# Class 9 RD Sharma Solutions- Chapter 20 Surface Area And Volume of A Right Circular Cone – Exercise 20.2

• Last Updated : 25 Jan, 2021

### Question 1. Find the volume of the right circular cone with:

(iii) Height is 21cm and slant height 28cm

Solution:

Height of Cone(h)=7cm

As we know that the Volume of a Right Circular Cone = 1/3 πr2h

By putting the values in formula we get,

= 1/3 x 3.14 x 62 x 7 = 264

Hence, Volume of a Right Circular Cone is 264 cm3

Height of Cone(h)=12cm

Volume of a Right Circular Cone = 1/3 πr2h

By putting the values in formula we get,

= 1/3 x 3.14 x 3.52 x 12 =154

Hence, Volume of a Right Circular Cone is 154 cm3

(iii) Height of Cone(h)=21 cm

Slant height of Cone(l) = 28 cm

As we know that, l2 = r2 + h2

282 = r2 + 212

r = 7√7

As we know that Volume of a Right Circular Cone = 1/3 πr2h

By putting the values in formula we get,

= 1/3 x 3.14 x (7√7)2 x 21 = 7546

Hence, Volume of a Right Circular Cone is 7546 cm3

### Question 2. Find the capacity in liters of a Conical Vessel with :

(i) radius 7 cm, slant height 25 cm

(ii) height 12 cm, slant height 13 cm.

Solution:

(i) Radius of the Cone(r) =7 cm

Slant height of the Cone (l) =25 cm

As we know that l2 = r2 + h2

252 = 72 + h2

h = 24

We know that Volume of a right circular cone = = 1/3 πr2h

By putting the values in formula we get,

= 1/3 x 3.14 x (7)2 x 24 = 1232

Hence Volume of a right circular cone is 1232 cm3 or 1.232 liters

(ii) Height of Cone(h)=12 cm

Slant height of Cone(l)=13 cm

As we know that l2 = r2 + h2

132 = r2 + 122

r = 5

We know that Volume of a Right Circular Cone = 1/3 πr2h

By putting the values in formula we get,

= 1/3 x 3.14 x (5)2 x 12 = 314.28

Hence, Volume of a right circular cone is 314.28 cm3 or 0.314 liters.

### Question 3. Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1. Find the ratio of their volumes.

Solution:

Let us assume that the heights of the cones is h and 3h and radii of their bases is 3r and r. Then, their volumes are

Volume of 1st Cone (V1) = 1/3 π(3r)2h

Volume of 2nd Cone (V2) = 1/3 πr2(3h)

V1/V2 = 3/1

Hence, Ratio of two volumes is 3:1.

### Question 4. The radius and the height of a right circular cone are in the ratio 5:12. If its volume is 314 cubic meter, find the slant height and the radius. (Use π = 3.14).

Solution:

Let us assume that the ratio of radius and the height of a Right Circular Cone to be x.

Then the radius be 5x and height be 12x

As we know that l2 = r2 + h2

= (5x) 2 + (12x)2

= 25 x2 + 144 x2

l = 13x

Hence, Slant Height is 13 m.

Now it is given that the Volume of Cone = 314 m3

= 1/3πr2h = 314

= 1/3 x 3.14 x (25x2 ) x (12x) = 314

=x3=1

x = 1

Hence, Radius = 5x 1 = 5 m

Therefore, Slant height = 13m

### Question 5. The radius and height of a right circular cone are in the ratio 5:12 and its volume is 2512 cubic cm. Find the slant height and radius of the cone. (Use π = 3.14).

Solution:

Let us assume that the ratio of radius and height of a right circular cone be y.

Height of Cone (h) =12y

As we know that, l2 = r2 + h2

= (5y) 2 + (12y)2

= 25 y2 + 144 y2

= l = 13y

Volume of Cone, given 2512cm3

= 1/3πr2h = 2512

= 1/3 x 3.14 x (5y)2 x 12y = 2512

= y3 = (2512 x 3)/(3.14 x 25 x 12) = 8

= y = 2

Therefore,

Radius of Cone = 5y = 5×2 = 10cm

Slant Height (l) =13y = 13×2 = 26cm

### Question 6. The ratio of volumes of two cones is 4:5 and the ratio of the radii of their bases is 2:3. Find the ratio of their vertical heights.

Solution:

Let us assume that the ratio of the Radius is x and Ratio of the Volume is y.

Radius of 1st Cone (r1) =2x

Radius of 2nd Cone (r2) =3x

Volume of 1st Cone (V1)= 4y

Volume of 2nd Cone (V2)= 5y

As we know that the formula for Volume of a Cone = 1/3πr2h

Let’s assume that h1 and h2 be the heights of respective cones.

V1/V2 = 4/5 = (1/3 x π x (r1)2 x h1) / (1/3 x π x (r2)2 x h2)

4/5 = 4h1/9h2 = 9/5

Hence, Heights are in the ratio of 9 : 5.

