\displaystyle \textbf{Question 1: }\text{A toy is in the shape of a right circular cylinder with a hemisphere on one end}
\displaystyle \text{and a cone on the other. The radius and height of the cylindrical part are }5\text{ cm and}
\displaystyle 13\text{ cm respectively. The radii of the hemispherical and conical parts are the same as}
\displaystyle \text{that of the cylindrical part. Find the surface area of the toy if the total height of the toy}
\displaystyle \text{is }30\text{ cm.}\hfill\text{[CBSE 2002]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cylindrical, hemispherical and conical parts}=5\text{ cm.}
\displaystyle \text{Height of the cylindrical part}=13\text{ cm.}
\displaystyle \text{Height of the hemispherical part}=\text{Radius}=5\text{ cm.}
\displaystyle \text{Height of the conical part}=30-13-5=12\text{ cm.}
\displaystyle \text{Let the slant height of the conical part be }l\text{ cm.}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{5^2+12^2}
\displaystyle =\sqrt{25+144}=\sqrt{169}=13\text{ cm.}
\displaystyle \text{Surface area of the toy}
\displaystyle =\text{Curved surface area of cylinder}+\text{Curved surface area of hemisphere}
\displaystyle \quad+\text{Curved surface area of cone}
\displaystyle =2\pi rh+2\pi r^2+\pi rl
\displaystyle =2\pi\times5\times13+2\pi\times5^2+\pi\times5\times13
\displaystyle =(130+50+65)\pi
\displaystyle =245\pi
\displaystyle =245\times\frac{22}{7}=770\text{ cm}^2.
\displaystyle \therefore \text{The surface area of the toy is }770\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A cylindrical tube of radius }5\text{ cm and length }9.8\text{ cm is full of water. A solid in}
\displaystyle \text{the form of a right circular cone mounted on a hemisphere is immersed in the tube. If the}
\displaystyle \text{radius of the hemisphere is }3.5\text{ cm and height of the cone outside the hemisphere is}
\displaystyle 5\text{ cm, find the volume of the water left in the tube. (Take }\pi=\frac{22}{7}\text{).}
\displaystyle \hfill\text{[CBSE 2000C]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cylindrical tube}=5\text{ cm},\qquad \text{length}=9.8\text{ cm.}
\displaystyle \text{Volume of the cylindrical tube}=\pi r^2h
\displaystyle =\pi\times5^2\times9.8=245\pi\text{ cm}^3.
\displaystyle \text{Radius of the hemisphere and cone}=3.5=\frac{7}{2}\text{ cm.}
\displaystyle \text{Height of the cone}=5\text{ cm.}
\displaystyle \text{Volume of the hemisphere}=\frac{2}{3}\pi\left(\frac{7}{2}\right)^3
\displaystyle =\frac{343}{12}\pi\text{ cm}^3.
\displaystyle \text{Volume of the cone}=\frac{1}{3}\pi\left(\frac{7}{2}\right)^2\times5
\displaystyle =\frac{245}{12}\pi\text{ cm}^3.
\displaystyle \text{Volume of the immersed solid}=\frac{343}{12}\pi+\frac{245}{12}\pi
\displaystyle =\frac{588}{12}\pi=49\pi\text{ cm}^3.
\displaystyle \text{Since the tube was initially full, water equal to the volume of the solid flows out.}
\displaystyle \text{Volume of water left}=245\pi-49\pi
\displaystyle =196\pi
\displaystyle =196\times\frac{22}{7}=616\text{ cm}^3.
\displaystyle \therefore \text{The volume of water left in the tube is }616\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{A solid is composed of a cylinder with hemispherical ends. If the}
\displaystyle \text{whole length of the solid is }104\text{ cm and the radius of each hemispherical end is }7\text{ cm,}
\displaystyle \text{find the cost of polishing its surface at the rate of Rs. }10\text{ per dm}^2.\hfill\text{[CBSE 2006C]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cylinder and hemispherical ends}=7\text{ cm.}
\displaystyle \text{Whole length of the solid}=104\text{ cm.}
\displaystyle \text{Length of the cylindrical part}=104-2\times7=90\text{ cm.}
\displaystyle \text{Surface area of the solid}
\displaystyle =\text{Curved surface area of cylinder}+\text{Surface area of two hemispheres}
\displaystyle =2\pi rh+4\pi r^2
\displaystyle =2\pi\times7\times90+4\pi\times7^2
\displaystyle =1260\pi+196\pi=1456\pi
\displaystyle =1456\times\frac{22}{7}=4576\text{ cm}^2.
\displaystyle 4576\text{ cm}^2=\frac{4576}{100}=45.76\text{ dm}^2.
\displaystyle \text{Cost of polishing}=45.76\times10=\text{Rs. }457.60.
\displaystyle \therefore \text{The cost of polishing the surface is Rs. }457.60.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The}
\displaystyle \text{diameter of the hemisphere is }14\text{ cm and the total height of the vessel is }13\text{ cm.}
\displaystyle \text{Find the inner surface area of the vessel.}\hfill\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the hemisphere}=\frac{14}{2}=7\text{ cm.}
\displaystyle \text{Height of the hemispherical part}=7\text{ cm.}
\displaystyle \text{Height of the cylindrical part}=13-7=6\text{ cm.}
\displaystyle \text{Inner surface area of the vessel}
\displaystyle =\text{Curved surface area of hemisphere}+\text{Curved surface area of cylinder}
\displaystyle =2\pi r^2+2\pi rh
\displaystyle =2\pi r(r+h)
\displaystyle =2\times\frac{22}{7}\times7\times(7+6)
\displaystyle =44\times13=572\text{ cm}^2.
\displaystyle \therefore \text{The inner surface area of the vessel is }572\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{A toy is in the form of a cone of radius }3.5\text{ cm mounted on a hemisphere of}
\displaystyle \text{the same radius. The total height of the toy is }15.5\text{ cm. Find the total surface area}
\displaystyle \text{of the toy.}\hfill\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cone and hemisphere}=3.5\text{ cm.}
\displaystyle \text{Height of the hemisphere}=\text{Radius}=3.5\text{ cm.}
\displaystyle \text{Height of the conical part}=15.5-3.5=12\text{ cm.}
\displaystyle \text{Let the slant height of the cone be }l\text{ cm.}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{(3.5)^2+12^2}
\displaystyle =\sqrt{12.25+144}=\sqrt{156.25}=12.5\text{ cm.}
\displaystyle \text{Total surface area of the toy}
\displaystyle =\text{Curved surface area of cone}+\text{Curved surface area of hemisphere}
\displaystyle =\pi rl+2\pi r^2
\displaystyle =\pi r(l+2r)
\displaystyle =\frac{22}{7}\times3.5\times(12.5+7)
\displaystyle =11\times19.5=214.5\text{ cm}^2.
