Notes: Important formulas to be kept in mind:

Parameters of a Cone: Radius of the base ($r$), Height of the cone ($h$) and Slant Height of a Cone ($l$)

Volume of a Cone $= \frac{1}{3} \pi r^2 h$

Curved Surface area of a Cone $= \pi r l$

Total Surface area of a Cone $= \pi r^2+ \pi r l$

Volume of a Sphere $= \frac{4}{3} \pi r^3$

Surface area of a Sphere $= 4 \pi r^2$

Question 1: The volume of a conical tent is $1232 \ m^3$ and the area of the base floor is $154 \ m^2$. Calculate the: (i) radius of the floor (ii) height of the tent (iii) length of the canvas required to cover this conical tent if its width is $2 \ m$.    [2008]

Volume $= 1232 \ m^3$

Area of the base $= 154 \ m^2$

(i) $\pi r^2 = 154 \Rightarrow r = \sqrt{\frac{154}{22} \times 7} = 7 \ m$

(ii) $\frac{1}{3} \times \pi \times (7)^2\times h = 1232 \Rightarrow h =$ $\frac{1232 \times 3}{\pi \times 7^2}$ $= 24 \ m$

(iii) Curved Surface Area $= \pi r l = \pi (7) \sqrt{24^2+7^2} = 550 \ m^2$

Length of canvas needed $=$ $\frac{550}{2}$ $=225 \ m$

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Question 2: A solid sphere of radius $15 \ cm$ is melted and recast into solid right circular cones of radius $2.5 \$ and height $8 \ cm$. Calculate the number of cones recast.   [2013]

Sphere: Radius $= 15 \ cm$

Cone: Radius $= 2.5 \ cm$ and Height $= 8 \ cm$

Therefore number of cones re-casted $= \frac{\frac{4}{3} \times \pi \times (15)^3}{\frac{1}{3} \times \pi \times (2.5)^2 \times 8}$ $= 270$

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Question 3: A hollow sphere of internal and external diameters $4 \ cm$ and $8 \ cm$ respectively is melted into a cone of base diameter $8 \ cm$. Find the height of the cone.   [2002]

Internal diameter $= 4 \ cm \Rightarrow$ Internal radius $= 2 cm$

External diameter $= 8 \ cm \Rightarrow$ External radius $= 4 cm$

Radius of the $cone = 4 \ cm$

$\therefore \ \ \frac{4}{3} \times \pi \times (3)^3 - \frac{4}{3} \times \pi \times (2)^3 = \frac{1}{3} \times \pi \times (4)^2 \times h$

$\Rightarrow h =$ $\frac{4 \times (64-8)}{16}$ $= 14 \ cm$

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Question 4: A hemispherical bowl of diameter $7.2 \ cm$ is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius $4.8 \ cm$. Find the height of the cone if it is completely filled.    [2010]

Hemisphere: Radius $= 3.6 \ cm$

Cone: Radius $= 4.8 \ cm$, Height $= h$

$\therefore \frac{1}{2} \frac{4}{3} \times \pi \times (3.6)^3 = \frac{1}{3} \times \pi \times (4.8)^2 \times h$

$\Rightarrow h =$ $\frac{2 \times 3.6^3}{4.8^2}$ $= 4.05 \ cm$

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Question 5: A solid cone of radius $5 \ cm$ and height $8 \ cm$ is melted and made into small spheres of radius $0.5 \ cm$. Find the number of spheres formed.    [2011]

Cone: Radius $= 5 \ cm$, Height $= 8 \ cm$

Sphere: Radius $= 0.5 \ cm$

$n \times \frac{4}{3} \times \pi \times (0.5)^3 = \frac{1}{3} \times \pi \times (5)^2 \times 8$

$\Rightarrow n =$ $\frac{5^2 \times 8}{4 \times 0.5^3}$ $= 400$

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Question 6: The total area of a solid metallic sphere is $1256 \ cm^2$. It is melted and recast into solid right circular cones of radius $2.5 \ cm$ and height $8 \ cm$. Calculate: (i) the radius of the solid sphere, (ii) the number of cones recast.   [2000]

Surface area $= 1256 \ cm^2$

(i) $\therefore 4 \pi r^2 = 1256$

$\Rightarrow r^2 =$ $\frac{1256}{4 \times 3.14}$ $= 100$

$\Rightarrow r = 10 \ cm$

(ii) Cone: Radius $= 2.5 \ cm$, Height $= 8 \ cm$

$\frac{4}{3} \times \pi \times (10)^3 = n \times \frac{1}{3} \times \pi \times (2.5)^2 \times 8$

$\Rightarrow n =$ $\frac{4 \times 10^3}{2.5^2 \times 8}$ $= 80$

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Question 7: A hollow sphere of internal and external $6 \ cm$ and $8 \ cm$ respectively is melted and recast into small cones of base radius $2 \ cm$ and height $8 \ cm$. Find the number of cones.    [2012]

