Class 10: Cones and Sphere

$\displaystyle \textbf{Question 1: }\text{From a solid right circular cylinder of height }10\text{ cm}$
$\displaystyle \text{and base radius }6\text{ cm, a right circular cone of the same height}$
$\displaystyle \text{and the same base is removed. Find the volume of the remaining solid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius }r=6\text{ cm, Height }h=10\text{ cm}$
$\displaystyle \text{Volume of the remaining solid}=\text{Volume of cylinder}-\text{Volume of cone}$
$\displaystyle =\pi r^2h-\frac{1}{3}\pi r^2h$
$\displaystyle =\frac{2}{3}\pi r^2h$
$\displaystyle =\frac{2}{3}\times\frac{22}{7}\times6^2\times10$
$\displaystyle =\frac{5280}{7}=\frac{7542}{7}=754\frac{2}{7}\text{ cm}^3$
$\displaystyle \therefore \text{Volume of the remaining solid}=754\frac{2}{7}\text{ cm}^3$
$\\$

$\displaystyle \textbf{Question 2: }\text{From a solid cylinder of height }16\text{ cm and radius }12\text{ cm},$
$\displaystyle \text{a conical cavity of height }8\text{ cm and base radius }6\text{ cm is hollowed out.}$
$\displaystyle \text{Find the volume and total surface area of the remaining solid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the cylinder, }R=12\text{ cm and }H=16\text{ cm}$
$\displaystyle \text{For the conical cavity, }r=6\text{ cm and }h=8\text{ cm}$
$\displaystyle \text{Volume of the remaining solid}$
$\displaystyle =\text{Volume of cylinder}-\text{Volume of cone}$
$\displaystyle =\pi R^2H-\frac{1}{3}\pi r^2h$
$\displaystyle =\frac{22}{7}\left(12^2\times16-\frac{1}{3}\times6^2\times8\right)$
$\displaystyle =\frac{22}{7}(2304-96)$
$\displaystyle =\frac{48576}{7}=6939.43\text{ cm}^3$
$\displaystyle \therefore \text{Volume of the remaining solid}=6939.43\text{ cm}^3$
$\displaystyle \text{Slant height of the conical cavity }l=\sqrt{r^2+h^2}$
$\displaystyle =\sqrt{6^2+8^2}=10\text{ cm}$
$\displaystyle \text{Total surface area of the remaining solid}$
$\displaystyle =\text{CSA of cylinder}+\text{two circular ends}-\text{circular opening}+\text{CSA of cone}$
$\displaystyle =2\pi RH+2\pi R^2-\pi r^2+\pi rl$
$\displaystyle =2\times\frac{22}{7}\times12\times16+2\times\frac{22}{7}\times12^2$
$\displaystyle \qquad-\frac{22}{7}\times6^2+\frac{22}{7}\times6\times10$
$\displaystyle =\frac{8448+6336-792+1320}{7}$
$\displaystyle =\frac{15312}{7}=2187.43\text{ cm}^2$
$\displaystyle \therefore \text{Total surface area of the remaining solid}=2187.43\text{ cm}^2$
$\\$

$\displaystyle \textbf{Question 3: }\text{A circus tent is cylindrical up to a height of }4\text{ m and conical above it.}$
$\displaystyle \text{Its diameter is }105\text{ m and the slant height of the conical part is }80\text{ m}.$
$\displaystyle \text{Calculate the total area of canvas required. Also, find its total cost at }\text{Rs. }15$
$\displaystyle \text{per metre, if the width of the canvas is }1.5\text{ m}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the cylindrical and conical parts }r=\frac{105}{2}=52.5\text{ m}$
$\displaystyle \text{Height of the cylindrical part }h=4\text{ m}$
$\displaystyle \text{Slant height of the conical part }l=80\text{ m}$
$\displaystyle \text{Area of canvas required}=\text{CSA of cylinder}+\text{CSA of cone}$
$\displaystyle =2\pi rh+\pi rl$
$\displaystyle =2\times\frac{22}{7}\times52.5\times4+\frac{22}{7}\times52.5\times80$
$\displaystyle =1320+13200$
$\displaystyle =14520\text{ m}^2$
$\displaystyle \therefore \text{Total area of canvas required}=14520\text{ m}^2$
$\displaystyle \text{Width of the canvas}=1.5\text{ m}$
$\displaystyle \text{Length of canvas required}=\frac{\text{Area of canvas}}{\text{Width of canvas}}$
$\displaystyle =\frac{14520}{1.5}=9680\text{ m}$
$\displaystyle \text{Total cost of canvas}=9680\times15$
$\displaystyle =\text{Rs. }145200$
$\displaystyle \therefore \text{Total cost of the canvas}=\text{Rs. }145200$
$\\$

