Surface Area And Volume Test

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Surface Area And Volume Test - 7

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Question 1
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A metallic right circular cone of height 18 cm and base radius 7 cm is melted into a cuboid whose two sides measure 11 cm and 6 cm. What is the length of the third side of the cuboid?

A
7 cm

B
6 cm

C
14 cm

D
10 cm

Solution

According to the given condition, Volume of cone = Volume of cuboid

Let the third side of the cuboid measure x cm.

x = 14 cm
Question 2
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A cylinder of length 14 cm and radius 4 cm is cut out of a cube of side 12 cm.

How many small cuboids of length, breadth and height of 4 cm, 4 cm and 2 cm, respectively, can be obtained from the remaining solid?

A
34

B
32

C
33

D
31

Solution

Given: Side of the cube = 12 cm
Therefore,
Volume of the cube = (Side)³
= 12 × 12 × 12
= 1,728 cm³
Volume of cylinder (cut out) of length 14 cm and radius 4 cm =
= π × 4² × 14

=
= 704 cm³
Now, total volume of the remaining solid = 1,728 – 704 = 1024 cm³
Volume of the cuboid of length, breadth and height of 4 cm, 4 cm and 2 cm, respectively = length × breadth × height
= 4 × 4 × 2 = 32 cm³
Number of such small cuboids that can be obtained from the remaining solid = = 32
Therefore, 32 such cuboids can be obtained from the remaining solid.
Question 3
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The dimensions of a plot of land are 20 m by 9 m. A pit 10 m long, 4.5 m wide and 3 m deep is dug in one corner of the plot and the earth removed is spread evenly over the remaining area. What is the rise in the height of this area as a result of this operation?

A
1 m

B
2 m

C
3 m

D
4 m

Solution

Volume of the earth removed = Area of the remaining rectangular field Rise in height (h)
10 4.5 3 = (180 – 45) sq. m h m
= h

h = 1 m

Thus, this is the correct answer.
Question 4
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A toy is in the shape of a cone surmounted on a hemisphere of same radius. If s, p and q, respectively is the total surface area of the toy, the curved surface area of the hemisphere and the curved surface area of the cone, then which of the following relationships is true?

A
s + p = q

B
p = s - q

C
q = p - s

D
s + 2q = p

Solution

Total surface area of the toy = s
Curved surface area of the cone = q
Curved surface area of the hemisphere = p
Total surface area of the toy = Curved surface area of the cone + Curved surface area of the hemisphere
s = q + p
p = s - q
Question 5
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A metallic cylinder with 70 cm height and base diameter 15 cm is surmounted by a hemisphere of same diameter. Find the mass of the solid; given that 1 cm³ of metal has 10 g of mass (approximately). (Take = )

A
135.98 kg

B
132.59 kg

C
133.58 kg

D
130.95 kg

Solution

Volume of the whole solid =, where h = 70 cm and r = 7.5 cm

= cm³
= cm³
= cm³
= cm³
= 13258.93 cm³
Mass of 1 cm³ metal = 10 gm
Mass of 13258.93 cm³ of metal = (10 13258.93) gm = (132589.3) gm
= kg = 132.5893 kg = 132.59 kg
Question 6
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A construction company wants to partition a hall in its office into two equal parts, so it makes a wall in the centre of the hall. The length and breadth of the wall are 12 m and 7.5 cm, respectively. The number of cement blocks required to construct the wall is 1600 and the dimensions of each cement block are 25 cm x 15 cm x 12 cm. Find the height of the wall (in metres).

A
8 m

B
12 m

C
15 m

D
17 m

Solution

Volume of one block = (25 × 15 × 12) cm³ = 4500 cm³Number of blocks = 1600
Volume of the wall = Number of blocks × Volume of one block
Length × Breadth × Height = 1600 × 4500
1200 × 7.5 × Height = 72,00,000
9000 × Height = 72,00,000

Height =

Height = 800 cm
= 8 m
Question 7
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Three solid metallic spheres of radii 6, 8 and 10 centimetres are melted to form a single solid sphere. The radius of the sphere so formed is

A
24 cm

B
16 cm

C
18 cm

D
12 cm

Solution

Let R be the radius of the sphere

Sum of volumes of 3 spheres = Sum of volume of single sphere

=

R³= 6³ + 8³+ 10³R³ = 216 + 512 + 1000

R³= 1728 cm

R = 12
Question 8
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Some spherical balls are made from scrap paper. The thickness of a spherical ball is 4 cm, where the outer radius is 1.5 times the inner radius. What will be the approximate volume of the scrap paper used?

