Surface Area and Volume Questions are available with answers here. Class 10 students can practise the questions based on surface area and volumes to prepare for the exams. These surface area and volume problems are prepared by our subject experts, as per the NCERT curriculum and latest CBSE syllabus (2022-2023). Learn more: Surface Areas and Volume.
Surface Area and Volume Formulas:
- Total surface area of a cuboid = 2[lb + bh + lh]
- Total surface area of a cube = 6(side)2
- Lateral surface area of a cuboid = Area of walls of a room = 2(l + b) × h
- Lateral surface area of a cube = 4a2
- Curved surface area of cylinder = 2πrh
- Total surface area of a cylinder = 2πr(r + h)
- Curved surface area of a cone = πrl
- Total surface area of a cone = πr(r + l)
- Surface area of a sphere = 4πr2
- Curved surface area of a hemisphere = 2πr2
- Total surface area of a hemisphere = 3πr2
- Volume of a cuboid = l × b × h
- Volume of a cube = (side)3
- Volume of a cylinder = πr2h
- Volume of a cone = ⅓ πr2h
- Volume of a sphere = 4/3 πr3
- Volume of a hemisphere = ⅔ πr3
Surface Area and Volume Questions and Solutions
Q.1: If the radius of a sphere is 3r, what is its volume?
Solution: Given,
Radius of sphere = 3r
Volume of a sphere = 4/3 πr3
= 4/3 π(3r)3 (given)
= 4/3 π27r3
= 36πr3 cu.units.
Q.2: What is the total surface area of a cuboid whose length = 5 cm, width = 2 cm and height = 3 cm?
Solution: Given, the dimensions of cuboid are:
Length, l = 5 cm
Width, w = 2 cm
Height, h = 3 cm
Surface area of cuboid = 2(lb + bh + lh)
= 2 (5.2 + 2.3 + 3.5)
= 2 (10 + 6 + 15)
= 62 sq.cm.
Q.3: If the radius of a sphere is doubled, find the ratio of their volumes.
Solution: Let the radius of a sphere is r.
Volume of a sphere, V1 = 4/3 πr3
New radius = 2r
New volume, V2 = 4/3 π(2r)3 = 4/3 π.23r3
Ratio of volumes = V1/V2
= (4/3 πr3)/(4/3 π.23r3)
= ⅛
Hence, 1:8 is the required volume ratio.
Q.4: Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid.
Solution: Given,
When two cubes are joined end to end, then;
Length of the cuboid = 6 + 6 = 12 cm
Breadth of the cuboid = 6 cm
Height of the cuboid = 6 cm
Total surface area of the cuboid = 2 (lb + bh + hi)
= 2(12 x 6 + 6 × 6 + 6 ×12)
= 2(72 + 36 + 72) = 2(180) = 360 cm2
Q.5: Find the area of the sheet required to make a closed cylindrical vessel of height 1 m and diameter 140 cm.
Solution: Given,
Height of cylindrical vessel = 1m
Diameter = 140 cm
Radius = 140/2 = 70 cm = 0.7 m
Total surface area of the closed cylindrical tank = 2 π r (r + h)
= 2 × 22/7 × 0.7 m × (0.7 m + 1 m)
= 4.4 m × 1.7 m
= 7.48 m²
Hence, 7.48 m² of the sheet is required.
Q.6: A 5m wide cloth is used to make a conical tent of base diameter 14m and height 24m. Find the cost of cloth used at the rate of Rs. 25 per metre.
Solution: Given,
Height of tent, h = 24 m
Diameter = 14 m, radius, r = 7m
Slant height, l = √(h2 + r2) = √(242 + 72) = √625 = 25 m
Curved surface area of cone = πrl
= 22/7 x 7 x 25
= 550 m2
Length of cloth = Area/width of cloth
= 550/5
= 110 m
Cost of cloth per meter = Rs.25
Therefore, total cost of cloth used = 25 x 110 = Rs. 2750/-
Q.7: 30 circular plates, each of radius 14 cm and thickness 3 cm are placed one above the other to form a cylindrical solid. Find the total surface area.
Solution: Given, radius 14 cm, thickness =3 cm, number of plates =30
Therefore, the height of cylinder will be 3×30=90 cm
We know, total surface area of cylinder =2πr(r+h) cm2
Therefore, total surface area =2π(14)(14+(90))=9152 cm2
Q.8: The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume.
Solution:
Inner radius of hollow sphere (r) = 10 cm
Outer radius of hollow sphere (R) = 12 cm
Volume of sphere = 4/3 π (R3 – r3)
= 4/3 x 22/7 (123 – 103)
= 4/3 x 22/7 (1728 – 1000)
= 4/3 x 22/7 x 728
= 3050.67 cm3
Q.9: The radius of the base of a right circular cylinder is 3 cm and height is 7 cm. Find the curved surface area.
Solution: Given,
Radius of cylinder, r=3 cm
Height of cylinder, h=7 cm
Curved Surface Area of a Cylinder =2πrh
CSA = 2 x 22/7 x 3 x 7 = 132 sq.cm.
Q.10: A sphere and a right circular cylinder of the same radius have equal volumes. By what percentage does the diameter of the cylinder exceed its height?
Solution: Let the radius of sphere be r
Height of right circular cylinder be h
According to the question;
4/3 πr3 = πr2h
4/3 r = h
⇒ r = ¾ h
Therefore,
Diameter of cylinder = 2 x r = 2(¾ h) = 3/2 h
Increase in height = (3/2 – 1)h = ½ h
Therefore, the required percentage is:
(½ h)/h x 100
= 50%
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