\displaystyle \therefore \text{The total surface area of the toy is }214.5\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A cylindrical vessel with internal diameter }10\text{ cm and height }10.5\text{ cm is full}
\displaystyle \text{of water. A solid cone of base diameter }7\text{ cm and height }6\text{ cm is completely}
\displaystyle \text{immersed in water. Find the volume of water (i) displaced out of the cylinder,}
\displaystyle \text{(ii) left in the cylinder. (Take }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cylindrical vessel}=\frac{10}{2}=5\text{ cm.}
\displaystyle \text{Height of the cylindrical vessel}=10.5\text{ cm.}
\displaystyle \text{Volume of water initially in the cylinder}=\pi r^2h
\displaystyle =\frac{22}{7}\times5^2\times10.5=825\text{ cm}^3.

\displaystyle \text{(i) Radius of the base of the cone}=\frac{7}{2}=3.5\text{ cm.}
\displaystyle \text{Height of the cone}=6\text{ cm.}
\displaystyle \text{Since the cone is completely immersed,}
\displaystyle \text{Volume of water displaced}=\text{Volume of the cone}
\displaystyle =\frac{1}{3}\pi r^2h
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times(3.5)^2\times6
\displaystyle =77\text{ cm}^3.
\displaystyle \therefore \text{The volume of water displaced is }77\text{ cm}^3.

\displaystyle \text{(ii) Volume of water left in the cylinder}
\displaystyle =\text{Initial volume of water}-\text{Volume of water displaced}
\displaystyle =825-77=748\text{ cm}^3.
\displaystyle \therefore \text{The volume of water left in the cylinder is }748\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A hemispherical depression is cut out from one face of a cubical}
\displaystyle \text{wooden block of edge }21\text{ cm, such that the diameter of the hemisphere is equal to the}
\displaystyle \text{edge of the cube. Determine the volume and total surface area of the remaining block.}
\displaystyle \hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Edge of the cubical block}=21\text{ cm.}
\displaystyle \text{Diameter of the hemispherical depression}=21\text{ cm.}
\displaystyle \therefore \text{Radius of the hemisphere}=\frac{21}{2}=10.5\text{ cm.}
\displaystyle \text{Volume of the cube}=21^3=9261\text{ cm}^3.
\displaystyle \text{Volume of the hemispherical depression}=\frac{2}{3}\pi r^3
\displaystyle =\frac{2}{3}\times\frac{22}{7}\times(10.5)^3
\displaystyle =2425.5\text{ cm}^3.
\displaystyle \text{Volume of the remaining block}=9261-2425.5
\displaystyle =6835.5\text{ cm}^3.
\displaystyle \therefore \text{The volume of the remaining block is }6835.5\text{ cm}^3.
\displaystyle \text{Surface area of the cube}=6(21)^2=2646\text{ cm}^2.
\displaystyle \text{Area of the circular portion removed}=\pi r^2
\displaystyle =\frac{22}{7}\times(10.5)^2=346.5\text{ cm}^2.
\displaystyle \text{Curved surface area of the hemispherical depression}=2\pi r^2
\displaystyle =2\times346.5=693\text{ cm}^2.
\displaystyle \text{Total surface area of the remaining block}
\displaystyle =\text{Surface area of cube}-\text{Area of circular portion removed}
\displaystyle \quad+\text{Curved surface area of hemispherical depression}
\displaystyle =2646-346.5+693
\displaystyle =2992.5\text{ cm}^2.
\displaystyle \therefore \text{The total surface area of the remaining block is }2992.5\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A toy is in the form of a hemisphere surmounted by a right circular}
\displaystyle \text{cone of the same base radius as that of the hemisphere. If the radius of the base of the}
\displaystyle \text{cone is }21\text{ cm and its volume is }\frac{2}{3}\text{ of the volume of the hemisphere, calculate the}
\displaystyle \text{height of the cone and the surface area of the toy. (Use }\pi=\frac{22}{7}\text{).}
\displaystyle \hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cone and hemisphere}=21\text{ cm.}
\displaystyle \text{Let the height of the cone be }h\text{ cm.}
\displaystyle \text{Volume of the cone}=\frac{1}{3}\pi r^2h.
\displaystyle \text{Volume of the hemisphere}=\frac{2}{3}\pi r^3.
\displaystyle \text{Given, volume of cone}=\frac{2}{3}\times\text{Volume of hemisphere}.
\displaystyle \frac{1}{3}\pi r^2h=\frac{2}{3}\times\frac{2}{3}\pi r^3
\displaystyle \frac{1}{3}\pi r^2h=\frac{4}{9}\pi r^3
\displaystyle h=\frac{4r}{3}
\displaystyle =\frac{4\times21}{3}=28\text{ cm.}
\displaystyle \therefore \text{The height of the cone is }28\text{ cm.}
\displaystyle \text{Let the slant height of the cone be }l\text{ cm.}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{21^2+28^2}
\displaystyle =\sqrt{441+784}=\sqrt{1225}=35\text{ cm.}
\displaystyle \text{Surface area of the toy}
\displaystyle =\text{Curved surface area of hemisphere}+\text{Curved surface area of cone}
\displaystyle =2\pi r^2+\pi rl
\displaystyle =\pi r(2r+l)
\displaystyle =\frac{22}{7}\times21\times(42+35)
\displaystyle =66\times77=5082\text{ cm}^2.
