Please refer to the following lecture notes for the formulas used in this exercise: Notes

\displaystyle \textbf{Question 1: }\text{A solid sphere of radius }15\text{ cm is melted and recast into}
\displaystyle \text{solid right circular cones of radius }2.5\text{ cm and height }8\text{ cm}.
\displaystyle \text{Calculate the number of cones formed.}\hfill\text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the sphere}=15\text{ cm}
\displaystyle \text{Radius of each cone}=2.5\text{ cm},\quad \text{Height}=8\text{ cm}
\displaystyle \text{Number of cones}=\frac{\text{Volume of the sphere}}{\text{Volume of one cone}}
\displaystyle =\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 =\frac{4\times3375}{6.25\times8}=\frac{13500}{50}=270
\displaystyle \therefore \text{The number of cones recast is }270.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A hollow sphere of internal and external diameters }4\text{ cm and }8\text{ cm respectively}
\displaystyle \text{is melted and recast 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 the hollow sphere}=\frac{4}{2}=2\text{ cm}
\displaystyle \text{External radius of the hollow sphere}=\frac{8}{2}=4\text{ cm}
\displaystyle \text{Radius of the cone}=\frac{8}{2}=4\text{ cm}
\displaystyle \text{Let the height of the cone be }h\text{ cm}
\displaystyle \text{Volume of the hollow sphere}=\text{Volume of the cone}
\displaystyle \frac{4}{3}\pi\left(4^3-2^3\right)=\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 \\

\displaystyle \textbf{Question 3: }\text{The radii of the internal and external surfaces of a metallic spherical shell}
\displaystyle \text{are }3\text{ cm and }5\text{ cm respectively. It is melted and recast into}
\displaystyle \text{a solid right circular cone of height }32\text{ cm. Find the diameter of the base of the cone.}
\displaystyle \text{Answer:}
\displaystyle \text{Internal radius of the spherical shell}=3\text{ cm}
\displaystyle \text{External radius of the spherical shell}=5\text{ cm}
\displaystyle \text{Height of the cone}=32\text{ cm}
\displaystyle \text{Let the radius of the base of the cone be }r\text{ cm}
\displaystyle \text{Volume of the spherical shell}=\text{Volume of the cone}
\displaystyle \frac{4}{3}\pi\left(5^3-3^3\right)=\frac{1}{3}\pi r^2\times32
\displaystyle 4(125-27)=32r^2
\displaystyle 392=32r^2
\displaystyle r^2=12.25\text{ cm}^2
\displaystyle r=3.5\text{ cm}
\displaystyle \therefore \text{Diameter of the base}=2r=2\times3.5=7\text{ cm}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The total volume of three identical cones is equal to the volume}
\displaystyle \text{of a bigger cone whose height is }9\text{ cm and diameter is }40\text{ cm}.
\displaystyle \text{Find the radius of the base of each smaller cone if its height is }108\text{ cm}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the base of each smaller cone be }r\text{ cm}
\displaystyle \text{Height of each smaller cone}=108\text{ cm}
\displaystyle \text{Radius of the bigger cone}=\frac{40}{2}=20\text{ cm}
\displaystyle \text{Height of the bigger cone}=9\text{ cm}
\displaystyle \text{Volume of three smaller cones}=\text{Volume of the bigger cone}
\displaystyle 3\left(\frac{1}{3}\pi r^2\times108\right)=\frac{1}{3}\pi(20)^2\times9
\displaystyle 3\times108r^2=20^2\times9
\displaystyle 324r^2=3600
\displaystyle r^2=\frac{3600}{324}=\frac{100}{9}
\displaystyle r=\frac{10}{3}\text{ cm}=3\frac{1}{3}\text{ cm}
