Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$512$$

B
$$500$$

C
$$400$$

D
$$345$$

Solution

Volume of the spherical ball of radius 8 cm
$$=\cfrac {4}{3}\pi \times 8^3 cm^3$$
Also, volume of each smaller spherical ball of radius 1 cm
$$=\cfrac {4}{3}\pi \times 1^3cm^3$$
Let n be the number of smaller balls that can be made. Then, the volume of the larger ball is equal to the sum of all the volumes of n smaller balls.
Hence, $$\cfrac {4}{3}\pi \times n=\cfrac {4}{3}\pi \times 8^3$$
$$\Rightarrow n=8^3=512$$
Hence, the required number of balls $$=512$$.
Question 2
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Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope. What would be the area of chart paper required by her, if she wanted to make a kaleidoscope of length $$25\space cm$$ with a $$3.5\space cm$$ radius? You may take $$\left (\pi = \dfrac{22}{7}\right )$$

A
$$540\space cm^2$$

B
$$520\space cm^2$$

C
$$550\space cm^2$$

D
$$560\space cm^2$$

Solution

Kaleidoscope is of cylindrical shape.
$$\therefore $$ Area of chart paper required $$=$$ CSA of cylinder
$$=$$ $$2\pi rh$$
$$ =$$ $$2\times \dfrac { 22 }{ 7 } \times 3.5\times 25$$
$$ =$$ $$550$$ $${ cm }^{ 2 }$$
Question 3
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Three solid spheres of copper, whose radii are $$3$$ cm, $$4$$ cm and $$5$$ cm respestively are melted into a single solid sphere of radius R. The value of R is

A
$$12$$ cm

B
$$8$$ cm

C
$$4$$ cm

D
$$6$$ cm

Solution

$$\frac {4}{3}\pi R^3=\frac {4}{3}\pi (3^3+4^3+5^3)\Rightarrow R^3=27+64+125=216\Rightarrow R=6$$
Question 4
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Find the difference between total surface area & curved surface area of a hemisphere of radius $$21\space cm$$.

A
$$1376\space cm^2$$

B
$$1386\space cm^2$$

C
$$1396\space cm^2$$

D
$$1404\space cm^2$$

Solution

Curved Surface Area of a hemisphere of radius 'r' $$ = 2\pi { r }^{ 2 } $$

Total Surface Area of a hemisphere of radius 'r' $$ = 3\pi { r }^{ 2 } $$

Hence, required difference $$ = 3\pi { r }^{ 2 } - 2\pi { r }^{ 2 } = \pi { r }^{ 2 } = \dfrac {22}{7} \times 21 \times 21 = 1386\ {cm}^{2} $$
Question 5
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The diameter of a garden roller is $$1.4$$ m and it its $$2$$ m long. How much area will it cover in one revolution?

A
$$6.8$$ $$\displaystyle m^{2}$$

B
$$8.8$$ $$\displaystyle m^{2}$$

C
$$3.8$$ $$\displaystyle m^{2}$$

D
$$5.8$$ $$\displaystyle m^{2}$$

Solution

$$\Rightarrow$$ Diameter of the roller $$(d)=1.4\,m$$.

$$\Rightarrow$$ Radius of the roller $$(r)=\dfrac{1.4}{2}=0.7\,m.$$

$$\Rightarrow$$ Length of the roller $$(h)=2\,m.$$

If the roller complete one revolution

Then the area covered $$=$$ Curved surface area of the roller $$=2\pi r h$$

$$\Rightarrow$$ Area covered in $$1$$ revolution $$=2\pi r h$$

$$=2\times \dfrac{22}{7}\times 0.7\times 2\\$$
$$=88\times 0.1\\$$
$$=8.8\,m^2$$
Question 6
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The diameter of the base of a right circular cylinder is $$28$$ cm and its height is $$21$$ cm Curved surface area of the cylinder is

A
$$1540$$ $$\displaystyle cm^{2}$$

B
$$1648$$ $$\displaystyle cm^{2}$$

C
$$1848$$ $$\displaystyle cm^{2}$$

D
$$1548$$ $$\displaystyle cm^{2}$$

Solution

$$Diameter=28\ cm$$
$$\Rightarrow r=14\ cm$$
and $$ h =21 cm$$
C.S.A of cylinder$$\displaystyle =2\pi rh=2\times \dfrac{22}{7}\times 14\times 21=1848\ cm^{2}$$
Question 7
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The hollow sphere, in which the circus motorcyclist performs his stunts, has diameter of $$7\ m$$. Find the area available to motorcyclist for riding.

