Class 10 RD Sharma Solutions - Chapter 16 Surface Areas and Volumes - Exercise 16.2 | Set 2

Chapter 16 of RD Sharma's Class 10 textbook "Surface Areas and Volumes" explores the mathematical principles used to calculate the surface areas and volumes of the various geometric shapes. This chapter is crucial for understanding how to measure and interpret the size of three-dimensional objects which has practical applications in fields such as engineering, architecture, and everyday problem-solving. The focus is on developing the skills to accurately compute these measurements for the different solids including spheres, cylinders, cones, and frustums.

Surface Areas and Volumes

Surface Areas: The Surface area refers to the total area that the surface of the three-dimensional object occupies. It is measured in square units. For example: the surface area of a cube is calculated by finding the area of one face and multiplying it by six while the surface area of a cylinder involves adding the areas of its two circular bases and the area of the side.

Volumes: The Volume represents the amount of space inside a three-dimensional object measured in cubic units. To find the volume of a cube we cube the length of the one side. For a cylinder, the volume is calculated by multiplying the area of the base by the height of the cylinder. Each solid has a specific formula for the volume that depends on its shape and dimensions.

Question 13. A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylinder is 14/3 and the diameter of the hemisphere is 3.5 m. Calculate the volume and the internal surface area of the solid.

Solution: 

According to the question 

Diameter of the hemisphere(d) = 3.5 m,

So, the radius of the hemisphere(r) = 1.75 m,

Height of the cylinder(h) = 14/3 m,

Now we find the volume of the cylinder 

V1 = πr² h1

= π(1.75)² x 14/3 m³

Now we find the volume of the hemisphere

V2= 2/3 × 22/7 × r³

V2 = 2/3 × 22/7 × 1.753 m³

So, the total volume of the vessel is

V = V1 + V2

V = π(1.75)² x 14/3 + 2/3 × 22/7 × 1.753 

V = 56 m³

Now we find the internal surface area of solid 

S = 2πrh1 + 2πr²

= 2 π(1.75)(143) + 2 π(1.75)² = 70.51 m³

Hence, the internal surface area of the solid is 70.51 m³ and the volume is 56 m³

Question 14. Consider a solid which is composed of a cylinder with hemispherical ends. If the complete length of the solid is 104 cm and the radius of each of the hemispherical ends is 7 cm. Find the cost of polishing its surface at the rate of Rs. 10 per dm².

Solution: 

According to the question 

Radius of the hemispherical end (r) = 7 cm,

Height of the solid = h + 2r = 104 cm, h = 90 cm

Now we find the curved surface area of the cylinder 

A1 = 2 πr h

= 2 π(7) h 

= 2 π(7)(90)

= 3948.40 cm²

Now we find the curved surface area of the two Hemisphere 

A2 = 2 (2πr²)

= 22π(7)² = 615.75 cm²

Hence, the total curved surface area of the solid is

A = A1 + A2

= 3948.40 + 615.75 = 4571.8 cm² = 45.718 dm²

So, he cost of polishing the 1 dm² surface of the solid is Rs. 15                               

Therefore, the cost of polishing the 45.718 dm² surface of the solid = 10 45.718 = Rs. 457.18

Hence, the cost of polishing is Rs. 457.18.

Question 15. A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16cm and height of 42 cm. The total space between the two vessels is filled with Cork dust for heat insulation purposes. Find how many cubic cms of the Cork dust will be required?

Solution: 

According to the question 

Depth of the vessel = Height of the vessel = h = 42 cm,

Inner diameter of the vessel = 14 cm,

So, inner radius of the vessel = r1 = 14/2 = 7 cm 

Outer diameter of the vessel = 16 cm,

So, the outer radius of the vessel = r2 = 16/2 = 8 cm

Now we find the volume of the vessel 

V = π(r₂² - r₁²)h 

= π(82 - 72) x 42 = 1980 cm³

Hence, the volume of the vessel is 1980 cm³ which is equal to the amount of cork dust required.

Question 16. A cylindrical road roller made of iron is 1 m long. Its internal diameter is 54 cm and the thickness of the iron sheet used in making roller is 9 cm. Find the mass of the road roller if 1 cm³ of the iron has 7.8 gm mass.