\displaystyle \therefore \text{The surface area of the toy is }5082\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A solid is in the shape of a cone surmounted on a hemisphere, the radius of each}
\displaystyle \text{of them being }3.5\text{ cm and the total height of the solid is }9.5\text{ cm. Find the volume}
\displaystyle \text{of the solid. (Use }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cone and hemisphere}=3.5=\frac{7}{2}\text{ cm.}
\displaystyle \text{Height of the hemispherical part}=3.5\text{ cm.}
\displaystyle \text{Height of the conical part}=9.5-3.5=6\text{ cm.}
\displaystyle \text{Volume of the solid}=\text{Volume of cone}+\text{Volume of hemisphere}
\displaystyle =\frac{1}{3}\pi r^2h+\frac{2}{3}\pi r^3
\displaystyle =\frac{1}{3}\pi\left(\frac{7}{2}\right)^2\times6+\frac{2}{3}\pi\left(\frac{7}{2}\right)^3
\displaystyle =\frac{49}{2}\pi+\frac{343}{12}\pi
\displaystyle =\left(\frac{294+343}{12}\right)\pi
\displaystyle =\frac{637}{12}\pi
\displaystyle =\frac{637}{12}\times\frac{22}{7}
\displaystyle =\frac{1001}{6}\text{ cm}^3
\displaystyle \approx166.83\text{ cm}^3.
\displaystyle \therefore \text{The volume of the solid is }\frac{1001}{6}\text{ cm}^3\text{ (approximately }166.83\text{ cm}^3\text{).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The largest possible sphere is carved out of a wooden solid cube of side }7\text{ cm.}
\displaystyle \text{Find the volume of the wood left. (Use }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Side of the cube}=7\text{ cm.}
\displaystyle \text{For the largest possible sphere, diameter of sphere}=\text{side of cube}.
\displaystyle \therefore \text{Radius of the sphere}=\frac{7}{2}=3.5\text{ cm.}
\displaystyle \text{Volume of the cube}=7^3=343\text{ cm}^3.
\displaystyle \text{Volume of the sphere}=\frac{4}{3}\pi r^3
\displaystyle =\frac{4}{3}\times\frac{22}{7}\times\left(\frac{7}{2}\right)^3
\displaystyle =\frac{539}{3}\text{ cm}^3.
\displaystyle \text{Volume of wood left}=\text{Volume of cube}-\text{Volume of sphere}
\displaystyle =343-\frac{539}{3}
\displaystyle =\frac{1029-539}{3}
\displaystyle =\frac{490}{3}\text{ cm}^3
\displaystyle \approx163.33\text{ cm}^3.
\displaystyle \therefore \text{The volume of the wood left is }\frac{490}{3}\text{ cm}^3\text{ (approximately }163.33\text{ cm}^3\text{).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{From a solid cylinder of height }2.8\text{ cm and diameter }4.2\text{ cm, a conical}
\displaystyle \text{cavity of the same height and same diameter is hollowed out. Find the total surface area}
\displaystyle \text{of the remaining solid. (Take }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cylinder and conical cavity}=\frac{4.2}{2}=2.1\text{ cm.}
\displaystyle \text{Height of the cylinder and conical cavity}=2.8\text{ cm.}
\displaystyle \text{Let the slant height of the conical cavity be }l\text{ cm.}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{(2.1)^2+(2.8)^2}
\displaystyle =\sqrt{4.41+7.84}=\sqrt{12.25}=3.5\text{ cm.}
\displaystyle \text{Total surface area of the remaining solid}
\displaystyle =\text{Curved surface area of cylinder}+\text{Area of its circular base}
\displaystyle \quad+\text{Curved surface area of conical cavity}
\displaystyle =2\pi rh+\pi r^2+\pi rl
\displaystyle =\pi r(2h+r+l)
\displaystyle =\frac{22}{7}\times2.1\times(2\times2.8+2.1+3.5)
\displaystyle =\frac{22}{7}\times2.1\times11.2
\displaystyle =73.92\text{ cm}^2.
\displaystyle \therefore \text{The total surface area of the remaining solid is }73.92\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The largest cone is carved out from a solid cube of side }21\text{ cm. Find the volume}
\displaystyle \text{of the remaining solid.}\hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Side of the cube}=21\text{ cm.}
\displaystyle \text{For the largest cone, diameter of its base}=\text{side of the cube}=21\text{ cm.}
\displaystyle \therefore \text{Radius of the base of the cone}=\frac{21}{2}=10.5\text{ cm.}
\displaystyle \text{Height of the cone}=21\text{ cm.}
\displaystyle \text{Volume of the cube}=21^3=9261\text{ cm}^3.
\displaystyle \text{Volume of the largest cone}=\frac{1}{3}\pi r^2h
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times(10.5)^2\times21
\displaystyle =2425.5\text{ cm}^3.
\displaystyle \text{Volume of the remaining solid}=\text{Volume of cube}-\text{Volume of cone}
\displaystyle =9261-2425.5
\displaystyle =6835.5\text{ cm}^3.
\displaystyle \therefore \text{The volume of the remaining solid is }6835.5\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{A pen stand made of wood is in the shape of a cuboid with four conical depressions}
\displaystyle \text{and a cubical depression to hold the pens and pins, respectively. The dimensions of the}
\displaystyle \text{cuboid are }10\text{ cm}\times5\text{ cm}\times4\text{ cm. The radius of each conical depression is }0.5\text{ cm}
\displaystyle \text{and the depth is }2.1\text{ cm. The edge of the cubical depression is }3\text{ cm. Find the volume}
\displaystyle \text{of the wood in the entire stand.}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of the cuboid}=10\times5\times4=200\text{ cm}^3.
\displaystyle \text{Radius of each conical depression}=0.5\text{ cm},\qquad \text{depth}=2.1\text{ cm.}
\displaystyle \text{Volume of one conical depression}=\frac{1}{3}\pi r^2h
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times(0.5)^2\times2.1
\displaystyle =0.55\text{ cm}^3.
\displaystyle \text{Volume of four conical depressions}=4\times0.55=2.2\text{ cm}^3.
\displaystyle \text{Edge of the cubical depression}=3\text{ cm.}
\displaystyle \text{Volume of the cubical depression}=3^3=27\text{ cm}^3.
\displaystyle \text{Volume of wood in the stand}
\displaystyle =\text{Volume of cuboid}-\text{Volume of four cones}-\text{Volume of cubical depression}
\displaystyle =200-2.2-27
\displaystyle =170.8\text{ cm}^3.
