.} Thus, N×825.50=74295

Q74. A tomb is in the form of a cylinder surmounted by a hemisphere. The radius and height of the cylindrical portion are 10.5 m and 9 m. calculate its volume.

Solution

Radius of the cylinder = Radius of hemisphere (r) = 10.5 m Height of cylinder ( h) = 9 m Volume of the tomb = Volume of hemispherical part + volume of cylindrical part                                    

Q75. The length, breadth and height of a room are 5 m, 4m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m².

Solution

Area to be whitewashed does not include the floor. Therefore, area to be whitewashed = Area of four walls + area of the ceiling = 2 (l + b) h + l b = [2 (5 + 4) 3 + 5 4] m =54 + 20 m² = 74 m² Cost of white washing = Rs 7.50 74 = Rs 555

Q76. The height of a solid cylinder is 16 cm and the radius of the base is 3.5 cm. Two equal conical holes each of radius 3.5 cm and height 3 cm are cut off from both ends. Calculate the volume of the remaining solid.

Solution

Given: Radius of the cylinder = Radius of cone (r) = 3.5 cm Height of cylinder (H) = 16 cm Height of cone (h) = 3 cm Volume of the remaining solid = Volume of cylinder (2 × volume of cone)

Q77. The lateral surface area of a cube is 100 m². The volume of the cube is:

• 1) 100 m³
• 2) 1000 m³
• 3) 125 m³
• 4) 25 m³

Solution

Lateral surface area = 4a² = 100 a² = 25 a = 5 cm Volume of the cube = a³ = (5)³ = 125 m³

Q78. A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the well is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm 12 cm 8 cm, how many bricks would be required?

Solution

Length of the wall = 10 cm = 1000 cm Height of the wall = 4 m = 400 cm Thickness of the wall = 24 cm Volume of the wall = 1000 400 24 cm³ Volume of each brick = 24 12 8 cm³ No. of bricks required = Since, the brick is to be taken as a whole, so to build the wall 4167 bricks are required.

Q79. 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 two cases.

Solution

Radius of the spherical balloon = r₁ = 7 cm Surface area S₁ of the balloon = 4 = 4(7)² Radius of the spherical balloon when air is pumped into it = r₂­ = 14 cm Surface area S₂ of the balloon = = 4(14)²

Q80. If 1 m² of sheet costs Rs. 400. Find the cost of sheet required to make a closed cylindrical tank of height 1.4 m and base diameter 210 cm.

Solution

Given: Base diameter of the tank = 210 cm Then, radius (r) = 105 cm Height (h) = 1.4 m = 140 cm Let S denote the total surface area of the tank.                           Cost of 1 m² of sheet = Rs. 400 Cost of 16.17 m² of sheet = Rs. (400 x16.17) = Rs. 6468

Q81. A rectangular block of iron metal of dimensions is recasted into a cylindrical vessel of height 14 cm. Calculate its radius.

Solution

Dimensions of the rectangular block are, Length (l) = 22 m, breadth (b) = 14 m and height (h) = 7 m Volume of tank (V) = lbh = 2156 m³ This rectangular block is recasted into a cylindrical vessel. Height of the cylindrical vessel (h) = 14 cm Let r be the radius of the cylindrical vessel and V₁ denote the volume of the vessel. Thus, volume of cylindrical wire = volume of cubical mass Thus, radius of the cylindrical vessel is 7 cm.

Q82. A cone of slant height 25 cm has a curved surface area 550 cm². Find the height of the cone and hence its volume. .

Solution

For a cone, l = 25 cm, CSA = 550 cm² Curved surface area =rl = 550 cm² h = 24 cm = 1232 cm³

Q83. A hemispherical dome of a fort needs to be polished. If the circumference of the base of the dome is 27.72 m, find the cost of polishing it at the rate of Rs 25 per 1000 cm².

Solution

Let r= radius of the base of a hemispherical tomb. Circumference of the base = 27.72 m=2772 cm   Cost of polishing 1000 cm² = Rs 25 Therefore, cost of polishing 1222452 cm² = Rs                                                             = Rs 30561.30

Q84. A joker’s cap is in the form of a right circular cone of base radius 3 dm and height 4 dm. Find the area of the sheet required to make 35 such caps.

