Mensuration
- The whole surface of a cube is 150 sq.cm. Then the volume of the cube is
-
View Hint View Answer Discuss in Forum
Edge of cube = a cm (let)
∴ Its total surface area = 6a²
∴ 6a² = 150⇒ a² = 150 = 25 6
⇒ a = 25 = 5 cm
∴ Volume of cube = a³ = (5 × 5 × 5) cu.cm = 125 cu.cmCorrect Option: A
Edge of cube = a cm (let)
∴ Its total surface area = 6a²
∴ 6a² = 150⇒ a² = 150 = 25 6
⇒ a = 25 = 5 cm
∴ Volume of cube = a³ = (5 × 5 × 5) cu.cm = 125 cu.cm
- A solid metallic spherical ball of diameter 6 cm is melted and recast into a cone with diameter of the base as 12 cm. The height of the cone is
-
View Hint View Answer Discuss in Forum
Volume of metallic sphere = 4 πr³ 3 Volume of metallic sphere = 4 π × 3 × 3 × 3 3
= 36π cu.cm.
∴ Volume of cone = 36π cu.cm.⇒ 1 πR³h = 36π 3
⇒ R²h = 108
⇒ 6 × 6 × h = 108&RaRR; h = 108 = 3 CM. 6 × 6 Correct Option: B
Volume of metallic sphere = 4 πr³ 3 Volume of metallic sphere = 4 π × 3 × 3 × 3 3
= 36π cu.cm.
∴ Volume of cone = 36π cu.cm.⇒ 1 πR³h = 36π 3
⇒ R²h = 108
⇒ 6 × 6 × h = 108&RaRR; h = 108 = 3 CM. 6 × 6
- A hemispherical bowl of internal radius 15 cm contains a liquid. The liquid is to be filled into cylindrical shaped bottles of diameter 5 cm and height 6 cm. The number of bottles required to empty the bowl is
-
View Hint View Answer Discuss in Forum
Volume of hemi-spherical bowl = 2 πr³ 3 = 
2 π × 15 × 15 × 15 
cu.cm. 3
Volume of a bottle = πR²h= π × 5 × 5 × 6 cu.cm. 2 2 ∴ Number of bottles = 2 × π × 15 × 15 × 15 = 60 3 × π × 5 × 5 × 6 3 3 Correct Option: D
Volume of hemi-spherical bowl = 2 πr³ 3 = 2 π × 15 × 15 × 15 cu.cm. 3
Volume of a bottle = πR²h= π × 5 × 5 × 6 cu.cm. 2 2 ∴ Number of bottles = 2 × π × 15 × 15 × 15 = 60 3 × π × 5 × 5 × 6 3 3
- If V1, V2 and V3 be the volumes of a right circular cone, a sphere and a right circular cylinder having the same radius and same height, then
-
View Hint View Answer Discuss in Forum
Volume of cone = V1 = 1 πr²h 3 Volume of cone = V1 = 1 πr²h 3 Volume of sphere = V2 = 4 πr³ 3
Volume of cylinder = V3 = πr²h = πr³∴ V1 : V2 : V3 = 1 : 1 : 1 = 1 : 4 : 3 3 3 ∴ V1 = V2 = V3 4 3 Correct Option: E
Volume of cone = V1 = 1 πr²h 3 Volume of cone = V1 = 1 πr²h 3 Volume of sphere = V2 = 4 πr³ 3
Volume of cylinder = V3 = πr²h = πr³∴ V1 : V2 : V3 = 1 : 1 : 1 = 1 : 4 : 3 3 3 ∴ V1 = V2 = V3 4 3
- If the radius of a shphere be doubled, then the percentage increase in volume is
-
View Hint View Answer Discuss in Forum
Volume of sphere = 4 πr³ cu.units 3
Case II,
R = 2r units∴Volume of sphere = 4 π(2r)³ 3 ∴Volume of sphere = 32 πr³ cu.units 3 Difference = 32 πr³ - 4 πr³ = 28 πr³ 3 3 3 ∴ Percentage increase = 28 πr³ × 100 3 4 πr³ 3
= 700%
OR
Single equivalent per cent increase for increases of 100% and 100%= 100 + 100 + 100 × 100 % 100
= 300%
Single equivalent per cent increase for increases of 300% and 100%= 300 + 100 + 300 × 100 % 100
= 700%Correct Option: B
Volume of sphere = 4 πr³ cu.units 3
Case II,
R = 2r units∴Volume of sphere = 4 π(2r)³ 3 ∴Volume of sphere = 32 πr³ cu.units 3 Difference = 32 πr³ - 4 πr³ = 28 πr³ 3 3 3 ∴ Percentage increase = 28 πr³ × 100 3 4 πr³ 3
= 700%
OR
Single equivalent per cent increase for increases of 100% and 100%= 100 + 100 + 100 × 100 % 100
= 300%
Single equivalent per cent increase for increases of 300% and 100%= 300 + 100 + 300 × 100 % 100
= 700%