\displaystyle \therefore \text{The volume of wood in the entire stand is }170.8\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{A solid toy is in the form of a hemisphere surmounted by a right circular cone. The}
\displaystyle \text{height of the cone is }4\text{ cm and the diameter of the base is }8\text{ cm. Determine the volume}
\displaystyle \text{of the toy. If a cube circumscribes the toy, find the difference of the volumes of the cube}
\displaystyle \text{and the toy. Also, find the total surface area of the toy.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the hemisphere and cone}=\frac{8}{2}=4\text{ cm.}
\displaystyle \text{Height of the cone}=4\text{ cm.}
\displaystyle \text{Volume of the toy}=\text{Volume of hemisphere}+\text{Volume of cone}
\displaystyle =\frac{2}{3}\pi r^3+\frac{1}{3}\pi r^2h
\displaystyle =\frac{2}{3}\pi(4)^3+\frac{1}{3}\pi(4)^2\times4
\displaystyle =\frac{128}{3}\pi+\frac{64}{3}\pi=64\pi\text{ cm}^3.
\displaystyle =64\times\frac{22}{7}=\frac{1408}{7}\text{ cm}^3.
\displaystyle \therefore \text{The volume of the toy is }\frac{1408}{7}\text{ cm}^3.
\displaystyle \text{Total height of the toy}=4+4=8\text{ cm.}
\displaystyle \text{Maximum width of the toy}=\text{Diameter}=8\text{ cm.}
\displaystyle \therefore \text{Side of the cube circumscribing the toy}=8\text{ cm.}
\displaystyle \text{Volume of the cube}=8^3=512\text{ cm}^3.
\displaystyle \text{Difference of the volumes}=512-\frac{1408}{7}
\displaystyle =\frac{3584-1408}{7}=\frac{2176}{7}\text{ cm}^3
\displaystyle \approx310.86\text{ cm}^3.
\displaystyle \therefore \text{The difference of the volumes is }\frac{2176}{7}\text{ cm}^3.
\displaystyle \text{Let the slant height of the cone be }l\text{ cm.}
\displaystyle l=\sqrt{r^2+h^2}=\sqrt{4^2+4^2}=4\sqrt{2}\text{ cm.}
\displaystyle \text{Total surface area of the toy}
\displaystyle =\text{Curved surface area of hemisphere}+\text{Curved surface area of cone}
\displaystyle =2\pi r^2+\pi rl
\displaystyle =2\pi(4)^2+\pi\times4\times4\sqrt{2}
\displaystyle =32\pi+16\sqrt{2}\pi
\displaystyle =16\pi(2+\sqrt{2})\text{ cm}^2
\displaystyle \approx171.69\text{ cm}^2.
\displaystyle \therefore \text{The total surface area of the toy is }16\pi(2+\sqrt{2})\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A circus tent is in the shape of a cylinder surmounted by a conical}
\displaystyle \text{top of the same diameter. If their common diameter is }56\text{ m, the height of the cylindrical}
\displaystyle \text{part is }6\text{ m and the total height of the tent above the ground is }27\text{ m, find the area}
\displaystyle \text{of the canvas used in making the tent.}\hfill\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cylindrical and conical parts}=\frac{56}{2}=28\text{ m.}
\displaystyle \text{Height of the cylindrical part}=6\text{ m.}
\displaystyle \text{Height of the conical part}=27-6=21\text{ m.}
\displaystyle \text{Let the slant height of the conical part be }l\text{ m.}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{28^2+21^2}
\displaystyle =\sqrt{784+441}=\sqrt{1225}=35\text{ m.}
\displaystyle \text{Area of canvas used}
\displaystyle =\text{Curved surface area of cylinder}+\text{Curved surface area of cone}
\displaystyle =2\pi rh+\pi rl
\displaystyle =2\times\frac{22}{7}\times28\times6+\frac{22}{7}\times28\times35
\displaystyle =1056+3080
\displaystyle =4136\text{ m}^2.
\displaystyle \therefore \text{The area of canvas used in making the tent is }4136\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{A cone of maximum size is carved out from a cube of edge }
\displaystyle 14\text{ cm. Find the surface} \ \text{area of the cone and the volume and surface area of the}
\displaystyle \text{remaining solid left out after the cone is carved out.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Edge of the cube}=14\text{ cm.}
\displaystyle \text{For the cone of maximum size, diameter of its base}=\text{edge of the cube}.
\displaystyle \therefore r=\frac{14}{2}=7\text{ cm},\qquad h=14\text{ cm.}
\displaystyle \text{Let the slant height of the cone be }l\text{ cm.}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{7^2+14^2}
\displaystyle =\sqrt{49+196}=\sqrt{245}=7\sqrt{5}\text{ cm.}
\displaystyle \text{Total surface area of the cone}=\pi r(l+r)
\displaystyle =\frac{22}{7}\times7(7\sqrt{5}+7)
\displaystyle =154(\sqrt{5}+1)\text{ cm}^2.
\displaystyle \therefore \text{The surface area of the cone is }154(\sqrt{5}+1)\text{ cm}^2.
\displaystyle \text{Volume of the cube}=14^3=2744\text{ cm}^3.
\displaystyle \text{Volume of the cone}=\frac{1}{3}\pi r^2h
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times7^2\times14
\displaystyle =\frac{2156}{3}\text{ cm}^3.
\displaystyle \text{Volume of the remaining solid}=2744-\frac{2156}{3}
\displaystyle =\frac{8232-2156}{3}=\frac{6076}{3}\text{ cm}^3.
\displaystyle \therefore \text{The volume of the remaining solid is }\frac{6076}{3}\text{ cm}^3.
\displaystyle \text{Surface area of the cube}=6(14)^2=1176\text{ cm}^2.
\displaystyle \text{Area of the circular portion removed}=\pi r^2
\displaystyle =\frac{22}{7}\times7^2=154\text{ cm}^2.
\displaystyle \text{Curved surface area of the conical cavity}=\pi rl
\displaystyle =\frac{22}{7}\times7\times7\sqrt{5}=154\sqrt{5}\text{ cm}^2.
\displaystyle \text{Surface area of the remaining solid}
\displaystyle =\text{Surface area of cube}-\text{Area of circular portion removed}
\displaystyle \quad+\text{Curved surface area of conical cavity}
\displaystyle =1176-154+154\sqrt{5}
\displaystyle =(1022+154\sqrt{5})\text{ cm}^2.
\displaystyle \therefore \text{The surface area of the remaining solid is }(1022+154\sqrt{5})\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Marbles of diameter }1.4\text{ cm are dropped into a cylindrical beaker of diameter}
\displaystyle 7\text{ cm containing some water. Find the number of marbles that should be dropped into the}
\displaystyle \text{beaker so that the water level rises by }5.6\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of each marble}=\frac{1.4}{2}=0.7\text{ cm.}
\displaystyle \text{Radius of the cylindrical beaker}=\frac{7}{2}=3.5\text{ cm.}
\displaystyle \text{Rise in the water level}=5.6\text{ cm.}
\displaystyle \text{Let the number of marbles dropped into the beaker be }n.
