\displaystyle \textbf{Question 1: }\text{How many balls, each of radius }1\text{ cm, can be made from a solid sphere}
\displaystyle \text{of lead of radius }8\text{ cm?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of balls formed }=n.
\displaystyle \text{Radius of the solid lead sphere }=8\text{ cm.}
\displaystyle \text{Radius of each small ball }=1\text{ cm.}
\displaystyle \text{Since the lead sphere is recast into small balls,}
\displaystyle \text{Volume of the solid sphere}=\text{Volume of }n\text{ small balls.}
\displaystyle \frac{4}{3}\pi(8)^3=n\times\frac{4}{3}\pi(1)^3
\displaystyle 8^3=n
\displaystyle n=512
\displaystyle \therefore \text{The number of balls that can be made is }512.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{How many spherical bullets, each of }5\text{ cm diameter, can be cast from a}
\displaystyle \text{rectangular block of metal }11\text{ dm}\times1\text{ m}\times5\text{ dm?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of spherical bullets formed be }n.
\displaystyle 11\text{ dm}=110\text{ cm},\qquad1\text{ m}=100\text{ cm},\qquad5\text{ dm}=50\text{ cm.}
\displaystyle \text{Volume of the rectangular block}=110\times100\times50
\displaystyle =550000\text{ cm}^3.
\displaystyle \text{Diameter of each spherical bullet}=5\text{ cm.}
\displaystyle \therefore \text{Radius of each spherical bullet}=\frac{5}{2}\text{ cm.}
\displaystyle \text{Volume of each spherical bullet}=\frac{4}{3}\pi\left(\frac{5}{2}\right)^3
\displaystyle =\frac{4}{3}\times\frac{22}{7}\times\frac{125}{8}\text{ cm}^3.
\displaystyle \text{Since the block is recast into spherical bullets,}
\displaystyle n\times\frac{4}{3}\times\frac{22}{7}\times\frac{125}{8}=550000
\displaystyle n=550000\times\frac{3}{4}\times\frac{7}{22}\times\frac{8}{125}
\displaystyle n=8400
\displaystyle \therefore \text{The number of spherical bullets that can be cast is }8400.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{A spherical ball of radius }3\text{ cm is melted and recast into three spherical}
\displaystyle \text{balls. The radii of two of the balls are }1.5\text{ cm and }2\text{ cm respectively. Determine}
\displaystyle \text{the diameter of the third ball.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the third ball be }r\text{ cm.}
\displaystyle \text{Since the original ball is recast into three spherical balls,}
\displaystyle \text{Volume of original ball}=\text{Sum of volumes of the three balls.}
\displaystyle \frac{4}{3}\pi(3)^3=\frac{4}{3}\pi(1.5)^3+\frac{4}{3}\pi(2)^3+\frac{4}{3}\pi r^3
\displaystyle 27=(1.5)^3+2^3+r^3
\displaystyle 27=3.375+8+r^3
\displaystyle r^3=27-11.375=15.625
\displaystyle r^3=(2.5)^3
\displaystyle \therefore r=2.5\text{ cm.}
\displaystyle \text{Diameter of the third ball}=2r=2\times2.5=5\text{ cm.}
\displaystyle \therefore \text{The diameter of the third ball is }5\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the number of metallic circular discs with }1.5\text{ cm base diameter and}
\displaystyle \text{height }0.2\text{ cm to be melted to form a right circular cylinder of height }10\text{ cm}
\displaystyle \text{and diameter }4.5\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of metallic circular discs be }n.
\displaystyle \text{Radius of each disc}=\frac{1.5}{2}=0.75\text{ cm},\qquad \text{height}=0.2\text{ cm.}
\displaystyle \text{Volume of each disc}=\pi(0.75)^2\times0.2\text{ cm}^3.
\displaystyle \text{Radius of the resulting cylinder}=\frac{4.5}{2}=2.25\text{ cm.}
\displaystyle \text{Volume of the resulting cylinder}=\pi(2.25)^2\times10\text{ cm}^3.
\displaystyle \text{Since the discs are melted to form the cylinder,}
\displaystyle n\times\pi(0.75)^2\times0.2=\pi(2.25)^2\times10
\displaystyle n=\frac{(2.25)^2\times10}{(0.75)^2\times0.2}
\displaystyle n=\left(\frac{2.25}{0.75}\right)^2\times\frac{10}{0.2}
\displaystyle n=3^2\times50=450
\displaystyle \therefore \text{The number of metallic circular discs required is }450.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{How many spherical lead shots of diameter }4\text{ cm can be made out of}
\displaystyle \text{a solid cube of lead whose edge measures }44\text{ cm?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of spherical lead shots be }n.
