Surface Areas and Volumes Test

Result

Surface Areas and Volumes Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

A spherical ball made of iron has diameter 6 cm. If density of iron 8g/$$\displaystyle cm^{3} $$ then mass of the ball is nearly (use $$\displaystyle \pi =3.142 $$)

A
0.9 kg

B
0.8 kg

C
0.7 kg

D
0.62 kg

Solution

Volume of ball
V = $$\displaystyle \dfrac{4}{3}\times 3.14\times 3\times 3\times 3cm^{3}=113.04m^{3}$$
Mass of ball = vol x density
= 113.04 x 8 g = 904.32 g = 0.9 kg
Question 2
1 / -0

Total surface area of a cube of 2 centimetre side is

A
$$\displaystyle 20cm^{2}$$

B
$$\displaystyle 24cm^{2}$$

C
$$\displaystyle 25cm^{2}$$

D
$$\displaystyle 30cm^{2}$$

Solution

Total surface area of the cube=$$6a^2$$
Here a=2 cm
$$\Rightarrow 6\times (2^2)=24 cm^2$$
Question 3
1 / -0

A right circular cylinder and a sphere are of equal volumes and their radii are also equal If h is the height of the cylinder and d is the diameter of the sphere then

A
$$\displaystyle \frac{h}{3}=\frac{d}{2} $$

B
$$\displaystyle \frac{h}{2}=\frac{d}{3} $$

C
$$2h = d$$

D
$$h = d$$

Solution

Volume of cylinder = Volume of sphere
$$\displaystyle \Rightarrow \pi \left ( \dfrac{d}{2} \right )^{2}h=\dfrac{4}{3}\pi \left ( \dfrac{d}{2} \right )^{3}$$
$$\displaystyle \Rightarrow h=\dfrac{2}{3}d\Rightarrow \dfrac{h}{2}=\dfrac{d}{3}$$
Question 4
1 / -0

Three solid spheres of a lead are melted into a single solid sphere If the radii of the three spheres be 1 cm, 6 cm and 8 cm respectively Then radius of the new sphere is :

A
$$2 cm$$

B
$$3 cm$$

C
$$5 cm$$

D
$$9 cm$$

Solution

Let r be the radius of the new sphere Then
$$\displaystyle \frac{4}{3}\pi r^{3}=\frac{4}{3}\pi \left ( 1 \right )^{3}+\frac{4}{3}\pi \left ( 6 \right )^{3}+\frac{4}{3}\pi \left ( 8 \right )^{3}$$
$$\displaystyle \frac{4}{3}\pi \left [ \left ( 1 \right )^{3}+\left ( 6 \right )^{3}+\left ( 8 \right )^{3} \right ]=\frac{4}{3}\pi \left [ 1+216+512 \right ]$$
or $$\displaystyle r^{3}=729=\left ( 9 \right )^{3}$$
$$\displaystyle \therefore $$ Radius of the new sphere (r) =9 cm
Question 5
1 / -0

The surface areas of the two spheres are in the ratio $$1:2$$. The ratio of their volumes is

A
$$1:2$$

B
$$\displaystyle 1:\sqrt{2}$$

C
$$\displaystyle 1:2\sqrt{2}$$

D
$$\displaystyle 1:3\sqrt{2}$$

Solution

Area ratio = 1 : 2 (given)
$$\displaystyle \therefore $$ radii ratio $$\displaystyle =1\sqrt{2}$$
Volume ratio $$\displaystyle =1^{3}:\left ( \sqrt{2} \right )^{3}=1:2\sqrt{2}$$
Question 6
1 / -0

The volume of a right circular cylinder is 1100 $$\displaystyle cm^{3} $$ and the radius of its base is 5 cm. The area of its curved surface is

A
$$\displaystyle \left ( 440+25\pi \right )cm^{2} $$

B
$$\displaystyle \left ( 440+50\pi \right )cm^{2} $$

C
$$\displaystyle 440 cm^{2} $$

D
None of these

Solution

Volume of a cylinder$$=2\times\pi \times r\times h$$
$$\Rightarrow 1100=\displaystyle \dfrac{22}{7}\times 5^{2}\times h$$ {$$\because volume = 1100 \,cm^3\, is\, given$$}

$$\Rightarrow h =14cm$$

$$\therefore \displaystyle C.S.A =2\times \dfrac{22}{7}\times 5\times 14$$
$$\therefore C.S.A=440cm^{2}$$
Question 7
1 / -0

If the radius of the base of a cylinder is $$2$$ cm and its height $$7$$ cm, then what is its curved surface area?

