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 cm^{2}

Volume = l × b × h

= 12 × 2 × 5

= 120 cm^{3}

Therefore, the total surface area of cuboid is 188 cm^{2} and volume is 120 cm^{3}.

**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 cm^{3}

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

= (6)^{3}

= 216 cm^{3}

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

= (1)^{3}

= 1 cm^{3}

The total volume of three cubes = 512 + 216 + 1 = 729 cm^{3}

After melting these cubes, a new cube is formed.

Thus, the volume of the cube formed = 729 cm^{3}

⇒ (edge)^{3} = 729

⇒ (edge)^{3} = (9)^{3}

⇒ Edge = 9 cm

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

= 6 × 9 × 9

= 486 cm^{2}

Hence, the total surface area of the new cube is 486 cm^{2}.

**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 cm^{2}

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 cm^{2}

Therefore, the total surface area of each equal part is 400 cm^{2}.

**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

√(l^{2} + b^{2} + h^{2}) = 11 cm

Squaring on both sides,

l^{2} + b^{2} + h^{2} = (11)^{2} = 121

We know that,

(l + b + h)2 = [l^{2} + b^{2} + h^{2}] + 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 cm^{2}

Hence, the total surface area of the cuboid is 240 cm^{2}.

**Question 7: A room with a square floor of side 6 m contains 180 m ^{3} 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 m^{3}

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 cm^{3}

Dimensions of the brick are:

Length = 22 cm

Breadth = 10 cm

Height = 17 cm

Volume of 1 brick = 22 × 10 × 17 cm^{3}

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 = √(l^{2} + b^{2} + h^{2})

= √(10^{2} + 8^{2} + 6^{2})

= √(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 cm^{2} (given)

2(lb + bh + hl) = 558

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

15x^{2} + 6x^{2} + 10x^{2} = 279

31x^{2} = 279

x^{2} = 279/31

x^{2} = 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 cm^{2}

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

= 132 + 2πr^{2}

= 132 + 2 × (22/7) × 3 × 3

= 132 + 56.57

= 188.57 cm^{2}

Volume of the cylinder = πr^{2}h

= (22/7) × 3 × 3 × 7

= 198 cm^{3}

**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 cm ^{2}.**

**Solution:**

Given

Height of the cylinder = h = 21 cm

Area of its one end = 154 cm^{2}

πr^{2} = 154

(22/7) × r^{2} = 154

r^{2} = (154 × 7)/22 = 49

r = √49

r = 7 cm

Volume of cylinder = πr^{2}h

= (22/7) × 7 × 7 × 21

= 22 × 7 × 21

= 3234 cm^{3}

Curved surface area = 2πrh

= 2 × (22/7) × 7 × 21

= 2 × 22 × 21

= 924 cm^{2}

Therefore, the volume of the cylinder is 3234 cm^{3} and the curved surface area is 924 cm^{2}.

**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 r_{1} and h_{1} be the radius and height of first the cylinder r_{2} and h_{2} be the radius and height of the second cylinder.

According to the given,

r_{1}/r_{2} = 2/3

h_{1}/h_{2} = 5/4

Curved surface area of the first cylinder S_{1} = 2πr_{1}h_{1}

Curved surface area of the second cylinder S_{2} = 2πr_{2}h_{2}

Ratio of the curved surface areas is:

S_{1}/S_{2} = (2πr_{1}h_{1})/(2πr_{2}h_{2})

= (r_{1}/r_{2}) × (h_{1}/h_{2})

= (2/3) × (5/4)

= 5/6

S_{1} : S_{2} = 5 : 6

Ratio of volumes is:

V_{1}/V_{2} = (πr_{1}^{2}h_{1}) / (πr_{2}^{2}h_{2})

= (r_{1}/r_{2})^{2} × (h_{1}/h_{2})

= (2/3)2 × (5/4)

= (4/9) × (5/4)

= 5/9

V_{1} : V_{2} = 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 cm ^{2}. 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 cm^{2}

⇒ 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

4πh^{2} = 154

4 × (22/7) × h^{2} = 154

h^{2} = (154 × 7)/(22 × 4)

h^{2} = 49/4

h = √(49/4) = 7/2 cm

Thus, r = 2 × (7/2) = 7 cm

Volume of the cylinder = πr^{2}h

= (22/7) × 7 × 7 × (7/2)

= 11 × 49

= 539 cm^{3}

**Question 15: Find the volume of the cylinder whose curved surface area is 660 cm ^{2} and height is 15 cm.**

**Solution:**

Let r be the radius of the cylinder.

