Class 10: Cones and Sphere: Exercise 22(f)

$\displaystyle \textbf{Question 1: }\text{What is the least number of solid metallic spheres, each of diameter }6\text{ cm,}$
$\displaystyle \text{that should be melted and recast to form a solid metal cone}$
$\displaystyle \text{of height }45\text{ cm and diameter }12\text{ cm?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of each sphere }(r_1)=3\text{ cm}$
$\displaystyle \text{Radius of cone }(r_2)=6\text{ cm}$
$\displaystyle \text{Height of cone }(h)=45\text{ cm}$
$\displaystyle \text{Let the number of spheres be }n.$
$\displaystyle n\times\frac{4}{3}\pi r_1^3=\frac{1}{3}\pi r_2^2h$
$\displaystyle n=\frac{\frac{1}{3}\pi\times6^2\times45}{\frac{4}{3}\pi\times3^3}$
$\displaystyle =\frac{36\times45}{4\times27}$
$\displaystyle =15$
$\displaystyle \therefore\text{ Least number of spheres }=15.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{A largest sphere is carved out of a right circular cylinder}$
$\displaystyle \text{of radius }7\text{ cm and height }14\text{ cm. Find the volume of the sphere,}$
$\displaystyle \text{correct to the nearest integer.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of cylinder}=7\text{ cm}$
$\displaystyle \text{Height of cylinder}=14\text{ cm}$
$\displaystyle \text{Diameter of largest sphere}=14\text{ cm}$
$\displaystyle \therefore\text{ Radius of sphere }(r)=7\text{ cm}$
$\displaystyle \text{Volume of sphere}=\frac{4}{3}\pi r^3$
$\displaystyle =\frac{4}{3}\times\frac{22}{7}\times7^3$
$\displaystyle =\frac{4312}{3}=1437.33\text{ cm}^3$
$\displaystyle \therefore\text{ Volume of sphere, to the nearest integer,}=1437\text{ cm}^3.$
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$\displaystyle \textbf{Question 3: }\text{A right circular cylinder of diameter }12\text{ cm and height }15\text{ cm}$
$\displaystyle \text{is full of ice-cream. The ice-cream is filled into identical cones}$
$\displaystyle \text{of height }12\text{ cm and diameter }6\text{ cm, each having a hemispherical top.}$
$\displaystyle \text{Find the number of ice-creams formed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of cylinder}=6\text{ cm},\quad\text{Height}=15\text{ cm}$
$\displaystyle \text{Radius of each cone and hemisphere}=3\text{ cm}$
$\displaystyle \text{Height of each cone}=12\text{ cm}$
$\displaystyle \text{Volume of one ice-cream}=\text{Volume of cone}+\text{Volume of hemisphere}$
$\displaystyle =\frac{1}{3}\pi r^2h+\frac{2}{3}\pi r^3$
$\displaystyle =\frac{1}{3}\pi\times3^2\times12+\frac{2}{3}\pi\times3^3$
$\displaystyle =36\pi+18\pi=54\pi\text{ cm}^3$
$\displaystyle \text{Volume of cylinder}=\pi r^2h$
$\displaystyle =\pi\times6^2\times15=540\pi\text{ cm}^3$
$\displaystyle \text{Number of ice-creams}=\frac{540\pi}{54\pi}=10$
$\displaystyle \therefore\text{ Number of ice-creams formed}=10.$
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$\displaystyle \textbf{Question 4: }\text{A solid is in the form of a cone standing on a hemisphere,}$
$\displaystyle \text{both having radius }8\text{ cm. The height of the cone is equal to its radius.}$
$\displaystyle \text{Find, in terms of }\pi,\text{ the volume of the solid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius }(r)=8\text{ cm}$
$\displaystyle \text{Height of cone }(h)=8\text{ cm}$
$\displaystyle \text{Volume of solid}=\text{Volume of hemisphere}+\text{Volume of cone}$
$\displaystyle =\frac{2}{3}\pi r^3+\frac{1}{3}\pi r^2h$
