Mensuration
- A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5cm × 2 cm. [Assuming π = 22/7] The percentage wood wasted in the process is :
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Volume of cone = 1 πr²h = 1 × 22 × 1 × 7 = 22 cu.cm. 3 3 7 3
Volume of cubical block = 10 × 5 × 2 cm³. = 100 cm³.Wastage of wood = 100 - 22 cm³ = 300 - 22 = 278 cm³ 3 3 3
´∴ % Wastage = 278 × 100 3 100 = 278 = 92 2 % 3 3 Correct Option: A
Volume of cone = 1 πr²h = 1 × 22 × 1 × 7 = 22 cu.cm. 3 3 7 3
Volume of cubical block = 10 × 5 × 2 cm³. = 100 cm³.Wastage of wood = 100 - 22 cm³ = 300 - 22 = 278 cm³ 3 3 3
´∴ % Wastage = 278 × 100 3 100 = 278 = 92 2 % 3 3
- If the radius of a cylinder is decreased by 50% and the height is increased by 50% to form a new cylinder, the volume will be decreased by
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Required percentage decrease = 100 - 50 × 50 × 150 = 100 - 37.5 = 62.5 % 100 × 100 Correct Option: C
Required percentage decrease = 100 - 50 × 50 × 150 = 100 - 37.5 = 62.5 % 100 × 100
- Each of the height and base-radius of a cone is increased by 100%. The percentage increase in the volume of the cone is
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Required per cent
∴ 200 × 200 × 200 - 100 = 800 - 100 = 700 % 100 × 100 Correct Option: A
Required per cent
∴ 200 × 200 × 200 - 100 = 800 - 100 = 700 % 100 × 100
- If both the radius and height of a right circular cone are increased by 20%, its volume will be increased by
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Let height and radius both of a cylinder change by x%, then volume changes by
3x + 3x² + x³ % = 3 × 20 + 3 × 20 × 20 + 20 × 20 × 20 % 100 100² 100 10000
= (60 + 12 + 0.8)% = 72.8%Correct Option: D
Let height and radius both of a cylinder change by x%, then volume changes by
3x + 3x² + x³ % = 3 × 20 + 3 × 20 × 20 + 20 × 20 × 20 % 100 100² 100 10000
= (60 + 12 + 0.8)% = 72.8%
- If the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, the volume of the
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Volume of original cone = 1 πr²h 3
Now, r1 = radius of new cone = r/2
h1 = height of new cone = 3h∴ V1 = 1 πr1²h1 = 1 π r² × 3h = 1 πr²h × 3 = 3 V 3 3 4 3 4 4 ∴ Decrease % = V - 3 V × 100 4 V
= 25%Correct Option: D
Volume of original cone = 1 πr²h 3
Now, r1 = radius of new cone = r/2
h1 = height of new cone = 3h∴ V1 = 1 πr1²h1 = 1 π r² × 3h = 1 πr²h × 3 = 3 V 3 3 4 3 4 4 ∴ Decrease % = V - 3 V × 100 4 V
= 25%