$\displaystyle \textbf{Question 4: }\text{A circus tent is cylindrical up to a height of }8\text{ m and is surmounted}$
$\displaystyle \text{by a conical part. The total height of the tent is }13\text{ m and its base diameter is }24\text{ m}.$
$\displaystyle \text{Calculate: (i) the total surface area of the tent,}$
$\displaystyle \text{(ii) the area of canvas required, allowing }10\%\text{ extra for folds and stitching.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the tent }r=\frac{24}{2}=12\text{ m}$
$\displaystyle \text{Height of the cylindrical part }h=8\text{ m}$
$\displaystyle \text{Height of the conical part }H=13-8=5\text{ m}$
$\displaystyle \text{Slant height of the conical part }l=\sqrt{r^2+H^2}$
$\displaystyle =\sqrt{12^2+5^2}=\sqrt{169}=13\text{ m}$
$\displaystyle \text{(i) Total surface area of the tent}$
$\displaystyle =\text{CSA of cylinder}+\text{CSA of cone}$
$\displaystyle =2\pi rh+\pi rl$
$\displaystyle =2\times\frac{22}{7}\times12\times8+\frac{22}{7}\times12\times13$
$\displaystyle =\frac{4224}{7}+\frac{3432}{7}$
$\displaystyle =\frac{7656}{7}=1093.71\text{ m}^2$
$\displaystyle \therefore \text{Total surface area of the tent}=1093.71\text{ m}^2$
$\displaystyle \text{(ii) Canvas required}=1093.71+\frac{10}{100}\times1093.71$
$\displaystyle =1093.71\times1.10$
$\displaystyle =1203.09\text{ m}^2$
$\displaystyle \therefore \text{Area of canvas required}=1203.09\text{ m}^2$
$\\$

$\displaystyle \textbf{Question 5: }\text{A cylindrical boiler, }2\text{ m high, has a diameter of }3.5\text{ m}.$
$\displaystyle \text{It has a hemispherical lid. Find its interior volume, including the part covered by the lid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the cylinder and hemisphere }r=\frac{3.5}{2}=1.75\text{ m}$
$\displaystyle \text{Height of the cylinder }h=2\text{ m}$
$\displaystyle \text{Interior volume}=\text{Volume of cylinder}+\text{Volume of hemisphere}$
$\displaystyle =\pi r^2h+\frac{2}{3}\pi r^3$
$\displaystyle =\frac{22}{7}\left((1.75)^2\times2+\frac{2}{3}(1.75)^3\right)$
$\displaystyle =19.25+11.229$
$\displaystyle =30.479\text{ m}^3$
$\displaystyle \therefore \text{Interior volume of the boiler}=30.48\text{ m}^3$
$\\$

$\displaystyle \textbf{Question 6: }\text{A vessel is a hollow cylinder fitted with a hemispherical bottom}$
$\displaystyle \text{of the same base. The depth of the cylindrical part is }4\frac{2}{3}\text{ m}$
$\displaystyle \text{and the diameter of the hemisphere is }3.5\text{ m}.$
$\displaystyle \text{Calculate the capacity and internal surface area of the vessel.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the cylinder and hemisphere }r=\frac{3.5}{2}=1.75\text{ m}$
$\displaystyle \text{Height of the cylindrical part }h=4\frac{2}{3}=\frac{14}{3}\text{ m}$
$\displaystyle \text{Capacity of the vessel}=\text{Volume of cylinder}+\text{Volume of hemisphere}$
$\displaystyle =\pi r^2h+\frac{2}{3}\pi r^3$
$\displaystyle =\frac{22}{7}\left((1.75)^2\times\frac{14}{3}+\frac{2}{3}(1.75)^3\right)$
$\displaystyle =56.1458\text{ m}^3$
$\displaystyle \therefore \text{Capacity of the vessel}=56.15\text{ m}^3$
$\displaystyle \text{Internal surface area}=\text{CSA of cylinder}+\text{CSA of hemisphere}$
$\displaystyle =2\pi rh+2\pi r^2$
$\displaystyle =2\times\frac{22}{7}\left(1.75\times\frac{14}{3}+(1.75)^2\right)$
$\displaystyle =70.5833\text{ m}^2$
$\displaystyle \therefore \text{Internal surface area of the vessel}=70.58\text{ m}^2$
$\displaystyle \$