A
$$154\ m^2$$

B
$$144\ m^2$$

C
$$38.5\ m^2$$

D
$$176\ m^2$$

Solution

Surface area of a sphere of radius 'r' $$ = 4 \pi { r }^{ 2 } $$

Radius of the sphere $$ = \dfrac {7}{2} m $$

Hence, the area available to motorcyclist for riding $$ = 4 \dfrac {22}{7} \times \dfrac {7}{2} \times \dfrac {7}{2} = 154 {m}^{2} $$
Question 8
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The diameter of a garden roller is $$1.4\space m$$ and it is $$2\space m$$ long. How much area will it cover in $$5$$ revolutions? (Use $$\pi = \dfrac{22}{7}$$)

A
$$40\space m^2$$

B
$$41.40\space m^2$$

C
$$44\space m^2$$

D
$$42.40\space m^2$$

Solution

Since the diameter of the roller is
$$ 0.7 m $$, its radius is $$ 0.35 m $$.
The roller is in the shape of a cylinder.
In one revolution, the roller covers the distance of one curved surface area.
Curved Surface Area of a Cylinder of Radius "$$R$$" and height "$$h$$" $$= 2\pi Rh$$
Radius of the roller $$ = 0.7 m $$
Curved Surface Area of the roller of radius $$ 0.35 m = 2 \times \dfrac {22}{7}\times 0.7 \times 2 = 8.8 m^2 $$
Area covered in $$5$$ revolutions $$ = 5 \times 8.8 m^2 = 44 m^2 $$
Question 9
1 / -0

A metal pipe is $$77$$ cm long The inner diameter of a cross section is $$4$$ cm then inner curved surface area is

A
$$900$$ $$\displaystyle cm^{2}$$

B
$$960$$ $$\displaystyle cm^{2}$$

C
$$968$$ $$\displaystyle cm^{2}$$

D
$$964$$ $$\displaystyle cm^{2}$$

Solution

$$\Rightarrow$$ Diameter of inner section of pipe $$(d)=4\,cm$$

$$\Rightarrow$$ Radius of inner section of pipe $$(r)=\dfrac{4}{2}=2\,cm$$

$$\Rightarrow$$ Height of a pipe $$(h)=77\,cm$$

$$\Rightarrow$$ Inner surface area of a pipe $$=2\pi rh$$

$$=2\times \dfrac{22}{7}\times 2\times 77$$

$$=968\,cm^2$$
Question 10
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Curved surface area of a right circular cylinder is $$4.4$$ $$\displaystyle m^{2}$$ If the radius of the base of the cylinder is $$0.7$$ m then its height is

A
$$2$$ m

B
$$3$$ m

C
$$4 $$m

D
$$1$$ m

Solution

Curved surface area $$=\displaystyle 2\pi rh=4.4$$

$$\Rightarrow 2\times \dfrac{22}{7}\times 0.7\times h=4.4\Rightarrow h=1m.$$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 30

Result

Surface Areas and Volumes Test - 30

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SHARING IS CARING

Question 1

$$512$$

$$500$$

$$400$$

$$345$$

Solution

Question 2

$$540\space cm^2$$

$$520\space cm^2$$

$$550\space cm^2$$

$$560\space cm^2$$

Solution

Question 3

$$12$$ cm

$$8$$ cm

$$4$$ cm

$$6$$ cm

Solution

Question 4

$$1376\space cm^2$$

$$1386\space cm^2$$

$$1396\space cm^2$$

$$1404\space cm^2$$

Solution

Question 5

$$6.8$$ $$\displaystyle m^{2}$$

$$8.8$$ $$\displaystyle m^{2}$$

$$3.8$$ $$\displaystyle m^{2}$$

$$5.8$$ $$\displaystyle m^{2}$$

Solution

Question 6

$$1540$$ $$\displaystyle cm^{2}$$

$$1648$$ $$\displaystyle cm^{2}$$

$$1848$$ $$\displaystyle cm^{2}$$

$$1548$$ $$\displaystyle cm^{2}$$

Solution

Question 7

$$154\ m^2$$

$$144\ m^2$$

$$38.5\ m^2$$

$$176\ m^2$$

Solution

Question 8

$$40\space m^2$$

$$41.40\space m^2$$

$$44\space m^2$$

$$42.40\space m^2$$

Solution

Question 9

$$900$$ $$\displaystyle cm^{2}$$

$$960$$ $$\displaystyle cm^{2}$$

$$968$$ $$\displaystyle cm^{2}$$

$$964$$ $$\displaystyle cm^{2}$$

Solution

Question 10

$$2$$ m

$$3$$ m

$$4 $$m

$$1$$ m

Solution

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