Just like the previous chapter, this one deals with surface areas and volumes of solids. This exercise consists of problems on the conversion of solids from one type to the other. Students can make use of the RD Sharma Solutions Class 10 to clarify any doubts regarding this chapter. Further, the RD Sharma Solutions for Class 10 Maths Chapter 16 Surface Areas and Volumes Exercise 16.1 PDF can be downloaded below.
RD Sharma Solutions for Class 10 Maths Chapter 16 Surface Areas and Volumes Exercise 16.1
Access RD Sharma Solutions for Class 10 Maths Chapter 16 Surface Areas and Volumes Exercise 16.1
1. How many balls, each of radius 1 cm, can be made from a solid sphere of lead of radius 8 cm?
Solution:
Given,
A solid sphere of radius, R = 8 cm
With this sphere, we have to make spherical balls of radius r = 1 cm
Let’s assume that the number of balls made as n.
Then, we know that
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.
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:
Given,
A metallic block of dimension 11dm x 1m x 5dm.
The diameter of each bullet = 5 cm
We know that,
Volume of the sphere = 4/3 πr3
Since, 1 dm = 10-1m = 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
Let the number of bullets made from the rectangular block be n.
Then, from the question,
The volume of the rectangular block = Sum of the volumes of the n spherical bullets
0.55 = n x 4/3 π(2.5)3
Solving for n, we have
n = 8400
Therefore, 8400 can be cast from the rectangular block of metal.
3. A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of the two balls are 2 cm and 1.5 cm, respectively. Determine the diameter of the third ball.
Solution:
Given,
The radius of the spherical ball = 3 cm
We know that
The volume of the sphere = 4/3 πr3
So, its volume (V) = 4/3 πr3
That the ball is melted and recast into 3 spherical balls.
The volume (V1) of first ball = 4/3 π 1.53
The volume (V2) of second ball = 4/3 π23
Let the radius of the third ball = r cm
The volume of the third ball (V3) = 4/3 πr3
The volume of the spherical ball is equal to the volume of the 3 small spherical balls.
Now,
Cancelling out the common part from both sides of the equation, we get
(3)3 = (2)3 + (1.5)3 + r3
r3 = 33– 23– 1.53 cm3
r3 = 15.6 cm3
r = (15.6)1/3 cm
r = 2.5 cm
As diameter = 2 x Radius = 2 x 2.5 cm
= 5.0 cm.
Thus, the diameter of the third ball is 5 cm.
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:
Given,
2.2 dm3 of brass is to be drawn into a cylindrical wire of Diameter = 0.25 cm
So, the radius of the wire (r) = d/2
= 0.25/2 = 0.125*10-2 cm
Now, 1 cm = 0.01 m
So, 0.1cm = 0.001 m
Let the length of the wire be (h).
We know that
The volume of the cylinder = πr2h
It’s understood that
The volume of cylindrical wire = The volume of brass of 2.2 dm3
h = 448 m
Therefore, the length of the cylindrical wire drawn is 448 m.
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:
Given,
The diameter of the solid cylinder = 2 cm
The length of hollow cylinder = 16 cm
The solid cylinder is recast into a hollow cylinder of length 16 cm, the external diameter of 20 cm and thickness of 2.5 mm = 0.25 cm
We know that
The volume of a cylinder = πr2h
The radius of the solid cylinder = 1 cm
So,
The volume of the solid cylinder = π12h = πh cm3
Let’s assume the length of the solid cylinder as h.
And,
The volume of the hollow cylinder = πh(R2– r2)
The thickness of the cylinder = (R – r)
0.25 = 10 – r
So, the internal radius of the cylinder is 9.75 cm.
Volume of the hollow cylinder = π × 16 (100 – 95.0625)
Hence, it’s understood that
The volume of the solid cylinder = The volume of the hollow cylinder
πh = π × 16(100 – 95.06)
h = 79.04 cm
Therefore, the length of the solid cylinder is 79.04 cm.
