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.

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.

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 $1.5 \ m$.

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.

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.

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.

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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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 $24 \ cm$, 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 $\pi = 3.14$)

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.

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 $5.6 \ cm$.

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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Question 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 $2.25 / m^2$.

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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Question 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.

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

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.

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.

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 $20 \ cm$.

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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