\displaystyle \text{Edge of the solid cube}=44\text{ cm.}
\displaystyle \text{Volume of the solid cube}=44^3\text{ cm}^3.
\displaystyle \text{Diameter of each spherical shot}=4\text{ cm.}
\displaystyle \therefore \text{Radius of each spherical shot}=2\text{ cm.}
\displaystyle \text{Volume of each spherical shot}=\frac{4}{3}\pi(2)^3
\displaystyle =\frac{32}{3}\pi\text{ cm}^3.
\displaystyle \text{Since the cube is melted to form spherical shots,}
\displaystyle n\times\frac{32}{3}\pi=44^3
\displaystyle n=\frac{44^3\times3}{32\pi}
\displaystyle =\frac{44^3\times3\times7}{32\times22}
\displaystyle =2541
\displaystyle \therefore \text{The number of spherical lead shots that can be made is }2541.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Three cubes of a metal whose edges are in the ratio }3:4:5\text{ are melted and}
\displaystyle \text{converted into a single cube whose diagonal is }12\sqrt{3}\text{ cm. Find the edges}
\displaystyle \text{of the three cubes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the edges of the three cubes be }3x,\ 4x\text{ and }5x\text{ cm.}
\displaystyle \text{Let the edge of the resulting cube be }a\text{ cm.}
\displaystyle \text{Diagonal of a cube}=a\sqrt{3}.
\displaystyle a\sqrt{3}=12\sqrt{3}
\displaystyle \therefore a=12\text{ cm.}
\displaystyle \text{Since the three cubes are melted to form a single cube,}
\displaystyle (3x)^3+(4x)^3+(5x)^3=12^3
\displaystyle 27x^3+64x^3+125x^3=1728
\displaystyle 216x^3=1728
\displaystyle x^3=8
\displaystyle x=2
\displaystyle \therefore 3x=6\text{ cm},\qquad4x=8\text{ cm},\qquad5x=10\text{ cm.}
\displaystyle \therefore \text{The edges of the three cubes are }6\text{ cm},\ 8\text{ cm and }10\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A solid metallic sphere of radius }10.5\text{ cm is melted and recast into a number}
\displaystyle \text{of smaller cones, each of radius }3.5\text{ cm and height }3\text{ cm. Find the number}
\displaystyle \text{of cones so formed.}\hfill\text{[CBSE 2014, 17]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of cones formed be }n.
\displaystyle \text{Radius of the metallic sphere}=10.5\text{ cm.}
\displaystyle \text{Volume of the sphere}=\frac{4}{3}\pi(10.5)^3\text{ cm}^3.
\displaystyle \text{Radius of each cone}=3.5\text{ cm},\qquad \text{height}=3\text{ cm.}
\displaystyle \text{Volume of each cone}=\frac{1}{3}\pi(3.5)^2\times3
\displaystyle =\pi(3.5)^2\text{ cm}^3.
\displaystyle \text{Since the sphere is melted and recast into cones,}
\displaystyle n\times\pi(3.5)^2=\frac{4}{3}\pi(10.5)^3
\displaystyle n=\frac{4(10.5)^3}{3(3.5)^2}
\displaystyle =\frac{4}{3}\times\left(\frac{10.5}{3.5}\right)^2\times10.5
\displaystyle =\frac{4}{3}\times9\times10.5=126
\displaystyle \therefore \text{The number of cones formed is }126.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A copper rod of diameter }1\text{ cm and length }8\text{ cm is drawn into a wire of}
\displaystyle \text{length }18\text{ m of uniform thickness. Find the thickness of the wire.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the copper rod}=\frac{1}{2}\text{ cm},\qquad \text{length}=8\text{ cm.}
\displaystyle \text{Length of the wire}=18\text{ m}=1800\text{ cm.}
\displaystyle \text{Let the radius of the wire be }r\text{ cm.}
\displaystyle \text{Since the rod is drawn into a wire, its volume remains unchanged.}
\displaystyle \pi\left(\frac{1}{2}\right)^2\times8=\pi r^2\times1800
\displaystyle 2=1800r^2
\displaystyle r^2=\frac{1}{900}
\displaystyle r=\frac{1}{30}\text{ cm.}
\displaystyle \text{Thickness of the wire}=\text{Diameter of the wire}=2r
\displaystyle =2\times\frac{1}{30}=\frac{1}{15}\text{ cm}.
\displaystyle \frac{1}{15}\text{ cm}=\frac{10}{15}\text{ mm}=\frac{2}{3}\text{ mm}.
