\displaystyle \textbf{Question 1: }\text{The radii of the bases of a cylinder and a cone are in the ratio }3:4
\displaystyle \text{and their heights are in the ratio }2:3.\text{ What is the ratio of their volumes?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radii of the cylinder and cone be }3r\text{ and }4r\text{ respectively.}
\displaystyle \text{Let their heights be }2h\text{ and }3h\text{ respectively.}
\displaystyle \frac{V_{\text{cylinder}}}{V_{\text{cone}}}=\frac{\pi(3r)^2(2h)}{\frac{1}{3}\pi(4r)^2(3h)}
\displaystyle =\frac{18\pi r^2h}{16\pi r^2h}=\frac{9}{8}.
\displaystyle \therefore V_{\text{cylinder}}:V_{\text{cone}}=9:8.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If the heights of two right circular cones are in the ratio }1:2\text{ and the}
\displaystyle \text{perimeters of their bases are in the ratio }3:4,\text{ what is the ratio of their volumes?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radii of the two cones be }r_1\text{ and }r_2.
\displaystyle \frac{2\pi r_1}{2\pi r_2}=\frac{3}{4}\quad\Rightarrow\quad r_1:r_2=3:4.
\displaystyle \text{Also, }h_1:h_2=1:2.
\displaystyle \frac{V_1}{V_2}=\frac{\frac{1}{3}\pi r_1^2h_1}{\frac{1}{3}\pi r_2^2h_2}
\displaystyle =\left(\frac{3}{4}\right)^2\times\frac{1}{2}=\frac{9}{32}.
\displaystyle \therefore V_1:V_2=9:32.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If a cone and a sphere have equal radii and equal volumes, what is the ratio}
\displaystyle \text{of the diameter of the sphere to the height of the cone?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the common radius be }r\text{ and the height of the cone be }h.
\displaystyle \frac{1}{3}\pi r^2h=\frac{4}{3}\pi r^3
\displaystyle \therefore h=4r.
\displaystyle \text{Diameter of the sphere}=2r.
\displaystyle \therefore \text{Diameter of sphere}:\text{Height of cone}=2r:4r=1:2.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A cone, a hemisphere and a cylinder stand on equal bases and have the}
\displaystyle \text{same height. What is the ratio of their volumes?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the common radius be }r.
\displaystyle \text{Since the height of a hemisphere is }r,\text{ the common height is }r.
\displaystyle V_{\text{cone}}=\frac{1}{3}\pi r^3,\qquad V_{\text{hemisphere}}=\frac{2}{3}\pi r^3,
\displaystyle V_{\text{cylinder}}=\pi r^3.
\displaystyle \therefore V_{\text{cone}}:V_{\text{hemisphere}}:V_{\text{cylinder}}=1:2:3.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The radii of two cylinders are in the ratio }3:5\text{ and their}
\displaystyle \text{heights are in the ratio }2:3.\text{ What is the ratio of their curved surface areas?}
\displaystyle \text{Answer:}
\displaystyle \text{Curved surface area of a cylinder}=2\pi rh.
\displaystyle \therefore \text{Ratio of curved surface areas}=(3\times2):(5\times3)
\displaystyle =6:15=2:5.
\displaystyle \therefore \text{The ratio of their curved surface areas is }2:5.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Two cubes have their volumes in the ratio }1:27.\text{ What is the ratio of}
\displaystyle \text{their surface areas?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the edges of the two cubes be }a_1\text{ and }a_2.
\displaystyle a_1^3:a_2^3=1:27
\displaystyle \therefore a_1:a_2=1:3.
\displaystyle \text{Surface areas are proportional to the squares of their edges.}
\displaystyle \therefore \text{Ratio of surface areas}=1^2:3^2=1:9.
\displaystyle \therefore \text{The ratio of their surface areas is }1:9.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Two right circular cylinders of equal volumes have their heights in the ratio}
\displaystyle 1:2.\text{ What is the ratio of their radii?}
\displaystyle \text{Answer:}
\displaystyle \text{Since the volumes are equal,}
\displaystyle \pi r_1^2h_1=\pi r_2^2h_2.
