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Q1. How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid of dimensions 11 cm 10 cm 7 cm? (take )

Solution

Diameter of the coin = 1.75 cm Radius, r = Height = 2 mm = 0.2 cm Volume of 1 coin = r²h = 0.2 cm³ Volume of cuboid = 11 10 7 cm³ Since, the coins are melted to form a cuboid, so the total volumes of all coins will be equal to the volume of the cuboid. Let the number of coins be n. So, volume of n coins = volume of cuboid n = = 1600

Q2. If the radius and slant height of a cone are equal, then the surface area of the cone will be 

• 1) 2πl²
• 2) A or B
• 3) 2πl²
• 4) 2πr²

Solution

The surface area of a cone = 𝜋r(l + r) According to the question, l = r The surface area of the cone = πr(l + l) = πr(r + r) = 2πr² or 2πl² 

Q3. The interior of a building is in the form of a right circular cylinder of diameter 4.2 m and height 4m surmounted by a cone of vertical height 2.1 m. Find the volume of the building.

Solution

Q4. A cylinder and a cone are of the same base radius, but the height of the cone is four times that of the cylinder. The ratio of the volume of the cylinder to that of the cone is 

• 1) 1:2
• 2) 3:5
• 3) 3:4
• 4) 3:2

Solution

Volume of a cylinder = πr²h According to the question,  For a cone: H = 4h, where H and h are the heights of the cone and cylinder, respectively. Volume of the cone = The ratio of the volume is  

Q5. If a solid right circular cone of height 24 cm and base radius 6 cm is melted and recast in the shape of a sphere, then the radius of the sphere is

• 1) 8 cm
• 2) 4 cm
• 3) 12 cm
• 4) 6 cm

Solution

Here the volume of the cone will be equal to the volume of the sphere as the sphere is formed by the melting the cone. Let the radius of sphere = R Radius of cone = r, Height of cone = h Thus, we have

Q6. The internal radii of the ends of a bucket, full of milk and of internal height 16 cm are 14 cm and 7 cm. If this milk is poured into a hemispherical vessel, find the internal diameter of the vessel.

Solution

For bucket, h = 16 cm r₁ = 14 cm r₂ = 7 cm V₁ = Volume of bucket =   Let R cm be the radius of the hemispherical vessel V₂ = volume of vessel = V₁ = V₂ R = 14 cm Internal diameter of vessel = 2R = 28 cm

Q7. A container is in the form of the frustum of a cone. If its height is 16 cm and the radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the slant height of the container and also the cost of milk that the container can hold, if the cost of milk is Rs. 30/litre ( = 3.14)

Solution

R = 20 cm, r = 8 cm, h = 16 cm Slant height = = Volume =  

Q8. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder of same radius. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Solution

Radius of hemisphere = Radius of cylinder = 7 cm Height of cylinder = 13 cm - 7 cm = 6 cm Inner S.A. of vessel = Curved S.A. of cylindrical part + Curved S.A of hemispherical part

Q9. The diameter of a sphere is 12 cm. It is melted and drawn into a wire of diameter 6 mm. The length of the wire is 

• 1) 3.2 cm
• 2) 32 mm
• 3) 0.32 m
• 4) None of these

Solution

For a sphere, d = 12 cm, r = 6 cm For a cylindrical wire, d = 0.6 cm, r = 0.3 cm Let the length of the wire be x cm. According to the question, π(0.3)²h = h = 3200 cm = 32 m

Q10.

Solution

Q11. A cylinder whose height is 4/3 of its radius has the same volume as a sphere of radius 8 cm. The radius of the base of the cylinder will be

• 1) 2 cm
• 2) 4 cm
• 3) 5 cm
• 4) 8 cm

Solution

For a cylinder: h = Volume of the cylinder = πr²h = = For a sphere: d = 8 cm, r = 4 cm Volume of the sphere =   According to the question, The radius of a sphere is 8 cm.

Q12. A solid is hemispherical at the bottom and conical above. If the curved surface area of the two parts are equal, then find the ratio of the radius and height of the conical part.

Solution

S.A. of hemisphere = S.A. of cone 2r² = rl 2r = l So, h = h = h = r r : h = r : r = 1 :

Q13. A right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled in cones of height 12 cm and diameter 6 cm having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Solution

 

Q14. The given figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other end. Find the curved surface area of the solid.