\displaystyle \therefore \text{The thickness of the wire is }\frac{2}{3}\text{ mm (approximately }0.67\text{ mm).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{How many coins }1.75\text{ cm in diameter and }2\text{ mm thick must be melted}
\displaystyle \text{to form a cuboid }11\text{ cm}\times10\text{ cm}\times7\text{ cm?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of coins required be }n.
\displaystyle \text{Diameter of each coin}=1.75\text{ cm}.
\displaystyle \therefore \text{Radius of each coin}=\frac{1.75}{2}=0.875=\frac{7}{8}\text{ cm.}
\displaystyle \text{Thickness of each coin}=2\text{ mm}=0.2=\frac{1}{5}\text{ cm.}
\displaystyle \text{Volume of each coin}=\pi\left(\frac{7}{8}\right)^2\times\frac{1}{5}
\displaystyle =\frac{49\pi}{320}\text{ cm}^3.
\displaystyle \text{Volume of the cuboid}=11\times10\times7=770\text{ cm}^3.
\displaystyle \text{Since the coins are melted to form the cuboid,}
\displaystyle n\times\frac{49\pi}{320}=770
\displaystyle n=\frac{770\times320}{49\pi}
\displaystyle =\frac{770\times320}{49\times\frac{22}{7}}
\displaystyle =\frac{770\times320\times7}{49\times22}
\displaystyle =1600
\displaystyle \therefore \text{The number of coins required is }1600.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The surface area of a solid metallic sphere is }616\text{ cm}^2.\text{ It is melted and}
\displaystyle \text{recast into a cone of height }28\text{ cm. Find the diameter of the base of the cone}
\displaystyle \text{so formed. (Use }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the metallic sphere be }r\text{ cm.}
\displaystyle \text{Surface area of the sphere}=4\pi r^2.
\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 \text{Let the radius of the base of the cone be }R\text{ cm.}
\displaystyle \text{Since the sphere is melted and recast into a cone,}
\displaystyle \text{Volume of the sphere}=\text{Volume of the cone.}
\displaystyle \frac{4}{3}\pi(7)^3=\frac{1}{3}\pi R^2\times28
\displaystyle 4\times343=28R^2
\displaystyle R^2=\frac{1372}{28}=49
\displaystyle R=7\text{ cm.}
\displaystyle \therefore \text{Diameter of the base of the cone}=2R=2\times7=14\text{ cm.}
\displaystyle \therefore \text{The diameter of the base of the cone is }14\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A cylindrical bucket, }32\text{ cm high and with radius of base }18\text{ cm, is filled}
\displaystyle \text{with sand. This bucket is emptied out on the ground and a conical heap of sand is}
\displaystyle \text{formed. If the height of the conical heap is }24\text{ cm, find the radius and slant height}
\displaystyle \text{of the heap.}\hfill\text{[CBSE 2012, 14]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cylindrical bucket}=18\text{ cm},\qquad \text{height}=32\text{ cm.}
\displaystyle \text{Let the radius of the conical heap be }r\text{ cm.}
\displaystyle \text{Height of the conical heap}=24\text{ cm.}
\displaystyle \text{Since the sand in the bucket forms the conical heap,}
\displaystyle \text{Volume of the cylinder}=\text{Volume of the cone.}
\displaystyle \pi(18)^2\times32=\frac{1}{3}\pi r^2\times24
\displaystyle 18^2\times32=8r^2
\displaystyle r^2=\frac{324\times32}{8}=1296
\displaystyle \therefore r=36\text{ cm.}
\displaystyle \text{Let the slant height of the conical heap be }l\text{ cm.}
\displaystyle l^2=r^2+h^2
\displaystyle l^2=36^2+24^2
\displaystyle =1296+576=1872
\displaystyle l=\sqrt{1872}=12\sqrt{13}\text{ cm.}
\displaystyle 12\sqrt{13}\approx43.27\text{ cm.}
\displaystyle \therefore \text{The radius of the heap is }36\text{ cm and its slant height is }12\sqrt{13}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A solid cuboid of iron with dimensions }33\text{ cm}\times40\text{ cm}\times15\text{ cm is melted}
\displaystyle \text{and recast into a cylindrical pipe. The outer and inner diameters of the pipe are }8\text{ cm}
\displaystyle \text{and }7\text{ cm respectively. Find the length of the pipe.}\hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of the iron cuboid}=33\times40\times15=19800\text{ cm}^3.
\displaystyle \text{Outer radius of the pipe }R=\frac{8}{2}=4\text{ cm.}
\displaystyle \text{Inner radius of the pipe }r=\frac{7}{2}=3.5\text{ cm.}
\displaystyle \text{Let the length of the pipe be }h\text{ cm.}
\displaystyle \text{Volume of material in the hollow cylindrical pipe}=\pi(R^2-r^2)h.
