RBSE Maths Class 10 Chapter 16: Surface Area and Volume Important Questions and Solutions

RBSE Class 10 Maths Chapter 16 – Surface Area and Volume Important questions and solutions are available here. All these important questions are given to help the students in clearing the exams with flying colours. The RBSE Class 10 important questions and solutions provided at BYJU’S contain detailed explanations for all questions.

Chapter 16 of RBSE Class 10 has four exercises; each of these exercises contains questions on three-dimensional figures, namely cube, cuboid, cylinder, cone, sphere and hemisphere. Enough questions are given here which cover problems related to the curved surface area, total surface area, volume and the combination of these components.

RBSE Maths Chapter 16: Exercise 16.1 Textbook Important Questions and Solutions

Question 1: The length of a cuboid is 12 cm, breadth is 2 cm and height is 5 cm. Find the total surface area of the cuboid and volume of the cuboid.

Solution:

Given dimensions of the cuboid are:

Length = l = 12 cm

Breadth = b = 2 cm

Height = h = 5 cm

Total surface area of the cuboid = 2(lb + bh + hl)

= 2[(12 × 2) + (2 × 5) +(5 × 12)]

= 2[24 + 10 + 60]

= 2 × 94

= 188 cm2

Volume = l × b × h

= 12 × 2 × 5

= 120 cm3

Therefore, the total surface area of cuboid is 188 cm2 and volume is 120 cm3.

Question 2: The edges of three cubes are 8 cm, 6 cm and 1 cm, respectively. After melting these cubes a new cube is formed. Find the total surface area of the new cube.

Solution:

Given,

Edges of three cubes are 8 cm, 6 cm and 1 cm respectively.

Volume of the cube with edge 8 cm = (edge)3

= (8)3

= 512 cm3

Volume of the cube with edge 6 cm = (edge)3

= (6)3

= 216 cm3

Volume of the cube with edge 1 cm = (edge)3

= (1)3

= 1 cm3

The total volume of three cubes = 512 + 216 + 1 = 729 cm3

After melting these cubes, a new cube is formed.

Thus, the volume of the cube formed = 729 cm3

⇒ (edge)3 = 729

⇒ (edge)3 = (9)3

⇒ Edge = 9 cm

Total surface area of new cube = 6(edge)2

= 6 × 9 × 9

= 486 cm2

Hence, the total surface area of the new cube is 486 cm2.

Question 3: The dimensions of a box are 50 cm × 36 cm × 25 cm. How much sq.cm cloth will be required for making the cover of this box?

Solution:

Given dimensions of the cuboidal box are:

Length = l = 50 cm

Breadth = b = 36 cm

Height = h = 25 cm

Required cloth to make its cover = Surface area of the box

= 2(lb + bh + hl)

= 2[(50 × 36) + (36 × 25) + (25 × 50)]

= 2[1800 + 900 + 1250]

= 2 × 3950

= 7900 sq.cm

Hence, 7900 sq.cm cloth is required to make the cover of the box.

Question 4: Each surface area of a cube is 100 cm2. The cube is cut into two equal parts by a plane which is parallel to the base, then find the total surface area of equal part.

Solution:

Given,

Area of each face of cube = 100 cm2

Side of the cube = √100 = 10 cm

When a cube is cut into two equal parts by a plane which is parallel to its base then two cuboids will form.

Thus, length of each cube = l = 10 cm

Breadth = b = 5 cm

Height = h = 10 cm

Surface area of each cuboid formed

= 2(lb + bh + hl)

= 2[(10 × 5) + (5 × 10) + (10 × 10)]

= 2[50 + 50 + 100]

= 2 × 200

= 400 cm2

Therefore, the total surface area of each equal part is 400 cm2.

Question 5: A box without a lid is made by wood of thickness 3 cm. Its outer length is 146 cm, breadth is 116 cm and height is 83 cm. Find the cost of painting the internal surface of the box at the rate of Rs. 2 per 1000 sq.cm.

