Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

If the volume and surface area of a sphere are numerically the same, then its diameter is

A
$$6$$ units

B
$$8$$ units

C
$$10$$ units

D
$$12$$ units

Solution

Let radius of sphere be 'r', then
Volume of sphere=surface area of sphere
$$\Rightarrow \cfrac { 4 }{ 3 } \pi { R }^{ 3 }=4\pi { r }^{ 2 }\Rightarrow \cfrac { { r }^{ 3 } }{ { r }^{ 2 } } =3\Rightarrow r=3$$
$$\therefore $$Diameter$$=2\times 3=6$$ units
Question 2
1 / -0

The radii of two sphere are in the ratio $$3:4$$. The ratio of their surface area is:

A
$$3:16$$

B
$$4:16$$

C
$$6:16$$

D
$$9:16$$

Solution

Given, the ratio of radii of two spheres is $$3:4$$.
We know, the surface area of a sphere $$=4 \pi r^2$$.

Let $$r_1$$ and $$r_2$$ be the radii of the two sphere,
i.e. $$r_1 : r_2=\dfrac{r_1}{r_2}=3:4=\dfrac{3}{4}$$.

Then, the areas of the spheres be $$4 \pi r_1^2$$ and $$4 \pi r_2^2$$.

Therefore, the ratio of their surface area
$$=\dfrac{4 \pi r_1^2}{4\pi r_2^2}=\dfrac{r_1^2}{r_2^2}=(\dfrac{r_1}{r_2})^2=(\dfrac{3}{4})^2=\dfrac{9}{16}=9:16$$.

Hence, option $$D$$ is correct.
Question 3
1 / -0

The volume of a sphere of radius $$\pi$$ cm is ________ cc.

A
$$\dfrac {4\pi^{4}}{3}$$

B
$$\dfrac {4}{3}\pi r^{3}$$

C
$$4\pi r^{2}$$

D
$$\dfrac {4}{3}\pi r^{2}$$

Solution

The volume of a sphere with radius $$r = \dfrac {4}{3}\pi r^{3}$$
$$\therefore$$ The volume of a sphere with radius $$\pi \ cm$$.
$$= \dfrac {4}{3}\pi (\pi)^{3}$$
$$= \dfrac {4\pi^{4}}{3}(cm)^{3}$$.
Question 4
1 / -0

The total surface area of a closed cylindrical petrol storage tank whose diameter $$4.2$$ m and height $$4.5$$ m.

A
$$87.12$$ sq. m

B
$$78.21$$ sq. m

C
$$69.23$$ sq. m

D
$$65.45$$ sq. m

Solution

Total surface area of cylindrical tank $$=2\pi r(r+h) \\ = 2\times \cfrac{22}{7}\times \cfrac{4.2}{2}\left(\cfrac{4.2}{2}+4.5\right) \\ =13.2\times 6.6 = 87.12 cm^2$$
Question 5
1 / -0

A closed water tank has an internal size of $$10m \times 15m \times 20m$$. It needs to be lined with waterproofing cement on all its internal sides. At the rate of Rs. $$250\ \text {per} \ m^2$$, the total cost of lining will be Rs:

A
$$237, 500$$

B
$$325,000$$

C
$$400,000$$

D
$$475,000$$

Solution

Total surface area of water tank
$$=2(lb+bh+hl)=2(10\times 15+15\times 20+20\times 10) $$

$$= 2\times (150+300+200)=2\times 650 $$

$$ =1300m^2$$

Cost of lining $$=Rs (250\times 1300)=Rs\; 325000$$

Hence option $$'B'$$ is the answer.
Question 6
1 / -0

A building has $$25$$ cylindrical shaped poles. Each has a radius of $$28$$ cm and a height of $$4$$ cm. Find the cost of painting curved surface of all poles at the rate of Rs. $$8$$ per $$m^2$$.