\displaystyle \text{Since the cuboid is melted and recast into the pipe,}
\displaystyle 19800=\pi\left(4^2-3.5^2\right)h
\displaystyle 19800=\pi(16-12.25)h
\displaystyle 19800=\frac{15}{4}\pi h
\displaystyle h=\frac{19800\times4}{15\pi}
\displaystyle =\frac{5280}{\pi}
\displaystyle =5280\times\frac{7}{22}=1680\text{ cm.}
\displaystyle 1680\text{ cm}=16.8\text{ m.}
\displaystyle \therefore \text{The length of the pipe is }16.8\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The }\frac{3}{4}\text{th part of a conical vessel of internal radius }5\text{ cm and height }24\text{ cm}
\displaystyle \text{is full of water. The water is emptied into a cylindrical vessel with internal radius }10\text{ cm.}
\displaystyle \text{Find the height of water in the cylindrical vessel.}\hfill\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the conical vessel}=5\text{ cm},\qquad \text{height}=24\text{ cm.}
\displaystyle \text{Volume of the conical vessel}=\frac{1}{3}\pi(5)^2\times24
\displaystyle =200\pi\text{ cm}^3.
\displaystyle \text{Volume of water}=\frac{3}{4}\times200\pi=150\pi\text{ cm}^3.
\displaystyle \text{Radius of the cylindrical vessel}=10\text{ cm.}
\displaystyle \text{Let the height of water in the cylindrical vessel be }h\text{ cm.}
\displaystyle \text{Volume of water in the cylindrical vessel}=\pi(10)^2h=100\pi h.
\displaystyle \text{Since the same water is transferred to the cylindrical vessel,}
\displaystyle 100\pi h=150\pi
\displaystyle h=\frac{150}{100}=1.5\text{ cm.}
\displaystyle \therefore \text{The height of water in the cylindrical vessel is }1.5\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{150 spherical marbles, each of diameter }1.4\text{ cm, are dropped in a cylindrical}
\displaystyle \text{vessel of diameter }7\text{ cm containing some water, which are completely immersed in}
\displaystyle \text{water. Find the rise in the level of water in the vessel.}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of each spherical marble}=\frac{1.4}{2}=0.7\text{ cm.}
\displaystyle \text{Let the rise in the water level be }h\text{ cm.}
\displaystyle \text{Radius of the cylindrical vessel}=\frac{7}{2}=3.5\text{ cm.}
\displaystyle \text{Since the marbles are completely immersed,}
\displaystyle \text{Volume of water displaced}=\text{Volume of }150\text{ marbles.}
\displaystyle \pi(3.5)^2h=150\times\frac{4}{3}\pi(0.7)^3
\displaystyle (3.5)^2h=200(0.7)^3
\displaystyle 12.25h=200\times0.343
\displaystyle 12.25h=68.6
\displaystyle h=\frac{68.6}{12.25}=5.6\text{ cm.}
\displaystyle \therefore \text{The rise in the level of water is }5.6\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Sushant has a vessel, of the form of an inverted cone, open at the top, of height}
\displaystyle 11\text{ cm and radius of top as }2.5\text{ cm and is full of water. Metallic spherical balls each}
\displaystyle \text{of diameter }0.5\text{ cm are put in the vessel due to which }\left(\frac{2}{5}\right)^{\text{th}}\text{ of the water}
\displaystyle \text{in the vessel flows out. Find how many balls were put in the vessel. Sushant made the}
\displaystyle \text{arrangement so that the water that flows out irrigates the flower beds. What value has}
\displaystyle \text{been shown by Sushant?}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the conical vessel}=2.5=\frac{5}{2}\text{ cm},\qquad \text{height}=11\text{ cm.}
\displaystyle \text{Volume of water in the full conical vessel}
\displaystyle =\frac{1}{3}\pi\left(\frac{5}{2}\right)^2\times11
\displaystyle =\frac{275\pi}{12}\text{ cm}^3.
\displaystyle \text{Volume of water flowing out}=\frac{2}{5}\times\frac{275\pi}{12}
\displaystyle =\frac{55\pi}{6}\text{ cm}^3.
\displaystyle \text{Radius of each metallic spherical ball}=\frac{0.5}{2}=0.25=\frac{1}{4}\text{ cm.}
\displaystyle \text{Volume of each ball}=\frac{4}{3}\pi\left(\frac{1}{4}\right)^3
\displaystyle =\frac{\pi}{48}\text{ cm}^3.
\displaystyle \text{Let the number of balls put in the vessel be }n.