Solution:

Given external dimensions of the box are:

Length = 146 cm

Breadth = 116 cm

Height = 83 cm

Width of the wood = 3 cm

Thus, the internal dimensions of the box are:

Length = l = (146 – 3 – 3) = 146 – 6 = 140 cm

Breadth = b = (116 – 3 – 3) = 116 – 6 = 110 cm

Height = h = 83 – 3 = 80 cm

Internal surface area of the box without a lid = 2(l + b) × h + l × b

= 2(140 + 110) × 80 + (140 × 110)

= 160 × 250 + 15400

= 55400 sq. cm

Cost of painting 1000 sq.cm = Rs. 2

Cost of painting 55400 sq.cm = (55400/1000) × Rs. 2

= Rs. 110.8

Question 6: The sum of length, breadth and height of cuboid is 19 cm and length of its diagonal is 11 cm. Find the total surface area of the cuboid.

Solution:

Given,

Sum of the length, breadth and height of cuboid = 19 cm

l + b + h = 19 cm

Length of its diagonal = 11 cm

√(l2 + b2 + h2) = 11 cm

Squaring on both sides,

l2 + b2 + h2 = (11)2 = 121

We know that,

(l + b + h)2 = [l2 + b2 + h2] + 2(lb + bh + hl)

(19)2 = 121 + 2(lb + bh + hl)

361 = 121 + 2(lb + bh + hl)

⇒ 2(lb + bh + hl) = 361 – 121

⇒ 2(lb + bh + hl) = 240 cm2

Hence, the total surface area of the cuboid is 240 cm2.

Question 7: A room with a square floor of side 6 m contains 180 m3 air. Find the height of the room.

Solution:

Let h be the height of the room.

Given,

Floor of the room is square shaped.

Thus, the length of the room = l = 6 m

Breadth of the room = b = 6 m

Volume of the room = 180 m3

l × b × h = 180

6 × 6 × h = 180

h = 180/(6 × 6) = 5 m

Therefore, the height of the room is 5 m.

Question 8: How many bricks are required to make a wall of dimensions 44 m long, 1.5 m height and 85 cm broad if dimensions of 1 brick are 22 m × 10 cm × 7 cm?

Solution:

Given,

Length of the wall = l = 44 = 44 × 100 cm = 4400 cm

Height of the wall = h = 1.5 m = 1.5 × 100 = 150 cm

Width of the wall = w = 85 cm

Volume of the wall = lwh

= 4400 × 85 × 150 cm3

Dimensions of the brick are:

Length = 22 cm

Breadth = 10 cm

Height = 17 cm

Volume of 1 brick = 22 × 10 × 17 cm3

Number of bricks = (Volume of the wall)/(Volume of 1 brick)

= (4400 × 85 × 150) / (22 × 10 × 17)

= 15000

Question 9: Find the maximum length of rod which can be kept in a room of size 10 m × 8 m × 6 m.

Solution:

Given dimensions of the room are:

Length = l = 10 m

Breadth = b = 8 m

Height = h = 6 m

The maximum length of the rod that can be placed in a room will be its length of the diagonal.

Diagonal = √(l2 + b2 + h2)

= √(102 + 82 + 62)

= √(100 + 64 + 36)

= √200

= 10√2 m

Hence, the required length of the rod is 10√2 m.

Question 10: The ratio of length, breadth and height of a cuboid are 5 : 3 : 2. If the total surface area of the cuboid is 558 cm2, then find its dimensions.

Solution:

Given,

Ratio of length, breadth and height of a cuboid = 5 : 3 : 2

Let 5x, 3x and 2x be its length, breadth and height.

i.e. l = 5x, b = 3x, h = 2x

Total surface area of cuboid = 558 cm2 (given)

2(lb + bh + hl) = 558

(5x × 3x) + (3x × 2x) + (2x × 5x) = 558/2

15x2 + 6x2 + 10x2 = 279

31x2 = 279

x2 = 279/31

x2 = 9

x = √9 = 3 cm

Hence, the dimensions of the cuboid are:

Length = 5x = 5(3) = 15 cm

Breadth = 3x = 3(3) = 9 cm

Height = 2x = 2(3) = 6 cm

RBSE Maths Chapter 16: Exercise 16.2 Textbook Important Questions and Solutions

Question 11: Find the curved surface area, total surface area and volume of a right circular cylinder having radius of the base 3 cm and height 7 cm.

