(1) Find the mass of 200 steel spherical ball bearings, each of which has radius 0.7 cm, given that the density of steel is 7.95 g/cm3. (Mass = Volume x Density).
(2) The outer and inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume.
(3) The volume of a solid hemisphere is 1152 Π cu.cm. Find its curved surface area.
(4) Find the volume of the largest right circular cone that can be cut of a cube whose edge is 14 cm.
(5) The radius of a spherical balloon increase from 7 cm to 14 cm as air is being pumped into it. Find the ratio of volumes of the balloon in the two cases.
(1) Solution :
radius of spherical ball = 0.7 cm
Volume of one spherical ball = (4/3) Π r3
= (4/3) (22/7) ⋅ 0.7 ⋅ 0.7 ⋅ 0.7
= 4.312/3
= 1.437
Volume of 200 steel spherical ball = 200 ⋅ 1.437
= 287.46 cm3
1 cm³ = 7.95 g
Therefore mass of 200 spherical ball bearings
= 287.46 (7.95)
= 2285.307 gram
1000 gram = 1 kg
= 2285.307/1000
= 2.29 kg
Volume of 200 spherical balls = 2.29 kg.
(2) Solution :
From this information we have to find the volume
Outer radius (R) = 12 cm
Inner radius (r) = 10 cm
Volume of hollow sphere = (4/3) Π (R3-r3)
= (4/3)(22/7) (123-103)
= (88/21) (1728-1000)
= (88/21) (728)
= (64064/21)
= 3050.67 cm3
Volume of hollow sphere is 3050.67 cm3.
(3) Solution :
Volume of hollow sphere = 1152 Π
(2/3) Π r3 = 1152 Π
r3 = 1152 Π (3/2Π)
r3 = (576 x 3)/2
r3 = 1728
r = ∛1728
r = 12 cm
Curved surface area = 2Πr2
= 2Π(12)2
= 2Π(144)
= 288Π cm3
Curved surface area = 288Π cm3
(4) Solution :
Since it is cube length of all sides will be equal that is 14 cm. Diameter and height of cone are 14 cm.
r = 14/2 ==> 7
h = 14 cm
Volume of cone = (1/3) Π r2 h
= (1/3) ⋅ (22/7) ⋅ 72 ⋅ 14
= (1/3) ⋅ 22 ⋅ 49 ⋅ 2
= (49 ⋅ 44)/3
= 2156/3
= 718.67 cm3
Volume of cone is 718.67 cm3.
(5) Solution :
Let r₁ and r₂ are the radii of two spherical balloon
r1 : r2 = 7 : 14
Volume of one spherical balloon = (4/3) Π r3
(4/3) Π (7)3 : (4/3) Π 143
73 : 143
7 ⋅ 7 ⋅ 7 : 14 ⋅ 14 ⋅ 14
1 : 8
So, the required ratio is 1 : 8.
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