Mensuration
- If the surface area of a sphere is 346.5 cm2, then its radius (taking π = 22/7) is
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Surface area of sphere = 4πr²
∴ 4 × 22 × r² = 346.5 7
⇒ 4 × 22 × r² = 346.5 × 7⇒ r² = 346.5 × 7 = 27.5625 4 × 22
∴ r = √27.5625 = 5.25 cmCorrect Option: C
Surface area of sphere = 4πr²
∴ 4 × 22 × r² = 346.5 7
⇒ 4 × 22 × r² = 346.5 × 7⇒ r² = 346.5 × 7 = 27.5625 4 × 22
∴ r = √27.5625 = 5.25 cm
- A sphere and a hemisphere have the same radius. Then the ratio of their respective total surface areas is
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Radius of sphere = r units
∴ Surface area of sphere Surface area of hemisphere = 4πr² = 4 = 4 : 3 3πr² 3 Correct Option: C
Radius of sphere = r units
∴ Surface area of sphere Surface area of hemisphere = 4πr² = 4 = 4 : 3 3πr² 3
- Area of the floor of a cubical room is 48 sq.m. The length of the longest rod that can be kept in that room is
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Length of room = √48 = 4√3 metre
∴ Diagonal = √3 × (4√3)²
= √3 × 16 × 3 = 12 metreCorrect Option: B
Length of room = √48 = 4√3 metre
∴ Diagonal = √3 × (4√3)²
= √3 × 16 × 3 = 12 metre
- A toy is in the form of a cone mounted on a hemisphere. The radius of the hemisphere and that of the cone is 3 cm and height of the cone is 4 cm. The total surface area of the toy (taking π = 22/7) is
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Total surface area of the toy = 2πr² + πrl
= πr (2r + √r² + h²)= 22 × 3 (2 × 3 + √3² + 4²) 7 = 22 × 3 (6 + 5) 7 = 22 × 3 × 11 = 103.71 sq.cm. 7 Correct Option: B
Total surface area of the toy = 2πr² + πrl
= πr (2r + √r² + h²)= 22 × 3 (2 × 3 + √3² + 4²) 7 = 22 × 3 (6 + 5) 7 = 22 × 3 × 11 = 103.71 sq.cm. 7
- The length of each edge of a regular tetrahedron is 12 cm. The area (in sq. cm) of the total surface of the tetrahedron is
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Total surface of the tetrahedron = 4 × √3 × 12² 4
= 144√3 sq.cm.Correct Option: D
Total surface of the tetrahedron = 4 × √3 × 12² 4
= 144√3 sq.cm.