$\displaystyle \textbf{Question 7: }\text{A wooden toy is in the shape of a cone mounted on a cylinder,}$
$\displaystyle \text{as shown alongside. The height of the cone is }24\text{ cm and the total height}$
$\displaystyle \text{of the toy is }60\text{ cm. The radius of the cylinder is }10\text{ cm and the radius}$
$\displaystyle \text{of the cone is twice the radius of the cylinder. Find the total surface area of the toy.}$
$\displaystyle \text{Take }\pi=3.14.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the cylinder }r=10\text{ cm}$
$\displaystyle \text{Radius of the cone }R=2\times10=20\text{ cm}$
$\displaystyle \text{Height of the cylinder }h=60-24=36\text{ cm}$
$\displaystyle \text{Slant height of the cone }l=\sqrt{R^2+24^2}$
$\displaystyle =\sqrt{20^2+24^2}=\sqrt{976}=4\sqrt{61}\text{ cm}$
$\displaystyle \text{Total surface area of the toy}$
$\displaystyle =\text{CSA of cylinder}+\text{CSA of cone}+\text{base of cylinder}+\text{exposed annular area}$
$\displaystyle =2\pi rh+\pi Rl+\pi r^2+\pi(R^2-r^2)$
$\displaystyle =2(3.14)(10)(36)+(3.14)(20)(4\sqrt{61})+(3.14)(10)^2$
$\displaystyle \qquad+3.14\left(20^2-10^2\right)$
$\displaystyle =2260.80+1961.94+314+942$
$\displaystyle =5478.74\text{ cm}^2$
$\displaystyle \therefore \text{Total surface area of the toy}=5478.74\text{ cm}^2$
$\\$

$\displaystyle \textbf{Question 8: }\text{A cylindrical container with base diameter }42\text{ cm contains sufficient water}$
$\displaystyle \text{to submerge a rectangular iron solid measuring }22\text{ cm}\times14\text{ cm}\times10.5\text{ cm}.$
$\displaystyle \text{Find the rise in the water level when the solid is completely submerged.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the cylindrical container }r=\frac{42}{2}=21\text{ cm}$
$\displaystyle \text{Let the rise in the water level be }h\text{ cm}$
$\displaystyle \text{Volume of water displaced}=\text{Volume of the iron solid}$
$\displaystyle \pi r^2h=22\times14\times10.5$
$\displaystyle \frac{22}{7}\times21^2\times h=22\times14\times10.5$
$\displaystyle h=\frac{22\times14\times10.5}{\frac{22}{7}\times21^2}$
$\displaystyle =\frac{7}{3}=2\frac{1}{3}\text{ cm}$
$\displaystyle \therefore \text{Rise in the water level}=2\frac{1}{3}\text{ cm}$
$\displaystyle \$

$\displaystyle \textbf{Question 9: }\text{Spherical marbles of diameter }1.4\text{ cm are dropped into a beaker}$
$\displaystyle \text{containing water and are fully submerged. The diameter of the beaker is }7\text{ cm}.$
$\displaystyle \text{Find the number of marbles dropped if the water level rises by }5.6\text{ cm}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of each marble }r=\frac{1.4}{2}=0.7\text{ cm}$
$\displaystyle \text{Radius of the beaker }R=\frac{7}{2}=3.5\text{ cm}$
$\displaystyle \text{Rise in the water level }h=5.6\text{ cm}$
$\displaystyle \text{Let the number of marbles be }n$
$\displaystyle \text{Volume of water displaced}=\text{Total volume of the marbles}$
$\displaystyle \pi R^2h=n\times\frac{4}{3}\pi r^3$
$\displaystyle \pi(3.5)^2(5.6)=n\times\frac{4}{3}\pi(0.7)^3$
$\displaystyle n=\frac{3(3.5)^2(5.6)}{4(0.7)^3}$
$\displaystyle =150$
$\displaystyle \therefore \text{Number of marbles dropped}=150$
$\\$

