Mensuration
- The volume of metallic cylindrical (hollow) pipe of uniform thickness is 748 c.c. Its length is 14 cm and its external radius is 9 cm. The thickness of the pipe is
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Let the internal radius of pipe be r cm.
External radius = R cm = 9 cm.
∴ Volume of the material of pipe = π(R² - r²)h⇒ 22 (9² - r²) × 14 = 784 7 ⇒(81 - r²) = 748 × 7 = 17 22 × 14
⇒ r² = 81 – 17 = 64
⇒ r = 64 = 8 cm
∴ Thickness of pipe = 9 – 8 = 1 cm.Correct Option: C
Let the internal radius of pipe be r cm.
External radius = R cm = 9 cm.
∴ Volume of the material of pipe = π(R² - r²)h⇒ 22 (9² - r²) × 14 = 784 7 ⇒(81 - r²) = 748 × 7 = 17 22 × 14
⇒ r² = 81 – 17 = 64
⇒ r = 64 = 8 cm
∴ Thickness of pipe = 9 – 8 = 1 cm.
- The diagonal of a cube is 192 cm. Its volume (in cm³) will be
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Diagonal of cube = √3 × edge
√3 × edge = √192
⇒ √3x = √64 × 3
Where x = edge of cube
⇒ x = 8 cm
∴ Volume of cube = (8)³ = 512 cu.cm.Correct Option: C
Diagonal of cube = √3 × edge
√3 × edge = √192
⇒ √3x = √64 × 3
Where x = edge of cube
⇒ x = 8 cm
∴ Volume of cube = (8)³ = 512 cu.cm.
- The radius of the base of a right circular cone is 6 cm and its slant height is 10 cm. Then its volume is Use π = 22/7)
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Slant height of cone = l cm. = 10 cm.
Radius of base = r = 6 cm.
∴ h = √l² - r²
= √10² - 6² = √(10 + 6)(10 - 6)
= √16 × 4 = 4 × 2 = 8 cm.∴ Volume of cone = 1 πr²h 3 = 
1 × 22 × 6 × 6 × 8 
cubic unit 3 7 = 6336 = 301.71 cu. cm. 21 Correct Option: A
Slant height of cone = l cm. = 10 cm.
Radius of base = r = 6 cm.
∴ h = √l² - r²
= √10² - 6² = √(10 + 6)(10 - 6)
= √16 × 4 = 4 × 2 = 8 cm.∴ Volume of cone = 1 πr²h 3 = 1 × 22 × 6 × 6 × 8 cubic unit 3 7 = 6336 = 301.71 cu. cm. 21
- Three solid spheres have their radii r1, r2 and r3. The spheres are melted to form a solid sphere of bigger radius. Then the radius of the new sphere is :
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In both cases, volume remains same.
If the radius of new sphere be R units, then4 πR³ = 4 πr1³ + 4 πr2³ + 4 πr3³ 3 3 3 3
⇒ R³ = r1³ + r2³ + r3³
∴ R = (r1³ + r2³ + r3³)¹/3 unitsCorrect Option: C
In both cases, volume remains same.
If the radius of new sphere be R units, then4 πR³ = 4 πr1³ + 4 πr2³ + 4 πr3³ 3 3 3 3
⇒ R³ = r1³ + r2³ + r3³
∴ R = (r1³ + r2³ + r3³)¹/3 units
- The ratio of the weights of two spheres is 8 : 27 and the ratio of weights per 1 cc of materials of two is 8 : 1. The ratio of the radii of the spheres is
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Mass = Volume × density
∴ Mass of sphere A Mass of sphere B = 4 πR³ × d1 3 4 πr³ × d2 3 ⇒ 8 = R³ × 8 27 r³ ⇒ R³ = 1 r³ 27 ⇒ R = ³√ 1 = 1 = 1 : 3 r 27 3 Correct Option: B
Mass = Volume × density
∴ Mass of sphere A Mass of sphere B = 4 πR³ × d1 3 4 πr³ × d2 3 ⇒ 8 = R³ × 8 27 r³ ⇒ R³ = 1 r³ 27 ⇒ R = ³√ 1 = 1 = 1 : 3 r 27 3
