Question 1: What is the least number of solid metallic spheres, each of $\displaystyle 6 \text{ cm }$ diameter, that should be melted and recast to form a solid metal cone whose height is $\displaystyle 45 \text{ cm }$ and diameter $\displaystyle 12 \text{ cm }$ ?

Radius of Sphere $\displaystyle (r_1) = 3 \text{ cm }$

Metal cone: Height $\displaystyle ( h) = 45 \text{ cm }$, Radius $\displaystyle ( r_2) = 6 \text{ cm }$

Let the number of cones melted $\displaystyle = n$

$\displaystyle \text{Therefore } \frac{4}{3} \pi {r_1}^3 \times n = \frac{1}{3} \pi {r_2}^2 h$

$\displaystyle \Rightarrow n = \frac{3}{4} \times \frac{1}{3} \times \frac{{r_2}^2}{{r_1}^3} \times h$

$\displaystyle \Rightarrow n = \frac{3}{4} \times \frac{1}{3} \times \frac{{6}^2}{{3}^3} \times 45$

$\displaystyle \Rightarrow n = 15$

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Question 2: A largest sphere is to be carved out of a right circular cylinder of radius $\displaystyle 7 \text{ cm }$ and height $\displaystyle 14 \text{ cm }$ . Find the volume of the sphere. (Answer correctly to the nearest integer).

Cylinder: Radius $\displaystyle ( r) = 7 \text{ cm }$, Height $\displaystyle (h) = 14 \text{ cm }$

Radius of sphere will be $\displaystyle 7 \text{ cm }$

$\displaystyle \text{Volume of the largest sphere } = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (7)^3 = 1437.33 \text{ cm}^3$

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Question 3: A right circular cylinder having diameter $\displaystyle 12 \text{ cm }$ and height $\displaystyle 15 \text{ cm }$ is full of ice-cream. The ice-cream is to be filled in identical cones of height $\displaystyle 12 \text{ cm }$ and diameter $\displaystyle 6 \text{ cm }$ having a hemi-spherical shape on the top. Find the number of cones required.

Cylinder: Radius $\displaystyle (r) = 6 \text{ cm }$, Height $\displaystyle (h) = 15 \text{ cm }$

$\displaystyle \text{Volume of ice-cream cone } = \frac{1}{3} \pi r^2 h + \frac{1}{2} \times \frac{4}{3} \pi r^3$

$\displaystyle = \frac{1}{3} \pi \times 3^2 \times 12 + \frac{1}{2} \times \frac{4}{3} \pi \times 3^3 = 54\pi$

Volume of cylinder $\displaystyle = \pi r^2 h = \pi (6)^2 \times 15 = 540 \pi$

$\displaystyle \text{Therefore the number of ice-cream cones } = \frac{540 \pi}{54 \pi} = 10$

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Question 4: A solid is in the form of a cone standing on a hemisphere with both their radii being equal to $\displaystyle 8 \text{ cm}$ and the height of the cone is equal to its radius. Find, in terms of $\displaystyle \pi$ , the volume of the solid.

Radius $\displaystyle (r) = 8 \text{ cm }$

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

$\displaystyle = \frac{1}{2} \times \frac{4}{3} \pi (8)^3 + \frac{1}{3} \pi (8)^2 \times 8 = 512 \pi \text{ cm}^3$

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Question 5: The diameter of a sphere is $\displaystyle 6 \text{ cm}$ . It is melted and drawn into a wire of diameter $\displaystyle 0.2 \text{ cm}$ . Find the length of the wire.

Sphere: Radius $\displaystyle (r_1)= 3 \text{ cm }$

Wire (cylinder): Radius $\displaystyle ( r_2) = 0.1 \text{ cm }$, Length $\displaystyle = l \text{ cm }$

Volume of sphere $\displaystyle =$ Volume of wire

$\displaystyle \Rightarrow \frac{4}{3} \pi (3)^3 = \pi (0.1)^2 \times l$

$\displaystyle \Rightarrow l = \frac{4 \times 3^2}{0.1 \times 0.1} = 3600 \text{ cm or } 36 \text{ m }$

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Question 6: Determine the ratio of the volume of a cube to that of a sphere that will exactly fit inside the cube.