Solution

Q15. If the radius of base of a cylinder is doubled and the height remains unchanged, its curved surface area becomes

• 1) Double
• 2) three times
• 3) Half
• 4) No change

Solution

Original radius = r (say) New radius = 2r Height = h Original curved surface area = New curved surface area = Thus, the curved surface area also becomes double.

Q16. How many solid spheres of diameter 6 cm are required to be melted to form a solid metal cylinder of height 45 cm and diameter 4 cm?

Solution

Let n be the required number of spheres. Since, the spheres are melted to form a cylinder. So, the volume of all the n spheres will be equal to the volume of the cylinder. n = 5 Thus, the required number of spheres which are melted to form the cylinder is 5.

Q17. A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted and cast into a cone of base diameter 8 cm. Find the height of the cone.

Solution

External diameter of sphere = 8 cm radius, R = 4 cm Internal diameter of sphere = 4 cm radius, r = 2 cm Diameter of the base of cone = 8 cm radius = 4 cm Now, Volume of the cone = Volume of the material in the sphere Hence, the height of the cone = 14 cm      

Q18. The diameter of a metallic solid sphere is 9 cm. It is melted and drawn into a wire having diameter of cross-section as 0.2 cm. Find the length of the wire.

Solution

For sphere: d = 9 cm r = cm Volume of sphere = r³ = cu. cm For wire/ cylinder: Diameter of cylinder = 0.2 cm R = 0.1 cm Volume of cylinder = r²h = (0.1)² h The sphere is melted to form a cylindrical wire. Volume of sphere = volume of cylinder = (0.1)²h h = = 12150 cm = 121.5 m Thus, the length of the wire is 121.5 m.

Q19. Radii of the cylinder and sphere are the same, and the height of the cylinder is twice its radius. Find the ratio of the volume of the cylinder to the sphere.

• 1) 3:5
• 2) 1:2
• 3) 3:4
• 4) 3:2

Solution

For a cylinder: h = 2r Volume of the cylinder = πr²h = πr²(2r) = 2πr³ For a sphere: Volume of the sphere =   The required ratio is  

Q20. Three cubes of volume 64 cm³ each are joined end to end to form a solid. Find the surface area of the cuboid so formed.

Solution

Volume of each cube = 64 cm³ Thus, side of each cube = 4 cm So, the dimensions of cuboid are 4 4 12. Surface area of the cuboid = 2[4 4 + 4 12 + 12 4] = 224 sq. cm

Q21. The slant height of the frustum of a cone is 5 cm. If the difference between the radii of its two circular ends is 4 cm, find the height of the frustum.

Solution

Let r and R be radii of the circular ends of the frustum of the cone. Then, R - r = 4, l = 5 We know l² = (R - r)² + h² 5² = 4² + h² h² = 25 - 16 = 9 h = 3 Thus, the height of the frustum is 3 cm.

Q22. If the radius of the base of a right circular cylinder is halved, keeping the height same, find the ratio of the volume of the reduced cylinder to that of the original cylinder.

Solution

Q23.

Solution

Q24. The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is

• 1) 120 cm²
• 2) 68 cm²
• 3) 136 cm²
• 4) 60 cm²

Solution

Height = 15 cm Diameter = 16 cm So, radius = 8 cm Slant height = l = Curved surface area =

Q25. A milk container is made of a metal sheet in the form of a frustum of a cone of height 16 cm, with radii of lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk which the container can hold when full at Rs. 20 per litre.

Solution

Height, h = 16 cm Radius of upper end = R = 20 cm Radius of lower end = r = 8 cm Volume of container = = 10449.92 cm³ = 10.45 litres Cost of milk = Rs. (20 10.45) = Rs. 209

Q26. A solid sphere of diameter 14 cm is cut into two halves by a plane passing through the centre. Find the combined surface area of the two hemispheres so formed.

Solution

Radius of each hemisphere = cm = 7 cm Surface area of one hemisphere = 3r² Surface area of two hemispheres (same radius) = 6r² = = 924 cm²

Q27. A vessel in the form of a hemispherical bowl is full of water. Its water is emptied in to a cylinder. The internal radii of bowl and the cylinder are cm and 7 cm respectively. Find the height of water in the cylinder.