6. A cylindrical vessel having a 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:
Given,
The diameter of the cylinder = The height of the cylinder
⇒ h = 2r, where h – the height of the cylinder and r – the radius of the cylinder
We know that
The volume of a cylinder = πr2h
So, the volume of the cylindrical vessel = πr22r = 2πr3 (as h = 2r)….. (i)
Now,
The volume of each identical vessel = πr2h
Diameter = 42 cm, so the radius = 21 cm
Height = 21 cm
So, the volume of two identical vessels = 2 x π 212 × 21 ….. (ii)
Since the volumes on equations (i) and (ii) are equal,
On equating both equations, we have
2πr3= 2 x π 212 × 21
r3 = (21)3
r = 21 cm
So, d = 42 cm
Therefore, the diameter of the cylindrical vessel is 42 cm.
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:
Given,
50 circular plates, each with diameter 14 cm. So,
The radius of circular plates = 7cm
The thickness of plates = 0.5 cm
As these plates are one above the other, the total thickness of all the plates = 0.5 x 50 = 25 cm
So, the 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
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:
Given,
250 circular plates, each with radius 10.5 cm and thickness of 1.6 cm. So,
As the plates are placed one above the other, the total height becomes = 1.6 x 25 = 40 cm
We know that
The curved surface area of a cylinder = 2πrh
= 2π × 10.5 × 40 = 2640 cm2
And, the volume of the cylinder = πr2h
= π × 10.52 × 40 = 13860 cm3
Therefore,
The curved surface area of the cylinder is 2640 cm2, and the volume of the cylinder is 13860 cm3
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:
Given,
Radius of each circular disc = r = 1.5/2 = 0.75 cm
Height of each circular disc = h = 0.2 cm
Radius of the cylinder = R = 4.5/ 2 = 2.25 cm
Height of the cylinder = H = 10 cm
So, the number of metallic discs required is given by n.
n = Volume of the 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.
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:
Given,
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
So, the volume of a spherical lead shot = 4/3 πr3
= 4/3 x 22/7 x 2.13
And the volume of the rectangular lead piece = 66 x 42 x 21
Thus,
The number of spherical lead shots = Volume of the rectangular lead piece/ Volume of the spherical lead shot
= 66 x 42 x 21/ (4/3 x 22/7 x 2.13)
= 1500
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:
Given,
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
Thus,
Number of the spherical lead shots = Volume of the cube/ Volume of each spherical lead shot
= 44 x 44 x 44/ (4/3 π 23)
= 2541
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 (in cm) be 3x, 4x and 5x, respectively.
So, the volume of the cube after melting will be = (3x)3 + (4x)3 + (5x)3
= 9x3 + 64x3 + 125x3 = 216x3
Now, let a be the edge of the new cube so formed after melting.
Then, we have
a3 = 216x3
a = 6x
We know that
Diagonal of the cube = √(a2 + a2 + a2) = a√3
So, 12√3 = a√3
a = 12 cm
x = 12/6 = 2
Thus, the edges of the three cubes are 6 cm, 8 cm and 10 cm, respectively.
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:
Given,
Radius of metallic sphere = R = 10.5 cm
So, its volume = 4/3 πR3 = 4/3 π(10.5)3
We also have,
Radius of each cone = r = 3.5 cm
Height of each cone = h = 3 cm
And, its volume = 1/3 πr2h = 1/3 π(3.5)2(3)
Thus,
The number of cones = Volume of the metallic sphere/ Volume of each cone
= 4/3 π(10.5)3 / 1/3 π(3.5)2(3)
= 126
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 a uniform cross-section. Find the length of the wire.
Solution:
Given,
The radius of the sphere = 9/2 cm
So, its volume = 4/3 πr3 = 4/3 π(9/2)3
And, the radius of the wire = 2 mm = 0.2 cm
Let the length of the wire = h cm
So, the volume of wire = πr2h = π(0.2)2h
Then, according to the question, we have
The volume of the wire = The volume of the sphere
π(0.2)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
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 together with that of the original ball.
Solution:
Let the radius of the big ball be x cm.
Then, the radius of the small ball = x/4 cm
And, let the number of balls = n
Then, according to the question, we have
The volume of n small balls = The 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
Next,
The surface area of all small balls/ The surface area of big ball = 64 x 4π(x/4)2/ 4π(x)2
= 64/16 = 4/1
Thus, the ratio of the surface area of the small balls to that of the original ball is 4:1.
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:
Given,
The radius of the copper sphere = 3 cm
We know that
The volume of the sphere = 4/3 π r3
= 4/3 π × 33 ….. (i)
Also, given that the copper sphere is melted and recasted into a right circular cone.