\displaystyle \therefore \text{The radius of the base of each smaller cone is }3\frac{1}{3}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{A solid rectangular block of metal measuring }49\text{ cm}\times44\text{ cm}\times18\text{ cm}
\displaystyle \text{is melted and formed into a solid sphere.}
\displaystyle \text{Calculate the radius of the sphere.}
\displaystyle \text{Answer:}
\displaystyle \text{Dimensions of the rectangular block}=49\text{ cm}\times44\text{ cm}\times18\text{ cm}
\displaystyle \text{Let the radius of the sphere be }r\text{ cm}
\displaystyle \text{Volume of the rectangular block}=\text{Volume of the sphere}
\displaystyle 49\times44\times18=\frac{4}{3}\pi r^3
\displaystyle r^3=\frac{49\times44\times18\times3}{4\times\frac{22}{7}}
\displaystyle =\frac{49\times44\times18\times21}{88}=9261
\displaystyle =21^3
\displaystyle r=21\text{ cm}
\displaystyle \therefore \text{The radius of the sphere is }21\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A hemispherical bowl of internal radius }9\text{ cm is full of liquid.}
\displaystyle \text{The liquid is poured into small conical containers, each of diameter }3\text{ cm}
\displaystyle \text{and height }4\text{ cm. How many containers are required to empty the bowl?}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the hemispherical bowl}=9\text{ cm}
\displaystyle \text{Radius of each conical container}=\frac{3}{2}=1.5\text{ cm}
\displaystyle \text{Height of each conical container}=4\text{ cm}
\displaystyle \text{Let the number of containers required be }n
\displaystyle \text{Volume of the hemispherical bowl}=n\times\text{Volume of one conical container}
\displaystyle \frac{2}{3}\pi(9)^3=n\left(\frac{1}{3}\pi(1.5)^2\times4\right)
\displaystyle 2\times9^3=n\times(1.5)^2\times4
\displaystyle n=\frac{2\times9^3}{(1.5)^2\times4}
\displaystyle =\frac{1458}{9}=162
\displaystyle \therefore \text{The number of containers required is }162.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A hemispherical bowl of diameter }7.2\text{ cm is filled completely}
\displaystyle \text{with chocolate sauce. The sauce is poured into an inverted cone}
\displaystyle \text{of radius }4.8\text{ cm. Find its height if it is completely filled.}\hfill\text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the hemisphere}=\frac{7.2}{2}=3.6\text{ cm}
\displaystyle \text{Radius of the cone}=4.8\text{ cm}
\displaystyle \text{Let the height of the cone be }h\text{ cm}
\displaystyle \text{Volume of the hemisphere}=\text{Volume of the cone}
\displaystyle \frac{2}{3}\pi(3.6)^3=\frac{1}{3}\pi(4.8)^2h
\displaystyle 2(3.6)^3=(4.8)^2h
\displaystyle h=\frac{2(3.6)^3}{(4.8)^2}
\displaystyle =4.05\text{ cm}
\displaystyle \therefore \text{The height of the cone is }4.05\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A solid cone of radius }5\text{ cm and height }8\text{ cm is melted}
\displaystyle \text{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 the cone}=5\text{ cm},\quad \text{Height}=8\text{ cm}
\displaystyle \text{Radius of each sphere}=0.5\text{ cm}
\displaystyle \text{Let the number of spheres formed be }n
\displaystyle \text{Volume of the cone}=n\times\text{Volume of one sphere}
\displaystyle \frac{1}{3}\pi(5)^2\times8=n\left(\frac{4}{3}\pi(0.5)^3\right)
\displaystyle n=\frac{5^2\times8}{4\times(0.5)^3}
\displaystyle =\frac{25\times8}{4\times0.125}
\displaystyle =\frac{200}{0.5}=400
\displaystyle \therefore \text{The number of spheres formed is }400.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The total surface area of a solid metallic sphere is }1256\text{ cm}^2.