Solution

Radius of hemispherical vessel = cm = cm Volume of water in hemispherical vessel = Let h be the height of water in the cylinder. Radius of cylinder = 7 cm Volume of water in cylinder = cm³ h = cm = 15.75 cm

Q28. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume is one- twenty seventh of the given cone, at what height above the base, the section is made.

Solution

Q29. A cylindrical copper rod of diameter 1 cm and length 8 cm is drawn into a cylindrical wire of length 18 m and uniform thickness. Find the thickness of the wire.

Solution

Volume of the rod = The cylindrical copper rod is drawn into a cylindrical wire. So, their volume is the same. Length of the new wire = 18 m = 1800 cm Let r be the radius of the cross-section of the wire. Volume of the wire = So, the diameter of the cross-section, i.e., the thickness of the wire is  

Q30. If the surface area of a sphere is 144 cm², then its radius is

• 1) 10 cm
• 2) 8 cm
• 3) 12 cm
• 4) 6 cm

Solution

Let r be the radius of the sphere. Surface area of a sphere = 4r² 144 = 4r² r² = 36 r = 6 cm

Q31. A wooden article was made by scooping out a hemisphere from each end of a cylinder. If the height of the cylinder is 15 cm, and the base is of diameter 14 cm, find the total surface area of the article when it is ready.

Solution

 

Q32. A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is filled in small cylindrical bottles of diameter 3 cm and height 4 cm. Determine the number of bottles used to empty the bowl.

Solution

 

Q33.

Solution

Q34. The surface area of a sphere is equal to the surface area of a cube whose radius is 21 cm. Find the side of the cube. 

• 1) 30.39 cm
• 2) 40.42 cm
• 3) 45.41 cm
• 4) 35.43 cm

Solution

Surface area of a sphere = 4𝜋r² Surface area of a cube = 6l² According to the question, 4𝜋r²= 6l²  l = 30.39 cm

Q35. If the radii of circular ends of frustum of a cone are 20 cm and 12 cm and its height is 6 cm, then the slant height of frustum (in cm) is:

• 1) 8
• 2) 15
• 3) 12
• 4) 10

Solution

Q36. The volume of a largest sphere than can be cut from cylindrical log of wood of base radius 1 m and height 4 m is:

• 1)
• 2)
• 3)
• 4)

Solution

Let r be the radius of the largest sphere which can be cut off from the cylindrical log of wood whose radius is 1 cm and height is 4 cm. From the given figure, we have: r = 1 cm Volume of the largest sphere that can be cut off =

Q37. The volume of a cube is half of its surface area. The side of the cube (in cm) is  

• 1) 9
• 2) 2
• 3) 3
• 4) 6

Solution

The surface area of a cube is 6s². The volume of the cube is s³. According to the question,   s ≠ 0  Hence, s = 3 cm.

Q38. A bucket is made up of a metal sheet in the form of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the bucket if the cost of the metal sheet is Rs. 15 per 100 cm².

Solution

Q39. How many lead shots, each 0.3 cm in diameter, can be made from a cuboid dimensions 9 cm 11 cm 12 cm

Solution

Volume of cuboid = Total volume of lead shots=no. of lead shots Volume of one lead shot radius of lead shots= Let n be the number of lead shots. 9 11 12 = n n = = 84000

Q40. 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 total surface area of the remaining solid.

Solution

         

Q41.

Solution

Q42. A shuttle cock used for playing badminton has the shape of the combination of :

• 1) A hemisphere and frustum cone
• 2) A cylinder and a hemisphere
• 3) A sphere and a cone
• 4) A cylinder and a sphere

Solution

A shuttle cock used for playing badminton has the shape of the combination of a hemisphere and frustum cone.

Q43. The volume of a sphere (in cu. cm) is equal to its surface area (in sq. cm). The diameter of the sphere (in cm) is

• 1) 4
• 2) 3
• 3) 6
• 4) 2

Solution

Volume of the sphere = Surface area of the sphere Thus, the diameter of the sphere is 6 cm.

