# 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 = πr

^{2}h= π 0.875

^{2}× 0.2Volume of the cuboid = 11 × 10 × 7 cm

^{3}Let the number of coins needed to be melted be n.

Volume of cuboid = n × Volume of cylinder

11 × 10 × 7 = π 0.875

^{2}× 0.2 x n11 × 10 × 7 = 22/7 x 0.875

^{2}× 0.2 x nn = 1600

Therefore, the number of coins required are 1600

**Question 20. The surface area of a solid metallic sphere is 616 cm**^{2}. It is melted and recast into a cone of height 28 cm. Find the diameter of the base of the cone so formed.

^{2}. 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 cm

^{3}Surface area of the sphere = 4πr

^{2}4πr

^{2}= 616r

^{2}= 49r = 7

Radius of the solid metallic sphere = 7 cm

Let radius of cone be x cm

Volume of the cone = 1/3 πx

^{2}h= 1/3 πx

^{2 }(28) ….. (i)Volume of the sphere = 4/3 πr

^{3}= 4/3 π7

^{3}………. (ii)(i) and (ii) are equal

1/3 πx

^{2 }(28) = 4/3 π7^{3}x

^{2 }(28) = 4 x 7^{3}x

^{2 }= 49r =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 × r

^{2}× hVolume of cone = 1/3 π × r

^{2}× hVolume of the conical heap = Volume of the cylindrical bucket

1/3 π × r

^{2}× 24 = π × 18^{2}× 32r

^{2}= 18^{2}x 4r = 18 x 2 = 36 cm

Let slant height of conical heap be l cm

l = √(h

^{2}+ r^{2})l = √(24

^{2}+ 36^{2}) = √1872l = 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 × r

^{3}= 4/3 π × 5.6

^{3}Volume of cone = 1/3 π × r

^{2}× h= 1/3 π × 2.8

^{2}× 3.2Volume of sphere = n * Volume of each cone

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

n = 4/3 π × 5.6

^{3}/(1/3 π × 2.8^{2}× 3.2)n = (4 x 5.6

^{3})/(2.8^{2}× 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) cm

^{3}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)

= πR

^{2}h – πr^{2}h= πh(R

^{2}– r^{2})= 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 × r

^{2}× h = 4/3 π × (R^{3 }– r^{3})π × 7

^{2}× h = 4/3 π × (5^{3 }– 3^{3})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 × (4

^{3}– 2^{3}) … (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 × 4

^{2}× (x) ….. (ii)(i) and (ii) are equal

4/3 π × (4

^{3}– 2^{3}) = 1/3 π × 4^{2}× h4 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 × (4

^{3}– 2^{3}) … (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 × 4

^{2}× h ….. (ii)4/3 π × (4

^{3}– 2^{3}) = 1/3 π × 4^{2}× h4 x (64 – 8) = 16 x h

h = 14

Let Slant height of the cone be l

l = √(h

^{2}+ r^{2})l = √(14

^{2}+ 4^{2}) = √212l = 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 πr

^{3}Volume = 4/3 π3

^{3}Volume of first ball = 4/3 π 1.5

^{3}Volume of second ball = 4/3 π2

^{3}Let the radius of the third ball = r cm

Volume of third ball = 4/3 πr

^{3}Volume of the spherical ball is equal to sum of volumes of three balls

4/3 π3

^{3 =}4/3 π 1.5^{3}+ 4/3 π2^{3 }+ 4/3 πr^{3}(3)

^{3}= (2)^{3}+ (1.5)^{3 }+ r^{3}r

^{3 }= 3^{3}– 2^{3}– 1.5^{3}cm^{3}r

^{3 }= 15.6 cm^{3}r = (15.625)

^{1/3}cmr = 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 (R

^{2}– r^{2}) × h= π (22

^{2}– 20^{2}) × 0.2= 52.8 m

^{3}Therefore, the volume of the hollow cylinder = 52.8 m

^{3}

**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= πr

^{2}h= π × 1.75

^{2}× 16The 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 r

^{2}h= π × 1

^{2}× 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 × r

^{2 }× h= π × [(1 + x)

^{2}– (1)^{2}]× 0.4 ….. (ii)(i) and (ii) are equal

π × 1

^{2}× 14 = π × [(1 + x)^{2}– (1)^{2}] x 0.414/0.4 = 1 + x

^{2 }+ 2x – 135 = x

^{2}+ 2xx

^{2}+ 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 r

^{2}h= π × 4

^{2 }× 14 …. (i)Width of the embankment = 3 m

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

Volume of the hollow embankment = π (R

^{2}– r^{2}) × h= π × (7

^{2}– 4^{2}) × h …… (ii)(i) and (ii) are equal

π × 4

^{2 }× 14 = π × (7^{2}– 4^{2}) × hh = 4

^{2 }× 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(R

^{2}– r^{2}) × h= π(5.5

^{2}– 1.5^{2}) × h ….. (i)Volume of earth dug out = π × r

^{2}× h= π × (3/2)

^{2}× 14 ….. (ii)(i) and (ii) are equal

π(5.5

^{2}– 1.5^{2}) × 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 × r

^{2}× h= 1/3 π × 4.5

^{2}× 9= 190.93 cm

^{3}Therefore, the volume of the largest cone to fit in the cube has a volume of 190.93 cm

^{3}

**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 × r

^{2}× hVolume of cone = 1/3 π × r

^{2}× hVolume of the conical heap = Volume of the cylindrical bucket

1/3 π × r

^{2}× 24 = π × 18^{2}× 32r

^{2}= 18^{2}X 4r = 18 × 2 = 36 cm

Slant height (l) = √(h

^{2}+ r^{2})l = √(24

^{2}+ 36^{2}) = √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 cm

^{3}= 240000 cm

^{3}…………….. (i)Radius of the cylindrical vessel = 20 cm

Let the height of the cylindrical vessel be h cm

Volume of the cylindrical vessel = π × r

^{2}× h= π × 20

^{2}× h ……….. (ii)(i) is equal to (ii)

240000 = π × 20

^{2}× hh = 190.9 cm

Therefore, the height of the cylindrical vessel = 191 cm