Topics covered: cube, cuboid, cylinder, cone, sphere, hemisphere. Each solution shows calculations on separate lines for clarity.
Formula Sheet
Sphere volume: \(\displaystyle V=\frac{4}{3}\pi r^3\)
Sphere surface area: \(\displaystyle A=4\pi r^2\)
Hemisphere volume (solid): \(\displaystyle V=\frac{2}{3}\pi r^3\)
Cone volume: \(\displaystyle V=\frac{1}{3}\pi r^2 h\)
Cone curved surface area: \(\displaystyle A=\pi r l\) where \(l=\sqrt{r^2+h^2}\)
Cylinder volume: \(\displaystyle V=\pi r^2 h\)
Cylinder curved surface area: \(\displaystyle A=2\pi r h\)
Cuboid volume: \(\displaystyle V=l b h\)
Cuboid base area (rectangular): \(\displaystyle A = l\times b\)
Cube volume: \(\displaystyle V=a^3\)
Cube surface area: \(\displaystyle A=6a^2\)
1. Half of a cuboidal water tank with length of 2.1 m and breadth of 1.5 m is filled with water. If 630 litre of water is poured into the tank, then calculate the increased height of water.
Convert litres to cubic metres: \(1\ \text{L} = 0.001\ \text{m}^3\) Volume added \(=630\ \text{L}=630\times 0.001=0.63\ \text{m}^3\) Base area \(=2.1\times 1.5=3.15\ \text{m}^2\) Increase in height \(=\dfrac{\text{volume}}{\text{base area}}=\dfrac{0.63}{3.15}=0.20\ \text{m}\) Answer: \(0.20\ \text{m} = 20\ \text{cm}\)
2. Height of a right circular cylinder is twice of its radius. If the height would be 6 times of its radius, then the volume of the cylinder would be greater by 539 cubic decimetres. Calculate the height of the cylinder.
Let radius \(=r\) (in dm). Original height \(=2r\). New height \(=6r\). Original volume \(V_1=\pi r^2(2r)=2\pi r^3\). New volume \(V_2=\pi r^2(6r)=6\pi r^3\). Difference \(V_2-V_1=4\pi r^3=539\). So \(r^3=\dfrac{539}{4\pi}\). Numeric: \(r^3\approx \dfrac{539}{4\pi}\approx 42.935\). \(r\approx 3.50047\ \text{dm}\). Original height \(=2r\approx 7.00094\ \text{dm}=0.70009\ \text{m}\). Answer: \( \approx 7.00\ \text{dm} \;(\approx 0.700\ \text{m})\)
3. In a right circular conical tent 11 persons can stay. For each person 4 sq. m. space in the base and 20 cu. m. air are necessary. Determine the height of the tent put up exactly for 11 persons.
Required base area \(=11\times 4=44\ \text{m}^2\). Required volume \(=11\times 20=220\ \text{m}^3\). For cone: \(V=\dfrac{1}{3}\times \text{base area}\times h\). So \(\dfrac{1}{3}\times 44\times h=220\). Solve: \(h=\dfrac{220\times 3}{44}=15\ \text{m}\). Answer: \(15\ \text{m}\)
4. Calculate how many spherical marbles with 1 cm radius each may be formed by melting a solid sphere of iron having 8 cm of radius.
Volume of large sphere \(=\dfrac{4}{3}\pi(8^3)=\dfrac{4}{3}\pi\times 512=\dfrac{2048\pi}{3}\ \text{cm}^3\). Volume of one marble \(=\dfrac{4}{3}\pi(1^3)=\dfrac{4\pi}{3}\ \text{cm}^3\). Number \(=\dfrac{2048\pi/3}{4\pi/3}=\dfrac{2048}{4}=512\). Answer: \(512\) marbles
5. The inner length, breadth and height of a tea box are 7.5 dcm, 6 dcm and 5.4 dcm respectively. If the weight of the box filled with tea is 52 kg 350 gm, but in empty state, its weight is 3.75 kg, then calculate the weight of 1 cubic dcm tea.
Volume \(=7.5\times 6\times 5.4=243\ \text{dm}^3\). Weight of filled box \(=52\ \text{kg}\ 350\ \text{g}=52.350\ \text{kg}\). Empty box weight \(=3.75\ \text{kg}\). Net weight of tea \(=52.350-3.750=48.600\ \text{kg}\). Weight per \(1\ \text{dm}^3=\dfrac{48.600}{243}=0.2\ \text{kg}=200\ \text{g}\). Answer: \(200\ \text{g per dm}^3\)
6. Two solid spheres with the radii of 1 cm and 6 cm lengths are melted and a hollow sphere with outer radius 9 cm is made. Determine the inner radius of the new hollow sphere.
