Coupon Accepted Successfully!


Solved Examples

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.
Description: n-16-15.tif
We can see in Fig. 1, that the AOA′P, BOB′Q and COC′R will be similar.
Description: n-16-16.tif
Fig. 1.
Description: n-16-17.tif
In Fig. 2, (r2 – r1)/(r2 + r1) = (r3 – r2)/(r3 + r2) = K

Using Componendo and Dividendo,
r2/r1 = r3/r2

Hence, the three radii are in a GP.

So, r2/20 = 5/r2, hence r2 = 10 units.

Result of this question can be used as a formula also.
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.
The diameter of the metallic sphere is 6 cm. Hence, radius of the sphere is 3 cm. Now, let the radius of the cross–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, Description: 2473.png × π × 33 = π × r2 × 3600
or, 4 × 9 π × 33 = 3600 πr2
This gives r2 = 0.01
i.e., r = 0.1
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 × 44 m. In what time the dam will be able to produce electricity?
The volume of water flowing through the cylindrical pipe in one hour at the rate of 5 km (5000 m) per hour
Description: 2483.png
[Since radius = 7 cm = Description: 2492.png m
= 77 m3
Thus, 77 m3 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., Description: 2501.png m,
Volume of the required quantity of water = 50 × 44
× Description: 2505.png m3 = 154 m3
Since 77 m3 of water falls into the tank in 1 h, therefore, 154 m3 of water will fall into the dam in Description: 2509.png h, i.e., 2 h.
So, the level of water will rise by 7 cm in 2 h.
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.
Let ∆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) 
Description: 2518.png cm = 25
Here, AO (and A’O) is the radius of the common base of the double cone formed by revolving the ∆ABC about BC.
Height of the cone BBA’ is BO and the slant height is 15 cm.
Height of the cone CAA’ is CO and the slant height is 20 cm
Using AA similarity,
Now, ∆AOB ∼ ∆CAB
Description: 16-19.tif
Therefore, Description: 2527.png
This gives Description: 2536.png 
Also, Description: 2545.png 
This gives Description: 2554.png
Thus, CO = 25 cm – 9 cm = 16 cm
Now, volume of the double cone
Description: 2564.png
Description: 2573.png 
= 3768 cm3
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?
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) = Description: 2582.png × 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 – volume of the solid
Description: 16-20.tif
Description: 2586.png 
= (16π − 8π) cm3
= 8π cm3
Hence, the right circular cylinder is having 8p cm3 more space than the solid.
A toy consists of a base that is the section of a sphere and a conical top. The volume of the conical top is 30π 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
  1. Description: 2595.png
  2. Description: 2604.png 
  3. Description: 2613.png
  4. Description: 2622.png
Height of the cone = 10 units
Volume of the cone = 30π cubic units
= diameter of the cross section from where the sphere has been sectioned = 6 units
Description: 2631.png 
= r = 5 units
= volume of the original sphere from which the base has been sectioned = Description: 2640.png

Test Your Skills Now!
Take a Quiz now
Reviewer Name