\displaystyle \text{Since the vessel was initially full, the volume of water flowing out equals}
\displaystyle \text{the total volume of the metallic balls immersed in the vessel.}
\displaystyle n\times\frac{\pi}{48}=\frac{55\pi}{6}
\displaystyle n=\frac{55}{6}\times48=440
\displaystyle \therefore \text{The number of metallic balls put in the vessel is }440.
\displaystyle \text{Sushant shows the values of water conservation, resourcefulness and environmental awareness}
\displaystyle \text{by reusing the water that flows out to irrigate the flower beds.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{16 glass spheres each of radius }2\text{ cm are packed into a cuboidal box of}
\displaystyle \text{internal dimensions }16\text{ cm}\times8\text{ cm}\times8\text{ cm and then the box is filled}
\displaystyle \text{with water. Find the volume of water filled in the box.}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of the cuboidal box}=16\times8\times8=1024\text{ cm}^3.
\displaystyle \text{Radius of each glass sphere}=2\text{ cm.}
\displaystyle \text{Volume of one glass sphere}=\frac{4}{3}\pi(2)^3=\frac{32}{3}\pi\text{ cm}^3.
\displaystyle \text{Volume of }16\text{ glass spheres}=16\times\frac{32}{3}\pi
\displaystyle =\frac{512}{3}\pi\text{ cm}^3.
\displaystyle \text{Volume of water}=\text{Volume of box}-\text{Volume of }16\text{ spheres}
\displaystyle =1024-\frac{512}{3}\pi
\displaystyle =1024-\frac{512}{3}\times\frac{22}{7}
\displaystyle =1024-\frac{11264}{21}
\displaystyle =\frac{21504-11264}{21}=\frac{10240}{21}\text{ cm}^3
\displaystyle \approx487.62\text{ cm}^3.
\displaystyle \therefore \text{The volume of water filled in the box is approximately }487.62\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The sum of the radius of base and height of a solid right circular cylinder is }
\displaystyle 37\text{ cm.} \ \text{If the total surface area of the solid cylinder is }1628\text{ cm}^2,\text{ find the volume of the}
\displaystyle \text{cylinder. (Use }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius and height of the cylinder be }r\text{ cm and }h\text{ cm respectively.}
\displaystyle r+h=37\text{ cm.}
\displaystyle \text{Total surface area of the cylinder}=2\pi r(r+h).
\displaystyle 2\times\frac{22}{7}\times r\times37=1628
\displaystyle \frac{44}{7}\times37r=1628
\displaystyle r=\frac{1628\times7}{44\times37}=7\text{ cm.}
\displaystyle \therefore h=37-r=37-7=30\text{ cm.}
\displaystyle \text{Volume of the cylinder}=\pi r^2h
\displaystyle =\frac{22}{7}\times7^2\times30
\displaystyle =4620\text{ cm}^3.
\displaystyle \therefore \text{The volume of the cylinder is }4620\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{A }5\text{ m wide cloth is used to make a conical tent of base diameter }
\displaystyle 14\text{ m and} \ \text{height }24\text{ m. Find the cost of cloth used at the rate of Rs. }25\text{ per metre.}
\displaystyle \text{(Use }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the base of the tent}=\frac{14}{2}=7\text{ m.}
\displaystyle \text{Height of the tent}=24\text{ m.}
\displaystyle \text{Let the slant height of the tent be }l\text{ m.}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{7^2+24^2}
\displaystyle =\sqrt{49+576}=\sqrt{625}=25\text{ m.}
\displaystyle \text{Cloth required}=\text{Curved surface area of the conical tent}
\displaystyle =\pi rl
\displaystyle =\frac{22}{7}\times7\times25=550\text{ m}^2.
\displaystyle \text{Width of the cloth}=5\text{ m.}
\displaystyle \text{Length of cloth required}=\frac{\text{Area of cloth}}{\text{Width of cloth}}
\displaystyle =\frac{550}{5}=110\text{ m.}
\displaystyle \text{Cost of cloth}=110\times25=\text{Rs. }2750.
\displaystyle \therefore \text{The cost of cloth used is Rs. }2750.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The volume of a hemisphere is }2425\frac{1}{2}\text{ cm}^3.\text{ Find its curved surface}
\displaystyle \text{area. (Use }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the hemisphere be }r\text{ cm.}
\displaystyle \text{Volume of the hemisphere}=\frac{2}{3}\pi r^3.