Solution:

Given,

Radius of the base of cylinder = r = 3 cm

Height of the cylinder = h = 7 cm

Curved surface area of cylinder = 2πrh

= 2 × (22/7) × 3 × 7

= 2 × 22 × 3

= 132 cm2

Total surface area = Curved surface area + 2 × Area of the base

= 132 + 2πr2

= 132 + 2 × (22/7) × 3 × 3

= 132 + 56.57

= 188.57 cm2

Volume of the cylinder = πr2h

= (22/7) × 3 × 3 × 7

= 198 cm3

Question 12: Find the curved surface area and volume of the cylinder whose height is 21 cm and area of its one end is 154 cm2.

Solution:

Given

Height of the cylinder = h = 21 cm

Area of its one end = 154 cm2

πr2 = 154

(22/7) × r2 = 154

r2 = (154 × 7)/22 = 49

r = √49

r = 7 cm

Volume of cylinder = πr2h

= (22/7) × 7 × 7 × 21

= 22 × 7 × 21

= 3234 cm3

Curved surface area = 2πrh

= 2 × (22/7) × 7 × 21

= 2 × 22 × 21

= 924 cm2

Therefore, the volume of the cylinder is 3234 cm3 and the curved surface area is 924 cm2.

Question 13: Find the ratio of curved surface area and volume of two right circular cylinders whose radii are in the ratio 2 : 3 and heights are in the ratio 5 : 4.

Solution:

Let r1 and h1 be the radius and height of first the cylinder r2 and h2 be the radius and height of the second cylinder.

According to the given,

r1/r2 = 2/3

h1/h2 = 5/4

Curved surface area of the first cylinder S1 = 2πr1h1

Curved surface area of the second cylinder S2 = 2πr2h2

Ratio of the curved surface areas is:

S1/S2 = (2πr1h1)/(2πr2h2)

= (r1/r2) × (h1/h2)

= (2/3) × (5/4)

= 5/6

S1 : S2 = 5 : 6

Ratio of volumes is:

V1/V2 = (πr12h1) / (πr22h2)

= (r1/r2)2 × (h1/h2)

= (2/3)2 × (5/4)

= (4/9) × (5/4)

= 5/9

V1 : V2 = 5 : 9

Hence, the required ratio of curved surface areas is 5 : 6 and volumes is 5 : 9.

Question 14: The total surface area of a solid cylinder is 462 cm2. Its curved surface area is one third of total surface area. Find the volume of the cylinder.

Solution:

Let r be the radius and h be the height of the cylinder.

Given,

Total surface area = 462 cm2

⇒ 2πr(h+r) = 462 ….(i)

According to the given,

Curved surface area = (1/3) × (total surface area)

= (1/3) × 462

2πrh = 154 ….(ii)

Dividing (i) by (ii),

2πr(h + r)/2πrh = 462/154

(h + r)/h = 3

h + r = 3h

r = 3h – h

r = 2h ….(iii)

Substituting (iii) in (ii),

2π(2h)h = 154

4πh2 = 154

4 × (22/7) × h2 = 154

h2 = (154 × 7)/(22 × 4)

h2 = 49/4

h = √(49/4) = 7/2 cm

Thus, r = 2 × (7/2) = 7 cm

Volume of the cylinder = πr2h

= (22/7) × 7 × 7 × (7/2)

= 11 × 49

= 539 cm3

Question 15: Find the volume of the cylinder whose curved surface area is 660 cm2 and height is 15 cm.

Solution:

Let r be the radius of the cylinder.

Given,

Height of cylinder = h = 15cm

Curved surface area = 660 cm2

2πrh = 660

2 × (22/7) × r × 15 = 660

r = (660 × 7)/(2 × 22 × 15)

r = 7 cm

Volume of the cylinder = πr2h

= (22/7) × 7 × 7 × 15

= 2310 cm3

Question 16: If the volume of a cylinder is 30π cm3 and the area of base is 6π cm2, then find the height of the cylinder.

Solution:

Let h be the height of the cylinder.

Given,

Volume of cylinder = 30π cm3

Base area = 6π cm2

Volume of cylinder = Base area × Height

30π = 6π × h

⇒ h = 30π/6π

⇒ h = 5 cm

Therefore, the height of the cylinder is 5 cm.

Question 17: 30800 cm3 water can be filled in a cylindrical vessel. If its internal radius is 14 cm, then find its internal curved surface area.

Solution:

Let h be the height of the vessel.