$\displaystyle =\frac{2}{3}\pi(8)^3+\frac{1}{3}\pi(8)^2(8)$
$\displaystyle =\frac{1024}{3}\pi+\frac{512}{3}\pi$
$\displaystyle =512\pi\text{ cm}^3$
$\displaystyle \therefore\text{ Volume of the solid}=512\pi\text{ cm}^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{The diameter of a sphere is }6\text{ cm. It is melted and drawn into a wire}$
$\displaystyle \text{of diameter }0.2\text{ cm. Find the length of the wire.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of sphere }(r_1)=3\text{ cm}$
$\displaystyle \text{Radius of wire }(r_2)=0.1\text{ cm}$
$\displaystyle \text{Let the length of the wire be }l\text{ cm}.$
$\displaystyle \text{Volume of sphere}=\text{Volume of wire}$
$\displaystyle \frac{4}{3}\pi(3)^3=\pi(0.1)^2l$
$\displaystyle l=\frac{4\times3^2}{(0.1)^2}=3600\text{ cm}$
$\displaystyle =36\text{ m}$
$\displaystyle \therefore\text{ Length of the wire}=3600\text{ cm}=36\text{ m}.$
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$\displaystyle \textbf{Question 6: }\text{Determine the ratio of the volume of a cube to the volume of a sphere}$
$\displaystyle \text{which exactly fits inside the cube.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the sphere be }r.$
$\displaystyle \therefore\text{ Side of the cube}=\text{Diameter of the sphere}=2r$
$\displaystyle \text{Volume of cube}=(2r)^3=8r^3$
$\displaystyle \text{Volume of sphere}=\frac{4}{3}\pi r^3$
$\displaystyle \text{Required ratio}=8r^3:\frac{4}{3}\pi r^3$
$\displaystyle =8:\frac{4\pi}{3}$
$\displaystyle =6:\pi$
$\displaystyle \text{Taking }\pi=\frac{22}{7},\text{ the ratio}=6:\frac{22}{7}=21:11$
$\displaystyle \therefore\text{ Required ratio}=6:\pi\text{ or approximately }21:11.$
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$\displaystyle \textbf{Question 7: }\text{An iron pole has a cylindrical portion }110\text{ cm high and }12\text{ cm in diameter.}$
$\displaystyle \text{It is surmounted by a cone }9\text{ cm high. Find the mass of the pole,}$
$\displaystyle \text{given that }1\text{ cm}^3\text{ of iron has a mass of }8\text{ g. Take }\pi=\frac{355}{113}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of cylinder and cone }(r)=\frac{12}{2}=6\text{ cm}$
$\displaystyle \text{Height of cylinder }(h_1)=110\text{ cm}$
$\displaystyle \text{Height of cone }(h_2)=9\text{ cm}$
$\displaystyle \text{Volume of pole}=\pi r^2h_1+\frac{1}{3}\pi r^2h_2$
$\displaystyle =\pi(6)^2(110)+\frac{1}{3}\pi(6)^2(9)$
$\displaystyle =3960\pi+108\pi=4068\pi\text{ cm}^3$
$\displaystyle \text{Mass of }1\text{ cm}^3\text{ of iron}=8\text{ g}$
$\displaystyle \text{Mass of pole}=4068\times\frac{355}{113}\times8$
$\displaystyle =102240\text{ g}=102.24\text{ kg}$
$\displaystyle \therefore\text{ Mass of the pole}=102.24\text{ kg approximately.}$
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$\displaystyle \textbf{Question 8: }\text{A metal cube is completely submerged in water contained in a cylindrical vessel}$
$\displaystyle \text{of diameter }30\text{ cm. The water level rises by }1\frac{41}{99}\text{ cm. Find:}$
$\displaystyle \text{(i) the length of an edge of the cube, (ii) the total surface area of the cube.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the edge of the cube be }a\text{ cm}.$
$\displaystyle \text{Radius of cylindrical vessel }(r)=\frac{30}{2}=15\text{ cm}$
$\displaystyle \text{Rise in water level}=1\frac{41}{99}=\frac{140}{99}\text{ cm}$
$\displaystyle \text{Volume of cube}=\text{Volume of water displaced}$