\displaystyle \frac{2}{3}\times\frac{22}{7}\times r^3=2425\frac{1}{2}
\displaystyle \frac{44}{21}r^3=\frac{4851}{2}
\displaystyle r^3=\frac{4851}{2}\times\frac{21}{44}
\displaystyle =\frac{9261}{8}=\left(\frac{21}{2}\right)^3
\displaystyle \therefore r=\frac{21}{2}=10.5\text{ cm.}
\displaystyle \text{Curved surface area of the hemisphere}=2\pi r^2
\displaystyle =2\times\frac{22}{7}\times\left(\frac{21}{2}\right)^2
\displaystyle =\frac{44}{7}\times\frac{441}{4}=693\text{ cm}^2.
\displaystyle \therefore \text{The curved surface area of the hemisphere is }693\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If the total surface area of a solid hemisphere is }462\text{ cm}^2,\text{ find its}
\displaystyle \text{volume. (Take }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the hemisphere be }r\text{ cm.}
\displaystyle \text{Total surface area of a solid hemisphere}=3\pi r^2.
\displaystyle 3\times\frac{22}{7}\times r^2=462
\displaystyle r^2=\frac{462\times7}{66}=49
\displaystyle \therefore r=7\text{ cm.}
\displaystyle \text{Volume of the hemisphere}=\frac{2}{3}\pi r^3
\displaystyle =\frac{2}{3}\times\frac{22}{7}\times7^3
\displaystyle =\frac{2}{3}\times22\times49
\displaystyle =\frac{2156}{3}\text{ cm}^3
\displaystyle \approx718.67\text{ cm}^3.
\displaystyle \therefore \text{The volume of the hemisphere is }\frac{2156}{3}\text{ cm}^3\text{ (approximately }718.67\text{ cm}^3\text{).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{A solid right circular cone of height }120\text{ cm and radius }60\text{ cm is placed in a}
\displaystyle \text{right circular cylinder full of water of height }180\text{ cm such that it touches the bottom.}
\displaystyle \text{Find the volume of water left in the cylinder, if the radius of the cylinder is equal to}
\displaystyle \text{the radius of the cone.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cylinder}=\text{Radius of the cone}=60\text{ cm.}
\displaystyle \text{Height of the cylinder}=180\text{ cm.}
\displaystyle \text{Height of the cone}=120\text{ cm.}
\displaystyle \text{Volume of the cylinder}=\pi r^2h
\displaystyle =\pi(60)^2\times180
\displaystyle =648000\pi\text{ cm}^3.
\displaystyle \text{Volume of the solid cone}=\frac{1}{3}\pi r^2h
\displaystyle =\frac{1}{3}\pi(60)^2\times120
\displaystyle =144000\pi\text{ cm}^3.
\displaystyle \text{Since the cylinder was initially full, water equal to the volume of the cone flows out.}
\displaystyle \text{Volume of water left}=\text{Volume of cylinder}-\text{Volume of cone}
\displaystyle =648000\pi-144000\pi
\displaystyle =504000\pi\text{ cm}^3.
\displaystyle =504000\times\frac{22}{7}=1584000\text{ cm}^3.
\displaystyle \therefore \text{The volume of water left in the cylinder is }1584000\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{A heap of rice is in the form of a cone of diameter }9\text{ m and height }
\displaystyle 3.5\text{ m.} \ \text{Find the volume of rice. How much canvas cloth is required to cover the heap?}
\displaystyle \hfill\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the conical heap}=\frac{9}{2}=4.5\text{ m.}
\displaystyle \text{Height of the conical heap}=3.5\text{ m.}
\displaystyle \text{Volume of the cone}=\frac{1}{3}\pi r^2h
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times(4.5)^2\times3.5
\displaystyle =74.25\text{ m}^3.
\displaystyle \therefore \text{The volume of rice is }74.25\text{ m}^3.
\displaystyle \text{Let the slant height of the conical heap be }l\text{ m.}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{(4.5)^2+(3.5)^2}
\displaystyle =\sqrt{20.25+12.25}=\sqrt{32.5}\text{ m.}
\displaystyle \text{Canvas required}=\text{Curved surface area of the conical heap}
\displaystyle =\pi rl
\displaystyle =\frac{22}{7}\times4.5\times\sqrt{32.5}
\displaystyle \approx80.63\text{ m}^2.
\displaystyle \therefore \text{The canvas cloth required is approximately }80.63\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{A factory manufactures }120000\text{ pencils daily. The pencils are cylindrical}
\displaystyle \text{in shape, each of length }25\text{ cm and circumference of base }1.5\text{ cm. Determine the cost }
\displaystyle \text{of colouring the curved surfaces of the pencils manufactured in one day at Rs. }0.05\text{ per dm}^2.
