Class 10 RD Sharma Solutions - Chapter 16 Surface Areas and Volumes - Exercise 16.1 | Set 2

Question 19. How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid 11 cm x 10 cm x 7 cm? 

Solution:

Diameter of the coin = 1.75 cm

Radius = 1.75/2 = 0.875 cm

Thickness or height = 2 mm = 0.2 cm

Volume of the cylinder = πr²h

                                    = π 0.875² × 0.2

Volume of the cuboid = 11 × 10 × 7 cm³

Let the number of coins needed to be melted be n.

Volume of cuboid = n × Volume of cylinder

11 × 10 × 7 = π 0.875² × 0.2 x n

11 × 10 × 7 = 22/7 x 0.875² × 0.2 x n

n = 1600

Therefore, the number of coins required are 1600

Question 20. The surface area of a solid metallic sphere is 616 cm². It is melted and recast into a cone of height 28 cm. Find the diameter of the base of the cone so formed.  

Solution:

Radius of sphere = r cm

Surface area of the solid metallic sphere = 616 cm³

Surface area of the sphere = 4πr²

4πr² = 616

r² = 49

r = 7

Radius of the solid metallic sphere = 7 cm

Let radius of cone be x cm

Volume of the cone = 1/3 πx²h

                                = 1/3 πx² (28) ….. (i)

Volume of the sphere = 4/3 πr³

                                   = 4/3 π7³  ……….  (ii)

(i) and (ii) are equal

1/3 πx² (28) = 4/3 π7³

x² (28) = 4 x 7³

x² = 49

r =7

Diameter of the cone = 7 x 2 = 14 cm

Therefore, the diameter of the base of the cone is 14 cm

Question 21.A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution:

Height of the cylindrical bucket = 32 cm

Radius of the cylindrical bucket = 18 cm

Volume of cylinder = π × r² × h

Volume of cone = 1/3 π × r² × h

Volume of the conical heap = Volume of the cylindrical bucket

1/3 π × r² × 24 = π × 18² × 32

r² = 18² x 4

r = 18 x 2 = 36 cm

Let slant height of conical heap be l cm

l = √(h² + r²)

l = √(24² + 36²) = √1872

l = 43.26 cm

Therefore, the radius = 36cm slant height of the conical heap = 43.26 cm 

Question 22. A solid metallic sphere of radius 5.6 cm is melted and solid cones each of radius 2.8 cm and height 3.2 cm are made. Find the number of such cones formed.

Solution:

Let the number of cones be n

Radius of metallic sphere = 5.6 cm

Radius of the cone = 2.8 cm

Height of the cone = 3.2 cm

Volume of a sphere = 4/3 π × r³

                                 = 4/3 π × 5.6³

Volume of cone = 1/3 π × r² × h

                          = 1/3 π × 2.8² × 3.2

Volume of sphere = n * Volume of each cone

Number of cones (n) = Volume of the sphere/ Volume of the cone

n = 4/3 π × 5.6³/(1/3 π × 2.8² × 3.2)

n = (4 x 5.6³)/(2.8² × 3.2)

n = 28

Therefore, 28 such cones can be formed.

Question 23. A solid cuboid of iron with dimensions 53 cm x 40 cm x 15 cm is melted and recast into a cylindrical pipe. The outer and inner diameters of pipe are 8 cm and 7 cm respectively. Find the length of pipe.

Solution:

Volume of cuboid = (53 x 40 x 15) cm³

Internal radius of the pipe = 7/2 cm = r

External radius of the pipe = 8/2 = 4 cm = R

Let h be length of pipe

Volume of iron in the pipe = (External Volume) – (Internal Volume)

                                           = πR²h – πr²h

                                           = πh(R²– r²)

                                           = πh(R – r) (R + r)

                                           = π(4 – 7/2) (4 + 7/2) x h

                                            = π(1/2) (15/2) x h

Volume of iron in the pipe = volume of iron in cuboid

π(1/2) (15/2) x h = 53 x 40 x 15

h = (53 x 40 x 15 x 7/22 x 2/15 x 2) cm

h = 2698.18 cm

Therefore, the length of the pipe is 2698.2 cm.

Question 24. The diameters of the internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, find the height of the cylinder.