$\displaystyle \textbf{Question 10: }\text{The cross-section of a railway tunnel is a rectangle }6\text{ m broad}$
$\displaystyle \text{and }8\text{ m high, surmounted by a semicircle, as shown in the figure.}$
$\displaystyle \text{The tunnel is }35\text{ m long. Find the cost of plastering its internal surface,}$
$\displaystyle \text{excluding the floor, at the rate of }\text{Rs. }2.25\text{ per m}^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Breadth of the tunnel}=6\text{ m}$
$\displaystyle \text{Height of each vertical wall}=8\text{ m}$
$\displaystyle \text{Radius of the semicircular roof }r=\frac{6}{2}=3\text{ m}$
$\displaystyle \text{Length of the tunnel}=35\text{ m}$
$\displaystyle \text{Area to be plastered}=\text{Area of two side walls}+\text{Curved area of roof}$
$\displaystyle =2\times8\times35+\pi r\times35$
$\displaystyle =2\times8\times35+\frac{22}{7}\times3\times35$
$\displaystyle =560+330=890\text{ m}^2$
$\displaystyle \text{Cost of plastering}=890\times\text{Rs. }2.25$
$\displaystyle =\text{Rs. }2002.50$
$\displaystyle \therefore \text{Cost of plastering the tunnel}=\text{Rs. }2002.50$
$\displaystyle \$

$\displaystyle \textbf{Question 11: }\text{The horizontal cross-section of a water tank is a rectangle}$
$\displaystyle \text{with a semicircle at one end, as shown in the figure. The water is }2.4\text{ m deep.}$
$\displaystyle \text{Calculate the volume of water in the tank in gallons.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of the rectangle}=21\text{ m}$
$\displaystyle \text{Width of the rectangle}=7\text{ m}$
$\displaystyle \text{Radius of the semicircle}=\frac{7}{2}=3.5\text{ m}$
$\displaystyle \text{Depth of water}=2.4\text{ m}$
$\displaystyle \text{Area of the horizontal cross-section}$
$\displaystyle =21\times7+\frac{1}{2}\times\frac{22}{7}\times(3.5)^2$
$\displaystyle =147+19.25=166.25\text{ m}^2$
$\displaystyle \text{Volume of water}=166.25\times2.4=399\text{ m}^3$
$\displaystyle \text{Since }1\text{ m}^3=1000\text{ litres and }1\text{ gallon}=4.5\text{ litres,}$
$\displaystyle \text{Volume in gallons}=\frac{399\times1000}{4.5}$
$\displaystyle =88666.67\text{ gallons}$
$\displaystyle \therefore \text{Volume of water}=88666.67\text{ gallons}$
$\displaystyle \$

$\displaystyle \textbf{Question 12: }\text{The figure shows the cross-section of a water channel consisting}$
$\displaystyle \text{of a rectangle and a semicircle. Assuming that the channel is always full, find the volume}$
$\displaystyle \text{of water discharged in one minute, if the water flows at }20\text{ cm/s}.$
$\displaystyle \text{Give your answer in cubic metres, correct to one decimal place.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of the rectangular part}=21\text{ cm}$
$\displaystyle \text{Height of the rectangular part}=7\text{ cm}$
$\displaystyle \text{Radius of the semicircular part}=\frac{21}{2}=10.5\text{ cm}$
$\displaystyle \text{Area of the cross-section}$
$\displaystyle =21\times7+\frac{1}{2}\times\frac{22}{7}\times(10.5)^2$
$\displaystyle =147+173.25$
$\displaystyle =320.25\text{ cm}^2$
$\displaystyle \text{Distance travelled by water in one minute}=20\times60=1200\text{ cm}$
$\displaystyle \text{Volume of water discharged}=320.25\times1200$
$\displaystyle =384300\text{ cm}^3$
$\displaystyle =\frac{384300}{1000000}\text{ m}^3$
$\displaystyle =0.3843\text{ m}^3$
$\displaystyle \therefore \text{Volume of water discharged}=0.4\text{ m}^3\text{, correct to one decimal place.}$
$\\$

$\displaystyle \textbf{Question 13: }\text{An open cylindrical vessel of internal diameter }7\text{ cm and height }8\text{ cm}$
$\displaystyle \text{stands on a horizontal table. A solid metallic right circular cone of base diameter }3.5\text{ cm}$
$\displaystyle \text{and height }8\text{ cm is placed inside it. Find the volume of water required to fill the vessel.}$
$\displaystyle \text{If this cone is replaced by another cone of height }1.75\text{ cm and base radius }2\text{ cm},$
$\displaystyle \text{find the drop in the water level.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the cylindrical vessel }R=\frac{7}{2}=3.5\text{ cm}$
$\displaystyle \text{Height of the cylindrical vessel }H=8\text{ cm}$
$\displaystyle \text{Radius of the first cone }r_1=\frac{3.5}{2}=1.75\text{ cm}$
$\displaystyle \text{Height of the first cone }h_1=8\text{ cm}$
$\displaystyle \text{Volume of water required}=\text{Volume of cylinder}-\text{Volume of first cone}$
$\displaystyle =\pi R^2H-\frac{1}{3}\pi r_1^2h_1$
$\displaystyle =\frac{22}{7}\left((3.5)^2\times8-\frac{1}{3}(1.75)^2\times8\right)$
$\displaystyle =282.33\text{ cm}^3$
$\displaystyle \therefore \text{Volume of water required}=282.33\text{ cm}^3$
$\displaystyle \text{For the second cone, }r_2=2\text{ cm and }h_2=1.75\text{ cm}$
$\displaystyle \text{Let the drop in the water level be }x\text{ cm}$
$\displaystyle \text{Volume corresponding to the drop}=\text{Difference in volumes of the cones}$
$\displaystyle \pi R^2x=\frac{1}{3}\pi\left(r_1^2h_1-r_2^2h_2\right)$
$\displaystyle (3.5)^2x=\frac{1}{3}\left((1.75)^2\times8-2^2\times1.75\right)$
$\displaystyle x=\frac{(1.75)^2\times8-2^2\times1.75}{3(3.5)^2}$
$\displaystyle =0.4762\text{ cm}$
$\displaystyle \therefore \text{Drop in the water level}=0.4762\text{ cm}$
$\\$