Q44. The rain water from a roof 20 m by 20 m drains out into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the vessel is just full, find the rainfall in cm.

Solution

Hence, the rainfall = 2.75 cm

Q45. A semi-circular sheet of paper of diameter 28 cm is bent into an open conical cup. Find the depth and capacity of the cup.

Solution

When the semi-circular sheet is bent into an open conical cup, the radius of the sheet becomes the slant height of the cup and the circumference of the sheet becomes the circumference of the base of the cone. l = Slant height of the conical cup = 14 cm Let r be the radius of and h be the height of the conical cup. Circumference of the base of the conical cup = circumference of the sheet 2r = 14 r = 7 cm h = = 12.12 cm Capacity of the cup = Volume of the cup = = 622.16 cm³

Q46.

Solution

Q47. If a cone is cut into two parts by a horizontal plane passing through the mid-points of its axis, the ratio of the volumes of the upper part and the cone is

• 1) 1:8
• 2) 1:2
• 3) 1:6
• 4) 1:4

Solution

Let the height and radius of the given cone be H and R respectively. Let the radius of smaller cone be r cm. In OCD and OAB, OCD = OAB = 90° COD = AOB (common) (AA similarity criterion) (Corresponding sides are proportional) Volume of smaller cone = Volume of given cone Ratio of volume of upper part to the cone is 1:8.

Q48. A rectangular sheet of paper 44 cm 18 cm is rolled a along its length (44 cm) and a cylinder is formed. Find the volume of the cylinder.

Solution

Given, a rectangular sheet of paper 44 cm 18 cm is rolled a along its length to form a cylinder. Height of cylinder = 18 cm Circumference = 44 cm r = 7 cm Volume of the cylinder = = 2772 cm³

Q49. A tent is of the shape of a right circular cylinder upto a height of 3 metres and conical above it. The total height of the tent is 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per square metre, if the radius of the base is 14 metres.

Solution

CSA of Tent = CSA of cylinder + CSA of cone = 2rh + rl = 2 (14) (3) + (14) = 264 + 14 =264 + 14 (17.5) = 264 + 770 = 1034 m². Cost of painting the inside of tent, i.e., 1034 m² at the rate of Rs. 2 per sq m = Rs 1034 x 2 = Rs. 2068.

Q50. The diameter of a sphere is 9 cm. It is melted and drawn into a wire of diameter 9 mm. The length of the wire is

• 1) 6 km
• 2) 6 mm
• 3) 6 cm
• 4) 6 m

Solution

For a sphere, d = 9 cm, r = 4.5 cm For a cylindrical wire, d = 0.9 cm, r = 0.45 cmLet the length of the wire be x cm. According to the question, Π(0.45)²h = h = 600 cm = 6 m 

Q51. A cylinder and a cone are of the same base radius, but the height of the cone is double that of the cylinder. The ratio of the volume of the cylinder to that of the cone is

• 1) 3:2
• 2) 1:2
• 3) 3:5
• 4) 3:4

Solution

Volume of the cylinder = πr²h According to the question, For a cone: H = 2h, where H and h are the heights of the cone and cylinder, respectively. Volume of the cone = The ratio of the volume is  

Q52. A solid sphere of radius 3 cm is melted and drawn into a long wire of uniform circular cross-section. If the length of the wire is 36 m, find its radius.

Solution

Wire is cylindrical in shape. Height of the wire = 36 m = 3600 cm Let r be the radius of the wire. Since, the sphere is melted to form a wire, so the volume of the sphere will be equal to the volume of the wire. r = = 0.1 cm = 1 mm Thus, the radius of the wire is 1 mm.

Q53. The radii of two cylinders are in the ratio 2:3. If their heights are in the ratio 3:5, then the ratio of their curved surface areas is 

• 1) 3:5
• 2) 1:2
• 3) 2:7
• 4) 2:5

Solution

Curved surface area of a cylinder = 2πrh  Let r₁ and r₂ be the radii of the cylinders and h₁ and h₂ be the heights of the cylinders. Ratio of curved surface areas =   

Q54. Water in a canal 6 m wide and 1.5 m deep is flowing with a speed of 10 km/h How much area will it irrigate in 30 minutes if 8 cm of standing water is needed?