Given,

Height of cylinder = h = 15cm

Curved surface area = 660 cm^{2}

2πrh = 660

2 × (22/7) × r × 15 = 660

r = (660 × 7)/(2 × 22 × 15)

r = 7 cm

Volume of the cylinder = πr^{2}h

= (22/7) × 7 × 7 × 15

= 2310 cm^{3}

**Question 16: If the volume of a cylinder is 30π cm ^{3} and the area of base is 6π cm^{2}, then find the height of the cylinder.**

**Solution:**

Let h be the height of the cylinder.

Given,

Volume of cylinder = 30π cm^{3}

Base area = 6π cm^{2}

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 cm ^{3} 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 cm^{3}

πr^{2}h = 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 cm^{2}

**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 cm^{2}.

**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(R^{2} – r^{2})h

= (22/7) × [(15)^{2} – (13)^{2}] × 20

= (22/7) × (225 – 169) × 20

= (22/7) × 56 × 20

= 3520

Therefore, the required volume is 3520 cm^{3}.

### 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 cm ^{3} 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 cm^{3}

(1/3)πr^{2}h = 1232

(1/3) × (22/7) × r^{2} × 24 = 1232

r^{2} = (1232 × 7 × 3) / (22 × 24)

r^{2} = 49

r = √49

r = 7 cm

Slant height of the cone = l = √(r^{2} + h^{2})

= √[(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 m^{2}

Hence, the total surface area of the cone is 704 m^{2}.

**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)πr^{2}h

= (1/3) × (22/7) × 6 × 6 × 8

= 2112/7

= 301.7 cm3

Therefore, the volume of the cone is 301.7 cm^{3}.

**Question 23: Find the radius of the base of a cone whose curved surface area is 1884.4 m ^{2} 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 m^{2}

∴ 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 cm ^{2}.**

**Solution:**

Given,

Slant height of the cone = l = 25 cm

Area of the base = 154 cm^{2}

πr^{2} = 154

(22/7) × r^{2} = 154

r^{2} = (154 × 7)/22

r^{2} = 7 × 7

r = √(7 × 7)

r = 7 cm

Let h be the height of the cone.

We know that,

l^{2} = r^{2} + h^{2}

(25)^{2} = (7)^{2} + h^{2}

⇒ h^{2} = 625 – 49

⇒ h^{2} = 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 cm ^{2}, 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 r_{1} and r_{2} be the radius of bigger and smaller cones, respectively.

⇒ r_{1} = r_{2}

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

l_{1}/l_{2} = 5/4 (given)

Curved surface area of smaller cone = 400 cm_{2}

Curved surface area of bigger cone/Curved surface area of smaller cone = πr_{1}l_{1}/πr_{2}l_{2}

Curved surface area of bigger cone = (l_{1}/l_{2}) × 400 [since r_{1} = r_{2}]

= (5/4) × 400

= 500

Therefore, the curved surface area of the bigger cone is 500 cm^{2}.

**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 m^{3}

**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 = πR^{2}H

= (1/3) × π × (10)^{2} × 18

= π × 100 × 6

= 600 π cm^{3}

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

πr^{2}h = 600 π

r^{2}h = 600

(5)^{2}h = 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

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

Height of the cone = h = 14 cm

Volume of the cone = (1/3)πr^{2}h

= (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 cm^{3}.

**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°

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

l^{2} = R^{2} + h^{2}

(12)^{2} = (4)^{2} + h^{2}

⇒ h^{2} = 144 – 16

⇒ h^{2} = 128

⇒ h = √128 = 11.31 cm

Volume of the cone = (1/3)πR^{2}h

= (1/3) × (22/7) × 4 × 4 × 11.31

= 3981.12/21

= 189.57 cm^{3}

### 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 = 4πr^{2}

= 4 × (22/7) × 1.4 × 1.4

= 24.64 cm^{2}

Volume = (4/3)πr^{3}

= (4/3) × (22/7) × 1.4 × 1.4 × 1.4

= 11.5 cm^{3}

Therefore, the surface of the sphere is 24.64 cm^{2} and volume is 11.5 cm^{3}.

**Question 31: Find the volume of the sphere whose surface area is 616 cm ^{2}.**

**Solution:**

Let r be the radius of the sphere.

Given,

Surface area of the sphere = 616 cm^{2}

4πr^{2} = 616

4 × (22/7) × r^{2} = 616

r^{2} = (616 × 7)/(22 × 4)

r^{2} = 49

r = √49

r = 7 cm

Volume = (4/3)πr^{3}

= (4/3) × (22/7) × 7 × 7 × 7

= 4312/3

= 1437.33

Hence, the volume of the sphere is 1437.33 cm^{3}.