$\displaystyle \textbf{Question 14: }\text{A cylindrical can with a horizontal base of radius }3.5\text{ cm contains}$
$\displaystyle \text{sufficient water so that, when a sphere is placed in it, the water just covers the sphere.}$
$\displaystyle \text{Given that the sphere just fits into the can, calculate:}$
$\displaystyle \text{(i) the total surface area of the can in contact with water when the sphere is in it,}$
$\displaystyle \text{(ii) the depth of water in the can before the sphere was put into it.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the cylinder and sphere }r=3.5\text{ cm}$
$\displaystyle \text{Depth of water after inserting the sphere}=2r=7\text{ cm}$
$\displaystyle \text{(i) Surface area of the can in contact with water}$
$\displaystyle =\text{CSA of cylinder}+\text{Area of its base}$
$\displaystyle =2\pi rh+\pi r^2$
$\displaystyle =2\times\frac{22}{7}\times3.5\times7+\frac{22}{7}\times(3.5)^2$
$\displaystyle =154+38.5$
$\displaystyle =192.5\text{ cm}^2$
$\displaystyle \therefore \text{Required surface area}=192.5\text{ cm}^2$
$\displaystyle \text{(ii) Let the initial depth of water be }x\text{ cm}$
$\displaystyle \text{Initial volume of water}=\text{Final volume of water}$
$\displaystyle \pi r^2x=\pi r^2(7)-\frac{4}{3}\pi r^3$
$\displaystyle x=7-\frac{4r}{3}$
$\displaystyle =7-\frac{4}{3}\times3.5$
$\displaystyle =7-\frac{14}{3}=\frac{7}{3}$
$\displaystyle =2\frac{1}{3}\text{ cm}$
$\displaystyle \therefore \text{Initial depth of water}=2\frac{1}{3}\text{ cm}$
$\\$

$\displaystyle \textbf{Question 15: }\text{A hollow cylinder has a solid hemisphere projecting inward from one end,}$
$\displaystyle \text{while the other end is closed by a flat circular plate. When the flat end is downward,}$
$\displaystyle \text{the water level is }10\text{ cm. Find the water level when the cylinder is inverted.}$
$\displaystyle \text{The common diameter is }7\text{ cm and the height of the cylinder is }20\text{ cm}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the cylinder and hemisphere }r=\frac{7}{2}=3.5\text{ cm}$
$\displaystyle \text{Initial depth of water}=10\text{ cm}$
$\displaystyle \text{Since the water does not reach the hemisphere initially,}$
$\displaystyle \text{Volume of water}=\pi r^2\times10$
$\displaystyle =\frac{22}{7}\times(3.5)^2\times10$
$\displaystyle =385\text{ cm}^3$
$\displaystyle \text{Let the water level after inversion be }h\text{ cm}$
$\displaystyle \text{Volume of water}=\text{Volume of cylinder up to height }h-\text{Volume of hemisphere}$
$\displaystyle \pi r^2h-\frac{2}{3}\pi r^3=385$
$\displaystyle \pi r^2h=385+\frac{2}{3}\pi r^3$
$\displaystyle h=\frac{385}{\pi r^2}+\frac{2r}{3}$
$\displaystyle =10+\frac{2}{3}\times3.5$
$\displaystyle =10+\frac{7}{3}=12\frac{1}{3}\text{ cm}$
$\displaystyle \therefore \text{Water level after inversion}=12\frac{1}{3}\text{ cm}$
$\\$