\displaystyle \text{It is melted and recast into solid right circular cones of radius }2.5\text{ cm}
\displaystyle \text{and height }8\text{ cm. Calculate: (i) the radius of the solid sphere}
\displaystyle \text{(ii) the number of cones formed.}\hfill\text{[ICSE 2000]}
\displaystyle \text{Answer:}
\displaystyle \text{Total surface area of the 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 \therefore \text{Radius of the solid sphere}=10\text{ cm}
\displaystyle \text{(ii) Radius of each cone}=2.5\text{ cm},\quad \text{Height}=8\text{ cm}
\displaystyle \text{Let the number of cones formed be }n
\displaystyle \text{Volume of the sphere}=n\times\text{Volume of one cone}
\displaystyle \frac{4}{3}\pi(10)^3=n\left(\frac{1}{3}\pi(2.5)^2\times8\right)
\displaystyle n=\frac{4\times10^3}{(2.5)^2\times8}
\displaystyle =\frac{4000}{6.25\times8}=\frac{4000}{50}=80
\displaystyle \therefore \text{The number of cones formed is }80.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A solid metallic cone of radius }6\text{ cm and height }10\text{ cm}
\displaystyle \text{is made of a heavy metal }A.\text{ To reduce its weight, a conical hole is made}
\displaystyle \text{in it and completely filled with a lighter metal }B.
\displaystyle \text{The conical hole has diameter }6\text{ cm and depth }4\text{ cm}.
\displaystyle \text{Find the ratio of the volume of metal }A\text{ to that of metal }B\text{ in the solid.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the original cone}=6\text{ cm},\quad \text{Height}=10\text{ cm}
\displaystyle \text{Radius of the conical hole}=\frac{6}{2}=3\text{ cm},\quad \text{Depth}=4\text{ cm}
\displaystyle \text{Volume of the original cone}=\frac{1}{3}\pi(6)^2\times10=120\pi\text{ cm}^3
\displaystyle \text{Volume of the conical hole}=\frac{1}{3}\pi(3)^2\times4=12\pi\text{ cm}^3
\displaystyle \text{Volume of metal }A=120\pi-12\pi=108\pi\text{ cm}^3
\displaystyle \text{Volume of metal }B=12\pi\text{ cm}^3
\displaystyle \text{Volume of metal }A:\text{Volume of metal }B
\displaystyle =108\pi:12\pi=9:1
\displaystyle \therefore \text{The required ratio is }9:1.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A hollow sphere of internal and external radii }6\text{ cm and }8\text{ cm}
\displaystyle \text{respectively is melted and recast into small cones of base radius }2\text{ cm}
\displaystyle \text{and height }8\text{ cm. Find the number of cones.}\hfill\text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Internal radius of the hollow sphere}=6\text{ cm}
\displaystyle \text{External radius of the hollow sphere}=8\text{ cm}
\displaystyle \text{Radius of each cone}=2\text{ cm},\quad \text{Height}=8\text{ cm}
\displaystyle \text{Let the number of cones be }n
\displaystyle \text{Volume of the hollow sphere}=n\times\text{Volume of one cone}
\displaystyle \frac{4}{3}\pi\left(8^3-6^3\right)=n\left(\frac{1}{3}\pi(2)^2\times8\right)
\displaystyle n=\frac{4(8^3-6^3)}{2^2\times8}
\displaystyle =\frac{4(512-216)}{4\times8}
\displaystyle =\frac{1184}{32}=37
\displaystyle \therefore \text{The number of cones formed is }37.
\displaystyle \\

\displaystyle \textbf{Question 12: }\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{(ii) the number of cones formed. Take }\pi=\frac{22}{7}.\hfill\text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Surface area of the sphere}=2464\text{ cm}^2
\displaystyle \text{Radius of each cone}=3.5\text{ cm},\quad \text{Height}=7\text{ cm}
\displaystyle \text{(i) }4\pi r^2=2464
\displaystyle r^2=\frac{2464\times7}{4\times22}=196
\displaystyle r=14\text{ cm}
\displaystyle \therefore \text{Radius of the sphere}=14\text{ cm}
\displaystyle \text{(ii) Let the number of cones formed be }n
\displaystyle \text{Volume of the sphere}=n\times\text{Volume of one cone}
\displaystyle \frac{4}{3}\pi(14)^3=n\left(\frac{1}{3}\pi(3.5)^2\times7\right)
\displaystyle n=\frac{4\times14^3}{(3.5)^2\times7}
\displaystyle =\frac{4\times2744}{12.25\times7}
\displaystyle =\frac{10976}{85.75}=128
\displaystyle \therefore \text{The number of cones formed is }128.
\displaystyle \\

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