Given,

The internal radius of vessel = r = 14 cm

Volume of the vessel = 30800 cm3

πr2h = 30800

(22/7) × 14 × 14 × h = 30800

h = (30800 × 7) / (22 × 14 × 14)

h = 50 cm

Internal surface area of the cylindrical vessel = 2πrh

= 2 × (22/7) × 14 × 50

= 4400 cm2

Question 18: If the width of the hollow cylinder is 2 cm, its internal diameter is 14 cm, height is 26 cm and both the ends of the cylinder are open, then find the total surface area of the hollow cylinder.

Solution:

Given,

Internal diameter of hollow cylinder = 14 cm

Height = h = 26 cm

∴ Internal radius = r = 7 cm

Width of the cylinder = 2 cm

∴ External radius = R = 7 + 2 = 9 cm

Total surface area of cylinder = 2π(R + r)(h + R – r)

= 2 × (22/7) × (9 + 7)(26 + 9 – 7)

= 2 × (22/7) × 16 × 28

= 44 × 16 × 4

= 2816

Hence, the total surface area of the cylinder is 2816 cm2.

Question 19: If both ends of a hollow cylinder are open, its height is 20 cm, internal and external radii are 26 cm and 30 cm, respectively, then find the volume of this hollow cylinder.

Solution:

Given,

Height of the cylinder = h = 20 cm

Internal diameter = 26 cm

External diameter = 30 cm

Internal radius = r = 26/2 = 13 cm

External radius = R = 30/2 = 15 cm

Volume the hollow cylinder = π(R2 – r2)h

= (22/7) × [(15)2 – (13)2] × 20

= (22/7) × (225 – 169) × 20

= (22/7) × 56 × 20

= 3520

Therefore, the required volume is 3520 cm3.

RBSE Maths Chapter 16: Exercise 16.3 Textbook Important Questions and Solutions

Question 20: Find the slant height of the right circular cone whose volume is 1232 cm3 and height is 24 cm.

Solution:

Let r be the radius of the cone.

Given,

Height of the cone = h = 24 cm

Volume of the cone = 1232 cm3

(1/3)πr2h = 1232

(1/3) × (22/7) × r2 × 24 = 1232

r2 = (1232 × 7 × 3) / (22 × 24)

r2 = 49

r = √49

r = 7 cm

Slant height of the cone = l = √(r2 + h2)

= √[(7)2 + (24)2]

= √(49 + 576)

= √625

= 25

Therefore, the radius of the cone is 7 cm and slant height is 25 cm.

Question 21: Find the total surface area of the cone whose diameter of base is 14 m, and slant height is 25 m.

Solution:

Given,

Slant height of the cone = l = 25 m

Diameter of the base of cone = 14 m

Radius of cone = r = 14/2 = 7 cm

Total surface area of the cone = πr(l+r)

= (22/7) × 7 × (25 + 7)

= (22/7) × 7 × 32

= 22 × 32

= 704 m2

Hence, the total surface area of the cone is 704 m2.

Question 22: Find the volume of the right circular cone whose radius of base is 6 cm and height is 8 cm.

Solution:

Given,

Radius of base of the cone = r = 6 cm

Height of the cone = h = 8 cm

Volume of the cone = (⅓)πr2h

= (1/3) × (22/7) × 6 × 6 × 8

= 2112/7

= 301.7 cm3

Therefore, the volume of the cone is 301.7 cm3.

Question 23: Find the radius of the base of a cone whose curved surface area is 1884.4 m2 and its slant height is 12 m.

Solution:

Let r be the radius of base of the cone.

Given,

Slant height of cone = l = 12 m

Lateral surface area = 1884.4 m2

∴ πrl = 1884.4

(22/7) × 12 × r = 1884.4

r = (1884.4 × 7)/(22 × 12)

= 50 m (approx)

Therefore, the radius of the cone is 50 m.

Question 24: Find the height of the right circular cone of slant height 25 cm and the area of its base is 154 cm2.

Solution:

Given,

Slant height of the cone = l = 25 cm

Area of the base = 154 cm2

πr2 = 154

(22/7) × r2 = 154

r2 = (154 × 7)/22

r2 = 7 × 7

r = √(7 × 7)

r = 7 cm

Let h be the height of the cone.

We know that,

l2 = r2 + h2

(25)2 = (7)2 + h2

⇒ h2 = 625 – 49

⇒ h2 = 576

⇒ h = √576

⇒ h = 24 cm

Therefore, the height of the cone is 24 cm.

