Question 1: From a solid right circular cylinder with height \displaystyle 10 \text{ cm } and radius of the base is \displaystyle 6 \text{ cm } , a right circular cone of the same height and the same base is removed. Find the volume of the remaining solid.

Answer:

\displaystyle \text{Cylinder: Height } = 10\text{ cm }\text{ Radius } = 6 \text{ cm }  

\displaystyle \text{Cone: Height } = 10\text{ cm }\text{ Radius } = 6 \text{ cm }  

Remaining Volume = Volume of Cylinder – Volume of Cone

\displaystyle = \pi r^2 h - \frac{1}{3} \pi r^2 h  

\displaystyle = \frac{2}{3} \times \frac{22}{7} \times 6^2 \times 10 = \frac{5290}{7} = 754 \frac{2}{7} \text{ cm}^2  

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Question 2: From a solid cylinder whose height is \displaystyle 16 \text{ cm } and radius is \displaystyle 12 \text{ cm } , a conical cavity of height \displaystyle 8 \text{ cm } and of base radius \displaystyle 6 \text{ cm } is hollowed out” Find the volume and total surface area of the remaining solid.

Answer:

Cylinder: Height \displaystyle = 16\text{ cm }\text{ Radius } = 12 \text{ cm }  

Cone: Height \displaystyle = 8\text{ cm }\text{ Radius } = 6 \text{ cm }  

Remaining Volume = Volume of Cylinder – Volume of Cone

\displaystyle = \pi r^2 h - \frac{1}{3} \pi r^2 h  

\displaystyle = \frac{22}{7} (12^2 \times 12 - \frac{1}{3} \times 6^2 \times 8)  

\displaystyle = \frac{22}{7} (2304 - 96) = 6939.43 \text{ cm}^2  

Surface Area calculations:

\displaystyle \text{Area of the top of the cylinder } = \pi r^2 = \frac{22}{7} \times 12^2 = \frac{3168}{7} \text{ cm}^2  

\displaystyle \text{Curved surface area of the cylinder } = 2 \pi r h = 2 \times \frac{22}{7} \times 12 \times 16 = \frac{8448}{7} \text{ cm}^2  

\displaystyle \text{Curved Surface Area of Cone } = \pi r l = \frac{22}{7} \times 6 \times \sqrt{6^2 + 8^2} = \frac{1320}{7} \text{ cm}^2  

\displaystyle \text{Area of the base of the cone } = \pi r^2 = \frac{22}{7} \times 6^2 = \frac{792}{7} \text{ cm}^2  

\displaystyle \text{Therefore the total surface area } = \frac{8448}{7} + 2 \times \frac{3168}{7} + \frac{1320}{7} - \frac{792}{7} = 2187.43 \text{ cm}^2  

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Question 3: A circus tent is cylindrical to a height of \displaystyle 4 \text{ m } and conical above it. If its diameter is \displaystyle 105 \text{ m } and its slant height is \displaystyle 80 \text{ m } , calculate the total area of canvas required. Also, find the total cost of the canvas at \displaystyle  \text{ Rs. }  15 / meter if the width if \displaystyle 1.5 \text{ m } .

Answer:

Cylinder: Height \displaystyle = 4 \text{ m,} \text{ Radius } = 52.5 \text{ m }  

Cone: Slant Height \displaystyle = 80 \text{ m,} \text{ Radius } = 52.5 \text{ cm }  

\displaystyle \text{Curved surface area of the cylinder } = 2 \pi r h = 2 \times \frac{22}{7} \times 52.5 \times 4 = \frac{9240}{7} 1320 \text{ m}^2  

\displaystyle \text{Curved Surface Area of Cone } = \pi r l = \frac{22}{7} \times 52.5 \times 80 = \frac{92400}{7} = 13200 \text{ m}^2  

Total Surface Area of the tent \displaystyle = 1320 + 13200 = 14520 \text{ m}^2  

\displaystyle \text{Length of the canvas needed } = \frac{14520}{1.5} = 9680 \text{ m }  

Cost of the total canvas \displaystyle = 9680 \times 15 = 145200  \text{ Rs. }   

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Question 4: A circus tent is cylindrical to a height of \displaystyle 8 \text{ m } surmounted by a conical part. If total height of the tent is \displaystyle 13 \text{ m } and the diameter of its base is \displaystyle 24 \text{ m } ; calculate: (i) total surface area of the tent, (ii) area of canvas, required to make this tent allowing \displaystyle 10\% of the canvas used for folds and stitching.