Height of the cone = 3 cm
We know that
The volume of the right circular cone = 1/3 π r2h
= 1/3 π × r2 × 3 ….. (ii)
On comparing equations (i) and (ii) we have
4/3 π × 33 = 1/3 π × r2 × 3
r2 = 36
r = 6 cm
Therefore, the radius of the base of the cone is 6 cm.
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:
Given,
Diameter of the copper wire = 1 cm
So, radius of the copper wire = 1/2 cm = 0.5 cm
Length of the copper rod = 8 cm
We know that
The volume of the cylinder = π r2h
= π × 0.52 × 8 ……. (i)
The length of the wire = 18 m = 1800 cm
The volume of the wire = π r2h
= π r2 × 1800 ….. (ii)
On equating both equations, we have
π × 0.52 × 8 = π r2 × 1800
r2 = 2 /1800 = 1/900
r = 1/30 cm
Therefore, the diameter of the wire is 1/15 cm, i.e., 0.67 mm, which is the thickness of the wire.
18. The diameters of the 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 the length of 8/3, find the diameter of the cylinder.
Solution:
Given,
The internal diameter of the hollow sphere = 6 cm
So, the internal radius of the hollow sphere = 6/2 cm = 3 cm = r
The external diameter of the hollow sphere = 10 cm
So, the external radius of the hollow sphere = 10/2 cm = 5 cm = R
We know that,
The volume of the hollow spherical shell = 4/3 π × (R3 – r3)
= 4/3 π × (53 – 33) ….. (i)
And given, the length of the solid cylinder = 8/3 cm
Let the radius of the solid cylinder be r cm.
We know that,
Volume of the cylinder = π × r2 × h
= π × r2 × 8/3 ….. (ii)
Now equating both (i) and (ii), we have
4/3 π × 53 – 33 = π × r2 × 8/3
4/3 x (125 – 27) = r2 × 8/3
98/2 = r2
r2 = 49
r = 7
So, d = 7 x 2 = 14 cm
Therefore, the diameter of the cylinder is 14 cm.
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:
Given,
The diameter of the coin = 1.75 cm
So, its radius = 1.74/2 = 0.875 cm
Thickness or the height = 2 mm = 0.2 cm
We know that,
The volume of the cylinder (V1) = πr2h
= π 0.8752 × 0.2
And, the volume of the cuboid (V2) = 11 × 10 × 7 cm3
Let the number of coins needed to be melted be n.
So, we have
V2 = V1 × n
11 × 10 × 7 = π 0.8752 × 0.2 x n
11 × 10 × 7 = 22/7 x 0.8752 × 0.2 x n
On solving, we get n = 1600
Therefore, the number of coins required is 1600.
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:
Given,
The height of the cone = 28 cm
The surface area of the solid metallic sphere = 616 cm3
We know that,
The surface area of the sphere = 4πr2
So, 4πr2 = 616
r2 = 49
r = 7
The radius of the solid metallic sphere = 7 cm
Let’s assume r to be the radius of the cone.
We know that,
The volume of the cone = 1/3 πr2h
= 1/3 πr2 (28) ….. (i)
The volume of the sphere = 4/3 πr3
= 4/3 π73 ………. (ii)
On equating equations (i) and (ii), we have
1/3 πr2 (28) = 4/3 π73
Eliminating the common terms, we get
r2 (28) = 4 x 73
r2 = 49
r =7
So, diameter of the cone = 7 x 2 = 14 cm
Therefore, the diameter of the base of the cone is 14 cm.
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:
Given,
The height of the cylindrical bucket = 32 cm
The radius of the cylindrical bucket = 18 cm
The height of conical heap = 24 cm
We know that,
The volume of the cylinder = π × r2 × h
And, the volume of the cone = 1/3 π × r2 × h
Then, from the question,
The volume of the conical heap = The volume of the cylindrical bucket
1/3 π × r2 × 24 = π × 182 × 32
r2 = 182 x 4
r = 18 x 2 = 36 cm
Now,
The slant height of the conical heap (l) is given by
l = √(h2 + r2)
l = √(242 + 362) = √1872
l = 43.26 cm
Therefore, the radius and slant height of the conical heap are 36 cm and 43.26 cm, respectively.