Question 25: The base of two cones are of the same diameter. Ratio of their slant height is 5 : 4. If the curved surface area of a smaller cone is 400 cm2, then find the curved surface area of the bigger one.

Solution:

Given that, two cones have the same base diameter.

Thus, their radii is equal.

Let r1 and r2 be the radius of bigger and smaller cones, respectively.

⇒ r1 = r2

Let l1 and l2 be the slant heights of two cones.

l1/l2 = 5/4 (given)

Curved surface area of smaller cone = 400 cm2

Curved surface area of bigger cone/Curved surface area of smaller cone = πr1l1/πr2l2

Curved surface area of bigger cone = (l1/l2) × 400 [since r1 = r2]

= (5/4) × 400

= 500

Therefore, the curved surface area of the bigger cone is 500 cm2.

Question 26: The circumference of the base of a conical tent is 9 m and height is 44 m. Find the volume of air inside it.

Solution:

Given,

Height of the conical tent = h = 9 m

Circumference of the base = 44 m

2πr = 44

2 × (22/7) × r = 44

r = (44 × 7)/(22 × 2)

r = 7 cm

Volume of the air inside the conical tent = (1/3)πr2h

= (1/3) × (22/7) × 7 × 7 × 9

= 22 × 21

= 462 m3

Question 27: The radius and height of a conical vessel are 10 cm and 18 cm, respectively, which is filled with water to the brim. It is poured in a cylindrical vessel of radius 5 cm. Find the height of the water level in a cylindrical vessel.

Solution:

Given,

Base radius of conical vessel = R = 10 cm

Height = H = 18 cm

Volume of the conical vessel = πR2H

= (1/3) × π × (10)2 × 18

= π × 100 × 6

= 600 π cm3

Let h be the water level in the cylindrical vessel.

Radius of cylindrical vessel = r = 5 cm

According to the given,

Volume of water in the cylindrical vessel = Volume of water in conical vessel

πr2h = 600 π

r2h = 600

(5)2h = 600

25h = 600

h = 600/25

h = 24 cm

Hence, the water level in the cylindrical vessel is 24 cm.

Question 28: A cone of maximum height is cut from a cube of edge 14 cm. Find the volume of the cone.

Solution:

Given,

Edge of the cube = a = 14 cm

Diameter of the largest cone can be cut off from the cube = 14 cm

RBSE class 10 maths chapter 16 imp que 28 sol

Radius of the cone = r = 14/2 = 7 cm

Height of the cone = h = 14 cm

Volume of the cone = (1/3)πr2h

= (1/3) × (22/7) × 7 × 7 × 14

= 2156/3

= 718.67

Therefore, the volume of the conical piece cut off from the cube is 718.67 cm3.

Question 29: The radius of a sector is 12 cm and angle is 120°. By coinciding its straight sides, a cone is formed. Find the volume of that cone.

Solution:

Radius of the sector = OB = OB = r = 12 cm

Angle of the sector = θ = 120°

RBSE class 10 maths chapter 16 imp que 29 sol

Length of the arc AB = πrθ/180°

= [π × 12 × 120°]/180°

= 8π cm

Let R be the radius and h be the height of the cone formed when straight lines of the sector are joined.

Thus, 2πR = 8π

R = 4 cm

Also, slant height of the cone = l = 12 cm

l2 = R2 + h2

(12)2 = (4)2 + h2

⇒ h2 = 144 – 16

⇒ h2 = 128

⇒ h = √128 = 11.31 cm

Volume of the cone = (1/3)πR2h

= (1/3) × (22/7) × 4 × 4 × 11.31

= 3981.12/21

= 189.57 cm3

RBSE Maths Chapter 16: Exercise 16.4 Textbook Important Questions and Solutions

Question 30: Find the surface area and volume of a sphere of radius 1.4 cm.

Solution:

Given,

Radius of sphere = r = 1.4 cm

Surface area = 4πr2

= 4 × (22/7) × 1.4 × 1.4

= 24.64 cm2

Volume = (4/3)πr3

= (4/3) × (22/7) × 1.4 × 1.4 × 1.4

= 11.5 cm3

Therefore, the surface of the sphere is 24.64 cm2 and volume is 11.5 cm3.