Answer:

Cylinder: \displaystyle \text{ Height } = 8 \text{ m,} \text{ Radius } = 12 \text{ m }  

Cone: \displaystyle \text{ Height } = 5 \text{ m,} \text{ Radius } = 12 \text{ cm }  

\displaystyle \text{(i) Curved surface area of the cylinder } = 2 \pi r h = 2 \times \frac{22}{7} \times 12 \times 8 = \frac{4224}{7} \text{ m}^2

\displaystyle \text{Curved Surface Area of Cone } = \pi r l = \frac{22}{7} \times 12 \times \sqrt{5^2+12^2} = \frac{3432}{7} \text{ m}^2

\displaystyle \text{Total Surface area = } \frac{4224}{7} + \frac{3432}{7} = 1093.71 \text{ m}^2

(ii) Area of canvas \displaystyle = 1093.71 \times 1.1 = 1203.08 \text{ m}^2

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Question 5: A cylindrical boiler, \displaystyle 2 \text{ m } high, is \displaystyle 3.5 \text{ m } in diameter. It has a hemispherical lid. Find the volume of its interior, including the part covered by the lid.

Answer:

Cylinder: \displaystyle \text{ Height } = 2 \text{ m,} \text{ Radius } = 1.75 \text{ m }  

Hemisphere: \displaystyle \text{ Radius } = 1.75 \text{ cm }  

\displaystyle \text{Volume } = \pi r^2 h + \frac{1}{2} . \frac{4}{3} \pi r^3

\displaystyle = \frac{22}{7} (1.75^2 \times 2 + \frac{2}{3} \times 1.75^3) = 30.48 \text{ m}^3

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Question 6: A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylindrical part is \displaystyle 4\frac{2}{3} \text{ m } and diameter of hemisphere is \displaystyle 3.5 \text{ m } . Calculate the capacity and the internal surface area of vessel.

Answer:

\displaystyle \text{Cylinder: } \text{ Height } = 4\frac{2}{3} \text{ m,} \text{ Radius } = 1.75 \text{ m }  

Hemisphere: \displaystyle \text{ Radius } = 1.75 \text{ m }  

\displaystyle \text{Volume } = \pi r^2 h + \frac{1}{2} . \frac{4}{3} \pi r^3

\displaystyle = \frac{22}{7} (1.75^2 \times 4\frac{2}{3} + \frac{2}{3} \times 1.75^3) = 56.15 \text{ m}^3

\displaystyle \text{Total Internal surface area } = 2 \pi r h + \frac{1}{3} . 4 \pi r^2

\displaystyle = 2 \times \frac{22}{7} (1.75 \times 4 \frac{2}{3} + 1.75^2) = 70.58 \text{ m}^2

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Question 7: A wooden toy is in the shape of a cone mounted on a cylinder as shown alongside. If the height of the cone is \displaystyle 24 \text{ cm } , the total height of the toy is \displaystyle 60 \text{ cm } and the radius of the base of the cone = twice the radius of the base of the cylinder \displaystyle = 10 \text{ cm } ; find the total surface area of the toy. [Take \displaystyle \pi = 3.14 )

Answer:

Cylinder: \displaystyle \text{ Height } = 36\text{ cm }\text{ Radius } = 10 \text{ cm }  

Cone: \displaystyle \text{ Height } = 24\text{ cm }\text{ Radius } = 20 \text{ cm }  

Curved surface area of the cylinder \displaystyle = 2 \pi r h

\displaystyle = 2 \times \frac{22}{7} \times 10 \times 36 = 2260.81 \text{ cm}^2

Curved Surface Area of Cone \displaystyle = \pi r l

\displaystyle = \frac{22}{7} \times 20 \times \sqrt{20^2+24^2} = 1961.93 \text{ cm}^2

Area of the base of the Cylinder \displaystyle = \pi r^2

\displaystyle = \frac{22}{7} \times 10^2 = 314 \text{ cm}^2

Area of the base of the cone \displaystyle = \pi 20^2 - \pi 10^2 = 942 \text{ cm}^2

Therefore the total surface area of the toy \displaystyle = 2260.81 + 1961.93 + 314 + 942 = 5478.74 \text{ cm}^2

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Question 8: A cylindrical container with diameter of base \displaystyle 42 \text{ cm } contains sufficient water to submerge a rectangular solid of iron with dimensions \displaystyle 22 cm \times 14 cm \times 10.5 \text{ cm } . Find the rise in level of the water when the solid is submerged.

