Mensuration
- The base 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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Let the radii of two cylinders are r1, r2 and length of the cylinders are h1, h2 respectively.
According to the questionr1 = 2 and h1 = 5 r2 3 h2 3
∴ Ratio of their volume = πr1²h1 : πr2²h2
= r1²h1 : r2²h2
= (2)² × 5 : (3)² × 3 = 4 × 5 : 9 × 3 = 20 : 27Correct Option: B
Let the radii of two cylinders are r1, r2 and length of the cylinders are h1, h2 respectively.
According to the questionr1 = 2 and h1 = 5 r2 3 h2 3
∴ Ratio of their volume = πr1²h1 : πr2²h2
= r1²h1 : r2²h2
= (2)² × 5 : (3)² × 3 = 4 × 5 : 9 × 3 = 20 : 27
- The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m³.
(Taking π = 22/7) . Find the ratio of its diameter to its height.
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If r be radius of base and h the height, then
Curved surface of cylindrical pillar = 2πrh.
and volume = πr²h.
∴ 2πrh = 264 m² ...(i)
πpr2h = 924 m³ ...(ii)
On dividing (ii) by (i), we getπr²h = 924 m. 2πrh 264 ⇒ r = 924 m. 2 264 ⇒ r = 324 × 2 m = 7 m 264
∴ Diameter = 2 × 7 = 14m From (i),h = 264 = 264 × 7 = 6m. π × d 22 × 14 ∴ Required ratio = 14 i.e, 7 : 3 6 Correct Option: D
If r be radius of base and h the height, then
Curved surface of cylindrical pillar = 2πrh.
and volume = πr²h.
∴ 2πrh = 264 m² ...(i)
πpr2h = 924 m³ ...(ii)
On dividing (ii) by (i), we getπr²h = 924 m. 2πrh 264 ⇒ r = 924 m. 2 264 ⇒ r = 324 × 2 m = 7 m 264
∴ Diameter = 2 × 7 = 14m From (i),h = 264 = 264 × 7 = 6m. π × d 22 × 14 ∴ Required ratio = 14 i.e, 7 : 3 6
- A hollow iron pipe is 21 cm long and its exterior diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm³, then the weight of the pipe is (Take π = 22/7)
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The pipe can be assumed as hollow cylinder.
External radius = 8 = 4 cm. 2
Thickness = 1 cm
∴ Internal radius = 4 – 1 = 3 cm.
Volume of the material = π h (R2 – r2)= 22 × 21 × (4² - 3²) 7 = 22 × 21 × 7 = 462 cm³ 7
Now, 1cm3 iron weighs = 8 gm
∴ 462 cm³ iron weighs
= 462 × 8 gm= 462 × 8 kg 1000
= 3.696 kgCorrect Option: A
The pipe can be assumed as hollow cylinder.
External radius = 8 = 4 cm. 2
Thickness = 1 cm
∴ Internal radius = 4 – 1 = 3 cm.
Volume of the material = π h (R2 – r2)= 22 × 21 × (4² - 3²) 7 = 22 × 21 × 7 = 462 cm³ 7
Now, 1cm3 iron weighs = 8 gm
∴ 462 cm³ iron weighs
= 462 × 8 gm= 462 × 8 kg 1000
= 3.696 kg
- The volume of a right circular cylinder, 14 cm in height, is equal to that of a cube whose edge is 11 cm. (Taking π = 22/7) the radius of the base of the cylinder is
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Volume of the cube = (edge)³ = (11 × 11 × 11)cm³
∴ Volume of cylinder = 11 × 11 × 11
⇒ πr² × 14 = 11 × 11 × 11⇒ r² = 11 × 11 × 11 × 7 = 11 × 11 22 × 14 4 ⇒ r = √ 11² = 11 = 5.5 cm. 4 2 Correct Option: B
Volume of the cube = (edge)³ = (11 × 11 × 11)cm³
∴ Volume of cylinder = 11 × 11 × 11
⇒ πr² × 14 = 11 × 11 × 11⇒ r² = 11 × 11 × 11 × 7 = 11 × 11 22 × 14 4 ⇒ r = √ 11² = 11 = 5.5 cm. 4 2
- If the volume of a right circular cylinder is 9πh m³, where h is its height (in metres) then the diameter of the base of the cylinder is equal to
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Let the radius of base be r metres.
∴ πr²h = 9πh
⇒ r⊃2 = 9
⇒ r = 3 m
∴ Diameter = 2 × 3 = 6 metres.Correct Option: B
Let the radius of base be r metres.
∴ πr²h = 9πh
⇒ r⊃2 = 9
⇒ r = 3 m
∴ Diameter = 2 × 3 = 6 metres.