Let radius of sphere be $\displaystyle r$. The the side of the cube would be $\displaystyle 2r$

$\displaystyle \text{Volume of sphere } = \frac{4}{3} \pi r^3$

Volume of cube $\displaystyle = ( 2r)^3 = 8r^3$

$\displaystyle \text{Hence Ratio } = \frac{8r^3}{\frac{4}{3} \pi r^3} = \frac{6}{\pi} = \frac{21}{11}$

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Question 7: An iron pole consisting of a cylindrical portion $\displaystyle 110 \text{ cm }$ high and of base diameter $\displaystyle 12 \text{ cm }$ is surmounted by a cone $\displaystyle 9 \text{ cm }$ high. Find the mass of the pole, given that $\displaystyle 1 \text{ cm}^3$ of iron 355 has $\displaystyle 8 \text{ gm }$ of mass (approx.) (Take $\displaystyle \pi = \frac{355}{113}$ )

Cylinder (Iron Pole): Height $\displaystyle (h_1) = 110 \text{ cm }$, Radius $\displaystyle (r) = 6 \text{ cm }$

Cone: Height $\displaystyle (h_2) = 9 \text{ cm }$

$\displaystyle \text{Volume of the structure } = \pi (6)^2 \times 110 + \frac{1}{3} \pi (6)^2 \times 9 = 3960\pi + 108 \pi = 4068 \pi$

Density $\displaystyle = 8 \text{ gm } / \text{ cm}^3$

$\displaystyle \text{Therefore Weight } = 4068 \times \frac{355}{113} \times 8 = 102240 \text{ gm } \text{ or } 102.24 \text{ kg }$

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Question 8: When a metal cube is completely submerged in water contained in a cylindrical vessel with diameter $\displaystyle 30 \text{ cm }$ , the level of water rises by $\displaystyle 1\frac{41}{99} \text{ cm }$ . Find : (i) the length of the edge of the cube, (ii) the total surface area of the cube.

Let side of the cube $\displaystyle = a$

Cylinder: Radius $\displaystyle (r) = 15 \text{ cm }$

$\displaystyle \text{Increase in height } = 1 \frac{41}{99} = \frac{140}{99} \text{ cm }$

$\displaystyle \text{ i) Therefore } \pi ( 15)^2 \times \frac{140}{99} = a^3$

$\displaystyle \Rightarrow a^3 = \frac{22}{7} \times 15^2 \times \frac{140}{99} = 2^3 \times 5^3$

$\displaystyle \text{Therefore } a = 2 \times 5 = 10 \text{ cm }$

$\displaystyle \text{ii) Surface are of the cube } = 6a^2 = 6 \times 10^2 = 600 \text{ cm}^2$

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Question 9: In the following diagram a rectangular platform with a semi-circular end on one side is $\displaystyle 22 \text{ meters }$ long from one end to the other end. If the length of the half circumference is $\displaystyle 11 \text{ meters }$ , find the cost of constructing the platform, $\displaystyle 1.5 \text{ meters }$ high at the rate of $\displaystyle Rs. 4 \text{ m}^3$ .

$\displaystyle \pi r = 11 \Rightarrow r = \frac{7 \times 11}{22} = 3.5 \text{ m }$

Therefore length of the rectangle $\displaystyle = 22 - 3.5 = 18.5 \text{ m }$

Volume of rectangle $\displaystyle = 18.5 \times ( 2 \times 3.5) \times 1.5 = 194.25 \text{ m}^3$

$\displaystyle \text{Volume of semi circle } = \frac{1}{2} \times \frac{22}{7} \times (3.5)^2 \times 1.5 = 28.875 \text{ m}^3$

Total volume $\displaystyle = 194.25 + 28.875 = 223.125 \text{ m}^3$

Total cost $\displaystyle = 223.125 \times 4 = 892.5$ Rs.