$\displaystyle a^3=\pi r^2h$
$\displaystyle a^3=\frac{22}{7}\times15^2\times\frac{140}{99}$
$\displaystyle a^3=1000=10^3$
$\displaystyle \therefore a=10\text{ cm}$
$\displaystyle \text{(i) Length of an edge of the cube}=10\text{ cm}$
$\displaystyle \text{Total surface area of cube}=6a^2$
$\displaystyle =6\times10^2=600\text{ cm}^2$
$\displaystyle \text{(ii) Total surface area of the cube}=600\text{ cm}^2.$
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$\displaystyle \textbf{Question 9: }\text{A rectangular platform with a semicircular end is }22\text{ m long}$
$\displaystyle \text{from one end to the other. The length of the semicircular arc is }11\text{ m.}$
$\displaystyle \text{Find the cost of constructing the platform, }1.5\text{ m high, at the rate of }\text{Rs. }4\text{ per m}^3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Please refer to the diagram.}$
$\displaystyle \text{Length of semicircular arc}=\pi r$
$\displaystyle \pi r=11$
$\displaystyle r=11\times\frac{7}{22}=3.5\text{ m}$
$\displaystyle \text{Width of rectangular part}=2r=7\text{ m}$
$\displaystyle \text{Length of rectangular part}=22-3.5=18.5\text{ m}$
$\displaystyle \text{Volume of rectangular part}=18.5\times7\times1.5$
$\displaystyle =194.25\text{ m}^3$
$\displaystyle \text{Volume of semicircular part}=\frac{1}{2}\pi r^2h$
$\displaystyle =\frac{1}{2}\times\frac{22}{7}\times(3.5)^2\times1.5$
$\displaystyle =28.875\text{ m}^3$
$\displaystyle \text{Total volume}=194.25+28.875=223.125\text{ m}^3$
$\displaystyle \text{Cost of construction}=223.125\times4$
$\displaystyle =\text{Rs. }892.50$
$\displaystyle \therefore\text{ Cost of constructing the platform}=\text{Rs. }892.50.$
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$\displaystyle \textbf{Question 10: }\text{The cross-section of a tunnel is a square of side }7\text{ m,}$
$\displaystyle \text{surmounted by a semicircle, as shown in the adjoining figure.}$
$\displaystyle \text{The tunnel is }80\text{ m long. Calculate: (i) its volume,}$
$\displaystyle \text{(ii) its surface area, excluding the floor, and (iii) its floor area.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Please refer to the diagram.}$
$\displaystyle \text{Side of square}=7\text{ m}$
$\displaystyle \text{Radius of semicircle}=\frac{7}{2}=3.5\text{ m}$
$\displaystyle \text{Length of tunnel}=80\text{ m}$
$\displaystyle \text{(i) Cross-sectional area}=7^2+\frac{1}{2}\pi(3.5)^2$
$\displaystyle =49+\frac{1}{2}\times\frac{22}{7}\times(3.5)^2$
$\displaystyle =49+19.25=68.25\text{ m}^2$
$\displaystyle \text{Volume of tunnel}=68.25\times80$
$\displaystyle =5460\text{ m}^3$
$\displaystyle \text{(ii) Surface area, excluding floor}$
$\displaystyle =(\text{semicircular arc}+2\text{ sides})\times\text{length}$
$\displaystyle =\left(\pi\times3.5+2\times7\right)\times80$
$\displaystyle =(11+14)\times80=2000\text{ m}^2$
$\displaystyle \text{(iii) Floor area}=7\times80=560\text{ m}^2$
$\displaystyle \therefore\text{ Volume}=5460\text{ m}^3,$
$\displaystyle \text{surface area}=2000\text{ m}^2\text{ and floor area}=560\text{ m}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{A cylindrical water tank of diameter }2.8\text{ m and height }4.2\text{ m}$
$\displaystyle \text{is filled through a pipe of diameter }7\text{ cm. Water flows through the pipe}$
$\displaystyle \text{at }4\text{ m/s. Calculate, in minutes, the time taken to fill the tank.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of tank }(R)=\frac{2.8}{2}=1.4\text{ m}$