Solution

Consider an area of cross section of canal be ABCD.   Area of cross section = 6 × 1.5 = 9 m²   Speed of water = 10 km/h =    Volume of water that flows in 1 minute from canal = = 1500 m³   Volume of water that flows in 30 minutes from canal = 30 × 1500 = 45000 m³   Let irrigated area be A. Volume of water irrigated in the required area will be equal to the volume of water flowed in 30 minutes from canal.   Volume of water that flows in 30 minutes from canal = Volume of water irrigated in the required area   So, area irrigated in 30 minutes is 562500 m².

Q55. A solid iron pole having cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that the mass of 1 cm³ of iron is 8 gm.

Solution

Q56. A hemispherical bowl of internal diameter 36 cm is full of liquid. This liquid is to be filled in cylindrical bottles of radius 3 cm and height 6 cm. How many such bottles are required to empty the bowl?

Solution

Volume of liquid in hemispherical bowl Volume of liquid filled in one cylindrical bottle = (3)² (6) cm³ Number of bottles required to empty the bowl = 72

Q57.

Solution

Q58. A well of diameter 3 m is dug 14 m deep. The earth taken out of it is spread evenly all around it to a width of 4 m, to form an embankment. Find the height of the embankment. (Use ).

Solution

Volume of earth dug out = = 99 m³ Let h be height of embankment. Now, volume of embankment = Volume of earth taken out

Q59. If the height of a right circular cylinder is doubled keeping the radius the same, then the ratio of the volume of the cylinder thus obtained to the volume of the original cylinder is 

• 1) 1:2
• 2) 1:3
• 3) 2:1
• 4) 1:4

Solution

Volume of the cylinder = πr²h The height is doubled keeping the radius the same. Volume of the new cone = πr²(2h) = 2πr²h The ratio of the volume of the new cylinder to the original one is 2:1.

Q60. A container, open from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at the rate of Rs.20/ litre. Also find the cost of metal sheet used to make the container if it costs Rs. 8 per 100 cm².

Solution

Q61. A girl fills a cylindrical bucket 32 cm in height and 18 cm in radius with sand. She empties the bucket on the ground and makes a conical heap of the sand. If the height of the conical heap is 24 cm, find the radius and the slant height of the heap. (Answer correct to one place of decimal)

Solution

Q62. A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm³ of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of the metal sheet used in its making.

Solution

Q63. A bucket made up of a metal sheet is in the form of a frustum of cone of height 16 cm with radii of its lower and upper ends 8 cm and 20 cm respectively. Find the cost of the bucket if the cost of metal sheet used is Rs. 15 per 100 cm². (use = 3.14)

Solution

Here r₁ = 8 cm, r₂ = 20 cm, h = 16 cm l = Surface area of the bucket = l (r₁ + r₂) + r₁² = 3.14 20 28 + 3.14 64 = 3.14 (560 + 64) = 1959.36 sq. cm Cost of 100 sq. cm metal = Rs. 15 Cost of 1 sq. cm metal = Rs. 0.15 Cost of 1959.36 sq. cm metal = Rs. 293.90

Q64. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has 8 gm mass.

Solution

V_pole = V_{small cylinder} + V_{big cylinder} V_pole = (4)² (60) + (12)² 220 = 102490 cm³ Mass of the pole = 102490 8 gms = 819920 gms 819.9 kg

Q65. A circus tent consists of a cylindrical base surmounted by a conical roof. The radius of the cylinder is 20 m. The height of the tent is 63 m and that of the cylindrical base is 42 m. Find the area of the canvas used for making it.

Solution

Q66. Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tank will rise by 21 cm.

Solution

For the tank, vol = l b h = 50 44 0.21 m³ 'r' of pipe = 0.07 m Vol of pipe = vol of tank of height 21 cm r²h₁ = 50 44 0.21 22/7 0.07 0.07 h₁ = 50 44 0.21 h₁ = 30,000 = 30 km Water flows @ 15 km/hr, so time taken to cover 30 km = 30/15 = 2 hours

Q67. A circus tent made of canvas is in the form of right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 21 m and 4 m respectively. The slant height of conical part is 12.5 m. Find the area of the canvas used for making the tent. Also, find total cost of the tent if the canvas used is Rs. 12/sq metre.