Question 31: Find the volume of the sphere whose surface area is 616 cm2.

Solution:

Let r be the radius of the sphere.

Given,

Surface area of the sphere = 616 cm2

4πr2 = 616

4 × (22/7) × r2 = 616

r2 = (616 × 7)/(22 × 4)

r2 = 49

r = √49

r = 7 cm

Volume = (4/3)πr3

= (4/3) × (22/7) × 7 × 7 × 7

= 4312/3

= 1437.33

Hence, the volume of the sphere is 1437.33 cm3.

Question 32: Find the surface area and volume of a hemisphere of radius 4.5 cm.

Solution:

Given,

Radius of hemisphere = r = 4.5 cm

Surface area = 3πr2

= 3 × (22/7) × 4.5 × 4.5

= 190.93 cm2

Volume = (2/3)πr3

= (2/3) × (22/7) × 4.5 × 4.5 × 4.5

= 4009.5/21

= 190.93 cm3

Therefore, the surface of the hemisphere is 190.93 cm2 and volume is 190.93 cm3.

Question 33: A cylinder is made of lead whose radius is 4 cm and height is 10 cm. By melting this, how many spheres of radius 2 cm can be formed?

Solution:

Given,

Radius of the cylinder = R = 4 cm

Height = H = 10 cm

Volume = πR2H

= π × 4 × 4 × 10 cm3

Radius of a sphere formed = r = 2 cm

Volume of sphere = (4/3)πr3

= (4/3) × π × (2)3

= (4/3) × π × 2 × 2 × 2

Let n be the number of spheres formed after melting.

Volume of the cylinder = n × Volume of sphere

⇒ n = [π × 4 × 4 × 10] / [(4/3) × π × 2 × 2 × 2]

= (4 × 4 × 10 × 3)/(4 × 4 × 2)

= 15

Hence, 15 spheres are formed.

Question 34: A hollow spherical shell is 2 cm thick. If its outer radius is 8 cm, then find the volume of metal used in it.

Solution:

Given,

External radius of the hollow spherical shell R = 8 cm

Thickness of spherical shell = 2 cm

Internal radius r = 8 – 2 = 6

Volume of the metal used = (4/3)π(R3 – r3)

= (4/3) × (22/7) × [(8)3 – (6)3]

= (4/3) × (22/7) × (512 – 216)

= (4/3) × (22/7) × 296

= 1240.38 cm3

Hence, the volume of the metal used in making the hollow spherical shell is 1240.38 cm3.

Question 35: How many cones of 3 cm radius and 6 cm height are formed by melting a metallic sphere of radius 9 cm?

Solution:

Given,

Radius of sphere = r = 9 cm

Volume of the sphere = (4/3)πr3

= (4/3) × (22/7) × 9 × 9 × 9 cm3

Radius of the cone = R = 3 cm

Height of the cone = h = 6 cm

Volume of the cone = (1/3)πR2h

= (1/3) × (22/7) × 3 × 3 × 6 cm3

Let n be the number of cones formed.

∴ Volume of sphere = n × Volume of a cone

n = Volume of sphere/Volume of cone

= [(4/3) × (22/7) × 9 × 9 × 9] / [(1/3) × (22/7) × 3 × 3 × 6]

= 54

Hence, the required number of cones is 54.

Question 36: Eight spheres of the same radius are formed from a metallic sphere of 10 cm radius. Find the surface area of each sphere so obtained.

Solution:

Given,

Radius of the metallic sphere = R = 10 cm

∴ Volume of sphere = (4/3)πR3

= (4/3) × π × (10)3

Let r be the radius of each sphere formed.

Volume of large sphere = 8 × volume of spheres with radius r

(4/3) × π × (10)3 = 8 × (4/3)πr3

(10)3 = 8r3

(2R)3 = (10)3

2R = 10

R = 10/2 = 5 cm

Thus, the radius of the sphere casted = R = 5 cm

Surface of the sphere = 4πR2

= 4 × π × (5)2

= 4 × π × 25

= 100π

Therefore, the surface area of the sphere formed is 100π cm2.

Question 37: The dimensions of a solid rectangular slab of lead is 66 cm, 42 cm and 21 cm, respectively. Find by melting this, how many spheres of diameter 4.2 cm can be formed?

