\displaystyle \textbf{Notes: Important formulas to be kept in mind:}
\displaystyle \text{Parameters of a cone: Radius of the base }(r),\ \text{Height }(h),
\displaystyle \text{and Slant Height }(l).
\displaystyle \text{Volume of a Cone}=\frac{1}{3}\pi r^2h
\displaystyle \text{Curved Surface Area of a Cone}=\pi rl
\displaystyle \text{Total Surface Area of a Cone}=\pi r^2+\pi rl
\displaystyle \text{Volume of a Sphere}=\frac{4}{3}\pi r^3
\displaystyle \text{Surface Area of a Sphere}=4\pi r^2
\\

—————————-

\displaystyle \textbf{Question 1: }\text{The volume of a conical tent is }1232\text{ m}^3
\displaystyle \text{and the area of the base floor is }154\text{ m}^2.\text{ Calculate:}
\displaystyle \text{(i) the radius of the floor (ii) the height of the tent}
\displaystyle \text{(iii) the length of the canvas required to cover the tent}
\displaystyle \text{if its width is }2\text{ m}.\hfill \text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, Volume}=1232\text{ m}^3,\quad \text{Area of base}=154\text{ m}^2
\displaystyle \text{(i) }\pi r^2=154
\displaystyle \frac{22}{7}r^2=154
\displaystyle r^2=\frac{154\times7}{22}=49
\displaystyle r=7\text{ m}
\displaystyle \text{(ii) }\frac{1}{3}\pi r^2h=1232
\displaystyle \frac{1}{3}\times\frac{22}{7}\times7^2\times h=1232
\displaystyle h=24\text{ m}
\displaystyle \text{Slant height }l=\sqrt{7^2+24^2}=\sqrt{625}=25\text{ m}
\displaystyle \text{Curved surface area}=\pi rl=\frac{22}{7}\times7\times25=550\text{ m}^2
\displaystyle \text{Length of canvas required}=\frac{550}{2}=275\text{ m}
\displaystyle \therefore \text{(i) Radius}=7\text{ m},\ \text{(ii) Height}=24\text{ m},\ \text{(iii) Length of canvas}=275\text{ m}.
\\

\displaystyle \textbf{Question 2: }\text{A solid sphere of radius }15\text{ cm is melted}
\displaystyle \text{and recast into solid right circular cones of radius }2.5\text{ cm}
\displaystyle \text{and height }8\text{ cm. Calculate the number of cones recast.}
\displaystyle \hfill \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of sphere}=15\text{ cm}
\displaystyle \text{Radius of cone}=2.5\text{ cm},\quad \text{Height of cone}=8\text{ cm}
\displaystyle \text{Number of cones}=\frac{\frac{4}{3}\pi(15)^3}{\frac{1}{3}\pi(2.5)^2\times8}
\displaystyle =\frac{4\times15^3}{(2.5)^2\times8}
\displaystyle =270
\displaystyle \therefore \text{The number of cones recast is }270.
\\

\displaystyle \textbf{Question 3: }\text{A hollow sphere of internal and external diameters}
\displaystyle 4\text{ cm and }8\text{ cm respectively is melted into a cone of base diameter }8\text{ cm.}
\displaystyle \text{Find the height of the cone.}\hfill \text{[ICSE 2002]}
\displaystyle \text{Answer:}
\displaystyle \text{Internal radius of hollow sphere}=\frac{4}{2}=2\text{ cm}
\displaystyle \text{External radius of hollow sphere}=\frac{8}{2}=4\text{ cm}
\displaystyle \text{Radius of cone}=\frac{8}{2}=4\text{ cm}
\displaystyle \text{Volume of material in hollow sphere}=\text{Volume of cone}
\displaystyle \frac{4}{3}\pi(4)^3-\frac{4}{3}\pi(2)^3=\frac{1}{3}\pi(4)^2h
\displaystyle 4(64-8)=16h
\displaystyle 224=16h
\displaystyle h=14\text{ cm}
\displaystyle \therefore \text{The height of the cone is }14\text{ cm}.
\\

\displaystyle \textbf{Question 4: }\text{A hemispherical bowl of diameter }7.2\text{ cm}
\displaystyle \text{is filled completely with chocolate sauce. This sauce is poured into}
\displaystyle \text{an inverted cone of radius }4.8\text{ cm. Find the height of the cone}
\displaystyle \text{if it is completely filled.}\hfill \text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of hemisphere}=\frac{7.2}{2}=3.6\text{ cm}
\displaystyle \text{Radius of cone}=4.8\text{ cm}
\displaystyle \text{Volume of hemisphere}=\text{Volume of cone}
\displaystyle \frac{2}{3}\pi(3.6)^3=\frac{1}{3}\pi(4.8)^2h
\displaystyle h=\frac{2\times(3.6)^3}{(4.8)^2}
\displaystyle h=4.05\text{ cm}
\displaystyle \therefore \text{The height of the cone is }4.05\text{ cm}.
\\