\displaystyle \therefore \frac{r_1^2}{r_2^2}=\frac{h_2}{h_1}=\frac{2}{1}.
\displaystyle \therefore \frac{r_1}{r_2}=\sqrt{2}.
\displaystyle \therefore r_1:r_2=\sqrt{2}:1.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If the volumes of two cones are in the ratio }1:4\text{ and their}
\displaystyle \text{diameters are in the ratio }4:5,\text{ then write the ratio of their heights.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the diameters are in the ratio }4:5,\text{ their radii are also in the ratio }4:5.
\displaystyle \text{Volume of a cone}=\frac{1}{3}\pi r^2h.
\displaystyle \therefore \frac{V_1}{V_2}=\frac{r_1^2h_1}{r_2^2h_2}.
\displaystyle \frac{1}{4}=\frac{4^2h_1}{5^2h_2}
\displaystyle \frac{1}{4}=\frac{16h_1}{25h_2}
\displaystyle \therefore \frac{h_1}{h_2}=\frac{25}{64}.
\displaystyle \therefore h_1:h_2=25:64.
\displaystyle \therefore \text{The required ratio is }25:64.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A sphere and a cube have equal surface areas. What is the ratio of the}
\displaystyle \text{volume of the sphere to that of the cube?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the sphere be }r\text{ and the edge of the cube be }a.
\displaystyle 4\pi r^2=6a^2
\displaystyle \therefore \frac{r^2}{a^2}=\frac{3}{2\pi}.
\displaystyle \frac{V_{\text{sphere}}}{V_{\text{cube}}}=\frac{\frac{4}{3}\pi r^3}{a^3}
\displaystyle =\frac{4\pi}{3}\left(\frac{r^2}{a^2}\right)^{\frac{3}{2}}
\displaystyle =\frac{4\pi}{3}\left(\frac{3}{2\pi}\right)^{\frac{3}{2}}
\displaystyle =\sqrt{\frac{6}{\pi}}.
\displaystyle \therefore V_{\text{sphere}}:V_{\text{cube}}=\sqrt{\frac{6}{\pi}}:1.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{What is the ratio of the volume of a cube to that of a sphere which will fit}
\displaystyle \text{inside it?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the edge of the cube be }a.
\displaystyle \text{Diameter of the sphere}=a\quad\Rightarrow\quad r=\frac{a}{2}.
\displaystyle V_{\text{cube}}=a^3.
\displaystyle V_{\text{sphere}}=\frac{4}{3}\pi\left(\frac{a}{2}\right)^3=\frac{\pi a^3}{6}.
\displaystyle \therefore V_{\text{cube}}:V_{\text{sphere}}=a^3:\frac{\pi a^3}{6}=6:\pi.
\displaystyle \text{Using }\pi=\frac{22}{7},\qquad 6:\pi=6:\frac{22}{7}=21:11.
\displaystyle \therefore \text{The required ratio is }21:11.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{What is the ratio of the volumes of a cylinder, a cone and a sphere, if}
\displaystyle \text{each has the same diameter and the same height?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the common radius be }r.
\displaystyle \text{Since the height of the sphere is its diameter, the common height}=2r.
\displaystyle V_{\text{cylinder}}=\pi r^2(2r)=2\pi r^3.
\displaystyle V_{\text{cone}}=\frac{1}{3}\pi r^2(2r)=\frac{2}{3}\pi r^3.
\displaystyle V_{\text{sphere}}=\frac{4}{3}\pi r^3.
\displaystyle \therefore V_{\text{cylinder}}:V_{\text{cone}}:V_{\text{sphere}}
\displaystyle =2:\frac{2}{3}:\frac{4}{3}=3:1:2.
