Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

A table lamp is in the form of a frustum of a cone whose area of the top and bottom circle bases are 212 and 103 $$m^2$$. Find the total surface area, if slant height is 10 m and top and bottom radii are 2 and 4 m ( Use $$\pi$$ = 3)

A
595 $$m^2$$

B
550 $$m^2$$

C
510 $$m^2$$

D
505 $$m^2$$
Question 2
1 / -0

A solid sphere of radius 6 mm is melted and then cast into small spherical balls each of radius 0.6 mm. Find the number of balls thus obtained.

A
1000

B
2000

C
4000

D
3000

Solution

Let number of small balls be $$x$$
Therefore Volume of $$ x $$ small balls = volume of sphere
$$\displaystyle x = \frac{6 \times 6 \times 6}{0.6 \times 0.6 \times 0.6} = \frac{216}{0.216}$$
$$x = 1000$$
$$\therefore$$ Number of balls = 1000
Question 3
1 / -0

Three cubes, whose edges are 12 cm, x cm and 10 cm respectively, are melted and recasted into a single cube of edge 14 cm. Find 'x'.

A
1.5 cm

B
2.5 cm

C
4 cm

D
3.1 cm

Solution

Edges of three cubes are 12 cm, $$x$$ cm and 10 cm.
$$\therefore$$ Volume of these cubes are $$12^3 , x^3 $$ and $$10^3$$
Edge of new cube formed = 14 cm
$$\therefore$$ Volume of new cube = $$(side)^3 = (14)^3 = 2744$$
According to the problem,
$$12^3 + x^3 + 10^3 = 2744$$
$$1728 + x^3 + 1000 = 2744$$
$$x^3 = 16 \Rightarrow x = 2.5 cm$$
$$Hence, the edge of cube = 2.5 cm$$
Question 4
1 / -0

A water jug is in the shape of a frustum of a cone of height $$21\ cm$$. The radii of its two circular ends are $$12\ cm$$ and $$7\ cm$$. Find the capacity of the water jug. (Use $$\pi$$ = 3).

A
$$5117\ cm^3$$

B
$$5217\ cm^3$$

C
$$5817\ cm^3$$

D
$$5917\ cm^3$$

Solution

Volume of a frustum cone $$= \displaystyle \dfrac{\pi h}{3} (R^2 + Rr + r^2) $$
$$=\displaystyle \dfrac{3 \times 21}{3} (12^2 + 12 \times 7 + 7^2)$$
$$= 21(144 + 84 + 49)$$
$$=21 \times 277$$
$$=5817 cm^3$$
Question 5
1 / -0

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 21 m and the height of the cone is 10 m. Find the volume of the solid.

A
24,024 $$m^3$$

B
23,024 $$m^3$$

C
22,024 $$m^3$$

D
21,024 $$m^3$$

Solution

Radius of the cone = radius of the hemisphere = 21m
Height of the cone = 10 m
Volume of the solid = Volume of the cone + Volume of the hemisphere
Volume of the cone = $$\displaystyle \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3$$
$$= \frac{1}{3} \times 22/7 \times 21^2 \times 10 + \frac{2}{3} \times 22/7 \times 21^3$$
= 4,620 + 19,404
= 24,024 $$m^3$$
Question 6
1 / -0

A metallic solid of volume 2200 $$cm^3$$ is melted and drawn into the form of a wire of height 7 cm. Find the radius of the wire.

A
12 cm

B
11 cm

C
10 cm

D
9 cm

Solution

Assume let the wire is of the shape of the right circular cylinder.
$$\therefore$$ Volume of the cylinder = $$\pi r^2 h$$
$$\displaystyle 2200 = \frac{22}{7} \times r^2 \times 7$$
$$r^2 = 100$$
$$r = 10 cm$$
$$\therefore$$ The radius of the wire is 10 cm.
Question 7
1 / -0

If the radii of the circular ends of a conical water glass are 25 and 12 cm whose height is 45 cm. Find the capacity of the water glass? (Use $$\pi$$ = 3.14)

A
50,349.9 $$cm^3$$

B
51,349.9 $$cm^3$$

C
52,349.9 $$cm^3$$

D
53,349.9 $$cm^3$$

Solution

Volume of a frustum cone = $$ \displaystyle \frac{\pi h}{3} (R^2 + Rr + r^2) $$
$$=\displaystyle \frac{3.14 \times 45}{3} (25^2 + 25 \times 12 + 12^2)$$
$$= 47.1 (625 + 300 + 144)$$
=$$47.1 \times 1069$$
= $$50349.9 cm^3$$
Question 8
1 / -0

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $$12$$ m and height of the cone is $$5$$ m. Find the volume of the solid.

