# Class 10 RD Sharma Solutions – Chapter 16 Surface Areas and Volumes – Exercise 16.1 | Set 1

**Question 1. How many balls, each of radius 1 cm, can be made from a solid sphere of lead of radius 8 cm? **

**Solution:**

R = 8 cm

r = 1 cm

Let the number of balls = n

Volume of the sphere = 4/3 Ï€r

^{3}The volume of the solid sphere = sum of the volumes of n spherical balls.

n x 4/3 Ï€r

^{3}= 4/3 Ï€R^{3}n x 4/3 Ï€(1)

^{3}= 4/3 Ï€(8)^{3}n = 8

^{3}= 512Therefore, 512 balls can be made of radius 1 cm each with a solid sphere of radius 8 cm.

**Question 2. How many spherical bullets each of 5 cm in diameter can be cast from a rectangular block of metal 11dm x 1 m x 5 dm?**

**Solution:**

A metallic block of dimension 11dm x 1m x 5dm

Diameter of each bullet = 5 cm

Volume of the sphere = 4/3 Ï€r

^{3}1 dm = 0.1 m

The volume of the rectangular block = 1.1 x 1 x 0.5 = 0.55 m

^{3}Radius of the bullet = 5/2 = 2.5 cm = 0.025cm

Let the number of bullets = n.

The volume of the rectangular block = sum of the volumes of the n spherical bullets

0.55 = n x 4/3 Ï€(0.025)

^{3}n = 8400

Therefore, 8400 can be cast from the rectangular block of metal.

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

**Solution:**

Radius of the spherical ball = 3 cm

Volume (V) = 4/3 Ï€3

^{3}Ball is melted and recast into 3 spherical balls.

Volume (V

_{1}) of first ball = 4/3 Ï€ 1.5^{3}Volume (V

_{2}) of second ball = 4/3 Ï€2^{3}Let the radius of the third ball = r cm

Volume of third ball (V

_{3}) = 4/3 Ï€r^{3}V = V

_{1}+ V_{2}+ V_{3}4/3 Ï€3

^{3}= 4/3 Ï€ 1.5^{3}+ 4/3 Ï€2^{3}+4/3 Ï€r^{3}On Solving

3

^{3 }= 1.5^{3 }+ 2^{3 }+ r^{3}27 = 3.375 + 8 + r

^{3}r

^{3}= 15.625r= 2.5

d = 2Ã—r

= 5cm

**Question 4. 2.2 cubic dm of brass is to be drawn into a cylindrical wire of 0.25 cm in diameter. Find the length of the wire? **

**Solution:**

Radius of the wire (r) = d/2

= 0.25/2 =

= 0.125cm

1dm = 10cm

2.2 dm = 22cm

Let the length of the wire be (h)

Volume of the cylinder = Ï€r

^{2}hVolume of cylindrical wire = Volume of brass

Ï€r

^{2}h = 22(22/7)*(0.125)

^{2}h = 22h = 7/(0.125)

^{2}h = 448

Therefore, the length of the cylindrical wire drawn is 448 m

**Question 5. What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm?**

**Solution:**

Diameter of the solid cylinder = 2 cm

Length of hollow cylinder = 16 cm

Volume of a cylinder = Ï€r

^{2}hRadius of the solid cylinder = 1 cm

Volume of the solid cylinder = Ï€1

^{2}h = Ï€h cm^{3}Volume of the hollow cylinder = Ï€h(R

^{2}â€“ r^{2})Thickness of the cylinder = (R â€“ r)

0.25 = 10 â€“ r

Internal radius of the cylinder = 9.75 cm

Volume of the hollow cylinder = Ï€ Ã— 16 (10

^{2}-9.75^{2}) [ (a^{2}-b^{2}) = (a+b)(a-b)]= Ï€ Ã— 16 (10+9.75)(10-9.75)

= 16Ï€(19.75)(0.25)

Volume of the solid cylinder = volume of the hollow cylinder

Ï€h = 16Ï€(19.75)(0.25)

h = 79cm

Therefore, the length of the solid cylinder = 79.04 cm.

**Question 6. A cylindrical vessel having diameter equal to its height is full of water which is poured into two identical cylindrical vessels with diameter 42 cm and height 21 cm which are filled completely. Find the diameter of the cylindrical vessel?**

**Solution:**

The diameter of the cylinder = the height of the cylinder

â‡’ h = 2r

Volume of a cylinder = Ï€r

^{2}hVolume of the cylindrical vessel = Ï€r

^{2}2r = 2Ï€r^{3}(as h = 2r)â€¦.. (i)Diameter = 42 cm, so the radius = 21 cm

Height = 21 cm

Volume of two identical vessels = 2 x Ï€ 21

^{2}Ã— 21 â€¦.. (ii)Volumes on equation (i) and (ii) are equal

2Ï€r

^{3}= 2 x Ï€ 21^{2}Ã— 21r

^{3}= (21)^{3}r = 21 cm

d = 42 cm

Therefore, the diameter of the cylindrical vessel is 42 cm.