A
$$44\:cm^2$$

B
$$22\:cm^2$$

C
$$88\:cm^2$$

D
$$56\:cm^2$$

Solution

Given, $$r=2$$ cm, $$h=7$$ cm

Then curved surface area $$=2 \pi r h$$

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

$$=88 cm^{2}$$
Question 8
1 / -0

A sphere of diameter $$10$$ cm weighs $$44$$ kg. The weight of a sphere of the same material whose diameter is $$6$$ cm is

A
$$2.64$$ kg

B
$$1.584$$ kg

C
$$0.9504$$ kg

D
$$\displaystyle\frac{4}{3}(0.9504)\:kg$$

Solution

Let the weight of the sphere of diameter $$6$$ cm be $$x$$ kg and density of the material be $$d$$ gm/c.c.

$$\displaystyle\therefore\frac{x}{4.4}=\frac{\displaystyle\frac{4}{3\pi}\times3^3\times d}{\displaystyle\frac{4}{3\pi}\times5^3\times d}$$

$$\displaystyle\implies x=\frac{4.4\times27}{125}=0.9504\:kg$$
Question 9
1 / -0

Find the volume of a hemisphere of radius $$6.3 \ cm$$ ($$\displaystyle \pi =22/7$$)

A
$$523.9 \ \displaystyle cm^{3}$$

B
$$520.91 \ \displaystyle cm^{3}$$

C
$$512.91 \ \displaystyle cm^{3}$$

D
$$510.91 \ \displaystyle cm^{3}$$

Solution

$$\Rightarrow$$ Volume of a hemisphere $$=\dfrac{2}{3}\pi r^3$$

$$=\dfrac{2}{3}\times \dfrac{22}{7}\times (6.3)^3$$

$$=\dfrac{2}{3}\times \dfrac{22}{7}\times 6.3\times 6.3 \times 6.3$$

$$=2\times 22\times 2.1\times 0.9\times 6.3\\$$
$$=523.9\,cm^3$$
Question 10
1 / -0

How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm if each bullet has radius 2 cm?

A
3000

B
2541

C
1779

D
2332

Solution

Volume of the cube of side 44 $$\displaystyle cm^{2}$$=$$\displaystyle 44\times 44\times 44cm^{3}$$
Let the number of bullets be N Radius is 2 cm Therefore
Total volume of N spheres=Volume of the cube
$$\displaystyle N\cdot \frac{4}{3}\times \frac{22}{7}\times 2\times 2\times 2=44\times 44\times 44$$
$$\displaystyle N=\frac{44\times 44\times 44\times 3\times 7}{4\times 22\times 2\times 2\times 2}=2541cm$$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 31

Result

Surface Areas and Volumes Test - 31

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SHARING IS CARING

Question 1

0.9 kg

0.8 kg

0.7 kg

0.62 kg

Solution

Question 2

$$\displaystyle 20cm^{2}$$

$$\displaystyle 24cm^{2}$$

$$\displaystyle 25cm^{2}$$

$$\displaystyle 30cm^{2}$$

Solution

Question 3

$$\displaystyle \frac{h}{3}=\frac{d}{2} $$

$$\displaystyle \frac{h}{2}=\frac{d}{3} $$

$$2h = d$$

$$h = d$$

Solution

Question 4

$$2 cm$$

$$3 cm$$

$$5 cm$$

$$9 cm$$

Solution

Question 5

$$1:2$$

$$\displaystyle 1:\sqrt{2}$$

$$\displaystyle 1:2\sqrt{2}$$

$$\displaystyle 1:3\sqrt{2}$$

Solution

Question 6

$$\displaystyle \left ( 440+25\pi \right )cm^{2} $$

$$\displaystyle \left ( 440+50\pi \right )cm^{2} $$

$$\displaystyle 440 cm^{2} $$

None of these

Solution

Question 7

$$44\:cm^2$$

$$22\:cm^2$$

$$88\:cm^2$$

$$56\:cm^2$$

Solution

Question 8

$$2.64$$ kg

$$1.584$$ kg

$$0.9504$$ kg

$$\displaystyle\frac{4}{3}(0.9504)\:kg$$

Solution

Question 9

$$523.9 \ \displaystyle cm^{3}$$

$$520.91 \ \displaystyle cm^{3}$$

$$512.91 \ \displaystyle cm^{3}$$

$$510.91 \ \displaystyle cm^{3}$$

Solution

Question 10

3000

2541

1779

2332

Solution

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