Match the following columns:
Column I | Column II |
(a) A solid metallic sphere of radius 8 cm is melted and the material is used to make solid right cones with height 4 cm and radius of the base 8 cm. How many cones are formed? | (p) 18 |
(b) A 20-m-deep well with diameter 14 m is dug up and the earth from digging is evenly spread out to form a platform 44 m by 14 m. the height of the platform is…..m. | (q) 8 |
(c) A sphere of radius 6 cm is melted and recast into the shape of a cylinder of radius 4 cm. then the height of the cylinder is ….. cm. | (r) 16:9 |
(d) The volumes of two sphere are in the ratio 64:27. The ratio of their surface areas is ….. . | (s) 5 |
a—(q), b—(s), c—(p), d—(r)
a) Given: A solid metallic sphere of radius 8 cm
Solid right cones with height 4 cm and radius of the base 8 cm.
Volume of a metallic sphere is given by: 
× π × r3
Volume of a Right cone is given by: 
× π × r2 × h
Let V1 be the Volume of the metallic sphere.
∴ V1 = 
× π × r3 = 
× π × (8)3
Let V2 be the Volume of the Solid right cone.
∴ V2 = 
× π × r2 × h = 
× π × (8)2 × 4
Let ‘n’ be the number of right circular cones that are made from melting the metallic sphere.
∴ V1 = n × V2
× π × (8)3 = n × 
× π × (8)2 × 4
n = 
= 8
∴ 8 cones are formed from melting the metallic sphere.
b) Given: A 20-m-deep well with diameter 14 m ⇒ radius = 7cm
A platform 44 m by 14 m
Volume of a cylinder is given by: π × r2 × h
Volume of a platform(cuboid) is given by: l × b × h (here l, b, h are length, breadth, height respectively)
Let V1 be the Volume of the Well.
∴ V1 = π × r2 × h = π × (7)2 × 20
Let V2 be the Volume of the platform
∴ V2 = l × b × h = 44 × 14 × h
Here V1 = V2
∴ 44 × 14 × h = π × (7)2 × 20
⇒ h = = 
= 5cm
∴ h = 5cm
That is height of the platform is 5cm.
c) Given: A sphere of radius 6 cm
A cylinder of radius 4 cm
Volume of a metallic sphere is given by: 
× π × r3
Volume of a Cylinder is given by: π × r2 × h
Let V1 be the Volume of the metallic sphere.
∴ V1 = 
× π × r3 = 
× π × (6)3
Let V2 be the Volume of the Solid Cylinder.
∴ V2 = π × r2 × h = π × (4)2 × h
Here V1 = V2
∴ π × (4)2 × h = 
× π × (6)3
⇒ h = 
= 18cm
∴ h = 18cm
That is height of the cylinder is 18 cm.
d) Given: Volume ratio of two Spheres is: 64:27
Volume of the Sphere is: 
× π × r3 (where r is radius of sphere)
Surface area of the sphere is: 4 × π × r2 (where r is radius of sphere)
Let V1 and V2 be the volumes of different spheres.
V1: V2 = 64:27
× π × (r1)3: 
× π × (r2)3 = 64:27 (here r1 and r2 are the radii of V1 and V2 respectively)
(r1)3: (r2)3 = 64:27
r1: r2 = ∛64:∛27
r1: r2 = 4:3
Now,
Let S1 and S2 be the Surface areas of the spheres.
∴ S1:S2 = 4 × π × (r1)2:4 × π × (r2)2 (here r1 and r2 are the radii of S1 and S2 respectively)
⇒ S1:S2 = (r1)2: (r2)2
⇒ S1:S2 = (4)2: (3)2
⇒ S1:S2 = 16:9
∴ The ratios of the Surface areas is: 16:9