\displaystyle \therefore \text{The required ratio is }3:1:2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The surface area of a sphere is }616\text{ cm}^2.\text{ Find its radius.}
\displaystyle \hfill\text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the sphere be }r\text{ cm.}
\displaystyle 4\pi r^2=616
\displaystyle 4\times\frac{22}{7}\times r^2=616
\displaystyle r^2=\frac{616\times7}{88}=49
\displaystyle \therefore r=7\text{ cm.}
\displaystyle \therefore \text{The radius of the sphere is }7\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{A cylinder and a cone are of the same base radius and of the same height.}
\displaystyle \text{Find the ratio of the volume of the cylinder to that of the cone.}\hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle V_{\text{cylinder}}=\pi r^2h.
\displaystyle V_{\text{cone}}=\frac{1}{3}\pi r^2h.
\displaystyle \therefore V_{\text{cylinder}}:V_{\text{cone}}=1:\frac{1}{3}=3:1.
\displaystyle \therefore \text{The required ratio is }3:1.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Volume and surface area of a solid hemisphere are numerically equal.}
\displaystyle \text{What is the diameter of the hemisphere?}\hfill\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the hemisphere be }r\text{ cm.}
\displaystyle \text{Volume of hemisphere}=\frac{2}{3}\pi r^3.
\displaystyle \text{Total surface area of solid hemisphere}=3\pi r^2.
\displaystyle \frac{2}{3}\pi r^3=3\pi r^2
\displaystyle \frac{2}{3}r=3
\displaystyle r=\frac{9}{2}=4.5\text{ cm.}
\displaystyle \therefore \text{Diameter}=2r=9\text{ cm.}
\displaystyle \therefore \text{The diameter of the hemisphere is }9\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A solid metallic sphere of radius }3\text{ cm is melted and recast into the}
\displaystyle \text{shape of a solid cylinder of radius }2\text{ cm. Find the height of the cylinder.}
\displaystyle \hfill\text{[CBSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of the cylinder be }h\text{ cm.}
\displaystyle \text{Since the sphere is melted and recast into the cylinder,}
\displaystyle \text{Volume of sphere}=\text{Volume of cylinder.}
\displaystyle \frac{4}{3}\pi(3)^3=\pi(2)^2h
\displaystyle 36\pi=4\pi h
\displaystyle h=9\text{ cm.}
\displaystyle \therefore \text{The height of the cylinder is }9\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The radii of two cones are in the ratio }2:1\text{ and their volumes are}
\displaystyle \text{equal. What is the ratio of their heights?}
\displaystyle \text{Answer:}
\displaystyle \text{Since the volumes of the two cones are equal,}
\displaystyle \frac{1}{3}\pi r_1^2h_1=\frac{1}{3}\pi r_2^2h_2.
\displaystyle \therefore \frac{h_1}{h_2}=\frac{r_2^2}{r_1^2}=\left(\frac{1}{2}\right)^2=\frac{1}{4}.
\displaystyle \therefore h_1:h_2=1:4.
\displaystyle \therefore \text{The required ratio is }1:4.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Two cones have their heights in the ratio }1:3\text{ and radii }3:1.\text{ What}
\displaystyle \text{is the ratio of their volumes?}\hfill\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of a cone}=\frac{1}{3}\pi r^2h.
\displaystyle \therefore V_1:V_2=r_1^2h_1:r_2^2h_2.
\displaystyle =(3)^2\times1:(1)^2\times3
\displaystyle =9:3=3:1.
\displaystyle \therefore \text{The required ratio is }3:1.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{A hemisphere and a cone have equal bases. If their heights are also equal,}
\displaystyle \text{then what is the ratio of their curved surfaces?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the common radius be }r.
\displaystyle \text{Height of the hemisphere}=r.
\displaystyle \therefore \text{Height of the cone}=r.
\displaystyle \text{Slant height of the cone}=\sqrt{r^2+r^2}=r\sqrt{2}.
\displaystyle \text{Curved surface area of hemisphere}=2\pi r^2.
\displaystyle \text{Curved surface area of cone}=\pi r(r\sqrt{2})=\sqrt{2}\pi r^2.
\displaystyle \therefore \text{Ratio of curved surfaces}=2:\sqrt{2}=\sqrt{2}:1.