A
5095.62 cm³

B
4200.05 cm³

C
4304.83 cm³

D
5406.63 cm³

Solution

Volume of a sphere = × π × r³
Now, thickness of the ball = outer radius - inner radius
Let the inner radius be x.
4 = 1.5x - x
4 = 0.5x
x = 8
Inner radius = 8 cm
Outer radius = 1.5 × 8 = 12 cm

Volume of scrap paper used = Outer volume of ball - Inner volume of ball

Volume of scrap used = × π × 12³ - × π × 8³= × π × [12³ - 8³]

= × π × [1728 - 512]

= × × 1216 = 5095.62 cm³
Question 9
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A toy is prepared using a hemisphere which is surmounted by a cone. The diameter of the hemisphere is d and the radius of the cone is half of the diameter of the hemisphere. The height of the toy is twice the height of the hemisphere. According to this information, which of the following expressions represents the volume of the toy?(Write the answer in cubic units)

A

B

C
2πd³

D

Solution

According to the question,
Radius of the hemisphere = (d/2) units
Height of the hemisphere = Radius of the hemisphere = (d/2) units
Radius of the cone is same as that of the hemisphere.
Height of the toy is double the height of the hemisphere, which means height of the cone is same as that of the hemisphere, i.e. (d/2) units.
Thus,
Volume of the given figure = Volume of the hemisphere + Volume of the cone

= πr³ + πr²h

=

=

=
Question 10
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The ratio of the base radii of two right circular cones is 7 : 4 and the slant height of these right circular cones is same. Which of the following represents the ratio of their curved surface areas?

A
4 : 7

B
49 : 16

C
8 : 14

D
7 : 4

Solution

Let l₁ and l₂ be the slant heights of the first and second cones, respectively.
So, l₁ = l₂ = l
Let r₁ and r₂ be the base radii of the first and second cones, respectively.

Also, curved surface area of a cone = πrl

Ratio of curved surface areas of the two cones = Curved surface area of the first cone ÷ Curved surface area of the second cone

=

=

= =

Ratio = 7 : 4
Question 11
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The volume of a cone increases by x cubic units after its radius is increased by 6 units. Increasing the height of the same cone by 9 units also increases the volume by x cubic units. If the original height is 3 units, then the original radius is

A
8 units

B
6 units

C
7 units

D
9 units

Solution

Let us say the initial volume of the cone be r²h.
After increasing the radius by 6 units, volume = (r + 6)²h
(r + 6)²h = r²h + x
h[r² + 36 + 12r - r²] = x
h[36 + 12r] = x ..... (i)
After increasing the height by 9 units, volume = r²(h + 9)
r²(h + 9) = r²h + x
r²h + r²(9) - r²h = x
3r²= x

Putting the value of x in equation (i), we get
h[36 + 12r] = 3r²
9r² = h[36 + 12r]

Putting the value of h in this equation, we get
9r² = 3[36 + 12r]
3r² - 12r - 36 = 0
r² - 4r - 12 = 0
r² - 6r + 2r - 12 = 0
r = 6 units
Question 12
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A cylinder of the maximum possible size is made out of a solid wooden cube with each side 20 cm. What percentage of material (approx.) is left in this process?(Take π = 3.14)

A
20%

B
25%

C
22%

D
30%

Solution

Volume of cube (V) = Side³ = 20³ cm³Volume of cylinder (v) =

% of Material left =

= (100 - 78.5) %
= 21.5 % ≈ 22

Hence, 22% is the correct answer.
Question 13
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What will be the area of a given cricket ground if a ground level machine in a cylindrical shape, having a radius of 49 cm and length 160 cm, needs 600 revolutions to complete its levelling?