$\displaystyle \text{Height of tank }(H)=4.2\text{ m}$
$\displaystyle \text{Radius of pipe }(r)=\frac{7}{2}=3.5\text{ cm}=0.035\text{ m}$
$\displaystyle \text{Speed of water}=4\text{ m/s}$
$\displaystyle \text{Volume of tank}=\pi R^2H$
$\displaystyle =\frac{22}{7}\times(1.4)^2\times4.2$
$\displaystyle =25.872\text{ m}^3$
$\displaystyle \text{Volume of water flowing per second}=\pi r^2\times4$
$\displaystyle =\frac{22}{7}\times(0.035)^2\times4$
$\displaystyle =0.0154\text{ m}^3$
$\displaystyle \text{Time taken}=\frac{25.872}{0.0154}$
$\displaystyle =1680\text{ seconds}$
$\displaystyle =\frac{1680}{60}=28\text{ minutes}$
$\displaystyle \therefore\text{ Time taken to fill the tank}=28\text{ minutes}.$
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$\displaystyle \textbf{Question 12: }\text{Water flows at }9\text{ km/h through a cylindrical pipe}$
$\displaystyle \text{of cross-sectional area }25\text{ cm}^2.\text{ The water is collected in a rectangular cistern}$
$\displaystyle \text{of dimensions }7.5\text{ m}\times5\text{ m}\times4\text{ m. Find the rise in water level}$
$\displaystyle \text{in }1\text{ hour }15\text{ minutes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Speed of water}=9\text{ km/h}$
$\displaystyle =\frac{9000}{3600}=2.5\text{ m/s}$
$\displaystyle \text{Cross-sectional area of pipe}=25\text{ cm}^2$
$\displaystyle =\frac{25}{10000}=0.0025\text{ m}^2$
$\displaystyle \text{Time}=1\text{ hour }15\text{ minutes}=4500\text{ s}$
$\displaystyle \text{Volume of water collected}=\text{area}\times\text{speed}\times\text{time}$
$\displaystyle =0.0025\times2.5\times4500$
$\displaystyle =28.125\text{ m}^3$
$\displaystyle \text{Let the rise in water level be }h\text{ m}.$
$\displaystyle 7.5\times5\times h=28.125$
$\displaystyle h=\frac{28.125}{37.5}=0.75\text{ m}$
$\displaystyle =75\text{ cm}$
$\displaystyle \therefore\text{ Rise in water level}=75\text{ cm}.$
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$\displaystyle \textbf{Question 13: }\text{The figure shows a cone, a cylinder and a hemisphere,}$
$\displaystyle \text{all having the same diameter }10\text{ cm. The other dimensions are as shown.}$
$\displaystyle \text{Calculate: (a) the total surface area, (b) the total volume of the solid}$
$\displaystyle \text{and (c) its density, if its total mass is }1.7\text{ kg.}$
$\displaystyle \text{Answer:}$ 2019-05-19_11-25-07
$\displaystyle \text{Radius of each solid }(r)=\frac{10}{2}=5\text{ cm}$
$\displaystyle \text{Height of cone}=12\text{ cm}$
$\displaystyle \text{Length of cylinder}=12\text{ cm}$
$\displaystyle \text{Slant height of cone }(l)=\sqrt{12^2+5^2}=13\text{ cm}$
$\displaystyle \text{(a) Total surface area}$
$\displaystyle =\text{CSA of cone}+\text{CSA of cylinder}+\text{CSA of hemisphere}$
$\displaystyle =\pi rl+2\pi rh+2\pi r^2$
$\displaystyle =\pi(5)(13)+2\pi(5)(12)+2\pi(5)^2$
$\displaystyle =65\pi+120\pi+50\pi$
$\displaystyle =235\pi\text{ cm}^2$
$\displaystyle =235\times\frac{22}{7}=738.57\text{ cm}^2$
$\displaystyle \therefore\text{ Total surface area}=235\pi\text{ cm}^2\approx738.57\text{ cm}^2.$
$\displaystyle \text{(b) Total volume}$
$\displaystyle =\text{Volume of cone}+\text{Volume of cylinder}+\text{Volume of hemisphere}$
$\displaystyle =\frac{1}{3}\pi r^2h+\pi r^2h+\frac{2}{3}\pi r^3$
$\displaystyle =\frac{1}{3}\pi(5)^2(12)+\pi(5)^2(12)+\frac{2}{3}\pi(5)^3$
