# Solved Examples

Example-1

Three spheres are kept inside a cone, as given in the figure. Spheres are touching both the slant sides of the cone and the adjacent spheres. If the radius of the 1st sphere and the 3rd sphere are 5 units and 20 units respectively, find the radius of the 2nd sphere.

Solution

We can see in Fig. 1, that the AOAâ€²P, BOBâ€²Q and COCâ€²R will be similar.

Fig. 1.

In Fig. 2, (r

Using Componendo and Dividendo,

r

Hence, the three radii are in a GP.

So, r

Result of this question can be used as a formula also.

_{2 }â€“ r_{1})/(r_{2 }+ r_{1}) = (r_{3 }â€“ r_{2})/(r_{3 }+ r_{2}) = KUsing Componendo and Dividendo,

r

_{2}/r_{1}= r_{3}/r_{2}Hence, the three radii are in a GP.

So, r

_{2}/20 = 5/r_{2}, hence r_{2}= 10 units.Result of this question can be used as a formula also.

Example-2

The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform circular cross-section. If the length of the wire is 36 m, find its radius.

Solution

The diameter of the metallic sphere is 6 cm. Hence, radius of the sphere is 3 cm. Now, let the radius of the crossâ€“section of the wire be r cm. And as we know, metallic sphere is converted into a cylindrical shaped wire, then their volumes will be equal.
So, Ã— Ï€ Ã— 3
or, 4 Ã— 9 Ï€ Ã— 3

This gives r
i.e., r = 0.1

^{3}= Ï€ Ã— r^{2}Ã— 3600^{3}= 3600 Ï€r^{2}This gives r

^{2}= 0.01Example-3

Sardar Sarovar Dam which is rectangular in shape, can produce electricity only if the height of the water level in it is atleast 7cm. Now the water is pumped in at the rate of 5 km per hour through a pipe of diameter 14 cm into the dam area of dimensions 50 m Ã— 44 m. In what time the dam will be able to produce electricity?

Solution

The volume of water flowing through the cylindrical pipe in one hour at the rate of 5 km (5000 m) per hour

[Since radius = 7 cm = m

= 77 m

Thus, 77 m

Volume of the required quantity of water = 50 Ã— 44

Ã— m

Since 77 m

So, the level of water will rise by 7 cm in 2 h.

[Since radius = 7 cm = m

= 77 m

^{3}Thus, 77 m

^{3}of water will fall into the tank in 1 h. Since the level of the water is required to be raised by 7 cm i.e., m,Volume of the required quantity of water = 50 Ã— 44

Ã— m

^{3 }= 154 m^{3}Since 77 m

^{3}of water falls into the tank in 1 h, therefore, 154 m^{3}of water will fall into the dam in h, i.e., 2 h.So, the level of water will rise by 7 cm in 2 h.

Example-4

A right angled triangle ABC, whose two sides other than the hypotonuse are 15 cm and 20 cm. The triangle is made to revolve about its hypotonuse. Find the volume of the double cone so formed.

Solution

Let âˆ†ABC be the right angled triangle right angled at A, whose sides AB and AC measures 15 cm and 20 cm, respectively.
The length of the side BC (hypotenuse)

= cm = 25
Here, AO (and Aâ€™O) is the radius of the common base of the double cone formed by revolving the âˆ†ABC about BC.
Height of the cone BBAâ€™ is BO and the slant height is 15 cm.
Height of the cone CAAâ€™ is CO and the slant height is 20 cm

Using AA similarity,

Now, âˆ†AOB âˆ¼ âˆ†CAB Therefore,

This gives

Also,

This gives

Thus, CO = 25 cm â€“ 9 cm = 16 cm
Now, volume of the double cone

= 3768 cm

= cm = 25

Using AA similarity,

Now, âˆ†AOB âˆ¼ âˆ†CAB Therefore,

This gives

Also,

This gives

Thus, CO = 25 cm â€“ 9 cm = 16 cm

= 3768 cm

^{3}Example-5

When shopping in Big Bazar, I saw a peculiar solid toy in the form of a hemisphere surmounted by a right circular cone. Height of the cone was 2 cm and the diameter of the base was 4 cm. If a right circular cylinder circumscribed the solid, find out how much more space will it have, provided the height of the cone was 2 cm and diameter of the base was 4 cm respectively?

Solution

See the figure below.
Assume BPC be the hemisphere and ABC is the cone standing on the base of the hemisphere.
Radius BO of the hemisphere (as well as of the cone) = Ã— 4 cm = 2 cm.
Now, let the right circular cylinder EFGH circumscribe the given solid.
Radius of the base of the right circular cylinder = HP = BO = 2 cm.
Height of the cylinder = AP = AO + OP = 2 cm + 2 cm = 4 cm
Now, volume of the right circular cylinder â€“ volume of the solid

= (16Ï€ âˆ’ 8Ï€) cm

^{3}

= 8Ï€ cm

^{3}

Hence, the right circular cylinder is having 8p cm

^{3}more space than the solid.

Example-6

A toy consists of a base that is the section of a sphere and a conical top. The volume of the conical top is 30Ï€ sq. units and its height is 10 units. The total height of the toy is 19 units. The volume of the sphere (in cubic units), from which the base has been extracted, is

Solution

Height of the cone = 10 units
Volume of the cone = 30Ï€ cubic units
= diameter of the cross section from where the sphere has been sectioned = 6 units

= r = 5 units

= volume of the original sphere from which the base has been sectioned =

= r = 5 units

= volume of the original sphere from which the base has been sectioned =