Mensuration
- 12 spheres of the same size are made by melting a solid cylinder of 16 cm diameter and 2 cm height. The diameter of each sphere is :
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Volume of cylinder = πr²h = π × (8)² × 2 = 128p cm³ Let the radius of each sphere be r cm.
∴ 12 × 4 πr³ = 128π 3
⇒ 16πr³ = 128π⇒ r³ = 128π 16π
⇒ r = ³√8 = 2 cm
∴ Diameter = 2 × 2 = 4cmCorrect Option: B
Volume of cylinder = πr²h = π × (8)² × 2 = 128p cm³ Let the radius of each sphere be r cm.
∴ 12 × 4 πr³ = 128π 3
⇒ 16πr³ = 128π⇒ r³ = 128π 16π
⇒ r = ³√8 = 2 cm
∴ Diameter = 2 × 2 = 4cm
- By melting a solid lead sphere of diameter 12 cm, three small spheres are made whose diameters are in the ratio 3 : 4 : 5. The radius (in cm) of the smallest sphere is
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Volume of original sphere = 4 π(6)³ = 128π cm³ 3
Let the radii of small spheres be 3x, 4x and 5x cm respectively∴ 4 π[(3x)³ + (4x)³ + (5x)³] = 288 π 3 ⇒ 4 π[27x³ + 64x³ + 125³] = 288 π 3 ⇒ 4 π× 216x³ = 288 π 3 ⇒ x³ = 288π × 3 = 1 4π × 216
⇒ x = 1
∴ Required radius = 3 × 1 = 3cm.Correct Option: A
Volume of original sphere = 4 π(6)³ = 128π cm³ 3
Let the radii of small spheres be 3x, 4x and 5x cm respectively∴ 4 π[(3x)³ + (4x)³ + (5x)³] = 288 π 3 ⇒ 4 π[27x³ + 64x³ + 125³] = 288 π 3 ⇒ 4 π× 216x³ = 288 π 3 ⇒ x³ = 288π × 3 = 1 4π × 216
⇒ x = 1
∴ Required radius = 3 × 1 = 3cm.
- A well 20 m in diameter is dug 14 m deep and the earth taken out is spread all around it to a width of 5 m to form an embankment. The height of the embankment is :
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Radius of the base of well = 20 = 10 m 2 Volume of the earth taken out = πr²h = 22 × 10² × 14 m³ 7
Let the height of embankment be x metres. Then,
Volume = π (R² – r²) × x, where R = 15 m, r = 10mVolume = 22 (15² – 10²) × x 7 = 22 × 25 × 5 × x 7
Clearly,= 22 × 25 × 5 × x 7 = 22 10² × 14 7 ⇒ x = 100 × 14 = 11.2 m 25 × 5 Correct Option: C
Radius of the base of well = 20 = 10 m 2 Volume of the earth taken out = πr²h = 22 × 10² × 14 m³ 7
Let the height of embankment be x metres. Then,
Volume = π (R² – r²) × x, where R = 15 m, r = 10mVolume = 22 (15² – 10²) × x 7 = 22 × 25 × 5 × x 7
Clearly,= 22 × 25 × 5 × x 7 = 22 10² × 14 7 ⇒ x = 100 × 14 = 11.2 m 25 × 5
- A copper wire of Iength 36 m and diameter 2 mm is melted to form a sphere. The radius of the sphere (in cm) is
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Volume of the wire = πr²h = π × 0.1 × 0.1 × 3600 cm³
= 36π cm³Volume of solid sphere = 4 πR³ = 36π 3 ⇒ R³ = 36 × 3 = 27 4
∴R = ³√27 = 3 cmCorrect Option: B
Volume of the wire = πr²h = π × 0.1 × 0.1 × 3600 cm³
= 36π cm³Volume of solid sphere = 4 πR³ = 36π 3 ⇒ R³ = 36 × 3 = 27 4
∴R = ³√27 = 3 cm
- A child reshapes a cone made up of clay of height 24cm and radius 6cm into a sphere. The radius (in cm) of the sphere is
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Volume of the cone = 1 πr²h 3 = π × 6 × 6 × 24 cm³ 3
= Volume of the sphere If the radius of the sphere be r cm, then4 πr³ = π × 6 × 6 × 24 3 3
⇒r³ = 6 × 6 × 6
∴ r = √6 × 6 × 6 = 6 cm.Correct Option: A
Volume of the cone = 1 πr²h 3 = π × 6 × 6 × 24 cm³ 3
= Volume of the sphere If the radius of the sphere be r cm, then4 πr³ = π × 6 × 6 × 24 3 3
⇒r³ = 6 × 6 × 6
∴ r = √6 × 6 × 6 = 6 cm.