\displaystyle \therefore \text{The required ratio is }\sqrt{2}:1.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{A sphere of maximum volume is cut out from a solid hemisphere of radius }r.
\displaystyle \text{What is the ratio of the volume of the hemisphere to that of the cut-out sphere?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the largest sphere be }R.
\displaystyle \text{For the largest sphere inside the hemisphere, }2R=r.
\displaystyle \therefore R=\frac{r}{2}.
\displaystyle V_{\text{hemisphere}}=\frac{2}{3}\pi r^3.
\displaystyle V_{\text{sphere}}=\frac{4}{3}\pi\left(\frac{r}{2}\right)^3=\frac{1}{6}\pi r^3.
\displaystyle \therefore V_{\text{hemisphere}}:V_{\text{sphere}}=\frac{2}{3}:\frac{1}{6}=4:1.
\displaystyle \therefore \text{The required ratio is }4:1.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{A metallic hemisphere is melted and recast in the shape of a cone with}
\displaystyle \text{the same base radius }R\text{ as that of the hemisphere. If }H\text{ is the height of the cone,}
\displaystyle \text{then write the value of }\frac{H}{R}.
\displaystyle \text{Answer:}
\displaystyle \text{Since the hemisphere is melted and recast into the cone,}
\displaystyle \text{Volume of hemisphere}=\text{Volume of cone.}
\displaystyle \frac{2}{3}\pi R^3=\frac{1}{3}\pi R^2H
\displaystyle \therefore H=2R.
\displaystyle \therefore \frac{H}{R}=2.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{A right circular cone and a right circular cylinder have equal base and}
\displaystyle \text{equal height. If the radius of the base and height are in the ratio }5:12,\text{ write the}
\displaystyle \text{ratio of the total surface area of the cylinder to that of the cone.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the common radius be }5x\text{ and the common height be }12x.
\displaystyle \text{Slant height of the cone}=\sqrt{(5x)^2+(12x)^2}=13x.
\displaystyle \text{Total surface area of cylinder}=2\pi r(r+h)
\displaystyle =2\pi(5x)(5x+12x)=170\pi x^2.
\displaystyle \text{Total surface area of cone}=\pi r(l+r)
\displaystyle =\pi(5x)(13x+5x)=90\pi x^2.
\displaystyle \therefore \text{Required ratio}=170:90=17:9.
\displaystyle \therefore \text{The required ratio is }17:9.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{A cylinder, a cone and a hemisphere are of equal base and have the}
\displaystyle \text{same height. What is the ratio of their volumes?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the common radius be }r.
\displaystyle \text{Since the height of a hemisphere is }r,\text{ the common height is }r.
\displaystyle V_{\text{cylinder}}=\pi r^3,\qquad V_{\text{cone}}=\frac{1}{3}\pi r^3,
\displaystyle V_{\text{hemisphere}}=\frac{2}{3}\pi r^3.
\displaystyle \therefore V_{\text{cylinder}}:V_{\text{cone}}:V_{\text{hemisphere}}=3:1:2.
\displaystyle \therefore \text{The required ratio is }3:1:2.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{A solid piece of metal in the form of a cuboid of dimensions }11\text{ cm}\times7\text{ cm}
\displaystyle \times7\text{ cm is melted to form }n\text{ solid spheres of radius }\frac{7}{2}\text{ cm each. Find the value}
\displaystyle \text{of }n.\hfill\text{[CBSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of the cuboid}=11\times7\times7=539\text{ cm}^3.
\displaystyle \text{Volume of one sphere}=\frac{4}{3}\pi\left(\frac{7}{2}\right)^3
\displaystyle =\frac{4}{3}\times\frac{22}{7}\times\frac{343}{8}
\displaystyle =\frac{539}{3}\text{ cm}^3.
\displaystyle \text{Since the cuboid is melted and recast into }n\text{ spheres,}
\displaystyle n\times\frac{539}{3}=539
\displaystyle \therefore n=3.
\displaystyle \therefore \text{The value of }n\text{ is }3.
\displaystyle \\

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