### Question 7. A cylinder and a cone have equal radii of their bases and equal heights. Show that their volumes are in the ratio 3:1.

Solution:

Given that,

Cylinder and Cone are having equal radii of their bases and heights.

Let’s assume that Radius of the Cone = Radius of the Cylinder = r &

Height of the Cone = Height of the Cylinder = h

Volume of Cylinder / Volume of Cone = (πr2h) / (1/3 x π x r2 x h) = 3:1

Hence, The Ratio of their Volumes is 3:1.

### Question 8. If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone?

Solution:

Let us assume that r be the radius and h be the height of the cone,

As we know that Volume of Cone = 1/3 πr2h

Put the values of radius and same height we get,

Volume = 1/3 π(r/2)2h = 1/3 x π x r2/2 x h

= 1/4 x (1/3 x π x r2 x h)

Ratio of two cone’s = 1/3 πr2h : 1/4 x (1/3 πr2h) = 1 : 1/4 = 4:1

Hence, the ratio between Cones are 1:4

### Question 9.  A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover theheap?

Solution:

Given that,

Diameter of conical heap of wheat = 9 m

Radius = 9/2m and height = 3.5m

As we know that volume of cone = 1/3 πr2h =  1/3 x 22/7 x (4.5)2 x 3.5 = 74.18 m3

As we know that Curved Surface area = πrl

l = √r2 + h2 = √(4.5)2 + (3.5)2 = √130/2 m

Curved Surface area =  π x 4.5 x √130/2 = 22/7 x 4.5 x √130/2 = 80.54 m2

### Question 10.  Find the weight of a solid cone whose base is of diameter 14 cm and vertical height 51 cm, supposing the material of which it is made weighs 10 grams per cubic cm.

Solution:

Given that,

Diameter of the base of solid cone = 14 cm and vertical height (h) = 51 cm

As we know that volume of cone =  1/3 πr2h =  1/3 x 22/7 x 72 x 51 = 2618 cm3

Weight of 1 cm3 = 10 grams

Then total weight will = 2618 x 10gm = 26180 gm = 26.180 kg

### Question 11. A right-angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Findthe volume of the solid, thus generated. Also, find its curved surface area.

Solution:

Given that,

Length of sides of a right-angled triangle are 6.3 cm and 10 cm

By turning around the longer side a Cone is formed in which radius (r) is = 6.3 cm and height (h) = 10 cm

As we know that l = √r2 + h2 = √(6.3)2 + (10)2 = 11.82 cm

As we know that volume of cone = 1/3 πr2h = 1/3 x 22/7 x (6.3)2 x10 = 415.8 cm3

We know that Curved Surface Area of Cone = πrl = 22/7 x 6.3 x 11.82 = 234.03 cm2

### Question 12. Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14 cm.

Solution:

Given that,

Side of Cube = 14 cm,

Radius(r) of the largest cone that can be fitted in Cube = Side / 2 = 14/2 = 7cm,

Height(h) = 14cm

As we know that Volume of Cone =  1/3 πr2h =  1/3 x 22/7 x 72 x 14 = 718.67 cm3

### Question 13. The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find:

(i) Height of the cone

(ii) Slant height of the cone

(iii) Curved surface area of the cone.

Solution:

Given that,

Volume of a Right Circular Cone = 9856 cm3,

Diameter of the base = 28 cm,

1. Height of cone(h) = Volume x 3/πr2 = (9856x3x7)/(22x14x14) = 48cm

2. Slant Height(l) = √r2 + h2 = √(14)2 + (48)2 = 50cm

3. Curved Surface area = πrl = 22/7 x 14 x 50 = 2200 cm2

### Question 14. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? [NCERT]

Solution:

Given that,

Diameter of the top of the conical pit = 3.5 m,

Radius(r) = 3.5/2 = 1.75 m and depth(h) = 12 m

As we know that volume of pit = 1/3 πr2h

= 1/3 x 22/7 x (1.75)2 x 12 = 38.5 m3

Volume in Kiloliters = (38.5×1000) / 1000 = 38.5 Kiloliters

### Question 15. Monica has a piece of Canvas whose area is 551 m2. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amounts to approximately 1 m2. Find the volume of the tent that can be made with it.

Solution:

Given that,

Area of Canvas = 551 m2,

Area of wastage = 1 m2,

Actual area = 551 – 1 = 550 m2,

Base radius of the conical tent = 7 m

Let us assume l is the slant height and h is the vertical height then the Slant Height(l) = Area / πr

= (550 x 7) / 22×7 = 25m

As we know that Vertical height (h) = √l2 – r

= √252 – 72 = √576 = 24 m

As we know that volume of Tent = 1/3 πr2

= 1/3 x 22/7 x 7 x 7 x 24 = 1232 m3

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