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Question 10: The cross-section of a tunnel is a square of side $\displaystyle 7 \text{ m }$ surmounted by a semicircle as shown in the adjoining figure. The tunnel ii $\displaystyle 80 \text{ m }$ long. Calculate (i) volume, (ii) the surface area of the tunnel (excluding the floor), and (iii) it’s floor area.

i) Volume $\displaystyle =$ Cross Section Area $\displaystyle \times$ Length

$\displaystyle = (7 \times 7 + \frac{1}{2} \times \frac{22}{7} \times (3.5)^2) \times 80 = ( 49 + 19.25) \times 80 = 5460 \text{ m}^3$

ii) Surface area $\displaystyle = ( \pi (3.5) + 2 \times 7) \times 80 = 25 \times 80 = 2000 \text{ m}^2$

iii) Floor area $\displaystyle = 7 \times 80 = 560 \text{ m}^2$

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Question 11: A cylindrical water tank of diameter $\displaystyle 2.8 \text{ m }$ and height $\displaystyle 4.2 \text{ m }$ is being fed by a Pipe of diameter $\displaystyle 7 \text{ cm }$ through which water flows at the rate of $\displaystyle 4 \text{ m/s }$ . Calculate, in minutes, the time it takes to fill the tank.

Water tank: Radius $\displaystyle (r) = 1.4 \text{ m}$ , Height $\displaystyle (h) = 4.2 \text{ m}$

Pipe: Radius $\displaystyle (r) = 3.5 \text{ cm }$

Flow of water $\displaystyle = 4 \text{ m/s}$

Volume of water tank $\displaystyle = \pi \times (.14)^2 \times 4.2 = 29.568 \text{ m}^3$

$\displaystyle \text{Time to fill the tank } = \frac{29.568}{\pi (\frac{3.5}{100})^2 \times 4} = 1920 \text{ seconds or } 32 \text{ minutes }$

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Question 12: Water flows, at 9 km per hour, through a cylindrical pipe of cross-sectional area $\displaystyle 25 \text{ cm}^2$ . If this water is collected into a rectangular cistern of dimensions $\displaystyle 7.5 \text{ m } \times 5 \text{ m } \times 4 \text{ m }$ : calculate the rise in level in the cistern in 1 hour 15 minutes.

$\displaystyle \text{Rate of flow } = 9 \text{ km/hr } = \frac{9000 m}{3600 s} = 2.5 \text{ m/s }$

Volume of Cistern $\displaystyle = 7.5 \times 5 \times 4 = 150 \text{ m}^3$

Total volume of flow in $\displaystyle 1 \text{ hour } 15$ minutes

$\displaystyle = \frac{25}{10000} m^2 \times 2.5 \frac{m}{s} \times 4500 s = 28.1250 \text{ m}^3$

Height of water $\displaystyle \Rightarrow 7.5 \times 5 \times h = 28.1250 \Rightarrow h = 0.75 \text{ m } = 75 \text{ cm }$

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Question 13: The given figure shows the cross-section of a cone, a cylinder, and a hemisphere all with the same diameter 10 cm, and the other dimensions are as shown. Calculate: (a) the total surface area, (b) the total volume of the solid and (c) the density of the material if its total weight is $\displaystyle 1.7 \text{ kg }$ .

i) Total Surface Area

$\displaystyle = \frac{1}{2} \times 4\pi (5)^2 + 2 \pi (5) \times 12 + \pi (5) (13)$

$\displaystyle = 50\pi + 120 \pi + 65 \pi ($ Since $\displaystyle l = \sqrt{12^2 + 5^2} = 13 )$

$\displaystyle = 230 \pi = 738.57 \text{ m}^2$

$\displaystyle \text{ii) Total volume } = \frac{1}{3} \pi (5)^2 \times 12 + \pi (5)^2 \times 12 + \frac{1}{2} \times \frac{4}{3} \pi (5)^2$

$\displaystyle = 100 \pi + 300 \pi + 83.33 \pi = 1519.047 \text{ cm}^3$

$\displaystyle \text{iii) Density } = \frac{1.7 \text{ kg }}{1519.047 \text{ cm}^3} = \frac{1.7 \times 1000 \text{ gm }}{1519.047 \text{ cm}^3} = 1.121 \frac{\text{ gm }}{\text{ cm}^3}$

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Question 14: A solid, consisting of a right circular cone standing on a hemisphere, is placed upright in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is $\displaystyle 3 \text{ cm }$ and its height is $\displaystyle 6 \text{ cm }$, the radius of the hemisphere is $\displaystyle 2 \text{ cm }$ and the height of the cone is $\displaystyle 4 \text{ cm }$. Give your answer to the nearest cubic centimeter.

