Chapter 20 - Cylinder, Cone and Sphere (Surface Area and Volume) Exercise Ex. 20(E)

Question 1

A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.

Solution 1

Height of cone = 15 cm

and radius of the base = cm

Therefore, volume of the solid = volume of the conical part + volume of hemispherical part.

Question 2

A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 m and its volume is two-third of the hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two decimal places.

Solution 2

Radius of hemispherical part (r) = 3.5 m = 

Therefore, Volume of hemisphere = 

Volume of conical part =  (2/3 of hemisphere)

Let height of the cone = h

Then,

Height of the cone = 4.67 m

Surface area of buoy = 

But 

Therefore, Surface area =

Surface Area = 141.17 m²

Question 3

From a rectangular solid of metal 42 cm by 30 cm by 20 cm, a conical cavity of diameter 14 cm and depth 24 cm is drilled out. Find:

(i) the surface area of the remaining solid

(ii)the volume of remaining solid

(iii) the weight of the material drilled out if it weighs 7 gm per cm³.

Solution 3

(i) Total surface area of cuboid = 2(b + bh + h)

=2(42  30 + 30 20 + 20 42)

=2(1260 + 600 + 840)

=2  2700

=5400 cm²

Diameter of the cone = 14 cm

Radius of the cone = 

Area of circular base

Area of curved surface area of cone

Surface area of remaining part 

(ii) Dimensions of rectangular solids = (42  30  20) cm

volume = (42  30  20) = 25200 cm³

Radius of conical cavity (r) =7 cm

height (h) = 24 cm

Volume of cone = 

Volume of remaining solid = (25200 - 1232) = 23968 cm³

(iii) Weight of material drilled out

=1232  7 g = 8624g = 8.624 kg

Question 4

The cubical block of side 7 cm is surmounted by a hemisphere of the largest size. Find the surface area of the resulting solid.

Solution 4

The diameter of the largest hemisphere that can be placed on a face of a cube of side 7 cm will be 7 cm.

Therefore, radius = 

Its curved surface area =

Surface area of the top of the resulting solid = Surface area of the top face of the cube - Area of the base of the hemisphere

Surface area of the cube = 

Total area of resulting solid = 245 + 10.5 + 77 = 332.5 cm²

Question 5

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of the top which is open is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution 5

Height of cone = 8 cm

Radius = 5 cm

Volume = 

Therefore, volume of water that flowed out =

 

Radius of each ball = 0.5 cm = 

Volume of a ball = 

Therefore, No. of balls = 

Hence, number of lead balls = 100

Question 6

A hemispherical bowl has negligible thickness and the length of its circumference is 198 cm. find the capacity of the bowl.

Solution 6

Let r be the radius of the bowl.

Capacity of the bowl =

Question 7

Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r cm.

Solution 7

For the volume of cone to be largest, h = r cm

Volume of the cone

 

Question 8

The radii of the bases of two solid right circular cones of same height are r₁ and r₂ respectively. The cones are melted and recast into a solid sphere of radius R. Find the height of each cone in terms r₁, r₂ and R.

Solution 8

Let the height of the solid cones be 'h'

Volume of solid circular cones

  

Volume of sphere

Volume of sphere = Volume of cone 1 + volume of cone 2

Question 9

A solid metallic hemisphere of diameter 28 cm is melted and recast into a number of identical solid cones, each of diameter 14 cm and height 8 cm. Find the number of cones so formed.

Solution 9

Volume of the solid hemisphere

 

Volume of 1 cone

No. of cones formed

 

Question 10

A cone and a hemisphere have the same base and same height. Find the ratio between their volumes.

Solution 10

Let the radius of base be 'r' and the height be 'h'

Volume of cone, V_c

 

Volume of hemisphere, V_h