\displaystyle \textbf{Question 5: }\text{A solid cone of radius }5\text{ cm and height }8\text{ cm}
\displaystyle \text{is melted and made into small spheres of radius }0.5\text{ cm.}
\displaystyle \text{Find the number of spheres formed.}\hfill \text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of cone}=5\text{ cm},\quad \text{Height of cone}=8\text{ cm}
\displaystyle \text{Radius of each sphere}=0.5\text{ cm}
\displaystyle \text{Volume of cone}=\text{Volume of all the spheres}
\displaystyle n\times\frac{4}{3}\pi(0.5)^3=\frac{1}{3}\pi(5)^2\times8
\displaystyle n=\frac{5^2\times8}{4\times(0.5)^3}
\displaystyle =400
\displaystyle \therefore \text{The number of spheres formed is }400.
\\

\displaystyle \textbf{Question 6: }\text{The total surface area of a solid metallic sphere}
\displaystyle \text{is }1256\text{ cm}^2.\text{ It is melted and recast into solid right}
\displaystyle \text{circular cones of radius }2.5\text{ cm and height }8\text{ cm.}
\displaystyle \text{Calculate (i) the radius of the sphere (ii) the number of cones recast.}
\displaystyle \hfill \text{[ICSE 2000]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, Surface area of sphere}=1256\text{ cm}^2
\displaystyle \text{(i) }4\pi r^2=1256
\displaystyle r^2=\frac{1256}{4\times3.14}=100
\displaystyle r=10\text{ cm}
\displaystyle \text{(ii) Radius of each cone}=2.5\text{ cm},\quad \text{Height}=8\text{ cm}
\displaystyle \text{Volume of sphere}=\text{Volume of all the cones}
\displaystyle \frac{4}{3}\pi(10)^3=n\times\frac{1}{3}\pi(2.5)^2\times8
\displaystyle n=\frac{4\times10^3}{(2.5)^2\times8}
\displaystyle =80
\displaystyle \therefore \text{(i) Radius of the sphere}=10\text{ cm}
\displaystyle \text{and (ii) the number of cones recast is }80.
\\

\displaystyle \textbf{Question 7: }\text{A hollow sphere of internal and external radii}
\displaystyle 6\text{ cm and }8\text{ cm respectively is melted and recast into small cones}
\displaystyle \text{of base radius }2\text{ cm and height }8\text{ cm. Find the number of cones.}
\displaystyle \hfill \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Internal radius of hollow sphere}=6\text{ cm}
\displaystyle \text{External radius of hollow sphere}=8\text{ cm}
\displaystyle \text{Radius of each cone}=2\text{ cm},\quad \text{Height}=8\text{ cm}
\displaystyle \text{Volume of material in hollow sphere}=\text{Volume of all the cones}
\displaystyle \frac{4}{3}\pi(8)^3-\frac{4}{3}\pi(6)^3=n\times\frac{1}{3}\pi(2)^2\times8
\displaystyle n=\frac{4(8^3-6^3)}{2^2\times8}
\displaystyle =\frac{4(512-216)}{32}=37
\displaystyle \therefore \text{The number of cones formed is }37.
\\

\displaystyle \textbf{Question 8: }\text{The surface area of a solid metallic sphere is }2464\text{ cm}^2.
\displaystyle \text{It is melted and recast into solid right circular cones of radius }3.5\text{ cm}
\displaystyle \text{and height }7\text{ cm. Calculate (i) the radius of the sphere}
\displaystyle \text{and (ii) the number of cones recast. Take }\pi=\frac{22}{7}.
\displaystyle \hfill \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, Surface area of sphere}=2464\text{ cm}^2
\displaystyle \text{(i) }4\pi R^2=2464
\displaystyle R^2=\frac{2464\times7}{4\times22}=196
\displaystyle R=14\text{ cm}
\displaystyle \text{(ii) Radius of each cone}=3.5\text{ cm},\quad \text{Height}=7\text{ cm}
\displaystyle \text{Volume of sphere}=\text{Volume of all the cones}
\displaystyle \frac{4}{3}\pi(14)^3=n\times\frac{1}{3}\pi(3.5)^2\times7
\displaystyle n=\frac{4\times14^3}{(3.5)^2\times7}
\displaystyle =128
\displaystyle \therefore \text{(i) Radius of the sphere}=14\text{ cm}
\displaystyle \text{and (ii) the number of cones recast is }128.
\\