Solution: 

According to the question 

Height of the road roller (h) = 1 m = 100 cm,

Internal Diameter of the road roller = 54 cm,

So, the internal radius of the road roller (r)= 27 cm 

Thickness of the road roller (T) = 9 cm,

Let us considered R be the outer radii of the road roller. So, 

T = R - r

9 = R - 27

R = 27 + 9 = 36 cm

Now we find the volume of the Iron Sheet 

V = π × (R² − r²) × h

= π × (36² − 27²) × 100 = 1780.38 cm³

Hence the mass of 1 cm³ of the iron sheet = 7.8 gm                   [Given]

Therefore, the mass of 1780.38 cm³ of the iron sheet = 1388696.4 gm = 1388.7 kg

Hence, the mass of the road roller is 1388.7 kg

Question 17. A vessel in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13cm. Find the inner surface area of the vessel.

Solution: 

According to the question 

Diameter of the hemisphere = 14 cm,

So, the radius of the hemisphere = 7 cm

And the total height of the vessel = = h + r = 13 cm 

Now we find the inner surface area of the vessel 

A = 2πr (h + r) 

= 2 x 22/7 x (13) x (7)

= 572 cm²

Hence, the inner surface area of the vessel is 572 cm²

Question 18. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution: 

According to the question 

Radius of the conical part of the toy(r) = 3.5 cm 

Total height of the toy (h) = 15.5 cm 

Slant height of the cone (L) = 15.5 – 3.5 = 12 cm

Now we find the curved surface area of the cone 

A1 = πrL 

= π(3.5)(12) = 131.94 cm²

Now we find the curved surface area of the hemisphere 

A2 = 2πr²

= 2π(3.5)² = 76.96 cm²

Hence, the total surface area of the toy 

A = A1 + A2

= 131.94 + 76.96 = 208.90 cm²

Hence, the total surface area of the toy is 209 cm²  

Question 19. The difference between outside and inside surface areas of the cylindrical metallic pipe 14 cm long is 44 dm². If pipe is made of 99 cm² of metal. Find outer and inner radii of the pipe.

Solution: 

According to the question 

Length of the cylinder (h) = 14 cm         

Difference between the outer and the inner surface area = 44 dm²

Volume of the metal used = 99 cm²,  

Let's assume that the inner radius of the pipe be r1 and r2 be the outer radius of the pipe

Now, the surface area of the cylinder is = 2π x 14 x (r2 - r1) = 44

(r2 − r1) = 1/2          ----------------(i)

Volume of the cylinder is 

πh(r₂² - r₁²) = 99

πh(r2 - r1) (r2 + r1) = 99

= 22/7 x 14 x 1/2 x (r2 + r1) = 99

(r2 + r1) = 9/2                --------------(ii)

On Solving eq(i) and (ii), we get,

r2 = 5/2 cm

r1 = 2 cm

Hence, the inner radius is 2cm and outer radius of the pipe is 5/2 cm.

Question 20. A right circular cylinder having diameter 12 cm and height 15 cm is full ice-cream. The ice-cream is to be filled in cones of height 12 cm and diameter 6 cm having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.

Solution: 

According to the question 

Radius of cylinder (r1) = 6 cm,

Radius of hemisphere (r2) = 3 cm,

Height of cylinder (h) = 15 cm,

slant height of the cones (l) = 12 cm,

Now we find the volume of cylinder 

V = πr₁²h

= π × 6² ×15              --------------(i)

The volume of each ice cream cone = Volume of cone + Volume of hemisphere

= 1/3πr₂²h + 2/3πr₂³

= 1/3π x 6² x 12 + 2/3π x 3³              ----------------(ii)

Let's assume that number of cones be 'n'

n(Volume of each ice cream cone) = Volume of cylinder

n(1/3 π x 3² x 12 + 2/3 π x 3³) = π(6)² x15

n = 50/5 = 10

Hence, the number of cones being filled with ice-cream is 10

Question 21. Consider a solid iron pole having cylindrical portion 110 cm high and the base diameter of 12 cm is surmounted by a cone of 9 cm height. Find the mass of the pole. Assume that the mass of 1 cm³ of iron pole is 8 gm.