A
2956.7 m²

B
2956.8 m²

C
2956.9 m²

D
2957.5 m²

Solution

Radius of machine roll = 49 cm
Length of the roller (h) = 160 cm
Curved surface area of the roller = 2πrh
= 49,280 cm²
∴ Area covered by the roller in 1 revolution = 49,280 cm²
∴ Area covered by the roller in 600 revolutions = 49,280 × 600 cm²
= 2,95,68,000 cm²
Hence, Area of the playground == 2956.8 m²
Question 14
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Cost of electroplating a spherical steel ball is Rs. 887.04. If 9 paise are charged per square cm, then what is the diameter of the spherical steel ball? (π = 22/7)

A
56 cm

B
63 cm

C
65 cm

D
67 cm

Solution

Total cost of electroplating = Rs. 887.04
Surface area of ball = 887.04 ÷ 0.09 = 88,704 ÷ 9
= 9856 cm²According to the question,
4πr² = 9856

r² =

r² = 784
r = 28 cm

Diameter = 28 × 2 = 56 cm
Question 15
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Raman while making a lamp post in spherical shape decides to reduce the diameter of the lamp post by a quarter. What effect will it have on curved surface area of the lamp post?

A
43.75% decrease

B
44.75% decrease

C
45.75% decrease

D
46.75% decrease

Solution

Let, originally, the diameter of the sphere be 2r.
Then, radius of the sphere = r
Surface area of the sphere = 4πr² ... (i)
New diameter of the sphere = 2r – 2r ×
∴ New radius of the sphere =
Surface area of the new sphere = 4π × =
Decrease in surface area = 4πr² –=
Percent decrease ==== 43.75%
Hence, the surface area decreases by 43.75%.
Question 16
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Mariya is an artist and is very precise about her inventory to be used in assignments. She undertakes the work of colouring the bricks, which are to be used for building a wall. She decides to use a bucket of paint which has sufficient paint to colour 8.9 square metre of area. How many bricks of dimensions 55 cm, 20 cm and 15 cm will she paint regardless the total bricks available, if she decides to use only the fixed amount of paint available with her?

A
17

B
18

C
19

D
20

Solution

Here, l = 55 cm, b = 20 cm, h = 15 cm.
Total surface area of one brick = 2(lb + bh + hl)
= 2(55 × 20 + 20 × 15 + 15 × 55) cm²
= 2(1100 + 300 + 825) cm² = 4450 cm²
⇒ In metres square == 0.445 m²
Therefore, number of bricks that can be painted === 20
Question 17
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Anju has used cloth on edges of a rectangular aquarium to make it look fancy. The length of the aquarium is 60 cm, whereas the width and height are 50 cm each.

(i) What is the total surface area of the aquarium?
(ii) How much length of cloth is used on all edges?

A
(i) 15,000 cm²(ii) 610 cm

B
(i) 16,000 cm²
(ii) 630 cm

C
(i) 17,000 cm²
(ii) 640 cm

D
(i) 18,000 cm²
(ii) 650 cm

Solution

Here, l = 60 cm, b = 50 cm, h = 50 cm

(i) Total surface area of the aquarium = 2(lb + bh + hl)
= 2(60 × 50 + 50 × 50 + 50 × 60) cm²
= 2(3000 + 2500 + 3000) cm²
= 2 × 8500 cm² = 17,000 cm²
Hence, total surface area of the aquarium = 17,000 cm²

(ii) A cuboid has 12 edges. These consist of 4 lengths, 4 breadths and 4 heights.
∴ Length of the cloth required = 4l + 4b + 4h
= (4 × 60 + 4 × 50 + 4 × 50) cm
= (240 + 200 + 200) cm = 640 cm
Question 18
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The decorative block shown in the figure is made up of two solids, a cube and a hemisphere. The base of the block is a cube with edge 5 cm and the hemisphere fixed on the top has a diameter of 4.2 cm. The total surface area of the block is

A
150 cm²

B
160.86 cm²

C
162.86 cm²

D
163.86 cm²

Solution

Total S.A. = S.A. of the cube + C.S.A. of the hemisphere - Bottom area of the hemisphere
= 6 × 5²+ 2 × (2.1)² -(2.1)²= 6 × 25 + × 2.1 × 2.1
= 163.86 cm²
Question 19
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In a stable, Ram prepares liquid diet for each horse in a cylinder shaped bowl of diameter 14 cm, filled up to a height of 8 cm. What quantity of liquid diet must be prepared to satisfy 500 horses?