\displaystyle \textbf{Question 9: }\text{A vessel in the form of an inverted cone is filled}
\displaystyle \text{with water to the brim. Its height is }20\text{ cm and diameter is }16.8\text{ cm}.
\displaystyle \text{Two equal solid cones are dropped in it so that they are fully submerged.}
\displaystyle \text{As a result, one-third of the water in the original cone overflows.}
\displaystyle \text{What is the volume of each of the solid cones submerged?}\hfill \text{[ICSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of conical vessel}=\frac{16.8}{2}=8.4\text{ cm}
\displaystyle \text{Height of conical vessel}=20\text{ cm}
\displaystyle \text{Volume of water in the vessel}=\frac{1}{3}\pi(8.4)^2\times20
\displaystyle \text{Volume of water overflowed}=\frac{1}{3}\times\frac{1}{3}\pi(8.4)^2\times20
\displaystyle \text{Volume of two solid cones}=\frac{1}{9}\pi(8.4)^2\times20
\displaystyle \text{Volume of each solid cone}=\frac{1}{2}\times\frac{1}{9}\pi(8.4)^2\times20
\displaystyle =\frac{1}{18}\times\frac{22}{7}\times(8.4)^2\times20
\displaystyle =246.4\text{ cm}^3
\displaystyle \therefore \text{The volume of each solid cone is }246.4\text{ cm}^3.
\\

\displaystyle \textbf{Question 10: }\text{A metallic sphere of radius }10.5\text{ cm is melted}
\displaystyle \text{and then recast into small cones, each of radius }3.5\text{ cm}
\displaystyle \text{and height }3\text{ cm. Find the number of cones thus formed.}
\displaystyle \hfill \text{[ICSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of metallic sphere}=\frac{4}{3}\pi(10.5)^3\text{ cm}^3
\displaystyle \text{Volume of each cone}=\frac{1}{3}\pi(3.5)^2\times3\text{ cm}^3
\displaystyle \text{Number of cones formed}
\displaystyle =\frac{\frac{4}{3}\pi(10.5)^3}{\frac{1}{3}\pi(3.5)^2\times3}
\displaystyle =126
\displaystyle \therefore \text{The number of cones formed is }126.
\\

\displaystyle \textbf{Question 11: }\text{A girl fills a cylindrical bucket }32\text{ cm in height}
\displaystyle \text{and }18\text{ cm in radius with sand. She empties the bucket on the ground}
\displaystyle \text{and makes a conical heap of the sand. If the height of the heap is }24\text{ cm,}
\displaystyle \text{find (i) the radius and (ii) the slant height of the heap.}
\displaystyle \text{Give your answer correct to one place of decimal.}\hfill \text{[ICSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of conical heap}=\text{Volume of cylindrical bucket}
\displaystyle \text{(i) }\frac{1}{3}\pi r^2\times24=\pi(18)^2\times32
\displaystyle 8r^2=18\times18\times32
\displaystyle r^2=1296
\displaystyle r=36\text{ cm}
\displaystyle \text{(ii) Slant height }l=\sqrt{r^2+h^2}
\displaystyle l=\sqrt{36^2+24^2}
\displaystyle l=\sqrt{1296+576}=\sqrt{1872}
\displaystyle l=43.3\text{ cm}
\displaystyle \therefore \text{(i) Radius}=36\text{ cm and (ii) Slant height}=43.3\text{ cm}.
\\

\displaystyle \textbf{Question 12: }\text{A vessel is in the form of an inverted cone.}
\displaystyle \text{Its height is }11\text{ cm and the radius of its top, which is open, is }2.5\text{ cm.}
\displaystyle \text{It is filled with water to the rim. When lead shots, each of which}
\displaystyle \text{is a sphere of radius }0.25\text{ cm, are dropped into the vessel,}
\displaystyle \frac{2}{5}\text{ of the water flows out. Find the number of lead shots dropped.}
\displaystyle \hfill \text{[ICSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of water in the cone}=\frac{1}{3}\pi(2.5)^2\times11
\displaystyle =\frac{68.75}{3}\pi\text{ cm}^3
\displaystyle \text{Volume of water that flows out}=\frac{2}{5}\times\frac{68.75}{3}\pi
\displaystyle \text{Volume of one lead shot}=\frac{4}{3}\pi(0.25)^3
\displaystyle =\frac{0.0625}{3}\pi\text{ cm}^3
\displaystyle \text{Number of lead shots}
\displaystyle =\frac{\frac{2}{5}\times\frac{68.75}{3}\pi}{\frac{0.0625}{3}\pi}
\displaystyle =440
\displaystyle \therefore \text{The number of lead shots dropped is }440.
\\