\displaystyle \text{Volume of water displaced}=\pi(3.5)^2\times5.6\text{ cm}^3.
\displaystyle \text{Volume of one marble}=\frac{4}{3}\pi(0.7)^3\text{ cm}^3.
\displaystyle \text{Since the marbles are completely immersed,}
\displaystyle n\times\frac{4}{3}\pi(0.7)^3=\pi(3.5)^2\times5.6
\displaystyle n=\frac{3(3.5)^2\times5.6}{4(0.7)^3}
\displaystyle =150
\displaystyle \therefore \text{The number of marbles required is }150.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Two cones with the same base radius }8\text{ cm and height }15\text{ cm are joined}
\displaystyle \text{together along their bases. Find the surface area of the shape formed.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of each cone}=8\text{ cm},\qquad \text{height}=15\text{ cm.}
\displaystyle \text{Let the slant height of each cone be }l\text{ cm.}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{8^2+15^2}
\displaystyle =\sqrt{64+225}=\sqrt{289}=17\text{ cm.}
\displaystyle \text{Since the cones are joined along their bases, the circular bases are not exposed.}
\displaystyle \text{Surface area of the shape}=2\times\text{Curved surface area of one cone}
\displaystyle =2\pi rl
\displaystyle =2\pi\times8\times17
\displaystyle =272\pi\text{ cm}^2
\displaystyle \approx854.51\text{ cm}^2.
\displaystyle \therefore \text{The surface area of the shape formed is }272\pi\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{From a solid cube of side }7\text{ cm, a conical cavity of height }7\text{ cm and radius}
\displaystyle 3\text{ cm is hollowed out. Find the volume of the remaining solid.}
\displaystyle \text{Answer:}
\displaystyle \text{Side of the cube}=7\text{ cm.}
\displaystyle \text{Volume of the cube}=7^3=343\text{ cm}^3.
\displaystyle \text{Radius of the conical cavity}=3\text{ cm},\qquad \text{height}=7\text{ cm.}
\displaystyle \text{Volume of the conical cavity}=\frac{1}{3}\pi r^2h
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times3^2\times7
\displaystyle =66\text{ cm}^3.
\displaystyle \text{Volume of the remaining solid}
\displaystyle =\text{Volume of cube}-\text{Volume of conical cavity}
\displaystyle =343-66
\displaystyle =277\text{ cm}^3.
\displaystyle \therefore \text{The volume of the remaining solid is }277\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{(i) The difference between outside and inside surface areas of a cylindrical metallic}
\displaystyle \text{pipe }14\text{ cm long is }44\text{ cm}^2.\text{ If the pipe is made of }99\text{ cm}^3\text{ of metal, find the outer}
\displaystyle \text{and inner radii of the pipe.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the outer and inner radii of the pipe be }R\text{ cm and }r\text{ cm respectively.}
\displaystyle \text{Length of the pipe}=14\text{ cm.}
\displaystyle \text{Difference between outer and inner curved surface areas}=44\text{ cm}^2.
\displaystyle 2\pi Rh-2\pi rh=44
\displaystyle 2\pi h(R-r)=44
\displaystyle 2\times\frac{22}{7}\times14(R-r)=44
\displaystyle 88(R-r)=44
\displaystyle R-r=\frac{1}{2}\qquad\ldots\text{(i)}
\displaystyle \text{Volume of metal used}=\pi h(R^2-r^2)=99.
\displaystyle \pi h(R-r)(R+r)=99
\displaystyle \frac{22}{7}\times14\times\frac{1}{2}(R+r)=99
\displaystyle 22(R+r)=99
\displaystyle R+r=\frac{9}{2}\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle 2R=\frac{1}{2}+\frac{9}{2}=5
\displaystyle R=\frac{5}{2}=2.5\text{ cm.}
\displaystyle \text{From (ii),}\qquad r=\frac{9}{2}-\frac{5}{2}=2\text{ cm.}
\displaystyle \therefore \text{The outer radius is }2.5\text{ cm and the inner radius is }2\text{ cm.}

\displaystyle \text{(ii) The difference between the outer and inner radii of a hollow cylinder of length }14\text{ cm}
\displaystyle \text{is }1\text{ cm. If the volume of the metal used in making the cylinder is }176\text{ cm}^3,\text{ find the}
\displaystyle \text{outer and inner radii of the cylinder.}\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the outer and inner radii of the cylinder be }R\text{ cm and }r\text{ cm respectively.}
\displaystyle R-r=1\qquad\ldots\text{(i)}
\displaystyle \text{Length of the cylinder}=14\text{ cm.}
\displaystyle \text{Volume of metal used}=\pi h(R^2-r^2)=176.
\displaystyle \pi h(R-r)(R+r)=176
\displaystyle \frac{22}{7}\times14\times1\times(R+r)=176
\displaystyle 44(R+r)=176
\displaystyle R+r=4\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle 2R=5
\displaystyle R=2.5\text{ cm.}
\displaystyle \text{From (ii),}\qquad r=4-2.5=1.5\text{ cm.}
\displaystyle \therefore \text{The outer radius is }2.5\text{ cm and the inner radius is }1.5\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{A solid wooden toy is in the form of a hemisphere surmounted by a cone of the same}
\displaystyle \text{radius. The radius of the hemisphere is }3.5\text{ cm and the total wood used in making the toy}
\displaystyle \text{is }166\frac{5}{6}\text{ cm}^3.\text{ Find the height of the toy. Also, find the cost of painting the}
\displaystyle \text{hemispherical part of the toy at the rate of Rs. }10\text{ per cm}^2.\text{ (Take }\pi=\frac{22}{7}\text{).}
\displaystyle \hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the hemisphere and cone}=3.5=\frac{7}{2}\text{ cm.}
\displaystyle \text{Total volume of the toy}=166\frac{5}{6}=\frac{1001}{6}\text{ cm}^3.
\displaystyle \text{Volume of the hemisphere}=\frac{2}{3}\pi r^3
\displaystyle =\frac{2}{3}\times\frac{22}{7}\times\left(\frac{7}{2}\right)^3
\displaystyle =\frac{539}{6}\text{ cm}^3.