A
6,17,000 cm³

B
6,16,000 cm³

C
6,15,000 cm³

D
6,14,000 cm³

Solution

Here,
d = 14 cm
r = 7 cm, h = 8 cm
Capacity of 1 cylindrical bowl = πr²h

Hence, liquid diet consumed by 500 horses per day would be:

500 × 1232 cm³ = 6,16,000 cm³
Question 20
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The curved surface areas of a right-circular cone and a hemisphere, having the same radius are in the ratio of . Find the ratio of the height of the cone to the radius of the hemisphere.

A
1 : 3

B
3 : 1

C
1 : 2

D
1 : 4

Solution

Here, 'l' is slant height, l =
As both have the same radius, therefore:

Squaring both sides:

Question 21

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Match the following.

Column - I	Column - II
(P) A cylinder has radius 6 cm and height 3 cm. It is melted to form small cones of radius 2 cm and height 3 cm. How many such cones can be made?	(i) 20
(Q) A cylinder of radius 7 cm and length 196 cm is melted to form small cubes of side 14 cm. How many such cubes can be made?	(ii) 11
(R) The radius of a hemisphere is half that of a spherical ball. How many such hemispheres can be made out of the large sphere?	(iii) 27
(S) An object is made up of two shapes - cuboid at the bottom and cube at the top. If cuboid's breadth and height are double of cube's length, and cuboid's length is the same as cube's length, then how many small cuboids of height and breadth equal to one-quarter of the height of the big cuboid can be obtained from the object? (The length of a small cuboid is the same as that of the cube.)	(iv) 16

(P) - (iii), (Q) - (ii), (R) - (i), (S) - (iv)

(P) - (iii), (Q) - (ii), (R) - (iv), (S) - (i)

(P) - (iii), (Q) - (iv), (R) - (ii), (S) - (i)

(P) - (iii), (Q) - (i), (R) - (iv), (S) - (ii)

Solution

(P) Volume of the cylinder is three times the volume of the cone with same radius and same height as that of the cylinder.
Volume of the cylinder = πr²h
Volume of the cone = πr²h
Three cones of radius and height same as cylinder can be produced.

Now, the radius of the cone is one third of the radius of the cylinder.
If the radius of the cone is reduced to one third, its volume will be reduced by 3² = 9 times, as volume of the cone is πr²h.
As the height of the cone is same as that of the cylinder and the radius is one third, total number of cone that can be obtained = 3 × 9 = 27

(Q) Volume of the cylinder = πr²h = × 7 × 7 × 196 = 30,184 cm³Volume of each small cube = 14 × 14 × 14 = 2744 cm³Number of cubes that can be formed = 30,184 ÷ 2744 = 11

(R) Radius of the spherical ball = 2 cm
Radius of the hemisphere = 2 ÷ 2 = 1
Volume of the spherical ball =
Volume of the hemisphere =
Number of small hemispheres that can be made out of the spherical ball = = = = 16
So, 16 such small hemispheres can be obtained from the sphere.
(S) Let cube's height = 2x
Given: Cuboid's height = Double of cube's height = 4x
And small cuboid's height = One-quarter of cuboid's height = x
Now, Volume of the cube = Side³
= 8x³
Volume of the cuboid = length × breadth × height = (2x) × (4x) × (4x) = 32x³
Total volume of the object = (8 + 32)x³ = 40x³
Volume of a small cuboid = 2x × x × x = 2x³
Number of small cuboids that can be made out of the object = = 20
Hence, 20 small cuboids can be made out of the object.

Question 22
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A sphere of diameter 21 cm is placed on a cuboid of dimensions 33 cm, 21 cm, and 49 cm. Both are melted to form a large sphere. Find the ratio of the radius of small sphere to the radius of large sphere. {Take pi = 22/7}

A
4 : 2

B
1 : 2

C
3 : 2

D
2 : 3

Solution

Radius of the small sphere = 21 ÷ 2 = 10.5 cm

Let r be the radius of small sphere.
Volume of the small sphere = = = 4851 cm³Volume of the cuboid = 33 × 21 × 49 = 33,957 cm³Volume of the large sphere = 4851 + 33,957 = 38,808 cm³Let R be the radius of the large sphere. = 38,808

R³ = = 9261
R = 21 cm
Required ratio = 10.5 : 21 = 1 : 2
Question 23
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Directions: Read the given statements and write 'T' for true and 'F' for false.