Solution:

Given dimensions of the cuboid are:

Length = l = 66 cm

Breadth = b = 42 cm

Height = h = 21 cm

Volume of cuboid = l × b × h

= 66 × 42 × 21 cm3

Diameter of sphere formed = 4.2 cm

Radius = r = 4.2/2 = 2.1 cm

Volume of sphere = (4/3)πr3

= (4/3) × (22/7) × (2.1)3 cm3

Let n be the number of spheres formed by melting the cuboid.

Volume of cuboid = n × volume of sphere

66 × 42 × 21 = n × (4/3) × (22/7) × (2.1)3

n = (66 × 42 × 21 × 7 × 3) / (4 × 22 × 2.1 × 2.1 × 2.1)

= 1500

Hence, the required number of spheres is 1500.

Question 38: A hemispherical bowl of internal radius 9 cm is full of liquid. This liquid is to be filled into cylindrical shaped small bottles each of diameter 3 cm and height 4 cm. How many bottles are necessary to empty the bowl?

Solution:

Given,

Radius of hemispherical bowl = R = 9 cm

Volume of the bowl = (2/3) × π × R3

= (2/3) × π × (9)3 cm3

Diameter of cylindrical bottle = 3 cm

Radius = r = 3/2 = 1.5 cm

Height = h = 4 cm

Volume of the bottle = πr2h

= π × 1.5 × 1.5 × 4 cm3

Let n be the number of bottles will be needed to fill the whole Liquid.

Volume of hemispherical bowl = n × volume of cylindrical bottles

(2/3) × π × (9)3 = n × π × 1.5 × 1.5 × 4

⇒ n = (2 × 9 × 9 × 9) / (3 × 1.5 × 1.5 × 4)

= 54

Hence, the required number of bottles is 54.

Question 39: The diameter of a sphere is 0.7 cm. If 3000 spheres completely filled with water are drawn out from a water tank, then find the volume of water drawn out.

Solution:

Given,

Diameter of sphere = 0.7 cm

Radius of the sphere = r = 0.7/2 cm

Volume of sphere V = (4/3)πr3

= (4/3) × π × (0.7/2)3

= (4/3) × (22/7) × (0.7/2)3 cm3

Volume of water drawn out = 3000 × Volume of sphere

= 3000 × (4/3) × (22/7) × (0.7/2)3

= (3000 × 4 × 22

× 0.7 × 0.7 × 0.7) / (3 × 7 × 2 × 2 × 2)

= 539 cm3

Hence, the volume of water drawn out from a water tank is 539 cm3.

Question 40: A hollow hemispherical vessel has external and internal radius as 43 cm and 42 cm, respectively. If the cost of colouring it is 7 paisa per sq.cm, then find the cost of painting the vessel.

Solution:

Given,

External diameter of hemispherical bowl = 43 cm

Internal diameter = 42 cm

∴ External radius = R = 43/2 cm

Internal radius = r = 42/2 = 21 cm

External surface area of bowl = 2πR2

Internal surface area = 2πr2

Total surface area of the bowl = 2π(R2 + r2)

= 2 × (22/7) × [(43/2)2 + (21)2]

= 2 × (22/7) × (462.25 + 441)

= 2 × (22/7) × 903.25 sq.cm

Cost of colouring 1 sq.cm = 7 paisa = Rs. 0.07

Total cost of painting the bowl = 2 × (22/7) × 903.25 × 0.07

= Rs. 397.43

RBSE Maths Chapter 16: Additional Important Questions and Solutions

Question 1: The total surface area of a cube is 486 cm2, edge of cube is

(a) 6 cm

(b) 8 cm

(c) 9 cm

(d) 7 cm

Solution:

Correct answer: (c)

Let a be the edge of the cube.

Given,

Total surface area of the cube = 486 cm2

6a2 = 486

a2 = 486/6

a2 = 81

a = √81

a = 9 cm

Question 2: The diameter of a sphere is 6 cm, the volume of sphere will be

(a) 16π cm3

(b) 20π cm3

(c) 36π cm3

(d) 30π cm3

Solution:

Correct answer: (c)

Given,

Diameter of sphere = 6 cm

Radius of sphere = r = 6/2 = 3 cm

Volume of sphere = (4/3)πr3

= (4/3) × π × (3)3

= (4/3) × π × 3 × 3 × 3

= 36π cm3

Question 3: The volume and height of a cone is 308 cm2 and 6 cm respectively. Radius of its base will be:

(a) 7 cm

(b) 8 cm

(c) 6 cm

(d) none of these

Solution:

Correct answer: (a)

Let r be the radius of the base of the cone.