\displaystyle \text{Answer:}
\displaystyle \text{Length of each pencil}=25\text{ cm.}
\displaystyle \text{Circumference of the base of each pencil}=1.5\text{ cm.}
\displaystyle \text{Curved surface area of one pencil}
\displaystyle =\text{Circumference of base}\times\text{Length}
\displaystyle =1.5\times25=37.5\text{ cm}^2.
\displaystyle \text{Curved surface area of }120000\text{ pencils}
\displaystyle =120000\times37.5=4500000\text{ cm}^2.
\displaystyle 4500000\text{ cm}^2=\frac{4500000}{100}=45000\text{ dm}^2.
\displaystyle \text{Cost of colouring}=45000\times0.05=\text{Rs. }2250.
\displaystyle \therefore \text{The cost of colouring the pencils manufactured in one day is Rs. }2250.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{A well of diameter }2\text{ m is dug }14\text{ m deep. The earth taken out of it is}
\displaystyle \text{spread evenly all around it to form an embankment of height }40\text{ cm. Find the width}
\displaystyle \text{of the embankment.}\hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the well}=\frac{2}{2}=1\text{ m},\qquad \text{depth}=14\text{ m.}
\displaystyle \text{Volume of earth dug out}=\pi(1)^2\times14=14\pi\text{ m}^3.
\displaystyle \text{Height of the embankment}=40\text{ cm}=0.4\text{ m.}
\displaystyle \text{Let the width of the embankment be }x\text{ m.}
\displaystyle \text{Inner radius of the embankment}=1\text{ m.}
\displaystyle \text{Outer radius of the embankment}=(1+x)\text{ m.}
\displaystyle \text{Volume of the embankment}=\pi\left[(1+x)^2-1^2\right]\times0.4.
\displaystyle \text{Since the earth dug out is used to form the embankment,}
\displaystyle 14\pi=\pi\left[(1+x)^2-1\right]\times0.4
\displaystyle 14=0.4\left[(1+x)^2-1\right]
\displaystyle 35=(1+x)^2-1
\displaystyle (1+x)^2=36
\displaystyle 1+x=6
\displaystyle x=5\text{ m.}
\displaystyle \therefore \text{The width of the embankment is }5\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Water flows through a cylindrical pipe, whose inner radius is }1\text{ cm, at}
\displaystyle \text{the rate of }80\text{ cm/sec into an empty cylindrical tank, the radius of whose base is }
\displaystyle 40\text{ cm.}  \ \text{What is the rise of water level in the tank in half an hour?}
\displaystyle \text{Answer:}
\displaystyle \text{Inner radius of the pipe}=1\text{ cm.}
\displaystyle \text{Rate of flow of water}=80\text{ cm/sec.}
\displaystyle \text{Time}=\frac{1}{2}\text{ hour}=30\text{ minutes}=1800\text{ seconds.}
\displaystyle \text{Length of water column flowing in }1800\text{ seconds}
\displaystyle =80\times1800=144000\text{ cm.}
\displaystyle \text{Volume of water flowing into the tank}=\pi(1)^2\times144000
\displaystyle =144000\pi\text{ cm}^3.
\displaystyle \text{Radius of the cylindrical tank}=40\text{ cm.}
\displaystyle \text{Let the rise in water level be }h\text{ cm.}
\displaystyle \text{Volume of water in the tank}=\pi(40)^2h=1600\pi h.
\displaystyle \text{Equating the volume of water flowing in with the volume of water in the tank,}
\displaystyle 1600\pi h=144000\pi
\displaystyle h=\frac{144000}{1600}=90\text{ cm.}
\displaystyle \therefore \text{The rise of water level in the tank is }90\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Water in a canal }1.5\text{ m wide and }6\text{ m deep is flowing with a speed}
\displaystyle \text{of }10\text{ km/hr. How much area will it irrigate in }30\text{ minutes if }8\text{ cm of standing}
\displaystyle \text{water is desired?}\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Width of the canal}=1.5\text{ m},\qquad \text{depth}=6\text{ m.}
\displaystyle \text{Speed of water}=10\text{ km/hr.}
\displaystyle \text{Time}=30\text{ minutes}=\frac{1}{2}\text{ hour.}
\displaystyle \text{Distance travelled by water in }30\text{ minutes}
\displaystyle =10\times\frac{1}{2}=5\text{ km}=5000\text{ m.}
\displaystyle \text{Volume of water flowing through the canal}
\displaystyle =\text{Width}\times\text{Depth}\times\text{Distance}
\displaystyle =1.5\times6\times5000=45000\text{ m}^3.
\displaystyle \text{Depth of standing water}=8\text{ cm}=0.08\text{ m.}
\displaystyle \text{Let the area irrigated be }A\text{ m}^2.