A
$$1392\pi~ m^{3}$$

B
$$4176\pi~ m^{3}$$

C
$$1028\pi~ m^{3}$$

D
$$1308\pi~ m^{3}$$

Solution

The volume of solid $$=$$ volume of cone + volume of a hemisphere

$$=$$ $$\dfrac { 1 }{ 3 } \pi { r }^{ 2 }h+\dfrac { 2 }{ 3 } \pi { r }^{ 3 }$$

$$ =$$ $$\dfrac { 1 }{ 3 } \pi { r }^{ 2 }\left( h+2r \right) $$

$$ =$$ $$\dfrac { 1 }{ 3 } \times \pi\times 12\times 12(5+2\times 12)$$
$$=\dfrac{1}{3}\times \pi \times 144\times (5+24)$$

$$ =$$ $$\dfrac{1}{3}\times \pi \times144\times 29=1392\pi~{ m }^{ 3 }$$
Question 9
1 / -0

A cylindrical powder tin of 15 cm of height and 14 cm of radius is filled with water. The powder tin is emptied to make a conical heap of water on the ground. If the height of the conical heap is 42 cm, what is approximate value of the radius? (Use $$\pi$$ = 3).

A
14 cm

B
16 cm

C
18 cm

D
20 cm

Solution

Volume of powered tin = Volume of the cone = $$ \pi r^2 h \Rightarrow 3 \times 14 \times 14 \times 15$$
Volume of conical heap = Volume of the cone = $$\displaystyle \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3 \times r^2 \times 42$$
On equating, we get
$$\displaystyle 3 \times 14 \times 14 \times 15 = \frac{1}{3} \times 3 \times r^2 \times 42$$
$$26,460/126 = r^2$$
$$210 = r^2$$
$$r = 14 cm$$
Question 10
1 / -0

A brick whose length, breadth and height are $$3$$m, $$6$$m and $$9$$m respectively which are cut into single cuboid. Find the surface area of the cuboid.

A
$$198 m^{2}$$

B
$$188 m^{2}$$

C
$$178 m^{2}$$

D
$$158 m^{2}$$

Solution

Surface area of the cuboid $$= 2(lb + bh + hl)$$
$$= 2 \times (3 \times 6 + 6 \times 9 + 9 \times 3)$$
$$= 2 \times (18 + 54 + 27)$$
$$= 2 \times (99)$$
$$= 198 m^{2}$$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 42

Result

Surface Areas and Volumes Test - 42

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SHARING IS CARING

Question 1

595 $$m^2$$

550 $$m^2$$

510 $$m^2$$

505 $$m^2$$

Question 2

1000

2000

4000

3000

Solution

Question 3

1.5 cm

2.5 cm

4 cm

3.1 cm

Solution

Question 4

$$5117\ cm^3$$

$$5217\ cm^3$$

$$5817\ cm^3$$

$$5917\ cm^3$$

Solution

Question 5

24,024 $$m^3$$

23,024 $$m^3$$

22,024 $$m^3$$

21,024 $$m^3$$

Solution

Question 6

12 cm

11 cm

10 cm

9 cm

Solution

Question 7

50,349.9 $$cm^3$$

51,349.9 $$cm^3$$

52,349.9 $$cm^3$$

53,349.9 $$cm^3$$

Solution

Question 8

$$1392\pi~ m^{3}$$

$$4176\pi~ m^{3}$$

$$1028\pi~ m^{3}$$

$$1308\pi~ m^{3}$$

Solution

Question 9

14 cm

16 cm

18 cm

20 cm

Solution

Question 10

$$198 m^{2}$$

$$188 m^{2}$$

$$178 m^{2}$$

$$158 m^{2}$$

Solution

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