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 made be n
Given,
The radius of the metallic sphere = 5.6 cm
The radius of the cone = 2.8 cm
The height of the cone = 3.2 cm
We know that
Volume of a sphere = 4/3 π × r3
So, V1 = 4/3 π × 5.63
And,
Volume of the cone = 1/3 π × r2 × h
V2 = 1/3 π × 2.82 × 3.2
Thus, the 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.
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 the pipe are 8 cm and 7 cm, respectively. Find the length of the pipe.
Solution:
Let the length of the pipe be h cm.
Then, Volume of the 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
So, the 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
Then, from the question, it’s understood that
The volume of iron in the pipe = The volume of iron in the 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 cm
Therefore, the length of the pipe is 2698 cm.
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:
Given,
The internal diameter of the hollow spherical shell = 6 cm
So, the internal radius of the hollow spherical shell = 6/2 = 3 cm = r
External diameter of hollow spherical shell = 10 cm
So, the external diameter of the hollow spherical shell = 10/2 = 5 cm = R
Diameter of the cylinder = 14 cm
So, the radius of cylinder = 14/2 = 7 cm
Let the height of the cylinder be taken as h cm.
Then, according to the question we have,
The volume of the cylinder = The volume of the 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
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:
Given,
The internal diameter of the hollow sphere = 4 cm
So, the internal radius of the hollow sphere = 2 cm
The external diameter of the hollow sphere = 8 cm
So, the external radius of the hollow sphere = 4 cm
We know that,
Volume of the hollow sphere 4/3 π × (43 – 23) … (i)
Also given,
Diameter of the cone = 8 cm
So, the 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)
As the volume of the hollow sphere and cone are equal, we can equate equations (i) and (ii).
So, we get
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 so obtained will have a height of 14 cm.
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:
Given,
The internal radius of the hollow sphere = 2 cm
The external radius of the hollow sphere = 4 cm
We know that,
The volume of the hollow sphere 4/3 π × (43 – 23) … (i)
Also given,
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)
As the volume of the hollow sphere and cone are equal, we can equate equations (i) and (ii).
So, we get
4/3 π × (43 – 23) = 1/3 π × 42 × h
4 x (64 – 8) = 16 x h
h = 14
Now,
The slant height of the cone (l) is given by
l = √(h2 + r2)
l = √(142 + 42) = √212
l = 14.56 cm
Therefore, the height and slant height of the conical heap are 14 cm and 14.56 cm, respectively.
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:
Given,
The radius of the spherical ball = 3 cm
We know that,
The volume of the sphere = 4/3 πr3
So, its volume (V) = 4/3 πr3
That the ball is melted and recast into 3 spherical balls.
Volume (V1) of the first ball = 4/3 π 1.53
Volume (V2) of the second ball = 4/3 π23
Let the radius of the third ball = r cm
Volume of the third ball (V3) = 4/3 πr3
The volume of the spherical ball is equal to the volume of the 3 small spherical balls.
Now,
Cancelling out the common part from both sides of the equation, we get
(3)3 = (2)3 + (1.5)3 + r3
r3 = 33– 23– 1.53 cm3
r3 = 15.6 cm3
r = (15.6)1/3 cm
r = 2.5 cm
As diameter = 2 x radius = 2 x 2.5 cm
= 5.0 cm.
Thus, the diameter of the third ball is 5 cm
28. A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic meters of gravel is required to grave the path to a depth of 20 cm?
Solution:
Given,
Diameter of the circular pond = 40 m
So, the radius of the pond = 40/2 = 20 m = r
Thickness (width of the path) = 2 m
The whole view of the pond looks like a hollow cylinder.
And the height will be 20 cm = 0.2 m
So,
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 is the required amount of sand needed to spread across to a depth of 20 m.
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:
Let us assume the well to be a solid right-circular cylinder.
Radius(r) of the cylinder = 3.5/2 m = 1.75 m
Depth of the well or height of the cylinder (h) = 16 m
We know that,
The volume of the cylinder (V1) = πr2h
= π × 1.752 × 16
Given,
The length of the platform (l) = 27.5 m
The breadth of the platform (b) =7 m
Now, let the height of the platform be x m.
We know that,
The volume of the rectangle = l*b*h
V2 = 27.5*7*x
As the earth dug up is spread evenly to form the platform,
The volumes of both the well and the platform should be the same.
