Surface Areas and Volumes Test

Result

Surface Areas and Volumes Test - 49

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The shape of a glass (tumbler) (see above figure) is usually in the form of a

A
cone

B
frustrum of a cone

C
cylinder

D
sphere

Solution

The radius of the lower circular part is smaller than the upper part. So, it is frustrum of a cone.
Question 2
1 / -0

A right circular cylinder of radius $$r$$ cm and height $$h$$ cm (where $$h> 2r$$) just encloses of sphere of diameter

A
$$r$$ cm

B
$$2r$$ cm

C
$$h$$ cm

D
$$2h$$ cm

Solution

As the cylinder just enclosed the sphere so the radius or diameter of cylinder and sphere are equal i.e., $$2r$$ and height $$h > 2r$$.
Hence, verifies the option B.
Question 3
1 / -0

A medicine-capsule is in the shape of a cylinder of diameter $$0.5cm$$ with two hemispheres stuck to each to its ends. The length of entire capsule is $$2cm$$. The capacity of the capsule is

A
$$0.36cm^{3}$$

B
$$0.35cm^{3}$$

C
$$0.34cm^{3}$$

D
$$0.33cm^{3}$$

Solution

Capsule consists of 2 Hemispheres and a Cylinder (as shown in figure below),
$$r=\frac{0.5}{2}cm=0.25cm$$
$$\Rightarrow r=0.25cm$$
Total length of capsule = $$r+h+r$$
$$\Rightarrow 2cm=2r+h$$
$$\Rightarrow 2=2\times 0.25+h\Rightarrow h=2-0.5=1.5cm$$
Volume of capsule = Volume of two hemispheres + Volume of cylinder
$$=2\times \left ( \dfrac{4}{3}\pi r^{3}\times \dfrac{1}{2} \right )+\pi r^{2}h=\dfrac{4}{3}\pi r^{3}+\pi r^{2}h$$
$$=\pi r^{2}\left [ \dfrac{4}{3}r+h \right ]=\dfrac{22}{7}\times 0.25\times 0.25\left [ \dfrac{4}{3}\times 0.25+\dfrac{15}{10} \right ]$$
$$=\dfrac{22}{7}\times 0.25\times 0.25\left [ \dfrac{4}{3}\times \frac{25}{100}+\dfrac{3}{2} \right ]$$
$$=\dfrac{22}{7}\times 0.25\times 0.25\left [ \dfrac{1}{3}+\frac{3}{2} \right ]$$
$$=\dfrac{22}{7}\times 0.25\times 0.25\left [ \dfrac{2+9}{6} \right ]$$
$$=\dfrac{22\times 25\times 25\times 11}{7\times 6\times 100\times 100}=\dfrac{121}{42\times 8}=\dfrac{121}{336}$$
$$\therefore$$ Volume of capsule = $$0.3601 cm^{3} = 0.36 cm^{3}$$
Hence, verifies the option A.
Question 4
1 / -0

Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is:

A
$$4\ cm$$

B
$$3\ cm$$

C
$$2\ cm$$

D
$$6\ cm$$

Solution

Solid cylinder is recasted into 12 spheres.
So, the volume of 12 spheres will be equal to the volume of the cylinder,
12 sphere as shown in below figure 1.
$$R$$=?
Cylinder is shown in below figure 2.
$$r=\dfrac{2}{2}=1cm$$
$$h=16cm$$
$$\therefore$$ Volume of 12 spheres = Volume of cylinder
$$\Rightarrow \dfrac{4}{3}\pi R^{3}\times12=\pi r^{2}h$$
$$\Rightarrow 12\times\dfrac{4}{3}R^{3}=r^{2}h$$
$$\Rightarrow R^{3}=\dfrac{3r^{2}h}{4\times12}=\dfrac{3\times1\times1\times16}{4\times12}=1$$
$$\Rightarrow R=1cm$$
Hence, diameter is ($$2R= 2\times1=2$$) $$cm$$. So, right option is C.
Question 5
1 / -0

A cone is cut through a plane parallel to its base and then the cone is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called

A
a frustrum of cone

B
cone

C
cylinder

D
sphere

Solution

The new part that is left over on the other side of the plane is called a frustrum of a cone.
Question 6
1 / -0

The volume of a frustum of a cone of height h and ends radii $$r_1$$ and $$r_2$$ is

A
$$\frac{1}{3} \pi h (r^2_1 + r^2_2 + r_1 r_2)$$

B
$$\frac{1}{3} \pi h (r^2_1 + r^2_2 - r_1 r_2)$$

C
$$\frac{1}{2} \pi h (r^2_1 + r^2_2 - r_1 r_2)$$

D
$$\pi h (r^2_1 + r^2_2 + r_1 r_2)$$

Solution

The volume of a frustum of a cone of height h and ends radii $$r_1$$ and $$r_2$$ is $$\frac{1}{3} \pi h (r^2_1 + r^2_2 + r_1 r_2)$$
Question 7
1 / -0