Solution

Slant height of conical part, l = 12.5 m Radius of conical part = = 412.5 m² Total Surface area = (264 + 412.5) m² = 676.5 m² Cost = Rs 12 676.5 = Rs. 8118

Q68. The rain water from a roof 22 m 20 m drains into a conical vessel having the diameter of base as 2 m and height 3.5 cm. If the vessel is just full, find the rainfall in mm.

Solution

Let the rainfall be h m. Volume of cuboid = (22 20 h) cu. m Volume of conical vessel = 1 3.5 cu m Volume of cuboid = Volume of conical vessel 22 20 h = 1 3.5

Q69. The rain water from a roof 22 m 20 m drain in to a conical vessel having diameter of base as 2 m and height 3.5 m. If the vessel if just full, find the rain fall in cm.

Solution

Let the rain fall be x cm. Volume of water collected on roof = Volume of water in conical vessel =

Q70. A solid is in the form of a right circular cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder and the hemispheres is 7 cm. Find the volume and total surface area of the solid.

Solution

Let r be radius and h be the height of cylinder. r = cm, h = 19 - 2 = 12 cm Volume of solid = Vol. of cylinder + Vol. of two hemispheres = = 641.66 cm³ T.S.A. = C.S.A. of cylinder + S.A. of two hemisphere. = 2rh + 2 (2 r²) = 2r (h + 2r) = = 22 19 cm² = 418 cm²

Q71. A toy is in the form of a cone of radius 3.5 cm surmounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy

Solution

Radius of hemisphere = 3.5 cm Total height of the toy = 15.5 cm Height of the cone = 15.5 c m - 3.5 cm = 12.0 cm Slant height (l) of the cone = Thus, Lateral surface of the conical part = = 137.5 sq. cm Curved surface of hemisphere = 2r² = Total surface of the toy = 137.5 + 77 = 214.5 cm²

Q72. A cylindrical container 2 m high and 3.5 m in diameter has a hemispherical lid. Find its volume.

Solution

 

Q73. A spherical copper shell, of external diameter 18 cm, is melted and recast into a solid cone of base radius 14 cm and height cm. Find the inner diameter of the shell.

Solution

External radius of copper shell, r_c = 9 cm Let the internal radius of copper shell be r_i. Radius of cone, r_c = 14 cm Height = h_c = cm = cm Since, the spherical copper shell is melted and recast into a solid cone, their volumes are the same. (r_c³ - r_i³) = r_c² h_c

Q74. A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy.

Solution

Q75. Find the volume of the largest right circular cone that can be out of a cube of side 4.2 cm.

Solution

Diameter of base of the cone = Side of the cube Radius of base of the cone = Volume of cone cut out = 19.404 cm³

Q76. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 such gulab jamuns, each shaped like a cylinder with two hemispherical ends with total length 5 cm and diameter 2.8 cm.

Solution

It is given that each gulab jamuns is shaped like a cylinder with two hemispherical ends. Radius of cylinder and hemispherical ends = 1.4 cm Height of cylinder = 5 cm - 1.4 cm - 1.4 cm = 2.2 cm Volume of 1 gulab jamun = = = = = = 28.7 cm³ Volume of 45 gulab jamuns = 45 28.7 cm³ 1292 cm³ Volume of syrup in 45 gulab jamuns cm³ 387.6 cm³

Q77. A solid is made of a cylinder with hemispherical ends. If the entire length of the solid is 108 cm and the diameter of the hemispherical ends is 36 cm, find the cost of polishing the surface of the solid at the rate of 7 paise per sq. cm.

Solution

Q78. The radius and slant height of a right circular cone are in the ratio of 7: 13 and its curved surface area is 286 cm². Find its radius. (Use  = )

Solution

For a given cone, r = 7a , l = 13a Thus, Curved Surface Area = = (7a)(13a) = 91a² = 286 or a² = = 1 a = 1 Radius of the cone = 7 cm.

Q79. Solid spheres of diameter 6 cm each are dropped into a cylindrical beaker containing some water and are fully submerged. The water in the beaker rises by 40 cm. Find the number of solid spheres dropped into the beaker if the diameter of the beaker is 18 cm.

