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• RD Sharma Class 10 Solutions

# 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 πr3

The volume of the solid sphere = sum of the volumes of n spherical balls.

n x 4/3 πr3 = 4/3 πR3

n x 4/3 π(1)3 = 4/3 π(8)3

n = 83 = 512

Therefore, 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 πr3

1 dm = 0.1 m

The volume of the rectangular block = 1.1 x 1 x 0.5 = 0.55 m3

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 π(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 π33

Ball is melted and recast into 3 spherical balls.

Volume (V1) of first ball = 4/3 π 1.53

Volume (V2) of second ball = 4/3 π23

Let the radius of the third ball = r cm

Volume of third ball (V3) = 4/3 πr3

V = V1 + V2 + V3

4/3 π33 = 4/3 π 1.53 + 4/3 π23 +4/3 πr3

On Solving

33 = 1.53 + 23 + r3

27 = 3.375 + 8 + r3

r3 = 15.625

r= 2.5

d = 2×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 = πr2h

Volume of cylindrical wire = Volume of brass

πr2h = 22

(22/7)*(0.125)2h = 22

h = 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 = πr2h

Radius of the solid cylinder = 1 cm

Volume of the solid cylinder = π12h = πh cm3

Volume of the hollow cylinder = πh(R2– r2)

Thickness of the cylinder = (R – r)

0.25 = 10 – r

Internal radius of the cylinder = 9.75 cm

Volume of the hollow cylinder = π × 16 (102-9.752)                [ (a2-b2) = (a+b)(a-b)]

= π × 16 (10+9.75)(10-9.75)

= 16π(19.75)(0.25)

Volume of the solid cylinder = volume of the hollow cylinder

πh = 16π(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

⇒ h = 2r

Volume of a cylinder = πr2h

Volume of the cylindrical vessel = πr22r = 2πr3   (as h = 2r)….. (i)

Diameter = 42 cm, so the radius = 21 cm

Height = 21 cm

Volume of two identical vessels = 2 x π 212 × 21 ….. (ii)

Volumes on equation (i) and (ii) are equal

2πr3= 2 x π 212 × 21

r3 = (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 = 2πr × h + 2πr2

= 2πr (h + r)

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

= 2 x 22 x 32 = 1408 cm2

Therefore, the total surface area of the cylinder is 1408 cm2

### 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 = 2πrh

= 2π × 10.5 × 40

= 2640 cm2

Volume of the cylinder = πr2h

= π × 10.5× 40

= 13860 cm3

Therefore, curved surface area of the cylinder = 2640 cm2 and the volume of the cylinder = 13860 cm3

### 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 = πR2H/ πr2h

n = (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 × 42 cm × 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 πr3

= 4/3 x 22/7 x 2.13

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 2.13)

= 1500

Therefore, 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 πr3 = 4/3 π 23 cm3

Edge of the cube = 44 cm

Volume of the cube = 443 cm3

Number of spherical lead shots = Volume of cube/ Volume of each spherical lead shot

= 44 x 44 x 44/ (4/3 π 23)

= 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 12√3 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

= 27x3 + 64x3 + 125x3

= 216x3

Let a be the edge of the new cube so formed after melting

a3 = 216x3

a = 6x

Diagonal of the cube = √(a2 + a2 + a2) = a√3

12√3 = a√3

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 πR3 = 4/3 π(10.5)3

Radius of each cone = r = 3.5 cm

Height of each cone = h = 3 cm

Volume = 1/3 πr2h = 1/3 π(3.5)2(3)

The number of cones = Volume of metallic sphere/ Volume of each cone

= 4/3 π(10.5)3 / 1/3 π(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 πr3 = 4/3 π(9/2)3

Radius of the wire = 1 mm = 0.1 cm

Let the length of the wire = h cm

Volume of wire = πr2h = π(0.1)2h

Volume of wire = Volume of sphere

π(0.1)2h = 4/3 π(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 π(x/4)3 = 4/3 πx3

n x (x3/ 64) = x3

n = 64

Therefore, the number of small balls = 64

Surface area of all small balls/ surface area of big ball = 64 x 4π(x/4)2/ 4π(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 π r3

= 4/3 π × 33 ….. (i)

Height of the cone = 3 cm

Volume of the right circular cone = 1/3 π r2h

= 1/3 π × r× 3 ….. (ii)

(i) and (ii) are equal

4/3 π × 33 = 1/3 π × r× 3

r2 = 36

r = 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 = π r2h

= π × 0.5× 8 ……. (i)

Length of the wire = 18 m = 1800 cm

Volume of the wire = π r2h

= π r2 × 1800  ….. (ii)

(i) and (ii) are equal

π × 0.5× 8 = π r2 × 1800

r2 = 2 /1800 = 1/900

r = 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 π × (R3 – r3)

= 4/3 π × (53 – 33)         …..  (i)

Let the radius of the solid cylinder be r cm

Volume of the cylinder = π × r2 × h

= π × r2 × 8/3       ….. (ii)

(i) and (ii) are equal

4/3 π × (53 – 33) = π × r2 × 8/3

4/3 x (125 – 27) = r2 × 8/3

98/2 = r2

r2 = 49

r = 7

d = 7 x 2 = 14 cm

Therefore, the diameter of the cylinder is 14 cm