\displaystyle \text{Volume of the cone}=\frac{1001}{6}-\frac{539}{6}
\displaystyle =\frac{462}{6}=77\text{ cm}^3.
\displaystyle \text{Let the height of the cone be }h\text{ cm.}
\displaystyle \frac{1}{3}\pi r^2h=77
\displaystyle \frac{1}{3}\times\frac{22}{7}\times\left(\frac{7}{2}\right)^2h=77
\displaystyle \frac{77}{6}h=77
\displaystyle h=6\text{ cm.}
\displaystyle \text{Height of the toy}=\text{Height of cone}+\text{Radius of hemisphere}
\displaystyle =6+3.5=9.5\text{ cm.}
\displaystyle \therefore \text{The height of the toy is }9.5\text{ cm.}
\displaystyle \text{Area of the hemispherical part to be painted}=2\pi r^2
\displaystyle =2\times\frac{22}{7}\times\left(\frac{7}{2}\right)^2
\displaystyle =77\text{ cm}^2.
\displaystyle \text{Cost of painting}=77\times10=\text{Rs. }770.
\displaystyle \therefore \text{The cost of painting the hemispherical part is Rs. }770.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{A wall }24\text{ m long, }0.4\text{ m thick and }6\text{ m high is constructed with bricks, each}
\displaystyle \text{of dimensions }25\text{ cm}\times16\text{ cm}\times10\text{ cm. If the mortar occupies }\frac{1}{10}\text{th of the volume}
\displaystyle \text{of the wall, find the number of bricks used in constructing the wall.}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of the wall}=24\times0.4\times6=57.6\text{ m}^3.
\displaystyle \text{Volume occupied by mortar}=\frac{1}{10}\times57.6=5.76\text{ m}^3.
\displaystyle \text{Volume occupied by bricks}=57.6-5.76=51.84\text{ m}^3.
\displaystyle 25\text{ cm}=0.25\text{ m},\qquad16\text{ cm}=0.16\text{ m},\qquad10\text{ cm}=0.10\text{ m.}
\displaystyle \text{Volume of one brick}=0.25\times0.16\times0.10
\displaystyle =0.004\text{ m}^3.
\displaystyle \text{Number of bricks}=\frac{\text{Volume occupied by bricks}}{\text{Volume of one brick}}
\displaystyle =\frac{51.84}{0.004}=12960.
\displaystyle \therefore \text{The number of bricks used in constructing the wall is }12960.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{(i) A building is in the form of a cylinder surmounted by a hemispherical vaulted dome}
\displaystyle \text{and contains }41\frac{19}{21}\text{ m}^3\text{ of air. If the internal diameter of the dome is equal to its total}
\displaystyle \text{height above the floor, find the height of the building?}\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the hemispherical dome be }r\text{ m and the height of the cylindrical part be }h\text{ m.}
\displaystyle \text{Total height of the building}=h+r.
\displaystyle \text{Internal diameter of the dome}=2r.
\displaystyle \text{Given, }2r=h+r
\displaystyle \therefore h=r.
\displaystyle \text{Volume of air in the building}
\displaystyle =\text{Volume of cylinder}+\text{Volume of hemisphere}
\displaystyle =\pi r^2h+\frac{2}{3}\pi r^3.
\displaystyle \text{Since }h=r,
\displaystyle \pi r^3+\frac{2}{3}\pi r^3=41\frac{19}{21}
\displaystyle \frac{5}{3}\pi r^3=\frac{880}{21}
\displaystyle \frac{5}{3}\times\frac{22}{7}r^3=\frac{880}{21}
\displaystyle \frac{110}{21}r^3=\frac{880}{21}
\displaystyle r^3=8
\displaystyle r=2\text{ m.}
\displaystyle \therefore h=2\text{ m.}
\displaystyle \text{Height of the building}=h+r=2+2=4\text{ m.}
\displaystyle \therefore \text{The height of the building is }4\text{ m.}

\displaystyle \text{(ii) A room is in the form of a cylinder surmounted by a hemispherical dome. The base radius}
\displaystyle \text{of the hemisphere is half of the height of the cylindrical part. If the room contains }\frac{1408}{21}\text{ m}^3
\displaystyle \text{of air, find the height of the cylindrical part. (Use }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the hemispherical dome be }r\text{ m and the height of the cylinder be }h\text{ m.}
\displaystyle \text{Given, }r=\frac{h}{2}
\displaystyle \therefore h=2r.
\displaystyle \text{Volume of air in the room}
\displaystyle =\text{Volume of cylinder}+\text{Volume of hemisphere}
\displaystyle =\pi r^2h+\frac{2}{3}\pi r^3.
\displaystyle =\pi r^2(2r)+\frac{2}{3}\pi r^3
\displaystyle =2\pi r^3+\frac{2}{3}\pi r^3
\displaystyle =\frac{8}{3}\pi r^3.
\displaystyle \frac{8}{3}\times\frac{22}{7}r^3=\frac{1408}{21}
\displaystyle \frac{176}{21}r^3=\frac{1408}{21}
\displaystyle r^3=8
\displaystyle r=2\text{ m.}
\displaystyle \therefore h=2r=2\times2=4\text{ m.}
\displaystyle \therefore \text{The height of the cylindrical part is }4\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{A building is in the form of a cylinder surmounted by a hemispherical dome. The base}
\displaystyle \text{diameter of the dome is equal to }\frac{2}{3}\text{ of the total height of the building. Find the height}
\displaystyle \text{of the building, if it contains }67\frac{1}{21}\text{ m}^3\text{ of air.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the total height of the building be }H\text{ m.}
\displaystyle \text{Let the radius of the hemispherical dome be }r\text{ m.}
\displaystyle \text{Given, diameter of the dome}=\frac{2}{3}H
\displaystyle 2r=\frac{2}{3}H
\displaystyle \therefore H=3r.
\displaystyle \text{Height of the hemispherical dome}=r.
\displaystyle \text{Height of the cylindrical part}=H-r=3r-r=2r.
\displaystyle \text{Volume of air in the building}
\displaystyle =\text{Volume of cylinder}+\text{Volume of hemisphere}
\displaystyle =\pi r^2(2r)+\frac{2}{3}\pi r^3
\displaystyle =2\pi r^3+\frac{2}{3}\pi r^3
\displaystyle =\frac{8}{3}\pi r^3.