Solution

Given: Radius (r) = 3 dm, height (h) = 4 dm. Let slant height = l            

Q85. A solid right circular cylinder of radius 8 cm and height 3 cm is melted to make a cone of height 3 times that of cylinder. Find the curved surface area of the cone. (use = 12)

Solution

Height of cone = 3 times height of cylinder = 3 3 = 9 cm Volume of cylinder = volume of cone r² = 8 8 r = 8 cm l² = h² + r² = (9)² + (8)² l = _{= 12 cm} C.S.A (cone) = _{= 301.71 cm²}

Q86. If the circumference of the base, a hemispherical dome of a building is 352 cm, find the lateral surface area of the dome.

Solution

Let r = radius of the base of a hemispherical tomb. Circumference of the base = 352 cm   Lateral surface area of dome is 19712 cm²

Q87. A hollow cylindrical pipe 21 dm long is made of copper. Its outer and inner diameters are 10 cm and 6 cm respectively. Find the volume of the copper used in making the pipe.

Solution

Let h = height of the pipe = 21 dm = 210 cm R = External Radius = r = Internal Radius = Volume of copper used in making the pipe = Volume of External cylinder - Volume of Internal cylinder

Q88. The diameter of a roller 1 m long is 56 cm. If it takes 200 revolutions to level a playground, find the cost of levelling the playground at the rate of 50 paisa per m².

Solution

Given: Diameter = 56 cm ⇒ radius (r) = 28 cm Length (h) = 1 m = 100 cm   Let S be the curved surface area of the roller. Area covered in 200 revolutions = 200×17600 cm² = 3520000 cm² = 352 m²

Q89. A conical tent is to accommodate 11 persons. Each person must have 4 m² of the space on the ground and 20 m³ of air to breathe. Find the height of the cone.

Solution

Let the radius of the conical tent be 'r' m and height be 'h' m. Space on the ground required by each person = 4 m² Area of base = 11 4 = 44 m² Air required by each person to breathe = 20 m³ Volume of cone = 11 20 = 220 m³ Area of the base of conical tent = Volume of cone = 220 _{Thus, height of the cone is 15 m.}

Q90. The cost of painting the total outside surface of a closed cylindrical oil tank at 60 paise per sq. dm is Rs. 237.60. The height of tank is 6 times the radius of the base of the tank. Find volume of cylindrical tank correct to two decimal places.

Solution

Q91. The curved surface area of a solid cylinder is 440 cm² and its height is 10 cm. Find the circumference of the base of the cylinder.

Solution

Given: Curved surface area(S) = 440 cm², height (h) = 10 cm Circumference of the base of the cylinder =   Thus circumference of the circle is 44 cm.

Q92. How many solid spheres of diameter 6 cm can be cast out of a solid cylinder with radius of the base 9 cm and height 40 cm? .

Solution

For a sphere, diameter = 6 cm radius (R) = 3 cm Now, Volume of sphere = For cylinder, radius (r) = 9 cm, height (h) = 40 cm Now, Volume of cylinder = No. of solid spheres = =

Q93. Find the height of the cuboidal tank with square base of area 169 m², if it can hold 253500 litres of water.

Solution

Given, area of the base of the tank = 169 m² Let height of the tank = h m Volume of the rectangular tank = Area of the base × height = 169 × h m³ = 169 h m³  A rectangular tank can hold 253500 litres of water = Thus we have 169 h = 253.5 Thus height of the tank is 1.5 m.

Q94. In a conical vessel of radius 8.4 m and vertical height 3.5 m, how many full bags of wheat can be emptied, if space for wheat in each bag is 1.96 m³? .

Solution

Volume of conical vessel = No. of bags =  

Q95. A lead pencil consists of cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of graphite is 1 mm. If the length of the pencil is 14 cm. Find the volume of wood and that of graphite.

Solution

Q96. How many m² of tin would be required to line the inside of a rectangular tank on the sides and bottom if the inside dimensions of the tank are 8 m × 5 m × 4 m.?