Given,

Height of the cone = h = 6 cm

Volume of cone = 308 cm3

(1/3)πr2h = 308

(1/3) × (22/7) × r2 × 6 = 308

r2 = (308 × 3 × 7)/(22 × 6)

r2 = 49

r = √49

r = 7 cm

Question 4: A solid metallic hemisphere has diameter 42 cm. Find the cost of polishing the total surface at the rate of 20 paisa per cm2.

Solution:

Given,

Diameter of the hemisphere = 42 cm

Radius of the hemisphere = r = 42/2 = 21 cm

Total surface area of hemisphere = 3πr2

= 3 × (22/7) × 21 × 21

= 3 × 22 × 3 × 21

= 4158 cm2

∵ Cost of polishing 1 cm2 = 20 paise = Rs. 0.20

∴ Cost of polishing 4158 cm2 = 4158 × Rs. 0.20

= Rs. 831.60

Question 5: A cone, a hemisphere and a cylinder are formed by the same radius and same height. Write ratio of their volumes.

Solution:

Given,

A cone, a hemisphere and a cylinder are formed by the same radius and same height.

Thus, for each of these, r = h.

Volume of cone = V1 = (1/3)πr2h

= (1/3)π(r2)r

= (1/3)πr3

Volume of hemisphere = V2 = (2/3)πr3

Volume of cylinder = V3 = πr2h

= π(r2)r

= πr3

Now,

V1 : V2 : V3 = (1/3)πr3 : (2/3)πr3 : πr3

= (1/3) : (2/3) : 1

= 1 : 2 : 3

Hence, the required ratio of volumes is 1 : 2 : 3.

Question 6: The left part of a solid body is cylindrical and the right part is conical. If the diameter of cylindrical is 14 cm and length is 40 cm and diameter of cone is 14 cm and height is 12 cm, then find the volume of solid.

Solution:

Given,

Diameter of cylindrical portion = 14 cm

Radius = r = 14/2 = 7 cm

Height = h = 40 cm

Volume of the cylindrical portion V1 = πr2h

= (22/7) × 7 × 7 × 40

= 6160 cm3

Given that diameter of conical portion = 14 cm

Radius of the conical part = R = 14/2 = 7 cm

Height = H = 12 cm

Volume of conical portion V2 = (1/3)πR2H

= (1/3) × (22/7) × 7 × 7 × 12

= 616 cm3

Volume of the solid = Volume cylinder + Volume of cone

= 6160 + 616

= 6776 cm3

Question 7: A sphere of 6 cm diameter is dropped into a cylindrical vessel of diameter 12 cm. Find the rise in water in the vessel.

Solution:

Given,

Diameter of sphere = 6 cm

Radius of sphere = r = 6/2 = 3 cm

Volume of sphere = (4/3)πr3

= (4/3) × (22/7) × (3)3 cm3

Diameter of cylindrical vessel = 12 cm (given)

Radius of cylindrical vessel = R = 12/2 = 6 cm

Let h be the rise in water level when the sphere is put into the cylindrical vessel.

∴ Volume of sphere = Volume of water rise in the vessel

(4/3)πr3 = πR2h

(4/3) × (22/7) × (3)3 = (22/7) × (6)2 × h

4 × 3 × 3 = 6 × 6 × h

⇒ h = (4 × 3 × 3)/(6 × 6)

⇒ h = 1 cm

Therefore, the rise in water level is 1 cm.

Question 8: The length and diameter of a roller are 2.5 m and 1.4 m, respectively. How much area will be planned by roller in 10 revolutions?

Solution:

Given,

Length of the roller = h = 2.5 m

Diameter of roller = 1.4 m

Radius of the roller = r = 1.4/2 = 0.7 m

Area of flat surface by roller in 1 revolution = Curved surface area of cylinder

= 2πrh

= 2 × (22/7) × 0.7 × 2.5

= 44 × 0.1 × 2.5

= 11 m2

Area planned by roller in 10 revolutions = 10 × 11 = 110 m2

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