Mensuration
- A cylindrical can whose base is horizontal and is of internal radius 3.5 cm contains sufficient water so that when a solid sphere is placed inside, water just covers the sphere. The sphere fits in the can exactly. The depth of water in the can before the sphere was put, is
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Increase in water level = Volume of sphere Area of base of cylinder = 4 πr³ 3 πr² = 4 r = 4 × 3.5 = 14 cm 3 3 3 ∴ Required water level = 7 - 14 = 7 cm. 3 3 Correct Option: C
Increase in water level = Volume of sphere Area of base of cylinder = 4 πr³ 3 πr² = 4 r = 4 × 3.5 = 14 cm 3 3 3 ∴ Required water level = 7 - 14 = 7 cm. 3 3
- If A denotes the volume of a right circular cylinder of same height as its diameter and B is the volume of a sphere of same radius, then A/B is :
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Volume of cylinder = πr²h
⇒ A = πr² (2r) = 2πr³ [∵ h = 2r]Volume of sphere = 4 πr³ 3 ⇒ A = 2πr³ = 6 = 3 B 4 πr³ 4 2 3 Correct Option: B
Volume of cylinder = πr²h
⇒ A = πr² (2r) = 2πr³ [∵ h = 2r]Volume of sphere = 4 πr³ 3 ⇒ A = 2πr³ = 6 = 3 B 4 πr³ 4 2 3
- The base of a right circular cone has the same radius a as that of a sphere. Both the sphere and the cone have the same volume. Height of the cone is
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1 πa²h = 4 πa³ 3 3
⇒ h = 4aCorrect Option: B
1 πa²h = 4 πa³ 3 3
⇒ h = 4a
- The total surface area of a sphere is 8π square unit. The volume of the sphere is
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Surface area of sphere = 4πr²
⇒ 4πr⊃2; = 8π
⇒ r² = 2 ⇒ r ⇒ √2 units∴ Volume of sphere = 4 πr³ = 4 π × (√2)³ 8 2 = 8√2 π cubic units 3 Correct Option: A
Surface area of sphere = 4πr²
⇒ 4πr⊃2; = 8π
⇒ r² = 2 ⇒ r ⇒ √2 units∴ Volume of sphere = 4 πr³ = 4 π × (√2)³ 8 2 = 8√2 π cubic units 3
- A hollow spherical metallic ball has an external diameter 6 cm and is 1 2 cm thick. The volume of the ball (in cm³) is (Take π = 22/7)
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External radius,
R = 6 = 3 cm 2
Internal radius, r= 3 - 1 = 5 cm. 2 2 ∴ Volume of hollow sphere (material) = 4 π (R³ - r³) 3 = 4 π 
3³ - 
5 
² 
cm² 3 2 = 4 × 22 27 - 125 cm³ 3 7 8 = 4 × 22 216 - 125 cm³ 3 7 8 = 4 × 22 × 91 3 7 8 = 143 = 47 2 cm³ 3 3 Correct Option: C
External radius,
R = 6 = 3 cm 2
Internal radius, r= 3 - 1 = 5 cm. 2 2 ∴ Volume of hollow sphere (material) = 4 π (R³ - r³) 3 = 4 π 3³ - 5 ² cm² 3 2 = 4 × 22 27 - 125 cm³ 3 7 8 = 4 × 22 216 - 125 cm³ 3 7 8 = 4 × 22 × 91 3 7 8 = 143 = 47 2 cm³ 3 3