Solution

Given: length (l)=8 m, breadth(b)=5 m and height(h)=4 m Area of four walls and base=                                                 =2(8+5)4+(8 x 5)                                                 =144 m²                                                                 Thus 144 m² of tin is required to line the inside of the tank.

Q97. How many cylindrical glasses of 3 cm base radius and height 8 cm can be refilled from a cylindrical vessel of base radius 15 cm which is filled up to a height of 32 cm?

Solution

Volume of a glass = ₌ Volume of the vessel = Number of glasses that can be filled = = 5 5 4 = 100

Q98. If the ratio of the radii of two sphere is 2 : 3, then the ratio of their volumes is:

Solution

r₁ : r₂ = 2 : 3 Or V₁ : V₂ = 8 : 27

Q99. The slant height and altitude of a conical tomb are 15 m and 12 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 70 per m².

Solution

Altitude (h) = 12 m, slant height (l) =15 m. Let radius = r   Thus, r = 9 m.  Cost of white washing 1 m² of canvas = Rs. 70 Therefore, cost of whitewashing  = Rs. (×70)                                                             = Rs. 29700

Q100. A hemispherical dome of a building is to be white washed and the total cost of white wash of dome building is Rs. 924 at the rate of Rs. 3 per m²_­ then find the (a) Surface area of hemisphere (b) Volume of air in the dome

Solution

(i) Total cost of white wash = Rs. 924 Cost per m²= Rs. 3 Total area to be white wash = = 308 m² Surface area of hemisphere = 308 m² (ii) Now, surface area of hemisphere = 308 m² 2r² = 308 m² r = 7 m Volume of hemisphere = 718.66 m² air is present inside the dome.

Q101. A building is cylindrical to a height of 9 m and conical above it. If the base radius is 15 m and height of the cone is 8 m, find the area of the building.

Solution

Diameter= 24 m ⇒ radius =12 m Radius of the conical part = Radius of the cylindrical part (r) = 15 m Height of cylindrical part (h) = 9 m, height of the cone (h) = 8 m For the conical part of the building, Thus, l = 17 m Surface area of the building = Curved surface area of the conical part + curved surface area of the cylindrical part   The surface of the building is 1650 m²

Q102. Find the length of 2 m wide tarpaulin required to make conical tent of height 6 m and base diameter 16 m?

Solution

Given: Diameter = 16 m  Radius (r) = 8 m, height (h) = 6 m Let slant height = l Breadth of tarpaulin (B) = 2 m, let length of tarpaulin = L _{Area of tarpaulin required = curved surface area of the tent}   Area of tarpaulin required is 125.7 m 

Q103. 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.

Solution

Outer radius R = 5 + 0.25 = 5.25 cm Outer curved surface area = 2R²

Q104. The surface area of a sphere is 5544 cm². Calculate its volume.

Solution

Let r = radius of the sphere. Surface area of the sphere = 5544 cm²   Let V be its volume.  

Q105. The radii of two right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 4. Find the ratio of their curved surface areas.

Solution

Let r₁ = 2x, r₂ = 3x h₁ = 5h, h₂ = 4h Curved surface area of cylinder = 2 rh Ratio of Curved Surface Areas =

Q106. Three spheres of radii 3 cm, 4 cm and 5 cm are melted together to form a single sphere. Find the radius of new sphere.

Solution

Let R be the radius of the new sphere. Volume of new sphere = sum of volumes of three spheres = (3 cm)³ + (4 cm)³ + (5 cm)³ = 27 cm³ + 64 cm³ + 125 cm³ = 216 cm³ = (6 cm)³ R = 6 cm Thus, the radius of the new sphere is 6 cm.

Q107. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square meters of metal sheet would be needed to make it? .

Solution

Height of the vessel = 1 m Capacity of the vessel = 15.4 litres . Volume of the cylindrical vessel = r²h. r²h = 0.0154 r = 0.07 m Metal sheet needed to make the cylindrical vessel. = Total surface area of the cylindrical vessel. = 2r(r + h) = 0.4708 m² Therefore, 0.4708 m² of sheet will be required to make the vessel.