Volume of spheres combined \(=\dfrac{4}{3}\pi(1^3)+\dfrac{4}{3}\pi(6^3)=\dfrac{4}{3}\pi(1+216)=\dfrac{4}{3}\pi(217)\). Volume of hollow sphere \(=\dfrac{4}{3}\pi(9^3-R^3)=\dfrac{4}{3}\pi(729-R^3)\) where \(R\) is inner radius. Equate: \(729-R^3=217\). So \(R^3=512\). \(R=8\ \text{cm}\). Answer: \(8\ \text{cm}\)
7. The height of a right circular cone is twice its radius of base. If the height be seven times of the diameter of the base then volume of the cone would have been 539 cu cm more. Find the height of the cone.
Let radius \(=r\) (cm). Original height \(=2r\). New height (seven times diameter) \(=7\times(2r)=14r\). Original volume \(V_1=\dfrac{1}{3}\pi r^2(2r)=\dfrac{2}{3}\pi r^3\). New volume \(V_2=\dfrac{1}{3}\pi r^2(14r)=\dfrac{14}{3}\pi r^3\). Difference \(V_2-V_1=\dfrac{12}{3}\pi r^3=4\pi r^3=539\). So \(r^3=\dfrac{539}{4\pi}\). Numeric: \(r^3\approx 42.935\), \(r\approx 3.50047\ \text{cm}\). Original height \(=2r\approx 7.00094\ \text{cm}\). Answer: \(\approx 7.00\ \text{cm}\)
8. Curved surface area of a right circular cylindrical wooden log of uniform density is 440 sq. decimeter. Weight of 1 cubic decimeter of wood is 3 kg and weight of the log is 18.48 quintal. Find the diameter of the log.
Total weight \(=18.48\) quintal \(=18.48\times 100=1848\ \text{kg}\). Density \(=3\ \text{kg per dm}^3\). Volume \(=\dfrac{1848}{3}=616\ \text{dm}^3\). Curved surface area \(=2\pi r h=440\) (dm²). Volume \(=\pi r^2 h=616\) (dm³). From CSA: \(h=\dfrac{440}{2\pi r}=\dfrac{220}{\pi r}\). Substitute in volume: \(\pi r^2\times\dfrac{220}{\pi r}=220r=616\). So \(r=\dfrac{616}{220}=2.8\ \text{dm}\). Diameter \(=2r=5.6\ \text{dm}=56\ \text{cm}\). Answer: \(56\ \text{cm}\)
9. The length of outer and inner radii of a hollow right circular pipe are 5 cm and 4 cm respectively. If the total surface area of the pipe is 1188 sq. cm., find the length of the pipe.
Let length \(=h\). Outer radius \(R=5\) cm, inner radius \(r=4\) cm. Total surface area \(=\) curved outer \(+\) curved inner \(+\) two ends \(=2\pi R h+2\pi r h+2\pi(R^2-r^2)\). So TSA \(=2\pi(R+r)h+2\pi(R^2-r^2)\). Substitute numbers: \(2\pi(5+4)h+2\pi(25-16)=18\pi h+18\pi\). Given \(18\pi h+18\pi=1188\). Divide by \(18\pi\): \(h+1=\dfrac{1188}{18\pi}=\dfrac{66}{\pi}\). So \(h=\dfrac{66}{\pi}-1\). Numeric: \(h\approx \dfrac{66}{3.14159}-1\approx 20.02-1\approx 19.02\ \text{cm}\). Answer: \(\approx 19.02\ \text{cm}\)
10. A hemispherical pot with internal radius of 9 cm is completely filled with water. If we fill this water in cylindrical bottles with a diameter of 3 cm and height of 4 cm, then find the number of bottles to be required to make the pot empty.
Volume of hemispherical pot \(=\dfrac{2}{3}\pi(9^3)=\dfrac{2}{3}\pi\times 729=486\pi\ \text{cm}^3\). Bottle radius \(=1.5\) cm, bottle volume \(=\pi(1.5^2)(4)=\pi\times 2.25\times 4=9\pi\ \text{cm}^3\). Number \(=\dfrac{486\pi}{9\pi}=54\). Answer: \(54\) bottles
11. A solid metallic cylinder of radius 7 cm and height 10 cm is melted and recast into spheres each of radius 3.5 cm. Find the number of spheres.
Volume of cylinder \(=\pi(7^2)(10)=490\pi\ \text{cm}^3\). Volume of one sphere of radius 3.5 cm \(=\dfrac{4}{3}\pi(3.5^3)=\dfrac{4}{3}\pi\times 42.875\approx 57.1667\pi\ \text{cm}^3\). Number \(=\dfrac{490\pi}{57.1667\pi}\approx 8.5714\). Only whole spheres possible: 8 full spheres, remainder metal left. Answer: \(8\) full spheres (with leftover metal)
12. A right circular cone of height 24 cm and base radius 7 cm is melted to form a cylinder of height 8 cm. Find the radius of the cylinder.
