Unit Test (Solutions): Surface Area & Volumes

Time: 1 hour
M.M. 30 
Attempt all questions. 
Question numbers 1 to 5 carry 1 mark each. 
Question numbers 6 to 8 carry 2 marks each. 
Question numbers 9 to 11 carry 3 marks each. 
Question numbers 12 & 13 carry 5 marks each.

Q1. What is the curved surface area of a right circular cone? (1 Mark)
(a) πr²
(b) πrl
(c) πr(l + r)
(d) 2πrl

Ans: (b)
Sol: Curved Surface Area (CSA) of a right circular cone = πrl

Q2. What is the total surface area of a hemisphere? (1 Mark)
(a) 2πr²
(b) 3πr²
(c) 4πr²
(d) πr²

Ans: (b)
Sol: Total Surface Area of a hemisphere = Curved Surface Area + Base Area
= 2πr² + πr² = 3πr²

Q3. The volume of a cone is one-third the volume of a cylinder with the same: (1 Mark)
(a) Height only
(b) Radius only
(c) Height and radius
(d) Slant height

Ans: (c)
Sol: Volume of cone = (1/3)πr²h
Volume of cylinder = πr²h
So, cone is one-third of a cylinder when both radius and height are the same.

Q4. The surface area of a sphere of radius r is: (1 Mark)
(a) 2πr²
(b) 3πr²
(c) 4πr²
(d) πr²

Ans: (c)
Sol: Surface Area of a sphere = 4πr²

Q5. The slant height of a cone is denoted by: (1 Mark)
(a) h
(b) r
(c) l 
(d) d

Ans: (c)
Sol: In a right circular cone, slant height is denoted by l.

Q6. Find the curved surface area of a cone with base diameter 7 cm and slant height 8 cm. (Use π = 22/7) (2 Marks)

Ans: Given:
Diameter = 7 cm ⇒ Radius (r) = 7/2 = 3.5 cm
Slant height (l) = 8 cm
CSA of cone = πrl
= (22/7) × 3.5 × 8
= 88 cm²

Q7. Calculate the surface area of a sphere with a diameter of 14 cm. (Use π = 22/7) (2 Marks)

Ans: Diameter = 14 cm ⇒ Radius (r) = 14/2 = 7 cm
Surface Area = 4πr²
= 4 × (22/7) × 7 × 7
= 616 cm²

Q8. Find the curved surface area of a hemisphere with radius 10.5 cm. (Use π = 22/7) (2 Marks)

Ans: 

Radius (r) = 10.5 cm
CSA of hemisphere = 2πr²
= 2 × (22/7) × 10.5 × 10.5
= 693 cm²

Q9. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. (3 Marks)

Ans: Given,
Diameter of the cone = 24 m 
Radius of the cone (r) = 24/2 = 12 m 
Slant height of the cone (l) = 21 m 
Total surface area of a cone = πr(l + r) 
= (22/7) × 12 × (21 + 12) 
= (22/7) × 12 × 33 
= 1244.57 m²

Q10. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 sq. m. (3 Marks)

Ans: Given,
Slant height of a cone (l) = 25 m  
Diameter of the base of cone = 2r = 14 m  
∴ Radius = r = 7 m  
Curved Surface Area = πrl 
= (22/7) x 7 x 25 
= 22 × 25  
= 550 sq.m  
Also, given that the cost of white-washing 100 sq.m = Rs. 210 
Hence, the total cost of white-washing for 550 sq.m = (Rs. 210 × 550)/100 = Rs. 1155

Q11. The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding. (3 Marks)

Ans: Given,
Diameter of the sphere = 7 m
Radius (r) = 7/2 = 3.5 m 
Now, the riding space available for the motorcyclist = Surface area of the sphere
= 4πr²
= 4 × (22/7) × 3.5 × 3.5
= 154 m² 

Q12. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite. (5 Marks)

Ans: Given,
Diameter of the pencil = 7  mm
Radius of the pencil (R) = 7/2 mm
Diameter of the graphite cylinder = 1 mm
Radius of the graphite (r) = 1/2 mm
Height (h) = 14 cm = 140 mm (since 1 cm = 10 mm)
Volume of a cylinder = πr²h
Volume of graphite cylinder = πr2h 
= (22/7) × (1/2) × (1/2) × 140
= 110 mm³
Volume of pencil = πR²h
= (22/7) × (7/2) × (7/2) × 140
= 490 × 11 
= 5390 mm²
Volume of wood = Volume of pencil - Volume of graphite
= 5390- 110 = 5280 mm³
= 5280/1000 (since 1 mm³ = 1/1000 cm³)
= 5.28 cm³

Q13. A spherical ball is divided into two equal halves. Given that the curved surface area of each half is 56.57 cm, what will be the volume of the spherical ball? (5 Marks)

Ans: ​​​​Given,
Curved surface area of of half of the spherical ball = 56.57 cm²
(1/2) 4πr² = 56.57
2 × 3.14 × r² = 56.57
r² = 56.57/6.28
r² = 9 (approx)
r = 3 cm
Now,
Volume of spherical ball = (4/3)πr³
= (4/3) × 3.14 × 3 × 3 × 3
= 113.04 cm³​​​