$\displaystyle =100\pi+300\pi+\frac{250}{3}\pi$
$\displaystyle =\frac{1450}{3}\pi\text{ cm}^3$
$\displaystyle =\frac{1450}{3}\times\frac{22}{7}=1519.05\text{ cm}^3$
$\displaystyle \therefore\text{ Total volume}=\frac{1450}{3}\pi\text{ cm}^3\approx1519.05\text{ cm}^3.$
$\displaystyle \text{(c) Mass of the solid}=1.7\text{ kg}=1700\text{ g}$
$\displaystyle \text{Density}=\frac{\text{Mass}}{\text{Volume}}$
$\displaystyle =\frac{1700}{1519.05}$
$\displaystyle =1.119\text{ g/cm}^3$
$\displaystyle \therefore\text{ Density of the material}\approx1.12\text{ g/cm}^3.$
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$\displaystyle \textbf{Question 14: }\text{A solid consisting of a right circular cone standing on a hemisphere}$
$\displaystyle \text{is placed upright in a cylinder full of water and touches its bottom.}$
$\displaystyle \text{The cylinder has radius }3\text{ cm and height }6\text{ cm. The hemisphere has radius }2\text{ cm,}$
$\displaystyle \text{and the cone has height }4\text{ cm. Find the volume of water left,}$
$\displaystyle \text{correct to the nearest cubic centimetre.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of cylinder }(R)=3\text{ cm}$
$\displaystyle \text{Height of cylinder }(H)=6\text{ cm}$
$\displaystyle \text{Radius of hemisphere and cone }(r)=2\text{ cm}$
$\displaystyle \text{Height of cone }(h)=4\text{ cm}$
$\displaystyle \text{Volume of cylinder}=\pi R^2H$
$\displaystyle =\pi(3)^2(6)=54\pi\text{ cm}^3$
$\displaystyle \text{Volume of solid}=\text{Volume of hemisphere}+\text{Volume of cone}$
$\displaystyle =\frac{2}{3}\pi r^3+\frac{1}{3}\pi r^2h$
$\displaystyle =\frac{2}{3}\pi(2)^3+\frac{1}{3}\pi(2)^2(4)$
$\displaystyle =\frac{16}{3}\pi+\frac{16}{3}\pi$
$\displaystyle =\frac{32}{3}\pi\text{ cm}^3$
$\displaystyle \text{Volume of water left}=54\pi-\frac{32}{3}\pi$
$\displaystyle =\frac{130}{3}\pi\text{ cm}^3$
$\displaystyle \approx136.14\text{ cm}^3$
$\displaystyle \therefore\text{ Volume of water left}=136\text{ cm}^3\text{, approximately.}$
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$\displaystyle \textbf{Question 15: }\text{A metal container is in the form of a cylinder surmounted by a hemisphere}$
$\displaystyle \text{of the same radius. The internal height of the cylinder is }7\text{ m,}$
$\displaystyle \text{and its internal radius is }3.5\text{ m. Calculate:}$
$\displaystyle \text{(i) the internal surface area, excluding the base, and (ii) the internal volume.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius }(r)=3.5\text{ m},\quad\text{Height of cylinder }(h)=7\text{ m}$
$\displaystyle \text{(i) Internal surface area}$
$\displaystyle =\text{CSA of cylinder}+\text{CSA of hemisphere}$
$\displaystyle =2\pi rh+2\pi r^2$
$\displaystyle =2\pi(3.5)(7)+2\pi(3.5)^2$
$\displaystyle =49\pi+24.5\pi$
$\displaystyle =73.5\pi$
$\displaystyle =73.5\times\frac{22}{7}=231\text{ m}^2$
$\displaystyle \therefore\text{ Internal surface area}=231\text{ m}^2.$
$\displaystyle \text{(ii) Internal volume}$
$\displaystyle =\text{Volume of cylinder}+\text{Volume of hemisphere}$
$\displaystyle =\pi r^2h+\frac{2}{3}\pi r^3$
$\displaystyle =\pi(3.5)^2(7)+\frac{2}{3}\pi(3.5)^3$
$\displaystyle =85.75\pi+\frac{85.75}{3}\pi$
$\displaystyle =\frac{343}{3}\pi$
$\displaystyle =\frac{343}{3}\times\frac{22}{7}$
$\displaystyle =359.33\text{ m}^3$
$\displaystyle \therefore\text{ Internal volume}=359.33\text{ m}^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{An exhibition tent is in the form of a cylinder surmounted by a cone.}$
