Class 10: Cone and Sphere – Exercise 22(d)

$\displaystyle \textbf{Question 1: }\text{A cone of height }15\text{ cm and diameter }7\text{ cm is mounted}$
$\displaystyle \text{on a hemisphere of the same diameter. Determine the volume of the solid thus formed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the cone and hemisphere }r=\frac{7}{2}=3.5\text{ cm}$
$\displaystyle \text{Volume of the solid}=\text{Volume of cone}+\text{Volume of hemisphere}$
$\displaystyle =\frac{1}{3}\pi r^2h+\frac{2}{3}\pi r^3$
$\displaystyle =\frac{1}{3}\times\frac{22}{7}\times(3.5)^2\times15+\frac{2}{3}\times\frac{22}{7}\times(3.5)^3$
$\displaystyle =192.5+89.83=282.33\text{ cm}^3$
$\displaystyle \therefore \text{Volume of the solid}=282.33\text{ cm}^3$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{A buoy is made in the form of a hemisphere surmounted by a right circular cone}$
$\displaystyle \text{whose base coincides with the plane surface of the hemisphere. The radius of the cone is }3.5\text{ m}.$
$\displaystyle \text{Its volume is two-thirds of the volume of the hemisphere. Calculate the height of the cone}$
$\displaystyle \text{and the surface area of the buoy, correct to two decimal places.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the cone and hemisphere }r=3.5\text{ m}$
$\displaystyle \text{Let the height of the cone be }h\text{ m}$
$\displaystyle \text{Volume of cone}=\frac{2}{3}\times\text{Volume of hemisphere}$
$\displaystyle \frac{1}{3}\pi r^2h=\frac{2}{3}\times\frac{2}{3}\pi r^3$
$\displaystyle \frac{1}{3}\pi r^2h=\frac{4}{9}\pi r^3$
$\displaystyle h=\frac{4r}{3}=\frac{4}{3}\times3.5=\frac{14}{3}$
$\displaystyle \therefore h=4.67\text{ m}$
$\displaystyle \text{Slant height of the cone }l=\sqrt{r^2+h^2}$
$\displaystyle =\sqrt{(3.5)^2+\left(\frac{14}{3}\right)^2}$
$\displaystyle =\sqrt{\frac{1225}{36}}=\frac{35}{6}\text{ m}$
$\displaystyle \text{Surface area of the buoy}=\pi rl+2\pi r^2$
$\displaystyle =\frac{22}{7}\times3.5\times\frac{35}{6}+2\times\frac{22}{7}\times(3.5)^2$
$\displaystyle =64.17+77=141.17\text{ m}^2$
$\displaystyle \therefore \text{Height of the cone}=4.67\text{ m}$
$\displaystyle \text{and surface area of the buoy}=141.17\text{ m}^2$
$\\$

$\displaystyle \textbf{Question 3: }\text{From a rectangular solid of metal measuring }$
$\displaystyle 42\text{ cm}\times30\text{ cm}\times20\text{ cm}, \ \text{a conical cavity of diameter }14\text{ cm and depth }24\text{ cm is drilled out. Find:}$
$\displaystyle \text{(i) the surface area of the remaining solid,}$
$\displaystyle \text{(ii) the volume of the remaining solid,}$
$\displaystyle \text{(iii) the weight of the material drilled out, if its density is }7\text{ g/cm}^3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Dimensions of the cuboid}=42\text{ cm}\times30\text{ cm}\times20\text{ cm}$
$\displaystyle \text{Radius of the conical cavity }r=\frac{14}{2}=7\text{ cm}$
$\displaystyle \text{Depth of the conical cavity }h=24\text{ cm}$
$\displaystyle \text{Slant height }l=\sqrt{r^2+h^2}$
$\displaystyle =\sqrt{7^2+24^2}=\sqrt{625}=25\text{ cm}$
$\displaystyle \text{(i) Surface area of the remaining solid}$
$\displaystyle =\text{Surface area of cuboid}-\text{Area of circular opening}+\text{Curved surface area of cone}$
$\displaystyle =2(lb+bh+hl)-\pi r^2+\pi rl$
$\displaystyle =2(42\times30+30\times20+20\times42)-\frac{22}{7}\times7^2+\frac{22}{7}\times7\times25$
$\displaystyle =5400-154+550$
$\displaystyle \therefore \text{Surface area of the remaining solid}=5796\text{ cm}^2$
$\displaystyle \text{(ii) Volume of the remaining solid}$
$\displaystyle =\text{Volume of cuboid}-\text{Volume of conical cavity}$
$\displaystyle =42\times30\times20-\frac{1}{3}\pi r^2h$
$\displaystyle =25200-\frac{1}{3}\times\frac{22}{7}\times7^2\times24$
$\displaystyle =25200-1232$
$\displaystyle \therefore \text{Volume of the remaining solid}=23968\text{ cm}^3$
$\displaystyle \text{(iii) Weight of the material drilled out}$
$\displaystyle =\text{Volume of conical cavity}\times\text{density}$
$\displaystyle =1232\times7=8624\text{ g}$
$\displaystyle =8.624\text{ kg}$
$\displaystyle \therefore \text{Weight of the material drilled out}=8.624\text{ kg}$
$\\$

$\displaystyle \textbf{Question 4: }\text{A cubical block of side }7\text{ cm is surmounted by a hemisphere}$
$\displaystyle \text{of the largest possible size. Find the surface area of the resulting solid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Side of the cube}=7\text{ cm}$
$\displaystyle \text{Radius of the hemisphere}=\frac{7}{2}=3.5\text{ cm}$
$\displaystyle \text{Surface area of the solid}=\text{Surface area of cube}-\text{Area of circular base}+\text{Curved surface area of hemisphere}$
$\displaystyle =6(7)^2-\pi(3.5)^2+2\pi(3.5)^2$
$\displaystyle =6(49)+\pi(3.5)^2$
$\displaystyle =294+\frac{22}{7}\times(3.5)^2$
$\displaystyle =294+38.5$
$\displaystyle =332.5\text{ cm}^2$
$\displaystyle \therefore \text{Surface area of the resulting solid}=332.5\text{ cm}^2$
$\\$