Solution:

Internal diameter of hollow spherical shell = 6 cm

Internal radius of hollow spherical shell = 6/2 = 3 cm = r

External diameter of hollow spherical shell = 10 cm

External diameter of hollow spherical shell = 10/2 = 5 cm = R

Diameter of the cylinder = 14 cm

Radius of cylinder = 14/2 = 7 cm

Let the height of cylinder be taken as h cm

Volume of cylinder = Volume of spherical shell

π × r² × h = 4/3 π × (R³ – r³)

π × 7² × h = 4/3 π × (5³ – 3³)

h = 4/3 x 2

h = 8/3 cm

Therefore, the height of the cylinder = 8/3 cm

Question 25. A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Calculate the height of the cone?

Solution:

Internal diameter of hollow sphere = 4 cm

Internal radius of hollow sphere = 2 cm

External diameter of hollow sphere = 8 cm

External radius of hollow sphere = 4 cm

Volume of the hollow sphere = 4/3 π × (4³ – 2³)          … (i)

Diameter of the cone = 8 cm

Radius of the cone = 4 cm

Let the height of the cone be x cm

Volume of the cone =1/3 π × 4² × (x)          ….. (ii)

(i) and (ii) are equal

4/3 π × (4³ – 2³) = 1/3 π × 4² × h

4 x (64 – 8) = 16 x h

h = 14

Therefore, the height of the cone = 14 cm

Question 26. A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.

Solution:

The internal radius of hollow sphere = 2 cm

The external radius of hollow sphere = 4 cm

Volume of the hollow sphere = 4/3 π × (4³ – 2³)          … (i)

The base radius of the cone = 4 cm

Let the height of the cone be x cm

Volume of the cone = 1/3 π × 4² × h          ….. (ii)

4/3 π × (4³ – 2³) = 1/3 π × 4² × h

4 x (64 – 8) = 16 x h

h = 14

Let Slant height of the cone be l 

l = √(h² + r²)

l = √(14² + 4²) = √212

l = 14.56 cm

Therefore, the height = 14 cm and slant height = 14.56 cm

Question 27. A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm. Find the diameter of the third ball. 

Solution:

Radius of the spherical ball = 3 cm

Volume of the sphere = 4/3 πr³

Volume = 4/3 π3³

Volume of first ball = 4/3 π 1.5³

Volume of second ball = 4/3 π2³

Let the radius of the third ball = r cm

Volume of third ball = 4/3 πr³

Volume of the spherical ball is equal to sum of volumes of three balls

 4/3 π3^{3 =} 4/3 π 1.5³+ 4/3 π2³ + 4/3 πr³

(3)³ = (2)³ + (1.5)³ + r³

r³ = 3³– 2³– 1.5³ cm³

r³ = 15.6 cm³

r = (15.625)^1/3 cm

r = 2.5 cm

Diameter = 2 x radius = 2 x 2.5 cm

                                   = 5.0 cm.

Therefore, the diameter of the third ball = 5 cm

Question 28. A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic meters of gravel are required to grave the path to a depth of 20 cm?

Solution:

Diameter of the circular pond = 40 m

Radius of the pond = 40/2 = 20 m = r

Thickness (width of the path) = 2 m

Height = 20 cm = 0.2 m

Thickness (t) = R – r

2 = R – 20

R = 22 m

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

                                                  = π (22²– 20²) × 0.2

                                                  = 52.8 m³

Therefore, the volume of the hollow cylinder = 52.8 m³

Question 29. A 16 m deep well with diameter 3.5 m is dug up and the earth from it is spread evenly to form a platform 27.5 m by 7m. Find the height of the platform?

Solution:

Radius(r) of the cylinder = 3.5/2 m = 1.75 m

Depth of the well or height of the cylinder (h) = 16 m

Volume of the cylinder= πr²h

                                    = π × 1.75² × 16

The length of the platform (l) = 27.5 m

Breadth of the platform (b) =7 m

Let the height of the platform be x m

Volume of the cuboid = l×b×h

                                    = 27.5×7×(x)

Volume of cylinder = Volume of cuboid

π × 1.75 × 1.75 × 16 = 27.5 × 7 × x

x = 0.8 m = 80 cm

Therefore, the height of the platform = 80 cm.

Question 30. A well of diameter 2 m is dug 14 m deep. The earth taken out of it is evenly spread all around it to form an embankment of height 40 cm. Find the width of the embankment?

Solution:

Radius of the circular cylinder (r) = 2/2 m = 1 m

Height of the well (h) = 14 m

Volume of the solid circular cylinder = π r²h

                                                           = π × 1²× 14 …. (i)

The height of the embankment = 40 cm = 0.4 m

Let the width of the embankment be (x) m.