$\displaystyle \text{The total height of the tent is }85\text{ m, and the cylindrical part is }50\text{ m high.}$
$\displaystyle \text{If the diameter of its base is }168\text{ m, find the canvas required to make the tent.}$
$\displaystyle \text{Allow }20\%\text{ extra for folds and stitching. Give the answer to the nearest m}^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the tent }(r)=\frac{168}{2}=84\text{ m}$
$\displaystyle \text{Height of cylindrical part }(h_1)=50\text{ m}$
$\displaystyle \text{Height of conical part }(h_2)=85-50=35\text{ m}$
$\displaystyle \text{Slant height of cone }(l)=\sqrt{r^2+h_2^2}$
$\displaystyle =\sqrt{84^2+35^2}$
$\displaystyle =\sqrt{8281}=91\text{ m}$
$\displaystyle \text{Canvas area}=\text{CSA of cylinder}+\text{CSA of cone}$
$\displaystyle =2\pi rh_1+\pi rl$
$\displaystyle =2\pi(84)(50)+\pi(84)(91)$
$\displaystyle =8400\pi+7644\pi$
$\displaystyle =16044\pi$
$\displaystyle =16044\times\frac{22}{7}=50424\text{ m}^2$
$\displaystyle \text{Canvas required, including }20\%\text{ extra}$
$\displaystyle =50424\times\frac{120}{100}$
$\displaystyle =60508.8\text{ m}^2$
$\displaystyle \therefore\text{ Canvas required}=60509\text{ m}^2\text{, to the nearest m}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{The total surface area of a hollow cylinder, open at both ends,}$
$\displaystyle \text{is }3575\text{ cm}^2.\text{ The area of each base ring is }357.5\text{ cm}^2,$
$\displaystyle \text{and its height is }14\text{ cm. Find the thickness of the cylinder.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the external radius be }R\text{ cm and internal radius be }r\text{ cm}.$
$\displaystyle \text{Height }(h)=14\text{ cm}$
$\displaystyle \text{Area of one base ring}=\pi(R^2-r^2)$
$\displaystyle \pi(R-r)(R+r)=357.5$
$\displaystyle (R-r)(R+r)=357.5\times\frac{7}{22}$
$\displaystyle (R-r)(R+r)=113.75\qquad\ldots\text{(i)}$
$\displaystyle \text{Total surface area}=2\pi Rh+2\pi rh+2(357.5)$
$\displaystyle 2\pi(14)(R+r)+715=3575$
$\displaystyle 28\pi(R+r)=2860$
$\displaystyle 28\times\frac{22}{7}(R+r)=2860$
$\displaystyle 88(R+r)=2860$
$\displaystyle R+r=32.5\qquad\ldots\text{(ii)}$
$\displaystyle \text{From (i) and (ii),}$
$\displaystyle (R-r)(32.5)=113.75$
$\displaystyle R-r=\frac{113.75}{32.5}=3.5$
$\displaystyle \therefore\text{ Thickness of the cylinder}=3.5\text{ cm}.$
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$\displaystyle \textbf{Question 18: }\text{A test tube consists of a hemisphere and a cylinder}$
$\displaystyle \text{of the same radius. The volume of water required to fill the tube is }\frac{5159}{6}\text{ cm}^3.$
$\displaystyle \text{The volume required to fill it to a level }4\text{ cm below the top is }\frac{4235}{6}\text{ cm}^3.$
$\displaystyle \text{Find the radius of the tube and the length of its cylindrical part.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius be }r\text{ cm and the cylindrical length be }h\text{ cm}.$
$\displaystyle \text{Volume of the top cylindrical portion of height }4\text{ cm}$
$\displaystyle =\frac{5159}{6}-\frac{4235}{6}$
$\displaystyle =\frac{924}{6}=154\text{ cm}^3$
$\displaystyle \therefore \pi r^2(4)=154$
$\displaystyle 4\times\frac{22}{7}\times r^2=154$
$\displaystyle r^2=154\times\frac{7}{88}=12.25$
$\displaystyle r=\sqrt{12.25}=3.5\text{ cm}$
$\displaystyle \text{Volume of the complete test tube}$