A
$$7.23 $$ Rs

B
$$8 $$ Rs

C
$$14.08 $$ Rs

D
$$56.32 $$ Rs

Solution

Given,
Radius of each pillar= 28 cm
Height of each pillar=4 cm
So, as we all know
CSA of a cylinder=$$ 2 \dfrac {22} {7} rh$$

CSA of one pillar
$$= 2 \dfrac {22} {7} (28) (4)$$

$$=704 \ cm^2$$ = $$704 \times 10^{-4} \ m^2$$

$$=704 \times 10^{-4} \times (25)$$

$$=17600 \times 10^{-4} \ m^2$$

Since, cost of painting 1 sq m= Rs. 8
∴Cost of 1.76 sq m =Rs. 8 ( 1.76)
= Rs. 14.08
Question 7
1 / -0

From a solid sphere of radius $$R$$, a concentric solid sphere of radius $$\dfrac{R}{2}$$ is removed. The total surface area increases by

A
$$0\%$$

B
$$25\%$$

C
$$50\%$$

D
$$10\%$$

Solution

Given,
From a solid sphere of radius $$R$$, a concentric solid sphere of radius $$\dfrac{R}{2}$$ is removed.

Initial surface area of the sphere $$ \left( {A}_{1} \right) = 4 \pi {R}^{2}$$

Now, the new surface area of sphere $$\left( {A}_{2} \right) = 4 \pi {R}^{2} + 4 \pi {\left( \cfrac{R}{2} \right)}^{2}$$
$$=4 \pi {R}^{2} + \pi {R}^{2} $$
$$ = 5 \pi {R}^{2}$$.

$$\therefore$$ Increase in area is $$= \cfrac{{A}_{2} - {A}_{1}}{{A}_{1}} \times 100\%$$
$$= \cfrac{5 \pi {R}^{2} - 4 \pi {R}^{2}}{4 \pi {R}^{2}} \times 100\%$$
$$= \cfrac{\pi {R}^{2}}{4 \pi {R}^{2}} \times 100\%$$
$$ = 25 \%$$.

Hence, the area will be increased by $$25 \%$$.
Therefore, option $$B$$ is correct.
Question 8
1 / -0

Area of the four walls of a big room measuring $$25\ m \times 18\ m\times 10\ m$$ is

A
$$1000\ m^{2}$$

B
$$940\ m^{2}$$

C
$$860\ m^{2}$$

D
None of these

Solution
Question 9
1 / -0

A sphere has volume $$36\pi\ cm^{3}$$, find the radius of the sphere

A
$$4cm$$

B
$$3cm$$

C
$$6cm$$

D
$$8cm$$

Solution

Volume of sphere$$=\dfrac{4}{3}\pi r^3=36\pi \ c m^3$$

$$\implies r^3=27cm^3$$

Hence, $$r=3cm$$
Question 10
1 / -0

Which of the following is the formula for the volume of the cubiod

A
$$V=lbh$$

B
$$V=lh$$

C
$$V=lb$$

D
$$V=l^2bh$$

Solution

We know,
Volume of cuboid = $$length \times breadth \times height$$
Therefore, $$V=lbh$$
Hence, option A is correct.

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Surface Areas and Volumes Test - 47

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Surface Areas and Volumes Test - 47

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SHARING IS CARING

Question 1

$$6$$ units

$$8$$ units

$$10$$ units

$$12$$ units

Solution

Question 2

$$3:16$$

$$4:16$$

$$6:16$$

$$9:16$$

Solution

Question 3

$$\dfrac {4\pi^{4}}{3}$$

$$\dfrac {4}{3}\pi r^{3}$$

$$4\pi r^{2}$$

$$\dfrac {4}{3}\pi r^{2}$$

Solution

Question 4

$$87.12$$ sq. m

$$78.21$$ sq. m

$$69.23$$ sq. m

$$65.45$$ sq. m

Solution

Question 5

$$237, 500$$

$$325,000$$

$$400,000$$

$$475,000$$

Solution

Question 6

$$7.23 $$ Rs

$$8 $$ Rs

$$14.08 $$ Rs

$$56.32 $$ Rs

Solution

Question 7

$$0\%$$

$$25\%$$

$$50\%$$

$$10\%$$

Solution

Question 8

$$1000\ m^{2}$$

$$940\ m^{2}$$

$$860\ m^{2}$$

None of these

Solution

Question 9

$$4cm$$

$$3cm$$

$$6cm$$

$$8cm$$

Solution

Question 10

$$V=lbh$$

$$V=lh$$

$$V=lb$$

$$V=l^2bh$$

Solution

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