Answer:

Cylinder: \displaystyle \text{ Radius } = 10 \text{ cm }  

Volume of iron solid \displaystyle = 22 \times 14 \times 10.5 \text{ cm}^3

\displaystyle \text{Rise in water level } = \frac{22 \times 14 \times 10.5}{\pi \times 21^2} = 2 \frac{1}{3} \text{ cm }

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Question 9: Spherical marbles of diameter \displaystyle 1.4 \text{ cm } are dropped into beaker containing some water and are fully submerged. The diameter of the beaker is \displaystyle 7 \text{ cm } . Find how many marbles have been dropped in it if the water rises by \displaystyle 5.6 \text{ cm } .

Answer:

Sphere: Radius \displaystyle = 0.7 \text{ cm }

Beaker: Radius \displaystyle = 3.5\text{ cm }Height = 5.6 \text{ cm }

\displaystyle \text{No of marbles } = \frac{\pi \times 3.5^2 \times 5.6}{\frac{4}{3} \pi (0.7)^3} = \frac{3.5 \times 3.5 \times 5.6 \times 3}{4 \times 0.7 \times 0.7 \times 0.7} = 150

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Question 10: The cross-section of a railway tunnel is a rectangle \displaystyle 6 \text{ m } broad and \displaystyle 8 \text{ m } high surmounted by a semi-circle as shown in: the figure. The tunnel is \displaystyle 35 \text{ m } long. Find the cost of plastering the internal surface of the tunnel (excluding the floor) at the rate of \displaystyle 2.25 / \text{ m}^2 .

Answer:

Rectangle: Breadth \displaystyle = 6 \text{ m,} \text{ Height } = 8 \text{ m }  

Semi-circle: Radius \displaystyle = 3 \text{ m }  

Length of the tunnel \displaystyle = 35 \text{ m }  

\displaystyle \text{Surface Area of the tunnel } = \frac{1}{2} 2 \pi \times 3 \times 35 + 8 \times 35 \times 2

\displaystyle = 330 + 560 = 890 \text{ m}^2  

Cost of plastering \displaystyle = 890 \times 2.25 = 2002.5  \text{ Rs. }   

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Question 11: The horizontal cross-section of a water tank is in the shape of a rectangle with a semi-circle at one end, as shown in the following figure. The water is \displaystyle 2.4 meters deep in the tank. Calculate the volume of the water in the tank in gallons.

Answer:

Rectangle: Width \displaystyle = 21 \text{ m,} \text{ Height } = 7 \text{ m }

Semi-circle: Radius \displaystyle = 3.5 \text{ m }

Depth of the water \displaystyle = 2.4 \text{ m }

\displaystyle \text{Cross section of the tank } = 21 \times 7 + \frac{1}{2} \times \pi \times 3.5^2 = 166.25 \text{ m}^3

Therefore, water in the tank \displaystyle = 166.25 \times 2.4 = 399 \text{ m}^3

\displaystyle \text{Hence the volume of water in gallons } = \frac{399 \times 1000}{4.5} = 88666.67 gallons

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Question 12: The given figure shows the cross-section of a water channel consisting of a rectangle and a semi-circle. Assuming that the channel is always full, find the volume of water discharged through it in one minute if water is flowing at the rate of \displaystyle 20 cm/ second. Give your answer in cubic meters correct to one place of decimal.

Answer:

Rectangle: Length \displaystyle = 21\text{ cm }Breadth = 7 \text{ cm }

Semi-circle: Radius \displaystyle = 10.5 \text{ cm }

\displaystyle \text{Cross Section area } = 21 \times 7 + \frac{1}{2} \times \frac{22}{7} \times 10.5^2

\displaystyle = 147 + 173.25 = 320.25 \text{ cm}^3

\displaystyle \text{Amount of water flowing } = 320.25 \times 20 \times 60 \frac{cm^3}{min} = 384300 \frac{cm^3}{min}  

\displaystyle = 0.384300 \frac{m^3}{min} \approx 0.4 \frac{m^3}{min}  

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Question 13: An open cylindrical vessel of internal diameter \displaystyle 7 \text{ cm } and height \displaystyle 8 \text{ cm } stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is \displaystyle 3.5 \text{ cm } and height \displaystyle 8 \text{ cm } . Find the volume of water required to fill the vessel. If this cone is replaced by another cone, whose height is \displaystyle 1.75 \text{ cm } and the radius of whose base is \displaystyle 2 \text{ cm } , find the drop in the water level.

