Mensuration
- A sphere is cut into two hemispheres. One of them is used as bowl. It takes 8 bowlfuls of this to fill a conical vessel of height 12 cm and radius 6 cm. The radius of the sphere (in centimetre) will be
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Volume of conical vessel = 1 πr²h 3 Volume of conical vessel = 1 × π × 6 × 6 × 12 = 144π cu. cm 3
If the radius of sphere be R cm, then= 8 × 2 π R³ = 144π 3 ⇒ R³ = 144 × 3 = 9 × 3 = 3 × 3 × 3 8 × 2
∴ R = ³√3 × 3 × 3 = 3 cm.Correct Option: A
Volume of conical vessel = 1 πr²h 3 Volume of conical vessel = 1 × π × 6 × 6 × 12 = 144π cu. cm 3
If the radius of sphere be R cm, then= 8 × 2 π R³ = 144π 3 ⇒ R³ = 144 × 3 = 9 × 3 = 3 × 3 × 3 8 × 2
∴ R = ³√3 × 3 × 3 = 3 cm.
- The volumes of a right circular cylinder and a sphere are equal. The radius of the cylinder and the diameter of the sphere are equal. The ratio of height and radius of the cylinder is
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Radius of cylinder = r units
Radius of sphere = r/2 units
Let the height of cylinder be h units,
∴ Volume of cylinder = Volume of sphere⇒ πr²h = 4 π 
r 
³ 3 2 ⇒ πr²h = 1 πr³ 6 ⇒ h = 1 r 6 ⇒ h = 1 r 6 Correct Option: D
Radius of cylinder = r units
Radius of sphere = r/2 units
Let the height of cylinder be h units,
∴ Volume of cylinder = Volume of sphere⇒ πr²h = 4 π r ³ 3 2 ⇒ πr²h = 1 πr³ 6 ⇒ h = 1 r 6 ⇒ h = 1 r 6
- Some bricks are arranged in an area measuring 20 cu. m. If the length, breadth and height of each brick is 25 cm, 12.5 cm and 8 cm respectively, then in that pile the number of bricks are (suppose there is no gap in between two bricks)
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Volume of pile = 20 cu. metre
= 20 × (100)³ cu.cm.
Volume of one brick = (25 × 12.5 × 8) cu.cm.∴ Required number of bricks = 20 × 100 × 100 × 100 = 8000 25 × 12.5 × 8 Correct Option: B
Volume of pile = 20 cu. metre
= 20 × (100)³ cu.cm.
Volume of one brick = (25 × 12.5 × 8) cu.cm.∴ Required number of bricks = 20 × 100 × 100 × 100 = 8000 25 × 12.5 × 8
- The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be 1/27 the of the volume of the given cone, at what height above the base is the sectionm ade ?
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Let H and R be the height and radius of bigger cone respectively and h and r that of smaller cone.
From triangles AOB and AMN. ∠A is common and MN || OB.
∴ Triangles AOB and AMN are similar,∴ AO = BO AM MN ⇒ 30 = R ..........(i) h r Volume of smaller cone = 1 πr²h 3 Volume of bigger cone = 1 πR²h 3
∴ According to the question,1 πr²h = 1 πR²h × 1 3 3 27 ⇒ r²h = r²H 27
→ 27r²h = R²H⇒ 27h = R² H r² ⇒ 27h = 30 ² ....[From(i)] H h ⇒ 27h = 900 H r²
⇒ 27h³ = 900H = 900 × 30⇒ h³ = 900 × 30 = 1000 27
⇒ h = ³√1000 = 10 cm
∴ Required height = 30 – 10 = 20 cmCorrect Option: B
Let H and R be the height and radius of bigger cone respectively and h and r that of smaller cone.
From triangles AOB and AMN. ∠A is common and MN || OB.
∴ Triangles AOB and AMN are similar,∴ AO = BO AM MN ⇒ 30 = R ..........(i) h r Volume of smaller cone = 1 πr²h 3 Volume of bigger cone = 1 πR²h 3
∴ According to the question,1 πr²h = 1 πR²h × 1 3 3 27 ⇒ r²h = r²H 27
→ 27r²h = R²H⇒ 27h = R² H r² ⇒ 27h = 30 ² ....[From(i)] H h ⇒ 27h = 900 H r²
⇒ 27h³ = 900H = 900 × 30⇒ h³ = 900 × 30 = 1000 27
⇒ h = ³√1000 = 10 cm
∴ Required height = 30 – 10 = 20 cm
- The height of the right pyramid whose area of the base is 30 m² and volume is 500 m³, is
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Volume of pyramid = 1 × area of base × height 3 ⇒ 500 × 1 × 30 × h 3
⇒ 10h = 500⇒ h = 500 = 50 metre 10
Correct Option: A
Volume of pyramid = 1 × area of base × height 3 ⇒ 500 × 1 × 30 × h 3
⇒ 10h = 500⇒ h = 500 = 50 metre 10
