Mensuration
- A metallic hemisphere is melted and recast in the shape of a cone with the same base radius (R) as that of the hemisphere. If H is the height of the cone, then :
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When we change shape of a solid figure, volume remains constant
∴ Volume of hemisphere = Volume of cone⇒ 2 πR³= 1 πR³H 3 3
∴ 2R = HCorrect Option: A
When we change shape of a solid figure, volume remains constant
∴ Volume of hemisphere = Volume of cone⇒ 2 πR³= 1 πR³H 3 3
∴ 2R = H
- Three solid metallic spheres of diameter 6 cm, 8 cm and 10 cm are melted and recast into a new solid sphere. The diameter of the new sphere is :
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According to the question, three solid metallic spheres are melted and recast into a new solid sphere. It means that the volume of new solid sphere will be equal to the sum of volume of three solid spheres.
∴ Volume of new solid sphere = 4 π 
6 
³ + 4 π 8 ³ + 4 π 10 ³ 3 2 3 2 3 2 4 πr³ = 4 π [(3)³ + (4)³ + (5)³] 3 3
= 27 + 64 + 125
⇒ r³ = 216
⇒ r³ = (6)³
⇒ r = 6 cm
∴ Diameter of the new sphere = 2 × 6 = 12 cmCorrect Option: D
According to the question, three solid metallic spheres are melted and recast into a new solid sphere. It means that the volume of new solid sphere will be equal to the sum of volume of three solid spheres.
∴ Volume of new solid sphere = 4 π 6 ³ + 4 π 8 ³ + 4 π 10 ³ 3 2 3 2 3 2 4 πr³ = 4 π [(3)³ + (4)³ + (5)³] 3 3
= 27 + 64 + 125
⇒ r³ = 216
⇒ r³ = (6)³
⇒ r = 6 cm
∴ Diameter of the new sphere = 2 × 6 = 12 cm
- Three solid metallic balls of radii 3 cm, 4 cm and 5 cm are melted and moulded into a single solid ball. The radius of the new ball is :
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Let the radius of new ball = R cm
then4 πR³ = 4 π (3³ + 4³ + 5³) 3 3
R³ = 27 + 64 + 125 = 216
⇒ R = ³√6 × 6 × 6 = 6 cmCorrect Option: D
Let the radius of new ball = R cm
then4 πR³ = 4 π (3³ + 4³ + 5³) 3 3
R³ = 27 + 64 + 125 = 216
⇒ R = ³√6 × 6 × 6 = 6 cm
- Three solid spheres of a metal whose radii are 1 cm, 6 cm and 8 cm are melted to form an other solid sphere. The radius of this new sphere is
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4 πr³ 3 = 4 π (1)³ + 4 π (6)³ + 4 π (8)³ 3 3 3 = 4 πr³ = 4 (1 + 216 + 512) 3 3
⇒ r³ = 729 ⇒ r = ³√729
⇒ r = 9 cmCorrect Option: D
4 πr³ 3 = 4 π (1)³ + 4 π (6)³ + 4 π (8)³ 3 3 3 = 4 πr³ = 4 (1 + 216 + 512) 3 3
⇒ r³ = 729 ⇒ r = ³√729
⇒ r = 9 cm
- A sphere of radius 2 cm is put into water contained in a cylinder of base-radius 4 cm. If the sphere is completely immersed in the water, the water level in the cylinder rises by
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According to question Volume of sphere = Volume of displaced water
⇒ 4 π × 2 × 2 × 2 = π × 4 × 4 × h 3 ∴ h = 2 cm 3 Correct Option: C
According to question Volume of sphere = Volume of displaced water
⇒ 4 π × 2 × 2 × 2 = π × 4 × 4 × h 3 ∴ h = 2 cm 3
