Class 10 RD Sharma Solutions Chapter 16 Surface Areas and Volumes Exercise 16.2 | Set 2 (original) (raw)
Last Updated : 23 Jul, 2025
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 v_olume of the cylinder
V1 = πr2 h1
= π(1.75)2 x 14/3 m3
__Now we find the v_olume 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 i_nternal 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 m 3 **and the volume is 56 m 3
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 c_urved surface area of the cylinder
A1 = 2 πr h
= 2 π(7) h
= 2 π(7)(90)
= 3948.40 cm2
__Now we find the c_urved 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 v_olume of the vessel
V = π(r22 - r12)h
= π(82 - 72) x 42 = 1980 cm3
**Hence, the **v**olume of the vessel is 1980 cm 3 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 v_olume 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 i_nner 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 cm 2
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 c_urved surface area of the cone
A1 = πrL
= π(3.5)(12) = 131.94 cm2
__Now we find the c_urved 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 cm 2
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 v_olume 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 v_olume of the cylinder
V1 = π × r2 × h
= π × 62 × 110 cm3
__Now we find the v_olume 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 v_olume of the cylinder
V1 = π × r2 × h
= π × 22 × 4 cm3
__Now we find the v_olume of the cone
V2 = 1/3 π r2l
= 1/3 × π × 22 × 2
= 1/3 × π x 4 × 2 cm3
__Now we find the v_olume 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 cm 3
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 v_olume of the circular cone
V1 = 1/3 × πr2l
= 1/3 × π × 602 × 120 = 452571.429 cm3
__Now we find the v_olume of the hemisphere
V2 = 2/3 × πr3
= 2/3 × π × 603 = 452571.429 cm3
__Now we find the v_olume 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 m 3
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 v_olume 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 cm 3
****(ii)** __First we find the v_olume of the cylindrical vessel is
V2 = π × r2 × h
= π × 52 × 10.5
= 824.6 cm3 = 825 cm3
__Now we find the v_olume 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 cm 3
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