\displaystyle \textbf{Question 13: }\text{A solid, consisting of a right circular cone standing}
\displaystyle \text{on a hemisphere, is placed upright in a right circular cylinder full of water,}
\displaystyle \text{and touches the bottom. Find the volume of water left in the cylinder,}
\displaystyle \text{given that the radius of cylinder is }3\text{ cm and height is }6\text{ cm;}
\displaystyle \text{the radius of hemisphere is }2\text{ cm and height of cone is }4\text{ cm.}
\displaystyle \text{Give your answer to the nearest cubic centimeter.}\hfill \text{[ICSE 1998]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of cylinder}=3\text{ cm},\quad \text{Height of cylinder}=6\text{ cm}
\displaystyle \text{Radius of cone}=2\text{ cm},\quad \text{Height of cone}=4\text{ cm}
\displaystyle \text{Radius of hemisphere}=2\text{ cm}
\displaystyle \text{Volume of cylinder}=\pi(3)^2\times6=54\pi\text{ cm}^3
\displaystyle \text{Volume of hemisphere}=\frac{2}{3}\pi(2)^3=\frac{16}{3}\pi\text{ cm}^3
\displaystyle \text{Volume of cone}=\frac{1}{3}\pi(2)^2\times4=\frac{16}{3}\pi\text{ cm}^3
\displaystyle \text{Volume of water left}
\displaystyle =54\pi-\frac{16}{3}\pi-\frac{16}{3}\pi
\displaystyle =\frac{130}{3}\pi=136.19\text{ cm}^3
\displaystyle \approx136\text{ cm}^3
\displaystyle \therefore \text{The volume of water left is }136\text{ cm}^3.
\\

\displaystyle \textbf{Question 14: }\text{A metal container in the form of a cylinder is surmounted}
\displaystyle \text{by a hemisphere of the same radius. The internal height of the cylinder}
\displaystyle \text{is }7\text{ m and the internal radius is }3.5\text{ m. Calculate:}
\displaystyle \text{(i) the total area of the internal surface, excluding the base;}
\displaystyle \text{(ii) the internal volume of the container in m}^3\text{.}\hfill \text{[ICSE 1999]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius}=3.5\text{ m},\quad \text{Height of cylinder}=7\text{ m}
\displaystyle \text{(i) Curved surface area of cylinder}=2\pi rh
\displaystyle =2\times\frac{22}{7}\times3.5\times7=154\text{ m}^2
\displaystyle \text{Curved surface area of hemisphere}=2\pi r^2
\displaystyle =2\times\frac{22}{7}\times(3.5)^2=77\text{ m}^2
\displaystyle \text{Total internal surface area}=154+77=231\text{ m}^2
\displaystyle \text{(ii) Internal volume}
\displaystyle =\text{Volume of cylinder}+\text{Volume of hemisphere}
\displaystyle =\pi r^2h+\frac{2}{3}\pi r^3
\displaystyle =\frac{22}{7}\times(3.5)^2\times7+\frac{2}{3}\times\frac{22}{7}\times(3.5)^3
\displaystyle =269.5+89.83=359.33\text{ m}^3
\displaystyle \therefore \text{(i) Total internal surface area}=231\text{ m}^2
\displaystyle \text{and (ii) internal volume}=359.33\text{ m}^3.
\\

\displaystyle \textbf{Question 15: }\text{An exhibition tent is in the form of a cylinder}
\displaystyle \text{surmounted by a cone. The height of the tent above the ground is }85\text{ m}
\displaystyle \text{and the height of the cylindrical part is }50\text{ m. If the diameter}
\displaystyle \text{of the base is }168\text{ m, find the quantity of canvas required.}
\displaystyle \text{Allow }20\%\text{ extra for folds and stitching. Give your answer}
\displaystyle \text{to the nearest m}^2\text{.}\hfill \text{[ICSE 2001]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of base}=\frac{168}{2}=84\text{ m}
\displaystyle \text{Height of cylindrical part}=50\text{ m}
\displaystyle \text{Height of conical part}=85-50=35\text{ m}
\displaystyle \text{Slant height of cone}=\sqrt{84^2+35^2}
\displaystyle =\sqrt{7056+1225}=\sqrt{8281}=91\text{ m}
\displaystyle \text{Curved surface area of cylinder}=2\pi rh
\displaystyle =2\times\frac{22}{7}\times84\times50=26400\text{ m}^2
\displaystyle \text{Curved surface area of cone}=\pi rl
\displaystyle =\frac{22}{7}\times84\times91=24024\text{ m}^2
\displaystyle \text{Total canvas area}=26400+24024=50424\text{ m}^2
\displaystyle \text{Canvas required including }20\%\text{ extra}=50424\times1.20
\displaystyle =60508.8\text{ m}^2\approx60509\text{ m}^2
\displaystyle \therefore \text{The quantity of canvas required is }60509\text{ m}^2.
\\