Sphere: Internal radius $= 6 \ cm$, External radius $= 8 \ cm$

Cone: Radius $= 2 \ cm$, Height $= 8 \ cm$

$\frac{4}{3} \times \pi \times (8)^3 - \frac{4}{3} \times \pi \times (6)^3 = n \times \frac{1}{3} \times \pi \times (2)^2 \times 8$

$\Rightarrow n =$ $\frac{4 \times (8^3-6^3)}{2^2 \times 8}$ $= 37$

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Question 8: The surface area of a solid metallic sphere is $2464\ cm^2$. It is melted and recast into solid right circular cones of radius $3.5 \ cm$ and height $7 \ cm$. Calculate: (i) the radius of the sphere (ii) the number of cones recast. (Take $\pi = \frac{22}{7}$   [2014]

Surface area of sphere $= 2464 \ cm^2$

Cone: Radius $= 3.5 \ cm$, Height $= 7 \ cm$

(i) $4 \pi r^2 = 2464 \Rightarrow r^2 =$ $\frac{2464 \times 7}{4 \times 22}$ $= 196$

Hence $R = 14 \ cm$

(ii) $\frac{4}{3} \times \pi \times (14)^3 = n \times \frac{1}{3} \times \pi \times (3.5)^2 \times 7$

$\Rightarrow n =$ $\frac{4 \times 14^3}{3.5^2 \times 7}$ $= 128$

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Question 9: A vessel in the form of an inverted cone, is filled with water to the brim. Its height is $20 \ cm$ and diameter is $16.8 \ cm$. Two equal solid cone are dropped in it so that they are fully submerged. As a result, one third of the water in original cone overflows. What is the volume of each of the solid cones submerged?     [2006]

Conical Vessel: Height $= 20 \ cm$, Radius $= 8.4 \ cm$

The amount of water overflow $=$ $\frac{1}{3} \times \frac{1}{3}$ $\pi (8.4)^2 \times 20$

Therefore the volume of one sphere $=$ $\frac{1}{2} \times \frac{1}{3} \times \frac{1}{3}$ $\pi (8.4)^2 \times 20 = 246.4 \ cm^3$

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Question 10: A metallic sphere of radius $10.5 \ cm$ is melted and then recast into small cones each of radius $3.5 \ cm$ and height $3 \ cm$. Find the number of cones thus formed.    [2005]

Volume of the sphere melted $=$ $\frac{4}{3}$ $\pi r^3 =$ $\frac{4}{3}$ $\pi (10.5)^3 \ cm^3$

Volume of the cones formed $=$ $\frac{1}{3}$ $\pi r^2 h =$ $\frac{1}{3}$ $\pi (3.5)^2 \times 3 \ cm^3$

Therefore the number of cones formed $=$ $\frac{\frac{4}{3} \pi (10.5)^3}{\frac{1}{3} \pi (3.5)^2 \times 3}$ $= 126$

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Question 11: A girl fills a cylindrical bucket $32 \ cm$ in height and $18 \ cm$ in radius with sand. She empties the bucket on the ground and makes a conical heap of the sand. If the height of the conical heap is $24 \ cm$, find:  (i) the radius and (ii) the slant height of the heap. Give your answer correct to one place of decimal.    [2004]

Volume of the conical heap = Volume of the bucket

(i) $\Rightarrow \frac{1}{3} \pi r^2 \times 24 = \pi \times (18)^2 \times (32)$

$\Rightarrow r^2 = 1296 \Rightarrow r = 36 \ cm$

(ii) $l^2 = 24^2 + 36^2$

$\Rightarrow l^2 = 1872 \Rightarrow l = 43.3 \ cm$

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Question 12: A vessel is in the form of inverted cone. Its height is $11 cm$ and the radius of its top, which is open, is $2.5 cm$. It is filled with water to the rim. When lead shots, each of which is a sphere of radius $0.25 cm$, are dropped into the vessel, $\frac{2}{5}$ of the water flows out. Find the number of lead shots dropped into the vessel.    [2003]

Volume of water in the cone $=$ $\frac{1}{3}$ $\pi r^2 h =$ $\frac{1}{3}$ $\pi .(2.5)^2 . 11 =$ $\frac{68.75}{3}$ $\pi$

Volume of the lead shots = volume of the water that over flows

$=$ $\frac{2}{5} \times \frac{68.75}{3}$ $\pi$

Volume of one lead shot $=$ $\frac{4}{3}$ $\pi (0.25)^3 =$ $\frac{0.625}{3}$ $\pi$

Therefore the number of lead shots $=$ $\frac{Volume \ of \ water \ displaced \ by \ lead \ shots}{Volume \ of \ one \ lead \ shot} = \frac{\frac{2}{5} \times \frac{68.75}{3} \pi}{\frac{0.625}{3} \pi}$ $= 440$