$\displaystyle \textbf{Question 5: }\text{A vessel is in the form of an inverted cone of height }8\text{ cm}$
$\displaystyle \text{and open-top radius }5\text{ cm. It is filled with water up to the rim.}$
$\displaystyle \text{Lead shots, each spherical with radius }0.5\text{ cm, are dropped into the vessel.}$
$\displaystyle \text{If one-fourth of the water flows out, find the number of lead shots dropped.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of lead shots be }n$
$\displaystyle \text{Volume of }n\text{ lead shots}=\frac{1}{4}\times\text{Volume of water in the cone}$
$\displaystyle n\times\frac{4}{3}\pi(0.5)^3=\frac{1}{4}\times\frac{1}{3}\pi(5)^2\times8$
$\displaystyle n=\frac{5^2\times8}{4\times4\times(0.5)^3}$
$\displaystyle =\frac{200}{16\times0.125}=100$
$\displaystyle \therefore \text{Number of lead shots}=100$
$\\$

$\displaystyle \textbf{Question 6: }\text{A hemispherical bowl of negligible thickness has a circumference}$
$\displaystyle \text{of }198\text{ cm. Find the capacity of the bowl.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Circumference of the bowl}=198\text{ cm}$
$\displaystyle 2\pi r=198$
$\displaystyle r=\frac{198}{2\times\frac{22}{7}}=31.5\text{ cm}$
$\displaystyle \text{Capacity of the bowl}=\frac{2}{3}\pi r^3$
$\displaystyle =\frac{2}{3}\times\frac{22}{7}\times(31.5)^3$
$\displaystyle =65488.5\text{ cm}^3$
$\displaystyle \therefore \text{Capacity of the bowl}=65488.5\text{ cm}^3$
$\\$

$\displaystyle \textbf{Question 7: }\text{Find the maximum volume of a cone that can be carved out}$
$\displaystyle \text{of a solid hemisphere of radius }r\text{ cm}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the largest cone, its base coincides with the circular base of the hemisphere}$
$\displaystyle \text{and its vertex lies at the lowest point of the hemisphere.}$
$\displaystyle \therefore \text{Radius of the cone}=r\text{ cm and height of the cone}=r\text{ cm}$
$\displaystyle \text{Maximum volume of the cone}=\frac{1}{3}\pi r^2h$
$\displaystyle =\frac{1}{3}\pi r^2\times r$
$\displaystyle =\frac{1}{3}\pi r^3\text{ cm}^3$
$\displaystyle \therefore \text{Maximum volume of the cone}=\frac{1}{3}\pi r^3\text{ cm}^3$
$\\$

$\displaystyle \textbf{Question 8: }\text{The radii of the bases of two solid right circular cones of equal height}$
$\displaystyle \text{are }r_1\text{ and }r_2\text{ respectively. The cones are melted and recast into a solid sphere}$
$\displaystyle \text{of radius }R.\text{ Find the height of each cone in terms of }r_1,\ r_2\text{ and }R.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the common height of the cones be }h$
$\displaystyle \text{Volume of the two cones}=\text{Volume of the sphere}$
$\displaystyle \frac{1}{3}\pi r_1^2h+\frac{1}{3}\pi r_2^2h=\frac{4}{3}\pi R^3$
$\displaystyle \frac{1}{3}\pi h(r_1^2+r_2^2)=\frac{4}{3}\pi R^3$
$\displaystyle h(r_1^2+r_2^2)=4R^3$
$\displaystyle \therefore h=\frac{4R^3}{r_1^2+r_2^2}$
$\\$

$\displaystyle \textbf{Question 9: }\text{A solid metallic hemisphere of diameter }28\text{ cm is melted and recast}$
$\displaystyle \text{into identical solid cones, each of diameter }14\text{ cm and height }8\text{ cm}.$
$\displaystyle \text{Find the number of cones so formed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the hemisphere}=\frac{28}{2}=14\text{ cm}$
$\displaystyle \text{Radius of each cone}=\frac{14}{2}=7\text{ cm}$
$\displaystyle \text{Height of each cone}=8\text{ cm}$
$\displaystyle \text{Let the number of cones formed be }n$
$\displaystyle n\times\text{Volume of one cone}=\text{Volume of the hemisphere}$
$\displaystyle n\times\frac{1}{3}\pi(7)^2\times8=\frac{2}{3}\pi(14)^3$
$\displaystyle n=\frac{2\times14^3}{7^2\times8}$
$\displaystyle =14$
$\displaystyle \therefore \text{Number of cones formed}=14$
$\\$

$\displaystyle \textbf{Question 10: }\text{A cone and a hemisphere have the same base and the same height.}$
$\displaystyle \text{Find the ratio of their volumes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the common base radius be }r$
$\displaystyle \text{Height of the hemisphere}=r$
$\displaystyle \therefore \text{Height of the cone}=r$
$\displaystyle \text{Volume of cone}:\text{Volume of hemisphere}$
$\displaystyle =\frac{1}{3}\pi r^2\times r:\frac{2}{3}\pi r^3$
$\displaystyle =\frac{1}{3}\pi r^3:\frac{2}{3}\pi r^3$
$\displaystyle =1:2$
$\displaystyle \therefore \text{The ratio of their volumes}=1:2$
$\\$