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VOLUME AND SURFACE AREA

Aptitude

VOLUME AND SURFACE AREA -> IMPORTANT FACTS AND FORMULAE

I. CUBIOD
Let length = l, breadth = b and height = h units. Then,
1. Volume = (l x b x h) cubic units.
2. Surface area = 2 (lb + bh + lh)
II. CUBE
Let each edge of a cube be of length a. Then, 1. Volume = a³ cubic units.
2. Surface area = 6a² sq. units.
3. Diagonal = √3 a units.
III. CYLINDER
Let radius of base = r and Height (or length) = h Then,
1. Volume = (∏r²h) cubic units.
2. Curved surface area = (2∏rh) sq. units.
3. Total surface area = (2∏rh + 2∏r² sq. units)
= 2∏r (h + r) sq. units.
IV. CONE
Let radius of base = r and Height = h. Then,
1. Slant height, l = √h² + r ² units.
2. Volume = [1/3 ∏r²h] cubic units.
3. Total surface area = (∏rl + ∏r²) sq.units.
V. SPHERE
Let the radius of the sphere be r. Then,
1. Volume = [4/3 ∏r3] cubic units.
2. Surface area = (4∏r²) sq. units.
VI. HEMISPHERE
Let the radius of a hemisphere be r. Then,
1. Volume = [2/3 ∏r3] cubic units.
2. Curved surface area = (3∏r²) sq. units.
3. Total surface area = (3∏r²) sq. units.
Remember : 1 litre = 1000 cm³.

VOLUME AND SURFACE AREA -> SOLVED EXAMPLES

1. The diagonal of a cube is 6√3 cm. Find its volume and surface area.
  Sol. Let the edge of the cube be a.
∴ √3a = 6 √3 ⇒ a = 6.
So, Volume = a³ = (6 * 6 * 6) cm³ = 216 cn³
Surface area = 6a² = (6 * 6 * 6) cm² = 216 cm²
2. Find the volume and surface area of a cuboid 16m long, 14m broad and 7m high.
  Sol. Volume = (16 * 14 * 7) m³ = 1568 m³
Surface area = [2 (16 * 14 + 14 * 7 + 16 * 7)] cm² = (2 * 434) cm²
= 868 cm².
3. If the capacity of a cylindrical tank is 1848 m³ and the diameter of its base is 14m, then find the depth of the tank.
  Sol.Let the depth of the tank be h metres. then,
∏ * (7)² * h = 1848 ⇔ h = [1848 * 7/22 * 1/7*7] = 12 m.
4. How many iron rods, each of length 7m and diameter 2cm can be made out of 0.88 cubic metre of iron?
  Sol. Volume of 1 rod = [22/7 * 1/100 * 1/100 * 7] cu. m = 11/5000 cu.m.
Volume of iron = 0.88 cu. m.
Number of rods = [0.88 * 5000/11] = 400.
5. A cube of edge 15 cm is imersed completely in a rectangular vessel containing water. If the dimensions of the base of vessel are 20 cm * 15 cm, find the rise in waer level.
  Sol. Increase in volume = Volume of the cube = (15 * 15 * 15) cm³.
∴ Rise in water level = [Volume/Area] = [15 * 15 * 15/20 * 15] cm = 11.25 cm.
6. How many spherical bullets can be made out of a lead cylinder 28 cm high and with base radius 6 cm, each bullet being 1.5 cm ub duaneter?
  Sol. Volume of cylinder = (∏ * 6 * 6 * 28) cm³ = (36 * 28) ∏ cm³.
Bolume of each bullet = [4/3 ∏ * 3/4 * 3/4 * 3/4] cm³ = 9∏/16 cm³.
Number of bullets = Volume of cylinder / Volume of each bullet
= [(36 * 28)∏ * 16/9∏] = 1792.

VOLUME AND SURFACE AREA -> EXERCISE

31. A cylindrical cube of radius 12 cm contains water upto a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. The radius of the ball is :
 
  • A. 7 cm
  • B. 8.5 cm
  • C. 9 cm
  • D. 9.5 cm
Ans: C.
Sol.
Let the radius of the ball be r cm.
Volume of ball = Volume of water displaced by it.
∴ 4/3 ∏r³ = ∏ * 12 * 12 * 6.75
⇒ r³ = 9 * 9 * 9 ⇒ r = 9 cm.
 
32. The volume of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is :
 
  • A. 3/4∏
  • B. 4/3∏
  • C. 5∏
  • D. 5/9∏
Ans: B.
Sol.
Required volume = Volume of a sphere of radius 1 cm
=[4/3 ∏ * 1 * 1 * 1]cm³ = 4/3 ∏cm³
 
 
33. If the radius of a sphere is increased by 2 cm, then its surface area increases by 352 cm². The radius of the sphere before the increase was :
 
  • A. 4 cm
  • B. 5.5 cm
  • C. 6 cm
  • D. 8 cm
Ans: C.
Sol.
4∏(r + 2)² - 4∏r² = 352 ⇔ (r+2)² - r²
=[352 * 7/22 * 1/4] = 28.
⇔ (r+2+r) (r+2-r) = 28
⇔ 2r + 2 = 14 ⇔ r = [14/2 - 1] = 16 cm.