Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

A solid cone of radius $$r$$ and height $$h$$ is placed over a solid cylinder having same base radius and height as that of a cone. The total surface area of the combined solid is

A
$$\pi r\left [ \sqrt{r^{2}+h^{2}}+r+2h \right ]$$

B
$$\pi r\left [ \sqrt{r^{2}+h^{2}}+2r+3h \right ]$$

C
$$\pi r\left [2 \sqrt{r^{2}+h^{2}}+3r+2h \right ]$$

D
none of these

Solution

The combined solid is represented in above fig.

TSA of new solid $$=$$ CSA of Cone $$+$$ CSA of cylinder $$+$$ Base area of cylinder

TSA of solid $$= \pi r l + 2\pi r h + \pi { r }^{ 2 }$$

TSA of solid $$= \pi r(l+2h+r)$$

TSA of solid $$ = \pi r (\sqrt{r^2+h^2} + 2h + r )$$
Question 2
1 / -0

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

A
$$0.36 cm^{3}$$

B
$$0.35 cm^{3}$$

C
$$0.34 cm^{3}$$

D
$$0.33 cm^{3}$$

Solution

The capacity of the capsule is = Volume of cylinder +2*Volume of hemisphere
$$V=\Pi { \times 0.5 }^{ 2 }\times 1+2\times \frac { 2 }{ 3 } \times \Pi { 0.5 }^{ 3 }$$
$$V=0.36$$
Question 3
1 / -0

A cone is $$8.4\ cm$$ high and the radius of its base is $$2.1\ cm$$. It is melted and recast into a sphere. The radius of the sphere is

A
$$4.2\ cm$$

B
$$2.1\ cm$$

C
$$2.4\ cm$$

D
$$1.6\ cm$$

Solution

Volume of the cone $$=\dfrac{1}{3} \pi r_c^2 h$$
Volume of the sphere $$=\dfrac{4}{3} \pi r_s^3$$

$$r_c$$ is the radius of cone $$= 2.1$$ cm ....(Given)
Let $$r_s$$ be the radius of the sphere

$$h$$ is the height of cone $$=8.4$$ cm ...(Given)
Volume of cone $$=$$ Volume of sphere ...(Given)

$$\Rightarrow\dfrac13 \pi r_c^2h = \dfrac43\pi r_s^3$$
$$\Rightarrow\dfrac{1}{3} r_c^2h =\dfrac{4}{3} r_s^3$$
$$\Rightarrow r_s^3=9.261$$
$$\Rightarrow r_s=2.1$$ cm
Question 4
1 / -0

A metallic spherical shell of internal and external diameters $$4 $$ cm and $$8 $$ cm, respectively is melted and recast into the form a cone of base diameter $$8 $$ cm. The height of the cone is

A
$$12 $$ cm

B
$$14 $$ cm

C
$$15 $$ cm

D
$$18$$ cm

Solution

Given that volume of cone $$=$$ volume of sphere
$$\dfrac{1}{3} \pi r_1^2 h=\dfrac{4}{3} \pi [R^3-r^3] $$
$$ 16 h = 4 [4^3-2^3] $$
$$ 4 h=56 $$
$$ h = 14$$ cm
Question 5
1 / -0

The diameters of the two circular ends of the bucket are $$44 $$ cm and $$24 $$ cm. The height of the bucket is $$35$$ cm. The capacity of the bucket is

A
$$32.7$$ litres

B
$$33.7$$ litres

C
$$34.7$$ litres

D
$$31.7$$ litres

Solution

Volume of frustum of cone $$\displaystyle =\frac{\pi h}{3} (R^2+Rr+r^2)$$
$$1$$ litre$$=1000$$ cu. cm
Volume of a frustum cone $$=\dfrac{\pi(35)(22^2+12^2+(22)(12))}{3}$$
$$=32706.67$$ cu.cm
$$=32.7$$ litres
Question 6
1 / -0

The diameter of a metallic sphere is $$6 cm$$. It is melted and drawn into a wire of diameter $$2 cm$$, then the length of the wire is

A
$$12 cm$$

B
$$18 cm$$

C
$$36 cm$$

D
$$66 cm$$

Solution

Diameter of sphere $$=6$$ cm
Radius of sphere $$=\cfrac { 6 }{ 2 } =3cm$$
Volume of sphere $$=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }\\ =\cfrac { 4 }{ 3 } \pi \times 3\times 3\times 3\\ =36\pi { cm }^{ 3 }$$

Let the length of wire be $$h$$ cm
Diameter of wire $$=2$$ cm
Radius of wire $$=\cfrac { 2 }{ 2 } =1 cm$$
As the wire is in the shape of cylinder .so that volume of cylinder $$=\pi { r }^{ 2 }h$$
Now, volume of metal sphere $$=$$ volume of wire
$$36\pi =\pi \times { 1 }^{ 2 }\times h\\ h=36 cm$$
Question 7
1 / -0

A frustum of a right circular cone of height $$16\text{ cm}$$ with radii of its circular ends as $$8\text{ cm}$$ and $$20\text{ cm}$$ has its slant height equal to:

A
$$18\text{ cm}$$

B
$$16\text{ cm}$$

C
$$20\text{ cm}$$

D
$$24\text{ cm}$$

Solution

Given:
$$H=16\text{ cm}$$
$$R=20\text{ cm}$$
$$r=8\text{ cm}$$

Let the slant height of the frustum of cone be $$l$$,
$$\begin{aligned}{}l &= \sqrt {{h^2} + {{(R - r)}^2}} \\ &= \sqrt {{{(16)}^2} + {{(20 - 8)}^2}} \\ &= \sqrt {256 + 144} \\ &= \sqrt {400} \\ & = 20{\text{ cm}}\end{aligned}$$
Question 8
1 / -0

The side of a square is $$2$$ cm and semicircles are constructed on each side of the square, then the area of the whole figure is

A
$$(4 + 2\pi ) \ \text{cm}^2$$

B
$$(4 + 4\pi ) \ \text{cm}^2$$

C
$$4\pi \ \text{cm}^2$$

D
$$8\pi \ \text{cm}^2$$

Solution

Area of whole figure $$=$$ Area of square + 4 $$\times $$ area of semicircle
$$={ \left( 2 \right) }^{ 2 }+4\times \dfrac { 1 }{ 2 } \times \pi \times { \left( \dfrac { 2 }{ 2 } \right) }^{ 2 }$$
$$=\left( 4+2\pi \right) { cm }^{ 2 }$$
Question 9
1 / -0

A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains $$41\dfrac{19}{21}\ m^3$$ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building.

A
$$4\ m$$

B
$$8\ m$$

C
$$6\ m$$

D
$$10\ m$$

Solution

Diameter of hemisphere $$(D) =$$ Total height of building $$(H)$$
Radius of hemisphere $$(R) = \dfrac H2=$$ Height of the hemisphere
Height of cylinder $$= H - \dfrac H2 = \dfrac H2$$
Volume of building $$=$$ Vol. of cylinder $$+$$ Vol. of hemisphere
$$\dfrac { 880 }{ 21 } = \pi { { r }^{ 2 } }H + \dfrac { 2 }{ 3 } \pi { { r }^{ 3 } }$$
$$\dfrac { 880 }{ 21 } = \pi \left (\dfrac { H }{ 2 } \right)^{ 3 }+\dfrac { 2 }{ 3 } \pi \left (\dfrac { H }{ 2 } \right)^{ 3 }$$
$$ H = 4$$ m
Question 10
1 / -0

Two solid cones $$A$$ and $$B$$ are placed in a cylindrical tube as shown in the figure. The ratio of their capacities is $$2:1$$. Find their heights.

A
$$14$$cm, $$14$$cm

B
$$15$$cm, $$6$$cm

C
$$14$$cm, $$7$$cm

D
$$10$$cm, $$11$$cm

Solution

The ratio of volume of cones $$A$$ and $$B$$ is $$2:1$$, that is:

$$\frac { { V }_{ A } }{ { V }_{ B } } =\frac { 2 }{ 1 }$$

Let the height of the cone $$A$$ be $$h$$ then the height of the cone $$A$$ will be $$(21-h)$$.

$$\frac { { V }_{ A } }{ { V }_{ B } } =\frac { 2 }{ 1 } \\ \Rightarrow \frac { \frac { 1 }{ 3 } \pi { r }^{ 2 }h }{ \frac { 1 }{ 3 } \pi { r }^{ 2 }(21-h) } =\frac { 2 }{ 1 } \\ \Rightarrow \frac { h }{ 21-h } =\frac { 2 }{ 1 } \\ \Rightarrow h=42-2h\\ \Rightarrow 3h=42\\ \Rightarrow h=14$$

Since the height of cone $$A$$ is $$14$$ cm, therefore, the height of cone $$B$$ is $$21-14=7$$ cm.

Hence, the heights are $$14$$ cm and $$7$$ cm.

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Surface Areas and Volumes Test - 19

Result

Surface Areas and Volumes Test - 19

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SHARING IS CARING

Question 1

$$\pi r\left [ \sqrt{r^{2}+h^{2}}+r+2h \right ]$$

$$\pi r\left [ \sqrt{r^{2}+h^{2}}+2r+3h \right ]$$

$$\pi r\left [2 \sqrt{r^{2}+h^{2}}+3r+2h \right ]$$

none of these

Solution

Question 2

$$0.36 cm^{3}$$

$$0.35 cm^{3}$$

$$0.34 cm^{3}$$

$$0.33 cm^{3}$$

Solution

Question 3

$$4.2\ cm$$

$$2.1\ cm$$

$$2.4\ cm$$

$$1.6\ cm$$

Solution

Question 4

$$12 $$ cm

$$14 $$ cm

$$15 $$ cm

$$18$$ cm

Solution

Question 5

$$32.7$$ litres

$$33.7$$ litres

$$34.7$$ litres

$$31.7$$ litres

Solution

Question 6

$$12 cm$$

$$18 cm$$

$$36 cm$$

$$66 cm$$

Solution

Question 7

$$18\text{ cm}$$

$$16\text{ cm}$$

$$20\text{ cm}$$

$$24\text{ cm}$$

Solution

Question 8

$$(4 + 2\pi ) \ \text{cm}^2$$

$$(4 + 4\pi ) \ \text{cm}^2$$

$$4\pi \ \text{cm}^2$$

$$8\pi \ \text{cm}^2$$

Solution

Question 9

$$4\ m$$

$$8\ m$$

$$6\ m$$

$$10\ m$$

Solution

Question 10

$$14$$cm, $$14$$cm

$$15$$cm, $$6$$cm

$$14$$cm, $$7$$cm

$$10$$cm, $$11$$cm

Solution

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