\displaystyle 67\frac{1}{21}=\frac{1408}{21}
\displaystyle \therefore \frac{8}{3}\times\frac{22}{7}r^3=\frac{1408}{21}
\displaystyle \frac{176}{21}r^3=\frac{1408}{21}
\displaystyle r^3=8
\displaystyle r=2\text{ m.}
\displaystyle \therefore H=3r=3\times2=6\text{ m.}
\displaystyle \therefore \text{The height of the building is }6\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Two solid cones }A\text{ and }B\text{ are placed in a cylindrical tube as shown in}
\displaystyle \text{figure. The ratio of their capacities is }2:1.\text{ Find the heights and capacities of the}
\displaystyle \text{cones. Also, find the volume of the remaining portion of the cylinder.} \displaystyle \text{Answer:}
\displaystyle \text{Diameter of the cylindrical tube}=6\text{ cm.}
\displaystyle \therefore \text{Radius of the tube and the bases of the cones}=3\text{ cm.}
\displaystyle \text{Length of the cylindrical tube}=21\text{ cm.}
\displaystyle \text{Let the heights of cones }A\text{ and }B\text{ be }h_1\text{ cm and }h_2\text{ cm respectively.}
\displaystyle \text{Since the cones have the same base radius, their volumes are proportional to their heights.}
\displaystyle \frac{V_A}{V_B}=\frac{h_1}{h_2}=\frac{2}{1}
\displaystyle \therefore h_1:h_2=2:1.
\displaystyle \text{Also, }h_1+h_2=21.
\displaystyle \text{Let }h_1=2x\text{ and }h_2=x.
\displaystyle 2x+x=21
\displaystyle 3x=21
\displaystyle x=7
\displaystyle \therefore h_1=14\text{ cm},\qquad h_2=7\text{ cm.}
\displaystyle \text{Volume of cone }A=\frac{1}{3}\pi r^2h_1
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times3^2\times14
\displaystyle =132\pi\text{ cm}^3=\frac{2904}{7}\text{ cm}^3.
\displaystyle \text{Volume of cone }B=\frac{1}{3}\pi r^2h_2
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times3^2\times7
\displaystyle =66\pi\text{ cm}^3=\frac{1452}{7}\text{ cm}^3.
\displaystyle \text{Volume of the cylindrical tube}=\pi r^2h
\displaystyle =\frac{22}{7}\times3^2\times21
\displaystyle =594\text{ cm}^3.
\displaystyle \text{Total volume occupied by the two cones}=132\pi+66\pi
\displaystyle =198\pi=622\frac{2}{7}\text{ cm}^3.
\displaystyle \text{This exceeds the volume of the cylinder, so this interpretation is geometrically impossible.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{An ice-cream cone full of ice-cream having radius }5\text{ cm and height }10\text{ cm is as}
\displaystyle \text{shown in figure. Calculate the volume of ice-cream, provided that its }\frac{1}{6}\text{ part is left}
\displaystyle \text{unfilled with ice-cream.} \displaystyle \text{Answer:}
\displaystyle \text{Radius of the cone and hemisphere}=5\text{ cm.}
\displaystyle \text{Total height}=10\text{ cm.}
\displaystyle \text{Height of the hemispherical part}=5\text{ cm.}
\displaystyle \therefore \text{Height of the conical part}=10-5=5\text{ cm.}
\displaystyle \text{Total capacity}=\text{Volume of cone}+\text{Volume of hemisphere}
\displaystyle =\frac{1}{3}\pi r^2h+\frac{2}{3}\pi r^3
\displaystyle =\frac{1}{3}\pi(5)^2(5)+\frac{2}{3}\pi(5)^3
\displaystyle =\frac{125\pi}{3}+\frac{250\pi}{3}
\displaystyle =125\pi\text{ cm}^3.
\displaystyle \text{Since }\frac{1}{6}\text{ part is unfilled, the filled part}=1-\frac{1}{6}=\frac{5}{6}.
\displaystyle \text{Volume of ice-cream}=\frac{5}{6}\times125\pi
\displaystyle =\frac{625\pi}{6}\text{ cm}^3
\displaystyle \approx327.38\text{ cm}^3.
\displaystyle \therefore \text{The volume of ice-cream is }\frac{625\pi}{6}\text{ cm}^3\text{ (approximately }327.38\text{ cm}^3\text{).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{In figure, from a cuboidal solid metallic block of dimensions }
\displaystyle 15\text{ cm}\times10\text{ cm}\times5\text{ cm, a cylindrical hole of diameter }7\text{ cm is drilled out. }
\displaystyle \text{Find the surface area of the remaining block. (Take }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2015]} \displaystyle \text{Answer:}
\displaystyle \text{Length of the cuboid}=15\text{ cm},\qquad \text{breadth}=10\text{ cm},\qquad \text{height}=5\text{ cm.}
\displaystyle \text{Total surface area of the cuboid}=2(lb+bh+lh)
\displaystyle =2(15\times10+10\times5+15\times5)
\displaystyle =2(150+50+75)=550\text{ cm}^2.
\displaystyle \text{Radius of the cylindrical hole}=\frac{7}{2}=3.5\text{ cm.}
\displaystyle \text{Height of the cylindrical hole}=5\text{ cm.}
\displaystyle \text{Area of the two circular portions removed}=2\pi r^2
\displaystyle =2\times\frac{22}{7}\times(3.5)^2
\displaystyle =77\text{ cm}^2.
\displaystyle \text{Curved surface area of the cylindrical hole}=2\pi rh
\displaystyle =2\times\frac{22}{7}\times3.5\times5
\displaystyle =110\text{ cm}^2.
\displaystyle \text{Surface area of the remaining block}
\displaystyle =\text{Surface area of cuboid}-\text{Area of two circles}+\text{Curved surface area of hole}
\displaystyle =550-77+110
\displaystyle =583\text{ cm}^2.
\displaystyle \therefore \text{The surface area of the remaining block is }583\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{An empty cone is of radius }3\text{ cm and height }12\text{ cm. Ice-cream is filled in it}
\displaystyle \text{so that the lower part of the cone, which is }\left(\frac{1}{6}\right)^{\text{th}}\text{ of the volume of the cone, is}
\displaystyle \text{unfilled but a hemisphere is formed on the top. Find the volume of ice-cream.}
\displaystyle \text{(Take }\pi=3.14\text{).}\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cone}=3\text{ cm},\qquad \text{height}=12\text{ cm.}
\displaystyle \text{Volume of the cone}=\frac{1}{3}\pi r^2h
\displaystyle =\frac{1}{3}\times3.14\times3^2\times12
\displaystyle =113.04\text{ cm}^3.