Q108. The volume of a right circular cylinder is 3850 cubic cm. Find its height if it's diameter is 14 cm.

Solution

Volume of cylinder = r² h = 3850cm³ = 25 cm

Q109. Find the curved surface area of a 44 m long metal pipe if diameter of its cross section is 4.2 cm.

Solution

  Given: Diameter=4.2 cm ⇒ radius (r) = 2.1 cm Length (h) =44 m= 4400 cm   Let S be the curved surface area of the roller    

Q110. Find the volume of metal used in making a hollow hemispherical bowl with internal and external diameters 4 m and 10 m respectively. .

Solution

External radius = R = 5 m Internal radius = r = 2 m Volume of metal =

Q111. An open cylindrical vessel has base diameter 14 cm and height 21 cm. Find the cost of tin plating its inner surface area at the rate of Rs 3 per 10 cm². .

Solution

Inner CSA = 2rh = = 924 cm² Cost @ Rs 3 per 10 cm² = Rs 3 = Rs 3 = Rs 277.20

Q112. The length of a room is double its breadth. Its height is 3 m. The area of four walls excluding a door of dimensions 4 m 2 m is 100 m². Find the volume.

Solution

l = 2b, h = 3m Area of four walls (including the door) = 100 + 8 = 108 m² 108 = 2(l + b)h 108 = 2(2b + b) 3 108 = 6(3b) = 3b 18 = 3b b = 6 m l = 12 m Volume = l b h = 12 6 3 = 216 m³

Q113. A cube and cuboid have same volume. The dimensions of the cuboid are in the ratio 1 : 2 : 4. If the difference between the cost of polishing the cube and cuboid at the rate of Rs. 5 per m² is Rs. 80, find their volumes.

Solution

Let the dimensions of cuboid be x, 2x, 4x Volume of cuboid = 8x³ Volume of cube = 8x³ = side³ So, side of cube = 2x TSA of cuboid = 2(lb + bh + hl) = 2(2x² + 8x² + 4x²) = 28x² Cost of polishing the cuboid = Rs. 5 28 x² = 140x² TSA of cube = 6(side)² = 6(2x)² = 24x² Cost of polishing the cube = Rs. 5 24 x² = 120x² Now, from the given information, we have 140x² - 120x² = 80 20x² = 80 x² = 4 x = 2 Volume of cuboid = Volume of cube = 8 (2)³ cubic cm= 64 cubic m

Q114. The dimensions of a match box is 5 cm × 3 cm × 2 cm. Find the volume of a cardboard box containing 10 such match boxes.

Solution

Dimensions of the match box are Length (l) = 5 cm, breadth (b) = 3 cm and height (h) = 2 cm Volume of the match box = Length x Breadth x Height = 5 x 3 x 2 = 30 cm³ Thus, Volume of cardboard box containing 10 such matchboxes = 10 x 30 cm³ = 300 cm³

Q115. The volume of a cylinder is 350 cm³ and its height is 14 cm. Find the curved surface area.

Solution

Volume of cylinder = 350 cm² Height (h) = 14 cm Volume of cylinder = r²h 350 = r² 14 r = r = 5 cm Curved surface area of cylinder = 2rh = 440 cm²

Q116. Two hemispherical bowls are formed by cutting a spherical ball of diameter 28 cm from the centre. Calculate the total surface area of each of the hemispherical bowls so formed.

Solution

Diameter = 28 cm⇒ radius (r) =14 cm. Let S be the total surface area of the hemispherical bowl.  

Q117. A solid spherical ball of diameter 4.2 cm is completely immersed in water. How much water is displaced?

Solution

Radius of ball (r) = = 2.1 cm Volume = r³ = = 38.808 cm³ Quantity of water displaced = volume of ball immersed = 38.808 cm³

Q118. A cubical block of edge 22 cm is melted into small spherical balls of radius 1 cm. Calculate the number of balls that can be made from it.

Solution

Volume of cubical block = a³ = cm³ = 10648 cm³ Volume of spherical ball = = = 4.19 cm³   Number of spherical balls = =

Q119. There are two cones. The C.S.A. of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of the radii.

Solution

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