Cylinder: Radius $\displaystyle (r) = 3 \text{ cm }$, Height $\displaystyle (h) = 6 \text{ cm }$

Volume of cylinder $\displaystyle = \pi (3)^2 \times 6 = 54 \pi$

$\displaystyle \text{Volume of solid } = \frac{1}{2} \times \frac{4}{3} \pi (2)^3 + \frac{1}{3} \times \pi (20)^2 \times 4 = \frac{16}{3} \pi + \frac{16}{3} \pi = \frac{32}{3} \pi$

$\displaystyle \text{Therefore volume of water left } = 54 \pi - \frac{32}{3} \pi = \frac{130}{3} \pi = 136.19 \text{ cm}^3$

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Question 15: A metal container in the form of a cylinder is Surmounted by a hemisphere of the same radius. The internal height of the cylinder is $\displaystyle 7 \text{ m }$ and the internal radius is $\displaystyle 3.5 \text{ m }$ Calculate: (i) the total area of the internal surface excluding the base; (ii) the internal volume of the container in $\displaystyle \text{ m}^3$

i) Total internal surface area

$\displaystyle = 2 \pi (3.5)\times 7 + \frac{1}{2} \times 4 \pi (3.5)^2 = 49 \pi + 24.5 \pi = 73.5 \pi = 231 \text{ m}^2$

ii) Internal volume

$\displaystyle = \pi (3.5)^2 \times 7 + \frac{1}{2} \times \frac{4}{3} \pi (3.5)^3 = 85.75 \pi + \frac{85.75}{3} \pi = 114.33 \pi = 359.33 \text{ m}^3$

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Question 16: An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is $\displaystyle85 \text{ m }$ and the height of the cylindrical part is $\displaystyle 50 \text{ m.}$  If the diameter of the base is $\displaystyle 168 \text{ m }$, find the quantity of canvas required to make the tent. Allow $\displaystyle 20\%$ extra for fold and for stitching. Give your answer to the nearest $\displaystyle \text{ m}^2$

Since $\displaystyle l = \sqrt{35^2 + 84^2} = 91$

Total Surface Area $\displaystyle = 2 \pi (84) \times 50 + \pi (84) \times 91 = 8400 \pi + 7644 \pi = 16044 \pi = 50424 \text{ m}^2$

Therefore canvas required $\displaystyle = 50424 \times 1.20 = 60508.8 \text{ m}^2$

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Question 17: The total surface area of a hollow cylinder, which is open from both sides, is $\displaystyle 3575 \text{ cm}^2$ ; area of the base ring is $\displaystyle 357.5 \text{ cm}^2$ and height is $\displaystyle 14 \text{ cm}$ . Find the thickness of the cylinder.

Total surface area $\displaystyle = 3575 \text{ cm}^2$

Area of the base $\displaystyle = 357.5 \text{ cm}^2$

Height $\displaystyle (h) = 14 \text{ cm}$

$\displaystyle \pi r^2 - \pi x^2 = 357.5$

$\displaystyle \pi ( r-x)(r+x)= 357.5$

$\displaystyle ( r-x)(r+x)= 113.75 .$.. … … … … i)

$\displaystyle 2 \pi r ( 14) + 2 \pi x ( 14) + 2 \times 357.5 = 3575$

$\displaystyle 28 \pi r + 28 \pi x = 2860$

$\displaystyle 28r + 28x = 910$

$\displaystyle (r+x) = 32.5 .$.. … … … … ii)

Therefore from i) we get $\displaystyle (r-x) = 3.5 .$.. … … … … iii)

Solving ii) and iii) we get

$\displaystyle 2r = 36 \Rightarrow r = 18$

$\displaystyle 2x = 29 \Rightarrow x = 14.5$

Hence thickness $\displaystyle = 3.5 \text{ cm}$

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Question 18: A test-tube consists of a hemisphere and a cylinder of the same radius. The volume of the water required to fill the whole tube is $\displaystyle \frac{5159}{6} \text{ cm}^3$ , and $\displaystyle \frac{4235}{6} \text{ cm}^3$ of water is required to fill the tube to a level which is $\displaystyle 4 \text{ cm }$ below the top of the tube. Find the radius of the tube and the length of its cylindrical part.