**Question 7. 50 circular plates each of diameter 14 cm and thickness 0.5 cm are placed one above the other to form a right circular cylinder. Find its total surface area.**

**Solution:**

Radius of circular plates = 7cm

Thickness of plates = 0.5 cm

Total thickness of all the plates = 0.5 x 50 = 25 cm (height of cylinder)

Total surface area of the right circular cylinder formed = 2Ï€r Ã— h + 2Ï€r

^{2}= 2Ï€r (h + r)

= 2(22/7) x 7 x (25 + 7)

= 2 x 22 x 32 = 1408 cm

^{2}Therefore, the total surface area of the cylinder is 1408 cm

^{2}

**Question 8. 25 circular plates, each of radius 10.5 cm and thickness 1.6 cm, are placed one above the other to form a solid circular cylinder. Find the curved surface area and the volume of the cylinder so formed.**

**Solution:**

Total height = 1.6 x 25 = 40 cm

Curved surface area of a cylinder = 2Ï€rh

= 2Ï€ Ã— 10.5 Ã— 40

= 2640 cm

^{2}Volume of the cylinder = Ï€r

^{2}h= Ï€ Ã— 10.5

^{2 }Ã— 40= 13860 cm

^{3}Therefore, curved surface area of the cylinder = 2640 cm

^{2}and the volume of the cylinder = 13860 cm^{3}

**Question 9. Find the number of metallic circular discs with 1.5 cm base diameter and of height 0.2 cm to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.**

**Solution:**

Radius of each circular disc = r = 1.5/2 = 0.75 cm

Height of each circular disc = h = 0.2 cm

Radius of cylinder = R = 4.5/ 2 = 2.25 cm

Height of cylinder = H = 10 cm

Let the number of metallic discs required is given by n

n = Volume of cylinder / volume of each circular disc

n = Ï€R

^{2}H/ Ï€r^{2}hn = (2.25)

^{2}(10)/ (0.75)^{2}(0.2)n = 3 x 3 x 50 = 450

Therefore, 450 metallic discs are required.

**Question 10. How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm Ã— 42 cm Ã— 21 cm.**

**Solution:**

Radius of each spherical lead shot = r = 4.2/ 2 = 2.1 cm

The dimensions of the rectangular lead piece = 66 cm x 42 cm x 21 cm

Volume of a spherical lead shot = 4/3 Ï€r

^{3}= 4/3 x 22/7 x 2.1

^{3}Volume of the rectangular lead piece = 66 x 42 x 21

The number of spherical lead shots = Volume of rectangular lead piece/ Volume of a spherical lead shot

= 66 x 42 x 21/ (4/3 x 22/7 x 2.1

^{3})

=1500Therefore, number of spherical lead shots = 1500

**Question 11. How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm.**

**Solution:**

The radius of each spherical lead shot = r = 4/2 = 2 cm

Volume of each spherical lead shot = 4/3 Ï€r

^{3}= 4/3 Ï€ 2^{3}cm^{3}Edge of the cube = 44 cm

Volume of the cube = 44

^{3}cm^{3}Number of spherical lead shots = Volume of cube/ Volume of each spherical lead shot

= 44 x 44 x 44/ (4/3 Ï€ 2

^{3})= 2541

**Question 12. Three cubes of a metal whose edges are in the ratio 3: 4: 5 are melted and converted into a single cube whose diagonal is 12âˆš3 cm. Find the edges of the three cubes.**

**Solution:**

Let the edges of three cubes be 3x, 4x and 5x respectively.

Volume of the cube after melting will be = (3x)

^{3}+ (4x)^{3}+ (5x)^{3}= 27x

^{3}+ 64x^{3}+ 125x^{3}= 216x

^{3}Let a be the edge of the new cube so formed after melting

a

^{3}= 216x^{3}a = 6x

Diagonal of the cube = âˆš(a

^{2}+ a^{2}+ a^{2}) = aâˆš312âˆš3 = aâˆš3

a = 12 cm

x = 12/6 = 2

Therefore, the edges of the three cubes are 6 cm, 8 cm and 10 cm respectively.