**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 = 3πr^{2}

= 3 × (22/7) × 4.5 × 4.5

= 190.93 cm^{2}

Volume = (2/3)πr^{3}

= (2/3) × (22/7) × 4.5 × 4.5 × 4.5

= 4009.5/21

= 190.93 cm^{3}

Therefore, the surface of the hemisphere is 190.93 cm^{2} and volume is 190.93 cm^{3}.

**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 = πR^{2}H

= π × 4 × 4 × 10 cm^{3}

Radius of a sphere formed = r = 2 cm

Volume of sphere = (4/3)πr^{3}

= (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(R^{3} – r^{3})

= (4/3) × (22/7) × [(8)^{3} – (6)^{3}]

= (4/3) × (22/7) × (512 – 216)

= (4/3) × (22/7) × 296

= 1240.38 cm^{3}

Hence, the volume of the metal used in making the hollow spherical shell is 1240.38 cm^{3}.

**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)πr^{3}

= (4/3) × (22/7) × 9 × 9 × 9 cm^{3}

Radius of the cone = R = 3 cm

Height of the cone = h = 6 cm

Volume of the cone = (1/3)πR^{2}h

= (1/3) × (22/7) × 3 × 3 × 6 cm^{3}

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)πR^{3}

= (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)πr^{3}

(10)^{3} = 8r^{3}

(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 = 4πR^{2}

= 4 × π × (5)^{2}

= 4 × π × 25

= 100π

Therefore, the surface area of the sphere formed is 100π cm^{2}.

**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 cm^{3}

Diameter of sphere formed = 4.2 cm

Radius = r = 4.2/2 = 2.1 cm

Volume of sphere = (4/3)πr^{3}

= (4/3) × (22/7) × (2.1)^{3} cm^{3}

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 × R^{3}

= (2/3) × π × (9)^{3} cm^{3}

Diameter of cylindrical bottle = 3 cm

Radius = r = 3/2 = 1.5 cm

Height = h = 4 cm

Volume of the bottle = πr^{2}h

= π × 1.5 × 1.5 × 4 cm^{3}

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)πr^{3}

= (4/3) × π × (0.7/2)^{3}

= (4/3) × (22/7) × (0.7/2)^{3} cm^{3}

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 cm^{3}

Hence, the volume of water drawn out from a water tank is 539 cm^{3}.

**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 = 2πR^{2}

Internal surface area = 2πr^{2}

Total surface area of the bowl = 2π(R^{2} + r^{2})

= 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 cm ^{2}, 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 cm^{2}

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 cm ^{3}**

**(b) 20π cm ^{3}**

**(c) 36π cm ^{3}**

**(d) 30π cm ^{3}**

**Solution:**

Correct answer: (c)

Given,

Diameter of sphere = 6 cm

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

Volume of sphere = (4/3)πr^{3}

= (4/3) × π × (3)^{3}

= (4/3) × π × 3 × 3 × 3

= 36π cm^{3}

**Question 3: The volume and height of a cone is 308 cm ^{2} 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 cm^{3}

(1/3)πr^{2}h = 308

(1/3) × (22/7) × r^{2} × 6 = 308

r^{2} = (308 × 3 × 7)/(22 × 6)

r^{2} = 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 cm ^{2}.**

**Solution:**

Given,

Diameter of the hemisphere = 42 cm

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

Total surface area of hemisphere = 3πr^{2}

= 3 × (22/7) × 21 × 21

= 3 × 22 × 3 × 21

= 4158 cm2

∵ Cost of polishing 1 cm^{2} = 20 paise = Rs. 0.20

∴ Cost of polishing 4158 cm^{2} = 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 = V_{1} = (1/3)πr^{2}h

= (1/3)π(r^{2})r

= (1/3)πr^{3}

Volume of hemisphere = V_{2} = (2/3)πr^{3}

Volume of cylinder = V_{3} = πr^{2}h

= π(r^{2})r

= πr^{3}

Now,

V_{1} : V_{2} : V_{3} = (1/3)πr^{3} : (2/3)πr^{3} : πr^{3}

= (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 = πr^{2}h

= (22/7) × 7 × 7 × 40

= 6160 cm^{3}

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)πR^{2}H

= (1/3) × (22/7) × 7 × 7 × 12

= 616 cm^{3}

Volume of the solid = Volume cylinder + Volume of cone

= 6160 + 616

= 6776 cm^{3}

**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)πr^{3}

= (4/3) × (22/7) × (3)^{3} cm^{3}

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)πr^{3} = πR^{2}h

(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 m^{2}

Area planned by roller in 10 revolutions = 10 × 11 = 110 m^{2}