Answer:

Cylinder: \displaystyle \text{ Height } = 8\text{ cm }\text{ Radius } = 3.5 \text{ cm }  

Cone: \displaystyle \text{ Height } = 8\text{ cm }\text{ Radius } = 1.75 \text{ cm }  

Volume of water required \displaystyle = Volume of Cylinder - Volume of Cone

\displaystyle = \frac{22}{7} (3.5^2 \times 8 - \frac{1}{3} \times 1.75^2 \times 8) = 282.33 \text{ cm}^3

If the cone was replaced by another cone with \displaystyle \text{ Height } = 1.75\text{ cm }\text{ Radius } = 2 \text{ cm } , then let the drop in water level \displaystyle = h . Therefore

\displaystyle \pi \times 3.5^2 \times h = \frac{1}{3} \pi (1.75^2 \times 8 - 2^2 \times 1.75)

\displaystyle \Rightarrow h = \frac{1.75^2 \times 8 - 2^2 \times 1.75}{3 \times 3.5^2} = 0.4762 \text{ cm }

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Question 14: A cylindrical can, whose base is horizontal and of radius \displaystyle 3.5 \text{ cm }, contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate: (i) the total surface area of the can in contact with water when the sphere is in it; (ii) the depth of water in the can before the sphere was Put into the can.

Answer:

Cylinder: \displaystyle \text{ Height } = 7\text{ cm }\text{ Radius } = 3.5 \text{ cm }  

Sphere: \displaystyle \text{ Radius } = 3.5 \text{ cm }  

Total Surface Area = Curved Surface area of Cylinder + Area of the base

\displaystyle = 2 \pi r h + \pi r^2

\displaystyle = 2 \times \frac{22}{7} \times 3.5 \times 7 + \frac{22}{7} \times 3.5^2

\displaystyle = 154+38.5 = 192.5 \text{ cm}^2

Let the depth of the water \displaystyle = h

\displaystyle \text{Therefore: } \pi \times 3.5^2 \times (7 - h) = \frac{4}{3} \pi \times 3.5^3

\displaystyle \Rightarrow 7 - h = \frac{4}{3} \times 3.5

\displaystyle \Rightarrow h = 7 - \frac{4}{3} \times 3.5 = 2 \frac{1}{3} \text{ cm }

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Question 15: A hollow cylinder has solid hemisphere inward at one end and on the other end it is closed with a flat circular plate. The height of water is \displaystyle 10 \text{ cm } when flat circular surface is downward. Find the level of water, when it is inverted upside down, common diameter is \displaystyle 7 \text{ cm } and height of the cylinder is \displaystyle 20 \text{ cm } .

Answer:

Cylinder: \displaystyle \text{ Height } = 20\text{ cm }\text{ Radius } = 3.5 \text{ cm }  

Sphere: \displaystyle \text{ Radius } = 3.5 \text{ cm }  

\displaystyle \text{Volume of water } = \pi r^2 h = \frac{22}{7} \times (\frac{7}{2}) ^2 \times 10 = 385 \text{ cm}^3

Let the height of water \displaystyle = h when cylinder is upside down

\displaystyle \Rightarrow \pi r^2 h = 385 + \frac{1}{2} \times \frac{4}{3} \pi \times 3.5^3

\displaystyle \Rightarrow \frac{22}{7} \times 3.5^2 \times h = 385 + \frac{1}{2} \times \frac{4}{3} \pi \times 3.5^3

\displaystyle \Rightarrow h = \frac{385 \times 7}{22 \times 3.5^2} + \frac{22 \times 3.5^3 \times 7}{3 \times 22 \times 3.5^2}  

\displaystyle \Rightarrow h = 10 + \frac{7}{3} = 12 \frac{1}{3} \text{ cm }