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

Answer:

Cylinder: Height = 10 \ cm, Radius = 6 \ cm

Cone: Height = 10 \ cm, Radius = 6 \ cm

Remaining Volume = Volume of Cylinder – Volume of Cone

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

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

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

Answer:

Cylinder: Height = 16 \ cm, Radius = 12 \ cm

Cone: Height = 8 \ cm, Radius = 6 \ cm

Remaining Volume = Volume of Cylinder – Volume of Cone

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

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

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

Surface Area calculations:

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

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

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

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

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

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

Answer:

Cylinder: Height = 4 \ m, Radius = 52.5 \ m

Cone: Slant Height = 80 \ m, Radius = 52.5 \ cm

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

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

Total Surface Area of the tent = 1320 + 13200 = 14520 \ m^2

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

Cost of the total canvas = 9680 \times 15 = 145200 \ Rs.

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

Answer:

Cylinder: Height = 8 \ m, Radius = 12 \ m

Cone: Height = 5 \ m, Radius = 12 \ cm

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

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

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

(ii) Area of canvas = 1093.71 \times 1.1 = 1203.08 \ m^2

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

Answer:

Cylinder: Height = 2 \ m, Radius = 1.75 \ m

Hemisphere: Radius = 1.75 \ cm

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

= \frac{22}{7} (1.75^2 \times 2 + \frac{2}{3} \times 1.75^3) = 30.48 \ 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 4\frac{2}{3} \ m and diameter of hemisphere is 3.5 \ m . Calculate the capacity and the internal surface area of vessel.

Answer:

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

Hemisphere: Radius = 1.75 \ m

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

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

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

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

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m1Question 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 24 \ cm , the total height of the toy is 60 \ cm and the radius of the base of the cone = twice the radius of the base of the cylinder = 10 \ cm ; find the total surface area of the toy. [Take \pi = 3.14 )

Answer:

Cylinder: Height = 36 \ cm, Radius = 10 \ cm

Cone: Height = 24 \ cm, Radius = 20 \ cm

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

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

Curved Surface Area of Cone = \pi r l

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

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

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

Area of the base of the cone = \pi 20^2  - \pi 10^2 = 942 \ cm^2

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

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

Answer:

Cylinder: Radius = 10 \ cm

Volume of iron solid = 22 \times 14 \times 10.5 \ cm^3

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

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

Answer:

Sphere: Radius = 0.7 \ cm

Beaker: Radius = 3.5 \ cm, Height = 5.6 \ cm

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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m2Question 10: The cross-section of a railway tunnel is a rectangle 6 \ m broad and 8 \ m high surmounted by a semi-circle as shown in: the figure. The tunnel is 35 \ m long. Find the cost of plastering the internal surface of the tunnel (excluding the floor) at the rate of 2.25 / m^2 .

Answer:

Rectangle: Breadth = 6 \ m, Height = 8 \ m

Semi-circle: Radius = 3 \ m

Length of the tunnel = 35 \ m

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

= 330 + 560 = 890 \ m^2

Cost of plastering = 890 \times 2.25 = 2002.5 \ Rs.

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m3Question 11: The horizontal cross-section of a water tank is in the shape of a rectangle with semi-circle at one end, as shown in the following figure. The water is 2.4 \  meters deep in the tank. Calculate the volume of the water in the tank in gallons.

Answer:

Rectangle: Width = 21 \ m, Height = 7 \ m

Semi-circle: Radius = 3.5 \ m

Depth of the water = 2.4 \ m

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

Therefore, water in the tank = 166.25 \times 2.4 = 399 \ m^3

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

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m4Question 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 20 \ cm/ second . Give your answer in cubic meters correct to one place of decimal.

Answer:

Rectangle: Length = 21 \ cm, Breadth = 7 \ cm

Semi-circle: Radius = 10.5 \ cm

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

= 147 + 173.25 = 320.25 \ cm^3

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

= 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 7 \ cm and height 8 \ cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is 3.5 \ cm and height 8 \ cm . Find the volume of water required to fill the vessel. If this cone is replaced by another cone, whose height is 1.75 \ cm and the radius of whose base is 2 \ cm , find the drop in the water level.

Answer:

Cylinder: Height = 8 \ cm, Radius = 3.5 \ cm

Cone: Height = 8 \ cm, Radius = 1.75 \ cm

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

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

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

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

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

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Question 14: A cylindrical can, whose base is horizontal and of radius 3.5 \ 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: Height = 7 \ cm, Radius = 3.5 \ cm

Sphere: Radius = 3.5 \ cm

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

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

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

= 154+38.5 = 192.5 \ cm^2

Let the depth of the water = h

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

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

\Rightarrow h = 7 - \frac{4}{3} \times 3.5 = 2 \frac{1}{3} \ 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 10 \ cm when flat circular surface is downward. Find the level of water, when it is inverted upside down, common diameter is 7 \ cm and height of the cylinder is 20 \ cm .

Answer:

Cylinder: Height = 20 \ cm, Radius = 3.5 \ cm

Sphere: Radius = 3.5 \ cm

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

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

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

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

\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}

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

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