\displaystyle \text{Volume of water required}=A\times0.08\text{ m}^3.
\displaystyle A\times0.08=45000
\displaystyle A=\frac{45000}{0.08}=562500\text{ m}^2.
\displaystyle 1\text{ hectare}=10000\text{ m}^2.
\displaystyle \therefore A=\frac{562500}{10000}=56.25\text{ hectares.}
\displaystyle \therefore \text{The canal will irrigate an area of }562500\text{ m}^2\text{ or }56.25\text{ hectares.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{A cylindrical tank full of water is emptied by a pipe at the rate of }
\displaystyle 225\text{ litres} \ \text{per minute. How much time will it take to empty half the tank,}
\displaystyle \text{if the diameter of its base is }3\text{ m and its height is }3.5\text{ m? (Use }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the cylindrical tank}=\frac{3}{2}=1.5\text{ m.}
\displaystyle \text{Height of the tank}=3.5\text{ m.}
\displaystyle \text{Volume of the tank}=\pi r^2h
\displaystyle =\frac{22}{7}\times(1.5)^2\times3.5
\displaystyle =24.75\text{ m}^3.
\displaystyle \text{Volume of half the tank}=\frac{24.75}{2}=12.375\text{ m}^3.
\displaystyle 1\text{ m}^3=1000\text{ litres.}
\displaystyle \therefore 12.375\text{ m}^3=12375\text{ litres.}
\displaystyle \text{Rate of emptying}=225\text{ litres per minute.}
\displaystyle \text{Time required}=\frac{\text{Volume of water}}{\text{Rate of emptying}}
\displaystyle =\frac{12375}{225}=55\text{ minutes.}
\displaystyle \therefore \text{The time required to empty half the tank is }55\text{ minutes.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Water is flowing at the rate of }2.52\text{ km/h through a cylindrical pipe into a}
\displaystyle \text{cylindrical tank, the radius of the base is }40\text{ cm. If the increase in the level of water}
\displaystyle \text{in the tank, in half an hour is }3.15\text{ m, find the internal diameter of the pipe.}
\displaystyle \hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Rate of flow of water}=2.52\text{ km/h.}
\displaystyle \text{Time}=\frac{1}{2}\text{ hour.}
\displaystyle \text{Length of water column flowing in half an hour}
\displaystyle =2.52\times\frac{1}{2}=1.26\text{ km}=126000\text{ cm.}
\displaystyle \text{Let the internal radius of the pipe be }r\text{ cm.}
\displaystyle \text{Volume of water flowing through the pipe}=\pi r^2\times126000\text{ cm}^3.
\displaystyle \text{Radius of the cylindrical tank}=40\text{ cm.}
\displaystyle \text{Rise in water level}=3.15\text{ m}=315\text{ cm.}
\displaystyle \text{Increase in volume of water in the tank}=\pi(40)^2\times315\text{ cm}^3.
\displaystyle \text{Equating the volume of water flowing in with the increase in volume of the tank,}
\displaystyle \pi r^2\times126000=\pi(40)^2\times315
\displaystyle r^2=\frac{1600\times315}{126000}=4
\displaystyle r=2\text{ cm.}
\displaystyle \therefore \text{Internal diameter of the pipe}=2r=2\times2=4\text{ cm.}
\displaystyle \therefore \text{The internal diameter of the pipe is }4\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{The difference between the outer and inner curved surface areas of}
\displaystyle \text{a hollow right circular cylinder }14\text{ cm long is }88\text{ cm}^2.\text{ If the volume of metal used in}
\displaystyle \text{making the cylinder is }176\text{ cm}^3,\text{ find the outer and inner diameters of the cylinder.}
\displaystyle \text{(Use }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the outer and inner radii of the cylinder be }R\text{ cm and }r\text{ cm respectively.}
\displaystyle \text{Height of the hollow cylinder}=14\text{ cm.}
\displaystyle \text{Difference between outer and inner curved surface areas}=88\text{ cm}^2.
\displaystyle 2\pi Rh-2\pi rh=88
\displaystyle 2\pi h(R-r)=88
\displaystyle 2\times\frac{22}{7}\times14(R-r)=88
\displaystyle 88(R-r)=88
\displaystyle R-r=1\qquad\ldots\text{(i)}
\displaystyle \text{Volume of metal in the hollow cylinder}=\pi h(R^2-r^2).
\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 \text{Outer diameter}=2R=2\times2.5=5\text{ cm.}
\displaystyle \text{Inner diameter}=2r=2\times1.5=3\text{ cm.}
\displaystyle \therefore \text{The outer and inner diameters are }5\text{ cm and }3\text{ cm respectively.}
\displaystyle \\

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