\displaystyle \text{Since }\frac{1}{6}\text{ of the cone is unfilled, the filled part}
\displaystyle =1-\frac{1}{6}=\frac{5}{6}.
\displaystyle \text{Volume of ice-cream inside the cone}=\frac{5}{6}\times113.04
\displaystyle =94.2\text{ cm}^3.
\displaystyle \text{Radius of the hemispherical part}=3\text{ cm.}
\displaystyle \text{Volume of the hemisphere}=\frac{2}{3}\pi r^3
\displaystyle =\frac{2}{3}\times3.14\times3^3
\displaystyle =56.52\text{ cm}^3.
\displaystyle \text{Total volume of ice-cream}=94.2+56.52
\displaystyle =150.72\text{ cm}^3.
\displaystyle \therefore \text{The volume of ice-cream is }150.72\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{A solid is in the shape of a right circular cone surmounted on a}
\displaystyle \text{hemisphere, the radius of each of them being }7\text{ cm and the height of the cone is equal}
\displaystyle \text{to its diameter. Find the volume of the solid.}\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cone and hemisphere}=7\text{ cm.}
\displaystyle \text{Height of the cone}=\text{Diameter}=2\times7=14\text{ cm.}
\displaystyle \text{Volume of the solid}=\text{Volume of cone}+\text{Volume of hemisphere}
\displaystyle =\frac{1}{3}\pi r^2h+\frac{2}{3}\pi r^3
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times7^2\times14+\frac{2}{3}\times\frac{22}{7}\times7^3
\displaystyle =\frac{2156}{3}+\frac{2156}{3}
\displaystyle =\frac{4312}{3}\text{ cm}^3
\displaystyle =1437\frac{1}{3}\text{ cm}^3.
\displaystyle \therefore \text{The volume of the solid is }\frac{4312}{3}\text{ cm}^3\text{ or }1437\frac{1}{3}\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If the radii of the bases of a cylinder and a cone are in the ratio }
\displaystyle 3:4\text{ and}  \ \text{their heights are in the ratio }2:3,\text{ find the ratio of their volumes.}\hfill\text{[CBSE 2025]}
\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 \text{Volume of the cylinder}=\pi(3r)^2(2h)
\displaystyle =18\pi r^2h.
\displaystyle \text{Volume of the cone}=\frac{1}{3}\pi(4r)^2(3h)
\displaystyle =16\pi r^2h.
\displaystyle \therefore \text{Volume of cylinder}:\text{Volume of cone}
\displaystyle =18\pi r^2h:16\pi r^2h
\displaystyle =18:16=9:8.
\displaystyle \therefore \text{The ratio of their volumes is }9:8.
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{A vessel is in the form of an inverted cone. Its height is }8\text{ cm and the radius of its}
\displaystyle \text{top, which is open, is }5\text{ cm. It is filled with water up to the brim. When lead shots,}
\displaystyle \text{each of which is a sphere of radius }0.5\text{ cm, are dropped into the vessel, one-fourth}
\displaystyle \text{of the water flows out. Find the number of lead shots dropped in the vessel.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the conical vessel}=5\text{ cm},\qquad \text{height}=8\text{ cm.}
\displaystyle \text{Volume of water in the full vessel}=\frac{1}{3}\pi r^2h
\displaystyle =\frac{1}{3}\pi(5)^2\times8
\displaystyle =\frac{200\pi}{3}\text{ cm}^3.
\displaystyle \text{Volume of water flowing out}=\frac{1}{4}\times\frac{200\pi}{3}
\displaystyle =\frac{50\pi}{3}\text{ cm}^3.
\displaystyle \text{Radius of each spherical lead shot}=0.5=\frac{1}{2}\text{ cm.}
\displaystyle \text{Volume of one lead shot}=\frac{4}{3}\pi\left(\frac{1}{2}\right)^3
\displaystyle =\frac{\pi}{6}\text{ cm}^3.
\displaystyle \text{Let the number of lead shots dropped be }n.
\displaystyle \text{Since the vessel was initially full, the volume of water flowing out equals}
\displaystyle \text{the total volume of the lead shots immersed in water.}
\displaystyle n\times\frac{\pi}{6}=\frac{50\pi}{3}
\displaystyle n=\frac{50\pi}{3}\times\frac{6}{\pi}
\displaystyle =100.
\displaystyle \therefore \text{The number of lead shots dropped in the vessel is }100.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{A solid toy is in the form of a hemisphere surmounted by a right circular cone. The}
\displaystyle \text{height of the cone is }2\text{ cm and the diameter of the base is }4\text{ cm. Determine the volume}
\displaystyle \text{of the toy. Also, find the surface area of the toy. (Take }\pi=3.14\text{).}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cone and hemisphere}=\frac{4}{2}=2\text{ cm.}
\displaystyle \text{Height of the cone}=2\text{ cm.}
\displaystyle \text{Volume of the toy}=\text{Volume of hemisphere}+\text{Volume of cone}
\displaystyle =\frac{2}{3}\pi r^3+\frac{1}{3}\pi r^2h
\displaystyle =\frac{2}{3}\times3.14\times2^3+\frac{1}{3}\times3.14\times2^2\times2
\displaystyle =\frac{16}{3}\times3.14+\frac{8}{3}\times3.14
\displaystyle =8\times3.14
\displaystyle =25.12\text{ cm}^3.
\displaystyle \therefore \text{The volume of the toy is }25.12\text{ cm}^3.
\displaystyle \text{Let the slant height of the cone be }l\text{ cm.}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{2^2+2^2}
\displaystyle =\sqrt{8}=2\sqrt{2}\text{ cm.}
\displaystyle \text{Surface area of the toy}
\displaystyle =\text{Curved surface area of hemisphere}+\text{Curved surface area of cone}
\displaystyle =2\pi r^2+\pi rl
\displaystyle =2\times3.14\times2^2+3.14\times2\times2\sqrt{2}
\displaystyle =25.12+12.56\sqrt{2}
\displaystyle \approx42.88\text{ cm}^2.
\displaystyle \therefore \text{The surface area of the toy is approximately }42.88\text{ cm}^2.
\displaystyle \\

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