Kerala SSLC · Mathematics · Class 10

Mensuration: Important Questions with Answers

These are 10 important multiple-choice questions from the Kerala SSLC Mathematics chapter “Mensuration” (Kerala SCERT syllabus). Each one shows the correct answer and a short explanation of why it is right. Try answering before you read the answer. ExamSummary has 20 practice questions on this chapter in total.

1. Q1.A right circular cylinder has a radius of 7 cm and height of 10 cm. What is its Curved Surface Area? (Use π = 22/7)

   • A)440 cm²
   • B)880 cm²
   • C)1540 cm²
   • D)3080 cm²

   Answer: A) 440 cm²

   The formula for Curved Surface Area (CSA) is 2πrh. Substituting the values gives 2 × (22/7) × 7 × 10 = 440 cm².

2. Q2.If the radius of a cone is 5 cm and its vertical height is 12 cm, find the slant height.

   • A)13 cm
   • B)15 cm
   • C)17 cm
   • D)10 cm

   Answer: A) 13 cm

   Using the Pythagorean theorem, slant height l = √(r² + h²). Calculating √(5² + 12²) = √(25 + 144) = √169 = 13 cm.

3. Q3.What is the total surface area of a hemisphere of radius r?

   • A)2πr²
   • B)3πr²
   • C)4πr²
   • D)πr²

   Answer: B) 3πr²

   Total Surface Area includes the curved surface (2πr²) plus the flat circular base (πr²). Adding these together results in 3πr².

4. Q4.A spherical ball of radius 3 cm is melted and recast into small spherical balls of radius 1 cm. How many small balls can be made?

   • A)9
   • B)18
   • C)27
   • D)81

   Answer: C) 27

   Since volume is conserved, the number of small balls equals the ratio of volumes. (R/r)³ = (3/1)³ = 27. Thus, 27 small balls are produced.

5. Q5.A toy is in the shape of a cone mounted on a hemisphere of the same radius. Which surfaces contribute to the Total Surface Area?

   • A)Base of cone and base of hemisphere
   • B)Curved surface of both
   • C)Base of cone only
   • D)Base of hemisphere only

   Answer: B) Curved surface of both

   When the shapes are joined, the common base is internal and not exposed. Therefore, only the curved surface area of the hemisphere and the cone count towards the total surface area.

6. Q6.What is the ratio of the volume of a cone to the volume of a cylinder having the same base radius and height?

   • A)1:1
   • B)1:2
   • C)1:3
   • D)3:1

   Answer: C) 1:3

   Volume of cone is (1/3)πr²h while cylinder is πr²h. Comparing the two shows the cone occupies one-third the space of the cylinder, giving a ratio of 1:3.

7. Q7.A hollow cylindrical pipe has an outer radius of 5 cm, inner radius of 4 cm, and length 20 cm. Find the volume of the material used.

   • A)180π cm³
   • B)360π cm³
   • C)90π cm³
   • D)45π cm³

   Answer: A) 180π cm³

   Volume of material is π(R² - r²)h. Substituting values gives π(25 - 16) × 20 = π(9) × 20 = 180π cm³.

8. Q8.The cost of painting the curved surface of a cylindrical pillar is Rs. 2000 at the rate of Rs. 10 per sq meter. What is the curved surface area?

   • A)100 m²
   • B)200 m²
   • C)150 m²
   • D)50 m²

   Answer: B) 200 m²

   Area is calculated by dividing Total Cost by Rate per unit area. 2000 / 10 = 200 square meters.

9. Q9.Water flows through a pipe of cross-sectional area 10 cm² at a speed of 2 m/s. How much water flows in 1 minute?

   • A)1200 liters
   • B)120 liters
   • C)12 liters
   • D)12000 liters

   Answer: B) 120 liters

   Convert speed to cm/s (200 cm/s). Volume = Area × Speed × Time = 10 × 200 × 60 = 120,000 cm³. Since 1000 cm³ = 1 liter, the volume is 120 liters.

10. Q10.If the radius of a sphere is doubled, how does its volume change?

    • A)Doubles
    • B)Quadruples
    • C)Becomes 8 times
    • D)Becomes 4 times

    Answer: C) Becomes 8 times

    Volume of a sphere is proportional to the cube of its radius (V ∝ r³). If the radius doubles, the volume increases by a factor of 2³, which is 8.

All Kerala SSLC Mathematics chapters →