Volume of cone \(=\dfrac{1}{3}\pi(7^2)(24)=\dfrac{1}{3}\pi\times 49\times 24=392\pi\ \text{cm}^3\). Let cylinder radius \(=R\). Cylinder volume \(=\pi R^2(8)\). Equate: \(\pi R^2\times 8=392\pi\). So \(R^2=\dfrac{392}{8}=49\). Hence \(R=7\ \text{cm}\). Answer: \(7\ \text{cm}\)
13. A metal hemisphere of radius 4.2 cm is melted to form spheres each of radius 0.7 cm. Find how many spheres are formed.
Volume of hemisphere \(=\dfrac{2}{3}\pi(4.2^3)=\dfrac{2}{3}\pi\times 74.088\approx 49.392\pi\ \text{cm}^3\). Volume of one small sphere \(=\dfrac{4}{3}\pi(0.7^3)=\dfrac{4}{3}\pi\times 0.343\approx 0.457333\pi\ \text{cm}^3\). Number \(=\dfrac{49.392\pi}{0.457333\pi}\approx 108.0\). Answer: \(108\) spheres
14. A solid metallic hemisphere of diameter 7 cm is melted and recast into small cubes of edge 1 cm each. Find the number of cubes.
Radius \(=3.5\) cm. Volume of hemisphere \(=\dfrac{2}{3}\pi(3.5^3)=\dfrac{2}{3}\pi\times 42.875\approx 28.583333\pi\ \text{cm}^3\). Numeric volume \(\approx 28.583333\times 3.14159\approx 89.7972\ \text{cm}^3\). Number of unit cubes \(=\lfloor 89.7972\rfloor = 89\) full 1 cm³ cubes (one cannot have fraction of a cube). Answer: \(89\) full cubes (with small leftover)
15. A cylindrical jar of internal diameter 14 cm contains water to a height of 6 cm. A solid sphere of radius 3.5 cm is dropped into the jar. Find the rise in water level.
Sphere radius \(=3.5\) cm. Volume of sphere \(=\dfrac{4}{3}\pi(3.5^3)=\dfrac{4}{3}\pi\times 42.875\approx 57.166667\pi\ \text{cm}^3\). Jar radius \(=7\) cm. Base area \(=\pi(7^2)=49\pi\ \text{cm}^2\). Rise in level \(=\dfrac{\text{sphere volume}}{\text{base area}}=\dfrac{57.166667\pi}{49\pi}= \dfrac{57.166667}{49}\approx 1.1666667\ \text{cm}\). Answer: \( \approx 1.17\ \text{cm}\)
16. A solid cone of height 24 cm and base radius 7 cm is melted and recast into small cones each of height 2.4 cm and base radius 0.7 cm. Find the number of small cones.
Big cone volume \(=\dfrac{1}{3}\pi(7^2)(24)=392\pi\ \text{cm}^3\). Small cone volume \(=\dfrac{1}{3}\pi(0.7^2)(2.4)=\dfrac{1}{3}\pi(0.49)(2.4)=0.392\pi\ \text{cm}^3\). Number \(=\dfrac{392\pi}{0.392\pi}=1000\). Answer: \(1000\) small cones
17. A well of diameter 14 m is dug to a depth of 15 m. The earth taken out is spread evenly to form an embankment 5 m wide around the well. Find the height of the embankment.
Well radius \(=7\) m. Volume of earth removed \(=\pi(7^2)(15)=735\pi\ \text{m}^3\). Outer radius of embankment \(=7+5=12\) m. Area of annulus \(=\pi(12^2-7^2)=\pi(144-49)=95\pi\ \text{m}^2\). Height \(=\dfrac{\text{volume}}{\text{area}}=\dfrac{735\pi}{95\pi}=\dfrac{735}{95}\approx 7.7368421\ \text{m}\). Answer: \(\approx 7.74\ \text{m}\)
18. A hemispherical bowl of radius 9 cm is completely filled with rice. This rice is poured into a cylindrical container of radius 6 cm. Find the height of the rice in the container.
Volume of hemisphere \(=\dfrac{2}{3}\pi(9^3)=486\pi\ \text{cm}^3\). Base area of cylinder \(=\pi(6^2)=36\pi\ \text{cm}^2\). Height \(=\dfrac{486\pi}{36\pi}=\dfrac{486}{36}=13.5\ \text{cm}\). Answer: \(13.5\ \text{cm}\)