(i) If the radius of a hemisphere and a cylinder is same, then volume of the hemisphere : volume of the cylinder = 3 (radius of the hemisphere or the cylinder) : 2 (height of the cylinder).
(ii) In a cylinder having equal radius and height, if both radius and height are doubled, the total surface area would become 9 times.
(iii) If the radius of a cylinder and a cone is same, and the height of the cylinder is equal to the slant height of the cone, then the ratio of the curved surface area of the cylinder to that of the cone is 2 : 1.

A
F, T and F

B
F, T and T

C
F, F and T

D
T, T and T

Solution

(i) Let 'r' be the radius of the hemisphere and the cylinder.
Volume of the hemisphere =
Volume of the cylinder =
Now,
The ratio of the volume of the hemisphere to the volume of the cylinder =
×
=
Hence, it is clear that the ratio of the volume of the hemisphere to the volume of the cylinder = 2r : 3h
The ratio of the volume of the hemisphere to the volume of the cylinder is, 2 times the radius of the hemisphere or the cylinder : 3 times the height of the cylinder, if their radii are same.
Hence, the given statement is false.

(ii) Surface area of cylinder = 2πrh + 2πr² As, r = h
Surface area of cylinder = 2πr² + 2πr² = 4πr²
So if 'r' would be doubled, the surface area would become 4 times because the square of double is 4 times.
Hence, the given statement is true.

(iii) Let 'r' be the height of the cylinder and the cone too.
Curved surface area of the cylinder = 2πrh
As the slant height of the cone = Height of the cylinder = h,
Curved surface area of the cone = πrh (l = h) Now, the ratio of the curved surface area of the cylinder to that of the cone with the same radius =
= 2 : 1
Hence, the given statement is true.
Question 24
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The volume of a cylinder is 5184 cm³. A cone of radius and height same as those of cylinder is carved out of it. The cone is then melted to form a cube and then cut into two equal cuboids. Both the small cuboids are then plated gold at the cost of Rs. 2.50 per cm². Find the total cost of gold plating.

A
Rs. 2880

B
Rs. 2740

C
Rs. 2625

D
Rs. 2500

Solution

Volume of a cylinder is 3 times the volume of a cone having same radius and height.
Thus, volume of the given cone = 5184 ÷ 3 = 1728 cm³Volume of the cube = Volume of the cone
Let the side of cube be x cm.
x³ = 1728
x = 12 cm
If the cube is cut into two cuboids, dimensions of each part would be 12 cm, 12 cm and 6 cm.
Surface area of each small cuboid = 2(lb + bh + hl) = 2(12 × 12 + 12 × 6 + 12 × 6) = 2 (144 + 72 + 72) = 576 cm²Surface area of two cuboids = 576 × 2 = 1152 cm²Total cost of gold plating = 1152 × 2.5 = Rs. 2880
Question 25
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Which of the following statements is CORRECT?

(i) The radius of the base of a cone is equal to its height. If the height of the cone is decreased by one-quarter without changing the radius, then the new slant height istimes its previous slant height.
(ii) If the radius of a hemisphere is double of the radius of a sphere, their volumes would be same.

A
Both statements are correct.

B
(i) is correct and (ii) is incorrect.

C
(i) is incorrect and (ii) is correct.

D
Both statements are incorrect.

Solution

(i) Let height of the cone = h and Radius of the base of the cone = r

According to the question,

h = r

Let slant height of the cone =

Therefore, slant height =

=

=
= … Eq. 1

Now, when the height is decreased by one-quarter but the radius is kept the same, then:

New height =

=

=

Let the new slant height be .

Therefore, new slant height =

=

=

=
= … Eq. 2

Dividing Eq. 1 by Eq. 2,
=
=
=

Hence, it is clear that the given statement is incorrect.

(ii) Let the radius of the hemisphere be 2r units and that of the of sphere be r units.

Volume of the hemisphere = = = Volume of the sphere =
Volume of the given hemisphere would be 4 times the volume of the given sphere.
So, the given statement is incorrect.