Solution: 

According to the question 

Base diameter of the cylinder = 12 cm,

So, the radius of the cylinder (r) = 6 cm

Height of the cylinder (h) = 110 cm,

Slant height of the cone (L) = 9 cm,

Now we find the volume of the cylinder 

V1 = π × r² × h 

= π × 6² × 110 cm³

Now we find the volume of the cone 

V2 = 1/3 × πr²L

= 1/3 × π x 6² x 12  = 108π cm³

Hence, the volume of the pole (V)

V = V1 + V2

= 108π + π(6)² 110 = 12785.14 cm³

So, the mass of 1 cm³ of the iron pole = 8 gm          [Given]      

Then, the mass of 12785.14 cm³ of the iron pole = 8

12785.14 = 102281.12 gm = 102.2 kg

Hence, the mass of the iron pole is 102.2 kg

Question 22. A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the toy, find how much more space it will cover?

Solution:

According to the question 

Radius of the cone, cylinder, and hemisphere (r) = 2 cm

Height of the cone (l) = 2 cm

Height of the cylinder (h) = 4 cm

Now we find the volume of the cylinder 

V1 = π × r² × h 

= π × 22 × 4 cm³

Now we find the volume of the cone 

V2 = 1/3 π r²l

= 1/3 × π × 2² × 2

= 1/3 × π x 4 × 2 cm³

Now we find the volume of the hemisphere 

V3 = 2/3 π r³

= 2/3 × π × 2³ cm³

= 2/3 π × 8 cm³

Therefore, the remaining volume of the cylinder when the toy is inserted to it is 

V = V1 - (V2 + V3)

= 16π - 8π = 8π cm³

Hence, remaining volume of the cylinder when toy is inserted into it is 8π cm³

Question 23. Consider a solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm, is placed upright in the right circular cylinder full of water such that it touches bottoms. Find the volume of the water left in the cylinder, if radius of the cylinder is 60 cm and its height is 180 cm.

Solution:

According to the question 

Radius of the circular cone (r) = 60 cm,

Height of the circular cone (L) = 120 cm,

Radius of the hemisphere (r) = 60 cm,

Radius of the cylinder (R) = 60 cm,

Height of the cylinder (H) = 180 cm,

Now we find the volume of the circular cone 

V1 = 1/3 × πr²l

= 1/3 × π × 602 × 120 = 452571.429 cm³

Now we find the volume of the hemisphere 

V2 = 2/3 × πr³

= 2/3 × π × 603 = 452571.429 cm³

Now we find the volume of the cylinder 

V3 = π × R² × H 

= π × 60² × 180 = 2036571.43 cm³

Hence, the volume of water left in the cylinder 

V = V3 - (V1 + V2)

= 2036571.43 – (452571.429 + 452571.429)

= 2036571.43 – 905142.858 = 1131428.57 cm³  = 1.1314 m³

Hence, the volume of the water left in the cylinder is 1.1314 m³

Question 24. Consider a cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 8 cm and height 6 cm is completely immersed in water. Find the value of water when:

(i) Displaced out of the cylinder

(ii) Left in the cylinder

Solution: 

According to the question 

Internal diameter of the cylindrical vessel (D)= 10 cm,

Radius of the cylindrical vessel (r) = 5 cm

Height of the cylindrical vessel (h) = 10.5 cm,

Base diameter of the solid cone = 7 cm,

Radius of the solid cone (R) = 3.5 cm

Height of the cone (L) = 6 cm,

(i)We find the volume of water displaced out from the cylinder which is equal to the volume of the cone 

So, V1 = 1/3 × πR²L

V1 = 1/3 × π3.5² × 6 = 77 cm³

Hence, the volume of the water displaced out of the cylinder is 77 cm³

(ii)First we find the volume of the cylindrical vessel is 

V2 = π × r² × h 

= π × 5² × 10.5

= 824.6 cm³ = 825 cm³

Now we find the volume of the water left in the cylinder 

V = V2 – V1

V = 825 - 77 = 748 cm³

Hence, the volume of the water left in cylinder is 748 cm³

Read More:

• Surface Areas and Volumes
• Practice Problems on Surface Area and Volume Formulas