$\displaystyle \frac{1}{2} \times \frac{4}{3} \pi r^3 + \pi r^2 h = \frac{5159}{6}$

$\displaystyle \frac{2}{3} \pi r^3 + \pi r^2 h = \frac{5159}{6}$

$\displaystyle 4 \pi r^2 + 6 \pi r^2 h = 5159$

$\displaystyle 4r^3 + 6 r^2 h = 1641.5 .$.. … … … … i)

$\displaystyle \frac{2}{3} \pi r^3 + \pi r^2 (h-4) = \frac{4235}{6}$

$\displaystyle 4 \pi r^3 + 6 \pi r^2 ( h-4) = 4235$

$\displaystyle 4r^3 + 6 r^2 ( h - 4) = 1347.5 .$.. … … … … ii)

$\displaystyle \text{From i) } h = \frac{1641.5 - 4r^3}{6r^2}$

Substituting in ii)

$\displaystyle 4r^3 + 6 r^2 ( \frac{1641.5 - 4r^3}{6r^2} - 4) = 1347.5$

$\displaystyle 4r^3 + ( 1641.5 - 4r^3 - 24r^2) = 1347.5$

$\displaystyle 1641.5 - 24 r^2 = 1347.5$

$\displaystyle 24r^2 = 294$

$\displaystyle r^2 = 12.25$

$\displaystyle r = 3.5$

$\displaystyle \text{Therefore } h = \frac{1641.5 - 4(3.5)^3}{6(3.5)^2} = \frac{1470}{73.5} = 20 \text{ cm }$

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Question 19: A solid is in the form of a right circular cone mounted on a hemisphere. The diameter of the base of the cone. which exactly coincides with hemisphere, is $\displaystyle 7 \text{ cm}$ and its height is $\displaystyle 8 \text{ cm}$ . The solid is placed in a cylindrical vessel of internal radius $\displaystyle 7 \text{ cm}$ and height $\displaystyle 10 \text{ cm}$ . How much water, in $\displaystyle \text{ cm}^3$, will be required to fill the vessel completely.

Volume of cylinder $\displaystyle = \pi (7)^2 (10) = 490 \pi$

$\displaystyle \text{Volume of solid } = \frac{1}{2} \times \frac{4}{3} \pi (3.5)^3 + \frac{1}{3} \pi (3.5)^2 (8) = \frac{343}{12} \pi + \frac{98}{3} \pi = \frac{735}{12} \pi = 61.25 \pi$

Therefore Quantity of water required $\displaystyle = 490 \pi - 61.25 \pi = 1347.5 \text{ cm}^3$

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Question 20: A cone and a cylinder have their heights in the ratio $\displaystyle 4 :5$ and their diameters are in the ratio $\displaystyle 3 : 2$ . Find the ratio between their volumes.

Let height of the cone $\displaystyle = 4x$

Therefore the height of the cylinder $\displaystyle = 5x$

Let diameter of the cone $\displaystyle = 3y$

Therefore diameter of the cylinder $\displaystyle = 2y$

$\displaystyle \text{Ratio of volumes } = \frac{\frac{1}{3} \pi \Big( \frac{3y}{2} \Big)^2 \times 4x}{\frac{1}{3} \pi \Big( \frac{2y}{2} \Big)^2 \times 5x} = \frac{9y^2 \times 4x}{4y^2 \times 5x} = 9:5$

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Question 21: A sphere just fits in a cylindrical vessel and the height of the cylindrical vessel is the same as the height of the sphere. Show that the curved surface area of the cylinder is the same as the curved surface area of the sphere.

Curved surface area of cylinder $\displaystyle = 2 \pi r ( 2r) = 4 \pi r^2$
Curved surface area of sphere $\displaystyle = 4 \pi r^2$