V1 = V2
π × 1.75 × 1.75 × 16 = 27.5 × 7 × x
x = 0.8 m = 80 cm
Therefore, the height of the platform is 80 cm.
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:
Given,
The radius of the circular cylinder (r) = 2/2 m = 1 m
The height of the well (h) = 14 m
We know that,
The volume of the solid circular cylinder = π r2h
= π × 12× 14 …. (i)
And,
The height of the embankment (h) = 40 cm = 0.4 m
Let the width of the embankment be (x) m.
The embankment is a hollow cylinder with an external radius = 1 + x and an internal radius = 1
Volume of the embankment = π × r2 × h
= π × [(1 + x)2 – (1)2]× 0.4 ….. (ii)
As the well is spread evenly to form an embankment, then the volumes will be the same.
So, on equating equations (i) and (ii), we get
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
Solving by factorisation method, we have
(x + 7) (x – 5) = 0
So, x = 5 m can only be the solution as it’s a positive value.
Therefore, the width of the embankment is 5 m.
31. A well with an 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 to form an embankment. Find the height of the embankment.
Solution:
Given,
The inner radius of the well = 4 m
The depth of the well = 14 m
We know that,
The volume of the cylinder = π r2h
= π × 42 × 14 …. (i)
From the question, it’s said that
The earth taken out from the well is evenly spread all around it to form an embankment.
And the width of the embankment = 3 m
So, the outer radius of the well = 3 + 4 m = 7 m
We know that,
Volume of the hollow embankment = π (R2 – r2) × h
= π × (72 – 42) × h …… (ii)
On equating both the equations (i) and (ii), we get
π × 42 × 14 = π × (72 – 42) × h
h = 42 × 14 / (33)
h = 6.78 m
Therefore, the height of the embankment so formed is 6.78 m.
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:
Given,
Diameter of the well = 3 m
So, the radius of the well = 3/2 m = 1.5 m
Depth of the well (h) = 14 m
Width of the embankment (thickness) = 4 m
So, the radius of the outer surface of the embankment = (4 + 1.5) m = 5.5 m
Let the height of the embankment be taken as h m.
We know that the embankment is a hollow cylinder.
Volume of the embankment = π (R2 – r2) × h
= π (5.52 – 1.52) × h ….. (i)
Volume of earth dug out = π × r2 × h
= π × 22 × 14 ….. (ii)
On equating both (i) and (ii), we get
π (5.52 – 1.52) × h = π × (3/2)2 × 14
(30.25 – 2.25) x h = 9 x 14/ 4
h = 9 x 14/ (4 x 28)
h = 9/8 m
Therefore, the height of the embankment is 9/8 m
33. Find the volume largest right circular cone that can be cut out of a cube whose edge is 9 cm.
Solution:
Given,
The side of the cube = 9 cm
The largest cone that can be cut from the cube will have the base diameter = Side of the cube
2r = 9
r = 9/2 cm = 4.5 cm
And,
Height of cone = Side of the cube
So, the height of cone (h) = 9 cm
Thus,
Volume of the largest cone to fit in = 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
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:
Given,
Height of the cylindrical bucket = 32 cm
Radius of the cylindrical bucket = 18 cm
Height of the conical heap = 24 cm
We know that,
Volume of the cylinder = π × r2 × h
And, volume of the cone = 1/3 π × r2 × h
Then, from the question
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
Now,
The slant height of the conical heap (l) is given by
l = √(h2 + r2)
l = √(242 + 362) = √1872
l = 43.26 cm
Therefore, the radius and slant height of the conical heap are 36 cm and 43.26 cm, respectively.
35. Rainwater, 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:
Given,
Length of the rectangular surface = 6 m = 600 cm
Breadth of the rectangular surface = 4 m = 400 cm
Height of the perceived rain = 1 cm
So,
Volume of the rectangular surface = Length * Breadth * Height
= 600*400*1 cm3
= 240000 cm3 …………….. (i)
Also given,
Radius of the cylindrical vessel = 20 cm
Let the height of the cylindrical vessel be taken as h cm.
We know that,
Volume of the cylindrical vessel = π × r2 × h
= π × 202 × h ……….. (ii)
As all the rainwater is transferred to the cylindrical vessel,
We can equate both (i) and (ii) for equal volumes.
240000 = π × 202 × h
h = 190.9 cm
Therefore, the height of the cylindrical vessel is nearly 191 cm.
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