\displaystyle \textbf{Question 16: }\text{An open cylindrical vessel of internal diameter }7\text{ cm}
\displaystyle \text{and height }8\text{ cm stands on a horizontal table. Inside this is placed}
\displaystyle \text{a solid metallic right circular cone, the diameter of whose base is }3.5\text{ cm}
\displaystyle \text{and height is }8\text{ cm. Find the volume of water required to fill the vessel.}
\displaystyle \text{If this cone is replaced by another cone whose height is }1.75\text{ cm}
\displaystyle \text{and radius of base is }2\text{ cm, find the drop in water level.}\hfill \text{[ICSE 1993]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of cylindrical vessel}=\frac{7}{2}=3.5\text{ cm}
\displaystyle \text{Height of cylindrical vessel}=8\text{ cm}
\displaystyle \text{Volume of cylinder}=\pi r^2h
\displaystyle =\frac{22}{7}\times(3.5)^2\times8=308\text{ cm}^3
\displaystyle \text{Radius of first cone}=\frac{3.5}{2}=1.75\text{ cm}
\displaystyle \text{Height of first cone}=8\text{ cm}
\displaystyle \text{Volume of first cone}=\frac{1}{3}\pi(1.75)^2\times8
\displaystyle =25.67\text{ cm}^3
\displaystyle \text{Volume of water required}=308-25.67=282.33\text{ cm}^3
\displaystyle \text{Radius of second cone}=2\text{ cm},\quad \text{Height}=1.75\text{ cm}
\displaystyle \text{Volume of second cone}=\frac{1}{3}\pi(2)^2\times1.75
\displaystyle =7.33\text{ cm}^3
\displaystyle \text{Decrease in displaced volume}=25.67-7.33=18.34\text{ cm}^3
\displaystyle \text{Drop in water level}=\frac{18.34}{\pi(3.5)^2}
\displaystyle =\frac{18.34\times7}{22\times(3.5)^2}=0.48\text{ cm}
\displaystyle \therefore \text{Volume of water required}=282.33\text{ cm}^3
\displaystyle \text{and drop in water level}=0.48\text{ cm.}
\\

\displaystyle \textbf{Question 17: }\text{A cylindrical can, whose base is horizontal and of radius }3.5\text{ cm,}
\displaystyle \text{contains sufficient water so that when a sphere is placed in the can,}
\displaystyle \text{the water just covers the sphere. Given that the sphere just fits into the can,}
\displaystyle \text{calculate: (i) the total surface area of the can in contact with water}
\displaystyle \text{when the sphere is in it (ii) the depth of water in the can before the sphere}
\displaystyle \text{was put into it.}\hfill \text{[ICSE 1997]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of cylindrical can}=3.5\text{ cm}
\displaystyle \text{Since the sphere just fits into the can, radius of sphere}=3.5\text{ cm}
\displaystyle \text{Depth of water when sphere is in it}=2\times3.5=7\text{ cm}
\displaystyle \text{(i) Surface area of can in contact with water}
\displaystyle =2\pi rh+\pi r^2
\displaystyle =2\times\frac{22}{7}\times3.5\times7+\frac{22}{7}\times(3.5)^2
\displaystyle =154+38.5=192.5\text{ cm}^2
\displaystyle \text{(ii) Volume of sphere}=\frac{4}{3}\pi r^3
\displaystyle =\frac{4}{3}\times\frac{22}{7}\times(3.5)^3=179.67\text{ cm}^3
\displaystyle \text{Volume of cylinder up to height }7\text{ cm}
\displaystyle =\pi r^2h=\frac{22}{7}\times(3.5)^2\times7=269.5\text{ cm}^3
\displaystyle \text{Original volume of water}=269.5-179.67=89.83\text{ cm}^3
\displaystyle \text{Original depth of water}=\frac{89.83}{\pi(3.5)^2}
\displaystyle =\frac{89.83}{38.5}=2.33\text{ cm}
\displaystyle \therefore \text{(i) Surface area in contact with water}=192.5\text{ cm}^2
\displaystyle \text{and (ii) original depth of water}=2.33\text{ cm}.
\\

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