If a rectangle revolves about one of its sides and completes a full rotation the solid formed is called

A
sphere

B
cylinder

C
hemisphere

D
cone

Solution

If a rectangle revolves about one of its side and completes a full rotation the solid formed is called cylinder
Question 8
1 / -0

If the radii of the circular ends of a bucket $$24$$ cm high are $$5$$ cm and $$15$$ cm respectively, find the inner surface area of the bucket (i.e., the area of the metal sheet required to make the bucket)
(Take $$\pi = 3.14$$)

A
$$1711.30\, cm^{2}$$

B
$$1812.30\, cm^{2}$$

C
$$1171.30\, cm^{2}$$

D
$$1279.30\, cm^{2}$$

Solution

Material used to make the bucket $$ = $$ Curved surface area $$ + $$ Area of the base
Curved Surface area of a frustum of a cone $$=\pi l({ r }_{ 1 }+{ r }_{ 2 })$$ where $$l$$ is the slant height $$ = \sqrt { { h }^{ 2 }+{ ({ r }_{ 1 }-{ r }_{ 2 }) }^{ 2 } } $$ $$h$$ is the height.
$$ { r }_{ 1 } $$ and $${ r }_{ 2 }$$ are the radii of the lower and upper ends of a frustum of a cone.
So, $$l = \sqrt { { 24 }^{ 2 }+{ (15 - 5) }^{ 2 } } $$
$$l = 26\text{ cm} $$
Hence, Curved Surface area of this frustum of a cone $$=3.14 \times 26 \times (15 + 5) = 1632.8 \text{ cm}^{2} $$
Area of the base $$ = \pi{ { r }_{ 1 } }^{2} = 3.14 \times 25 = 78.5 \text{ sq cm} $$
So, the inner surface area of the bucket $$ = 1632.8 + 78.5 = 1711.30 \text{ cm}^2 $$
Question 9
1 / -0

A circus tent is cylindrical upto a height of 3 cm and conical above it. If the diameter of the base is 105 m and the slant height of the conical part is 53 m, find the total cost of the canvas used to make the tent when the cost per square metre of the canvas is Rs. 10. (Take $$\pi =22/7$$)

A
Rs. 97350

B
Rs. 97450

C
Rs. 97550

D
Rs. 97650

Solution

The amount of canvas used to make the tent $$ = $$ Curved surface area of cylindrical part $$ + $$ Curved surface area if the conical part.
Curved Surface Area of a Cylinder of Radius $$"R"$$ and height $$"h"$$ $$ = 2\pi Rh$$
Radius of the cylindrical part $$ = \dfrac {Diameter}{2} = \dfrac {105}{2} $$ m
Curved surface area of a cone $$= \pi rl$$ where $$r$$ is the radius of the cone and $$l$$ is the slant height.
Radius of the conical part $$ = \dfrac {Diameter}{2} = \dfrac {105}{2} $$
Hence, area of canvas = $$2\pi rh+\pi rl $$
$$ = 2 \times \dfrac {22}{7} \times \dfrac {105}{2} \times 3 + \dfrac {22}{7} \times \dfrac {105}{2} \times 53 = 9735 {m}^{2} $$

Cost of canvas at $$ Rs 10 $$ per sq m $$ = 9735 \times 10 = Rs.\ 97350 $$
Question 10
1 / -0

A solid is in the form of a cone mounted on a right circular cylinder both having same radii of their bases. Base of the cone is placed on the top base of the cylinder. If the radius of the base and height of the cone be 4 cm and 7 cm, respectively, and the height of the cylindrical part of the solid is 3.5 cm, the volume of the solid is equal to

A
$$\displaystyle 293\frac{1}{3}\, cm^{3}$$

B
$$\displaystyle 293\frac{1}{5}\, cm^{3}$$

C
$$\displaystyle 298\frac{1}{3}\, cm^{3}$$

D
$$\displaystyle 298\frac{1}{5}\, cm^{3}$$

Solution

Volume of the solid $$ = $$ Volume of the cylindrical part $$ + $$ Volume of conical part

Volume of a Cylinder of Radius $$R$$ and height $$h$$ $$ = \pi { R }^{2}h $$

Volumeof a cone $$ = \dfrac { 1 }{ 3 } \pi { r }^{ 2 }h $$ where $$r$$ is radius and $$h$$ is height of the cone.

Hence, Volume of the solid
$$\displaystyle =\frac { 22 }{ 7 } \times 4\times 4\times 3.5 + \frac{ 1 }{ 3 } \times \frac {22}{7} \times { 4 }^{ 2 }\times 7 \\ = 176 + \dfrac {352}{3} = 293 \dfrac {1}{3} {cm}^{2} $$