$\displaystyle =\frac{2}{3}\pi r^3+\pi r^2h$
$\displaystyle \frac{2}{3}\times\frac{22}{7}\times(3.5)^3+\frac{22}{7}\times(3.5)^2h=\frac{5159}{6}$
$\displaystyle \frac{539}{6}+\frac{77}{2}h=\frac{5159}{6}$
$\displaystyle \frac{77}{2}h=\frac{4620}{6}=770$
$\displaystyle h=770\times\frac{2}{77}=20\text{ cm}$
$\displaystyle \therefore\text{ Radius of the tube}=3.5\text{ cm}$
$\displaystyle \text{and length of its cylindrical part}=20\text{ cm}.$
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$\displaystyle \textbf{Question 19: }\text{A solid is in the form of a right circular cone mounted on a hemisphere.}$
$\displaystyle \text{The diameter of the common base is }7\text{ cm and the height of the cone is }8\text{ cm.}$
$\displaystyle \text{The solid is placed in a cylindrical vessel of internal radius }7\text{ cm}$
$\displaystyle \text{and height }10\text{ cm. Find the volume of water required to fill the vessel completely.}$
$\displaystyle \text{Answer:}$ 2019-05-19_11-12-47
$\displaystyle \text{Radius of cylinder}=7\text{ cm},\quad\text{Height}=10\text{ cm}$
$\displaystyle \text{Radius of hemisphere and cone}=\frac{7}{2}=3.5\text{ cm}$
$\displaystyle \text{Height of cone}=8\text{ cm}$
$\displaystyle \text{Volume of cylinder}=\pi r^2h$
$\displaystyle =\pi(7)^2(10)=490\pi\text{ cm}^3$
$\displaystyle \text{Volume of solid}=\text{Volume of hemisphere}+\text{Volume of cone}$
$\displaystyle =\frac{2}{3}\pi(3.5)^3+\frac{1}{3}\pi(3.5)^2(8)$
$\displaystyle =\frac{343}{12}\pi+\frac{98}{3}\pi$
$\displaystyle =\frac{735}{12}\pi=61.25\pi\text{ cm}^3$
$\displaystyle \text{Volume of water required}=490\pi-61.25\pi$
$\displaystyle =428.75\pi\text{ cm}^3$
$\displaystyle =428.75\times\frac{22}{7}=1347.5\text{ cm}^3$
$\displaystyle \therefore\text{ Volume of water required}=1347.5\text{ cm}^3.$
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$\displaystyle \textbf{Question 20: }\text{A cone and a cylinder have their heights in the ratio }4:5,$
$\displaystyle \text{and their diameters are in the ratio }3:2.\text{ Find the ratio of their volumes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the heights of the cone and cylinder be }4x\text{ and }5x\text{ respectively.}$
$\displaystyle \text{Let their diameters be }3y\text{ and }2y\text{ respectively.}$
$\displaystyle \therefore\text{ Their radii are }\frac{3y}{2}\text{ and }y\text{ respectively.}$
$\displaystyle \text{Required ratio}=\frac{\frac{1}{3}\pi\left(\frac{3y}{2}\right)^2(4x)}{\pi(y)^2(5x)}$
$\displaystyle =\frac{\frac{1}{3}\times\frac{9y^2}{4}\times4x}{5xy^2}$
$\displaystyle =\frac{3}{5}$
$\displaystyle \therefore\text{ Ratio of the volumes of the cone and cylinder}=3:5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{A sphere just fits inside a cylindrical vessel, and the height}$
$\displaystyle \text{of the cylinder is equal to the height of the sphere. Show that their curved}$
$\displaystyle \text{surface areas are equal.}$ 2019-05-19_11-00-33
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the sphere and cylinder be }r.$
$\displaystyle \therefore\text{ Height of cylinder}=\text{Diameter of sphere}=2r$
$\displaystyle \text{Curved surface area of cylinder}=2\pi rh$
$\displaystyle =2\pi r(2r)=4\pi r^2$
$\displaystyle \text{Curved surface area of sphere}=4\pi r^2$
$\displaystyle \therefore\text{ Curved surface area of cylinder}$
$\displaystyle =\text{Curved surface area of sphere.}$
$\displaystyle \\$