Q. 745.0( 1 Vote )

# Match the followi

Answer :

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 × r^{3}

Volume of a Right cone is given by: × π × r^{2} × h

Let V_{1} be the Volume of the metallic sphere.

∴ V_{1} = × π × r^{3} = × π × (8)^{3}

Let V_{2} be the Volume of the Solid right cone.

∴ V_{2} = × π × r^{2} × h = × π × (8)^{2} × 4

Let ‘n’ be the number of right circular cones that are made from melting the metallic sphere.

∴ V_{1} = n × V_{2}

× 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 × r^{2} × h

Volume of a platform(cuboid) is given by: l × b × h (here l, b, h are length, breadth, height respectively)

Let V_{1} be the Volume of the Well.

∴ V_{1} = π × r^{2} × h = π × (7)^{2} × 20

Let V_{2} be the Volume of the platform

∴ V_{2} = l × b × h = 44 × 14 × h

Here V_{1} = V_{2}

∴ 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 × r^{3}

Volume of a Cylinder is given by: π × r^{2} × h

Let V_{1} be the Volume of the metallic sphere.

∴ V_{1} = × π × r^{3} = × π × (6)^{3}

Let V_{2} be the Volume of the Solid Cylinder.

∴ V_{2} = π × r^{2} × h = π × (4)^{2} × h

Here V_{1} = V_{2}

∴ 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 × r^{3} (where r is radius of sphere)

Surface area of the sphere is: 4 × π × r^{2} (where r is radius of sphere)

Let V_{1} and V_{2} be the volumes of different spheres.

V_{1}: V_{2} = 64:27

× π × (r_{1})^{3}: × π × (r_{2})^{3} = 64:27 (here r_{1} and r_{2} are the radii of V_{1} and V_{2} respectively)

(r_{1})^{3}: (r_{2})^{3} = 64:27

r_{1}: r_{2} = ∛64:∛27

r_{1}: r_{2} = 4:3

Now,

Let S_{1} and S_{2} be the Surface areas of the spheres.

∴ S_{1}:S_{2} = 4 × π × (r_{1})^{2}:4 × π × (r_{2})^{2} (here r_{1} and r_{2} are the radii of S_{1} and S_{2} respectively)

⇒ S_{1}:S_{2} = (r_{1})^{2}: (r_{2})^{2}

⇒ S_{1}:S_{2} = (4)^{2}: (3)^{2}

⇒ S_{1}:S_{2} = 16:9

∴ The ratios of the Surface areas is: 16:9

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