Mensuration — Important Questions
Tamil Nadu Board · Class 10 · Mathematics
44 important questions from Mensuration for Tamil Nadu Board Class 10 Mathematics, with answers. Includes multiple choice questions.
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Important Questions from Mensuration
The curved surface area of a hemisphere of radius 14 cm is:
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1232 cm²
Step 1: Given r = 14 cm. We need CSA of hemisphere. Step 2: CSA of hemisphere = 2πr² Step 3: CSA = 2 × (22/7) × 14 × 14 Step 4: CSA = 2 × (22/7) × 196 = 2 × 22 × 28 = 1232 cm² Common mistake: Using 4πr² (full sphere) instead of 2πr² for hemisphere CSA.
A metallic sphere of radius 6 cm is melted and recast into small spheres of radius 1 cm each. How many small spheres are formed?
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216
Step 1: Large sphere radius R = 6 cm, small sphere radius r = 1 cm. Step 2: Number of small spheres = Volume of large sphere / Volume of small sphere Step 3: n = [(4/3)πR³] / [(4/3)πr³] = R³/r³ Step 4: n = 6³/1³ = 216/1 = 216 Common mistake: Using ratios of radii directly (6/1 = 6) instead of cubing the ratio.
The volume of a frustum of a cone with height 12 cm and radii 8 cm and 4 cm is:
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1408 cm³
Step 1: Given h = 12 cm, R = 8 cm, r = 4 cm. Step 2: Volume of frustum = (πh/3)(R² + r² + Rr) Step 3: = (22/7 × 12/3)(64 + 16 + 32) Step 4: = (22/7 × 4)(112) = (22 × 448)/7 = 9856/7 = 1408 cm³ Common mistake: Using the cone volume formula instead of the frustum formula.
A cylinder and a cone have the same base radius and height. The ratio of the volume of the cylinder to the volume of the cone is:
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3:1
Step 1: Let radius = r and height = h for both solids. Step 2: Volume of cylinder = πr²h Step 3: Volume of cone = (1/3)πr²h Step 4: Ratio = πr²h : (1/3)πr²h = 1 : 1/3 = 3 : 1 This is a fundamental relationship: a cylinder holds exactly 3 times the volume of a cone with the same base and height.
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