Mensuration
- A cylindrical vessel of height 5 cm and radius 4 cm is completely filled with sand. When this sand is poured out it forms a right circular cone of radius 6 cm. What will be the height of this cone? (Take π = 22/7)
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Volume of sand in cylindrical vessel
= πr²h = π × (4)² × 5 cu. cm. = 80p cu. cm.
According to the question volume of conical shape
= 80π cu. cm.⇒ 1 πR²H = 80π 3 ⇒ 1 × 6 × 6 × H = 80 3
⇒ 12 H = 80⇒ H = 80 = 6.67 cm. 12 Correct Option: A
Volume of sand in cylindrical vessel
= πr²h = π × (4)² × 5 cu. cm. = 80p cu. cm.
According to the question volume of conical shape
= 80π cu. cm.⇒ 1 πR²H = 80π 3 ⇒ 1 × 6 × 6 × H = 80 3
⇒ 12 H = 80⇒ H = 80 = 6.67 cm. 12
- The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is
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V1 = πr1²h1 = 
r1 
² h1 V2 πr2²h2 r2 h2 = 2 ² 5 = 20 = 20 : 27 3 3 27 Correct Option: B
V1 = πr1²h1 = r1 ² h1 V2 πr2²h2 r2 h2 = 2 ² 5 = 20 = 20 : 27 3 3 27
- Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm respectively are melted and formed into a single cube. The edge of the new cube formed is
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Volume of larger cube = x1³ + x2³ + x3³
= (6³ + 8³ + 10³) cu. cm.
= (216 + 512 + 1000) cu. cm.
= 1728 cu. cm.
∴ Its edge = ³√1728
= ³√12 × 12 × 12 = 12 cm.Correct Option: A
Volume of larger cube = x1³ + x2³ + x3³
= (6³ + 8³ + 10³) cu. cm.
= (216 + 512 + 1000) cu. cm.
= 1728 cu. cm.
∴ Its edge = ³√1728
= ³√12 × 12 × 12 = 12 cm.
- The radius of a sphere is 6 cm. It is melted and drawn into a wire of radius 0.2 cm. The length of the wire is
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In both cases, volume will remain same.
Volume of sphere = 4 πr³ 3 = 4 π(6)³ 3
= 288π cu. cm.
If the length of wire be h cm., then
⇒ πR²h = 288π
⇒ (0.2)² × h = 288⇒ h = 288 = 7200 cm. 0.04
= 72 metreCorrect Option: D
In both cases, volume will remain same.
Volume of sphere = 4 πr³ 3 = 4 π(6)³ 3
= 288π cu. cm.
If the length of wire be h cm., then
⇒ πR²h = 288π
⇒ (0.2)² × h = 288⇒ h = 288 = 7200 cm. 0.04
= 72 metre
- The radius of a wire is decreased to one-third. If volume remains the same, length will increase by
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Volume of wire (V1) = πr&sub2;h
Case II,Volume of wire (V1) = π r h1 3
∵ V1 = V1∴ πr²h = πr²h1 ⇒ h1 = 9h 9 Correct Option: D
Volume of wire (V1) = πr&sub2;h
Case II,Volume of wire (V1) = π r h1 3
∵ V1 = V1∴ πr²h = πr²h1 ⇒ h1 = 9h 9
