Class 10: Cones and Sphere – Exercise 22(e) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

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

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

$\displaystyle \$

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:

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

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

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

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

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

$\displaystyle \$

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 $\displaystyle = 9680 \times 15 = 145200 \text{ Rs. }$

$\displaystyle \$

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 $\displaystyle = 1093.71 \times 1.1 = 1203.08 \text{ m}^2$

$\displaystyle \$

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

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

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

$\displaystyle \$

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

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

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

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

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

$\displaystyle \$

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 $\displaystyle = 2260.81 + 1961.93 + 314 + 942 = 5478.74 \text{ cm}^2$

$\displaystyle \$

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 $\displaystyle = 22 \times 14 \times 10.5 \text{ cm}^3$

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

$\displaystyle \$

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 $\displaystyle = 3.5\text{ cm }Height = 5.6 \text{ cm }$

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

$\displaystyle \$

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 $\displaystyle = 890 \times 2.25 = 2002.5 \text{ Rs. }$

$\displaystyle \$

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 $\displaystyle = 166.25 \times 2.4 = 399 \text{ m}^3$

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

$\displaystyle \$

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 $\displaystyle = 10.5 \text{ cm }$

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

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

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

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

$\displaystyle \$

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

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

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

$\displaystyle \$

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 $\displaystyle = h$

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

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

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

$\displaystyle \$

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 $\displaystyle 20 \text{ cm } .$