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Question 13: 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 $3 \ cm$ and its height is $6 \ cm$; the radius of the hemisphere is $2 \ cm$ and the height of cone is $4 \ cm$. Give your answer to the nearest cubic centimeter.    [1998]

Cylinder: Radius $= 3 \ cm$, Height $= 6 \ cm$

Cone: Radius $= 2 \ cm$, Height $= 4 \ cm$

Hemisphere: Radius $= 2 \ cm$

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

Volume of hemisphere $=$ $\frac{1}{2} \times \frac{4}{3}$ $\pi r^3 =$ $\frac{2}{3}$ $\pi (2)3 =$ $\frac{16}{3}$ $\pi$

Volume of cone $=$ $\frac{1}{3}$ $\pi r^2 h =$ $\frac{1}{3}$ $\pi (2)^2 \times 4 =$ $\frac{16}{3}$ $\pi$

Therefore the volume of water left in the cylinder $= 54 \pi -$ $\frac{16}{3}$ $\pi -$ $\frac{16}{3}$ $\pi =$ $\frac{130}{3}$ $\pi = 136.19 \ cm^2 \approx 136 \ cm^2$

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Question 14: 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 $7 \ m$ and the internal radius is $3.5 \ m$. Calculate: (i) the total area of the internal surface, excluding the base;  (ii) the internal volume of the container in $m^3$   [1999]

Cylinder: Radius $= 3.5 \ m$, Height $= 7 \ m$

Hemisphere: Radius $= 3.5 \ cm$

(i) Internal surface area of cylinder $= 2 \pi r h = 2 \times \frac{22}{7} \times 3.5 \times 7 = 154 \ m^2$

Internal surface are of hemisphere $= \frac{1}{2} \times 4 \pi r^2 = \frac{1}{2} \times 4 \times \frac{22}{7} \times 3.5^2 = 77 \ m^2$

Therefore the total surface area $= 154 + 77 = 231 \ m^2$

(ii) Internal volume = Volume of the cylinder + volume of the hemisphere

$= \frac{22}{7} \times 3.5^2 \times 7 + \frac{1}{2} \times \frac{4}{3} \times \frac{22}{7} \times (3.5)^3 = 269.5 + 89.33 = 359.33 \ m^3$

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Question 15: An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is $85 \ m$ and the height of the cylindrical part is $50 \ m$. If the diameter of the base is $168 \ m$, find the quantity of the canvas required to make the tent. Allow $20\%$ extra for folds and for stitching. Give your answer to the nearest $m^2$.    [2001]

Cylinder: Radius $= 84 \ m$, Height $= 50 \ m$

Cone: Radius $= 84 \ m$, Height $= 35 \ m$

Surface area of the cylinder $= 2 \pi r h = 2 \times \frac{22}{7} \times 84 \times 50 = 26400 \ m^2$

Curved surface area of the cone $= \pi r l = \frac{22}{7} \times 84 \times \sqrt{84^2 + 35^2} = 24024 \ m^2$

Total surface area $= 26400 + 24024 = 50424 \ m^2$

Total cloth required (including 20%) $= 50420 \times 1.20 = 60508.8 \ m^2 \approx 60509 \ m^2$

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Question 16:  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 is $8 \ cm$. Find the volume of the 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.    [1993]

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

Cone: Radius $= 1.75 \ m$, Height $= 35 \ m$

(i) Volume of cylinder $= \pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 8 = 308 \ cm^3$

Volume of the cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (1.75)^2 (8) = 25.67 \ cm^3$

Therefore the volume of the water $= 308 - 25.67 = 282.33 \ cm^3$

(ii) New Cone: Radius $= 2 cm, Height = 1.75 cm$

Volume of new cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2)^2 (1.75) = 7.33 \ cm^3$

Therefore change in volume $= 25.67 - 7.33 = 18.34 \ cm^3$

Change in height: $\pi r^2 h_1 = 18.34 \Rightarrow h_1 = \frac{18.34 \times 7}{22 \times 3.5^2} = 0.4736 \ cm$

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Question 17: 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 is contact with water when the sphere is in it (ii) the depth of water in the can before the sphere was put into it.    [1997]

Cylinder: Radius $= 3.5 \ m$, Height $= 7 \ m$

Hemisphere: Radius $= 3.5 \ cm$

(i) Surface Area $= 2\pi r h + \pi r^2 = 2 \times \frac{22}{7} \times 3.5 \times 7 + \frac{22}{7} \times 3.5^2 = 154 + 38.5 = 192.5 \ m^2$

(ii) Volume of sphere $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (3.5)^2 = 179.66 \ cm^3$

Decrease in height:

$\pi r^2 h_1 = 179.66 \Rightarrow h_1 = \frac{179.66 \times 7}{22 \times 3.5^2} = 4.67 \ cm$

Original height $= 7 - 4.67 = 2.33 \ cm$

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