**Question 13. A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed.**

**Solution:**

Radius of metallic sphere = R = 10.5 cm

Volume = 4/3 Ï€R

^{3}= 4/3 Ï€(10.5)^{3}Radius of each cone = r = 3.5 cm

Height of each cone = h = 3 cm

Volume = 1/3 Ï€r

^{2}h = 1/3 Ï€(3.5)^{2}(3)The number of cones = Volume of metallic sphere/ Volume of each cone

= 4/3 Ï€(10.5)

^{3}/ 1/3 Ï€(3.5)^{2}(3)= 126

**Question 14. The diameter of a metallic sphere is equal to 9 cm. It is melted and drawn into a long wire of diameter 2 mm having uniform cross-section. Find the length of the wire.**

**Solution:**

Radius of the sphere = 9/2 cm

Volume = 4/3 Ï€r

^{3}= 4/3 Ï€(9/2)^{3}Radius of the wire = 1 mm = 0.1 cm

Let the length of the wire = h cm

Volume of wire = Ï€r

^{2}h = Ï€(0.1)^{2}hVolume of wire = Volume of sphere

Ï€(0.1)

^{2}h = 4/3 Ï€(9/2)^{3}h = 4 x 729/ (3 x 8 x 0.01) = 12150 cm

Therefore, the length of the wire = 12150 cm

**Question 15. An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is 1/4 of the radius of the original ball, how many such balls are made? Compare the surface area, of all the smaller balls combined with that of the original ball.**

**Solution:**

Let the radius of the big ball be x cm

Radius of the small ball = x/4 cm

Let the number of balls = n

Volume of n small balls = Volume of the big ball

n x 4/3 Ï€(x/4)

^{3}= 4/3 Ï€x^{3}n x (x

^{3}/ 64) = x^{3}n = 64

Therefore, the number of small balls = 64

Surface area of all small balls/ surface area of big ball = 64 x 4Ï€(x/4)

^{2}/ 4Ï€(x)^{2}= 64/16 = 4/1

Therefore, the ratio of the surface area of the small balls to that of the original ball is 4:1

**Question 16. A copper sphere of radius 3 cm is melted and recast into a right circular cone of height 3 cm. Find the radius of the base of the cone?**

**Solution:**

Radius of the copper sphere = 3 cm

Volume of the sphere = 4/3 Ï€ r

^{3}= 4/3 Ï€ Ã— 3

^{3}â€¦.. (i)Height of the cone = 3 cm

Volume of the right circular cone = 1/3 Ï€ r

^{2}h= 1/3 Ï€ Ã— r

^{2 }Ã— 3 â€¦.. (ii)(i) and (ii) are equal

4/3 Ï€ Ã— 3

^{3}= 1/3 Ï€ Ã— r^{2 }Ã— 3r

^{2}= 36r = 6 cm

Therefore, the radius of the base of the cone is 6 cm.

**Question 17. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire? **

**Solution:**

Diameter of the copper rod = 1 cm

Radius of the copper wire = 1/2 cm = 0.5 cm

Length of the copper rod = 8 cm

Volume of the cylinder = Ï€ r

^{2}h= Ï€ Ã— 0.5

^{2 }Ã— 8 â€¦â€¦. (i)Length of the wire = 18 m = 1800 cm

Volume of the wire = Ï€ r

^{2}h= Ï€ r

^{2}Ã— 1800 â€¦.. (ii)(i) and (ii) are equal

Ï€ Ã— 0.5

^{2 }Ã— 8 = Ï€ r^{2}Ã— 1800r

^{2}= 2 /1800 = 1/900r = 1/30 cm

Therefore, the diameter of the wire is 1/15 cm = 0.67 mm

**Question 18. The diameters of internal and external surfaces of a hollow spherical shell are 10cm and 6 cm respectively. If it is melted and recast into a solid cylinder of length of 8/3, find the diameter of the cylinder?**

**Solution:**

Internal diameter of the hollow sphere = 6 cm

Internal radius of the hollow sphere = 6/2 cm = 3 cm = r

External diameter of the hollow sphere = 10 cm

External radius of the hollow sphere = 10/2 cm = 5 cm = R

Volume of the hollow spherical shell = 4/3 Ï€ Ã— (R

^{3}â€“ r^{3})= 4/3 Ï€ Ã— (5

^{3}â€“ 3^{3}) â€¦.. (i)Let the radius of the solid cylinder be r cm

Volume of the cylinder = Ï€ Ã— r

^{2}Ã— h= Ï€ Ã— r

^{2}Ã— 8/3 â€¦.. (ii)(i) and (ii) are equal

4/3 Ï€ Ã— (5

^{3}â€“ 3^{3)}= Ï€ Ã— r^{2}Ã— 8/34/3 x (125 â€“ 27) = r

^{2}Ã— 8/398/2 = r

^{2}r

^{2}= 49r = 7

d = 7 x 2 = 14 cm

Therefore, the diameter of the cylinder is 14 cm

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