Solution

Radius r of sphere = = 3 cm Volume of sphere = r³ If R is the radius and H is the height of the cylinder, then Volume of water in the cylindrical beaker = R²H Volume of rise in level of water = R² (40) If 'N' is the number of spheres dropped in the beaker then volume of spheres = Volume of water rose in the beaker N r³ = R²(40) N (3)³ = 9 9 40 N = 90

Q80. A lead pencil consists of cylinder of wood with a solid cylinder of graphite fitted into it. The diameter of pencil is 7 mm. The diameter of graphite is 1 mm and length of pencil is 10 cm. Calculate the weight of the whole pencil in grams if the density of wood is 0.6 gm/cm³and of graphite 2.3 gm/cm³.

Solution

Q81. A bucket is 18 cm in diameter at the top and 6 cm in diameter at the bottom. If it is 8 cm high, find its capacity. Also, find the area of sheet used in making the bucket.

Solution

Here, R = 9 cm r = 3 cm h = 8 cm Capacity of the bucket = 8 (9² + 3² + 9 3 ) = 312 cm³ Area of the metallic sheet used in making the bucket = Curved surface area of frustum + r² = [ℓ (R + r)] + r² = = 120 + 9 = 129 cm²

Q82. The radius of the base and the height of a right circular cylinder are in the ratio of 2: 3 and its volume is 1617 cu. cm. Find the curved surface area of the cylinder.

Solution

Let the radius of the base of the cylinder be 2x and its height be 3x. _{Volume of a cylinder =} (2x)² (3x) = 1617 x = = 3.5 cm r = 7 cm, h = 10.5 cm Curved surface area of the cylinder = 2rh

Q83. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Solution

Consider an area of cross section of canal be ABCD. Area of cross section = 6 × 1.5 = 9 m² Speed of water = 10 km/h = Volume of water that flows in 1 minute from canal = =1500m³ Volume of water that flows in 30 minutes from canal = 30 × 1500 = 45000 m³ Let irrigated area be A. Volume of water irrigated in the required area will be equal to the volume of water flowed in 30 minutes from canal. Volume of water that flows in 30 minutes from canal = Volume of water irrigated in the required area A = 562500 m² So, area irrigated in 30 minutes is 562500 m².

Q84. A shuttle cock used for playing badminton has the shape of a frustum of  a cone mounted on a hemisphere. The external diameters of the frustum are 5 cm and 2 cm and the height of the entire shuttle cock is 7 cm. Find its external surface area.

Solution

   

Q85. The internal and external radii of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted to form a solid cylinder of height 10 cm, find the diameter of the cylinder.

Solution

Volume of metal in spherical shell = Let r be the radius of solid cylinder. The spherical shell is melted to form a solid cylinder. So, their volume remains the same. Diameter of the solid cylinder = 7 cm

Q86. Find the number of coins 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

Solution

                              

Q87. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.

Solution

Volume of the rod = r²h = 8 cm³ = 2 cm³ The length of the new wire of the same volume is = 18 m = 1800 cm Volume of the wire = volume of the rod r² 1800 = 2 r² = r =  cm So, the diameter of cross-section i.e. thickness of the wire is  cm.

Q88. The internal radii of the ends of a bucket, full of milk and of internal height 16 cm are 14 cm and 7 cm. If this milk is poured into a hemispherical vessel, find the internal diameter of the vessel.

Solution

For bucket, h = 16 cm r₁ = 14 cm r₂ = 7 cm V₁ = Volume of bucket =   Let R cm be the radius of the hemispherical vessel V₂ = volume of vessel = V₁ = V₂ R = 14 cm Internal diameter of vessel = 2R = 28 cm

Q89. A cylindrical pipe has inner diameter of 4 cm and water flows through it at the rate of 20 m per minute. How long would it take to fill a conical tank, with diameter of base as 80 cm and depth 72 cm?

Solution

Let the time to fill the conical flask be x minutes. In 1 minute, 20 m water flows. So in x minutes, 20x m water will flow. Volume of water flowed through cylindrical pipe = r² (20 x) 100 cm³ = (2)² (20x) 100 cm³ Volume of water filled in conical tank = Volume of water flowed through pipe = Volume of water filled in conical tank. (2)² (20x) 100 =