External radius = (1 + x)m

Internal radius = 1 m

Volume of the embankment = π × r² × h

                                              = π × [(1 + x)² – (1)²]× 0.4 ….. (ii)

(i) and (ii) are equal

π × 1² × 14 = π × [(1 + x)² – (1)²] x 0.4

14/0.4 = 1 + x² + 2x – 1

35 = x² + 2x

x² + 2x – 35 = 0

(x + 7) (x – 5) = 0

x = 5 or -7

Therefore, the width of the embankment = 5 m.

Question 31. A well with inner radius 4 m is dug up and 14 m deep. Earth taken out of it has spread evenly all around a width of 3 m it to form an embankment. Find the height of the embankment?

Solution:

Inner radius of the well = 4 m

Depth of the well = 14 m

Volume of the cylinder = π r²h

                                      = π × 4² × 14 …. (i)

Width of the embankment = 3 m

Outer radius of the well = 3 + 4 m = 7 m

Volume of the hollow embankment = π (R² – r²) × h

                                                        = π × (7² – 4²) × h …… (ii)

(i) and (ii) are equal

π × 4² × 14 = π × (7² – 4²) × h

h = 4² × 14 / (33)

h = 6.78 m

Therefore, the height of the embankment = 6.78 m.

Question 32. A well of diameter 3 m is dug up to 14 m deep. The earth taken out of it has been spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.

Solution:

Diameter of the well = 3 m

Radius of the well = 3/2 m = 1.5 m

Depth of the well (h) = 14 m

Width of the embankment (thickness) = 4 m

Radius of the outer surface of the embankment = (4 + 1.5) m = 5.5 m

Volume of the embankment = π(R² – r²) × h

= π(5.5² – 1.5²) × h            ….. (i)

Volume of earth dug out = π × r² × h

= π × (3/2)² × 14    ….. (ii)

(i) and (ii) are equal

π(5.5² – 1.5²) × h = π × (3/2)² × 14

(5.5+1.5)(5.5-1.5) x h = 9 × 14/ 4

h = 9 × 14/ (4 × 28)

h = 9/8 m

   = 1.125m

Therefore, the height of the embankment =1.125 m

Question 33. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm. 

Solution:

The side of the cube = 9 cm

Diameter of the cone = Side of cube

2r = 9

Radius of cone = 9/2 cm = 4.5 cm

Height of cone = side of cube

Height of cone (h) = 9 cm

Volume of the largest cone = 1/3 π × r² × h

                                            = 1/3 π × 4.5² × 9

                                            = 190.93 cm³

Therefore, the volume of the largest cone to fit in the cube has a volume of 190.93 cm³

Question 34. A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution:

Height of the cylindrical bucket = 32 cm

Radius of the cylindrical bucket = 18 cm

Height of conical heap = 24 cm

Volume of cylinder = π × r² × h

Volume of cone = 1/3 π × r² × h

Volume of the conical heap = Volume of the cylindrical bucket

1/3 π × r² × 24 = π × 18² × 32

r² = 18² X 4

r = 18 × 2 = 36 cm

Slant height (l) = √(h² + r²)

l = √(24² + 36²) = √576+1296

                           = √1872

                           = 43.26 cm

Therefore, the radius =36 cm and slant height = 43.26 cm 

Question 35. Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm . What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen? 

Solution:

Length of the rectangular surface = 6 m = 600 cm

Breadth of the rectangular surface = 4 m = 400 cm

Height of the rain = 1 cm

Volume of the rectangular surface = length * breadth * height

                                                       = 600×400×1 cm³

                                                      = 240000 cm³  …………….. (i)

Radius of the cylindrical vessel = 20 cm

Let the height of the cylindrical vessel be  h cm

Volume of the cylindrical vessel = π × r² × h

                                                   = π × 20² × h  ……….. (ii)

(i) is equal to (ii)

240000 = π × 20² × h

h = 190.9 cm

Therefore, the height of the cylindrical vessel = 191

Summary

Exercise 16.1 Set 2 focuses on problems related to cuboids and cubes. It covers calculations involving surface area (lateral and total) and volume of these three-dimensional shapes. The questions range from straightforward applications of formulas to more complex scenarios involving real-world situations. Students are required to apply their knowledge of geometry, use appropriate formulas, and demonstrate problem-solving skills.