Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

If the ratio of the radii of the circular ends of a conical bucket whose height is $$60$$cm is $$2:1$$ and addition of the areas is $$770$$ sq.cm. Find the capacity of the bucket in litres.

A
$$21.56$$ litres

B
$$215.6$$ litres

C
$$21.560$$ litres

D
$$2156$$ litres

Solution

The conical bucket is same as frustum of a cone.
Let $$r, R$$ be the radii of lower base and upper base of bucket respectively.
$$\therefore R=2r$$
Area of circular region $$=\pi r^2$$
$$\because$$ Addition of the areas is $$770$$ sq. cm
$$\therefore\ \pi R^2+\pi r^2=770\ cm^2$$
$$\implies\ \pi(2r)^2+\pi r^2=770\ cm^2$$
$$\implies\ 5r^2 \pi=770\ cm^2$$
$$\implies\ r^2=49\ cm^2$$
$$\therefore\ r=7\ cm$$
Capacity of bucket $$=$$ Volume of frustum $$=\dfrac{\pi h}{3}(R^2+Rr+r^2)$$
$$=\dfrac{\pi h}{3}(4r^2+2r^2+r^2)$$

$$=\dfrac{22}{7}\times \dfrac{60}{3}\times 7(r^2)$$

$$=22\times 20\times 49$$

$$=21560\ cm^3=21.56$$ litres

Hence, capacity of a bucket is $$21.56$$ litres.
Question 2
1 / -0

A shuttle cock used for playing badminton has the shape of the combination of

A
a cylinder and a sphere

B
a cylinder and a hemisphere

C
a sphere and a cone

D
frustum of a cone and a hemisphere

Solution

Frustum of a cone:A cone is cut through a plane parallel to its base and then the cone that 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 frustum of a cone
Hemisphere:In geometry it is an exact half of a sphere.
Question 3
1 / -0

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

A
a cone

B
frustum of a cone

C
a cylinder

D
a sphere

Solution

From definition, A cone is cut through a plane parallel to its base and that cone 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 frustum of a cone.
Question 4
1 / -0

A cone is cut through a plane parallel to its base and then the cone that 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 frustum of a cone

B
cone

C
cylinder

D
sphere

Solution

A cone is cut through a plane parallel to its base and then the cone that 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 frustum of a cone.
Question 5
1 / -0

During conversion of a solid from one shape to another, the volume of the new shape will

A
increase

B
decrease

C
remain unaltered

D
be doubled

Solution

Volume:The amount of space that a substance or object occupies, or that is enclosed within a container.
From the definition of volume ,Volume remain unchanged.
Question 6
1 / -0

The shape formed by rotating a right triangle about its height is

A
Sphere

B
Cylinder

C
Cone

D
Cuboid

Solution

The shape formed by rotating a right triangle about its height is cone
Question 7
1 / -0

Consider a frustum of cone with larger and smaller radius equal to $$r_1 \ and\ r_2$$, respectively. Then, volume of frustum of a cone is $$\frac {1}{3}{\pi}h(r_{1}^{2}+r_{2}^{2}+r_1r_2)$$.
State whether the statement is True or False.

A
True

B
False

C
Insufficient Data

D
None of the above

Solution

Area $$=\pi(R+r)\sqrt{(R-r)^2+h^2}$$

Volume = $$\dfrac{\pi}{3}h(R^2+r^2+R*r)$$
Question 8
1 / -0

An oxygen cylinder is in the form of a right circular cylinder with a cone at one end and a hemisphere at the other end. Their common radius is r and height of cylindrical part is H cm and that of conical part is h cm. Express the volume V in terms of H, h & r

A
$$V=\frac {\pi r^2}{3}(h+H+2r)$$

B
$$V=\frac {\pi r^2}{3}(h+2H+2r)$$

C
$$V=\frac {\pi r^2}{3}(h+3H+4r)$$

D
$$V=\frac {\pi r^2}{3}(h+3H+2r)$$

Solution

Reqd volume $$=$$ volume of (cone+cylinder+ hemisphere)
$$=\frac {1}{3}\pi r^2h+\pi r^2H+\frac {2}{3}\pi r^3$$
$$=\frac {\pi r^2}{3}(h+3H+2r)$$
Question 9
1 / -0

If $$ \displaystyle S_{1} $$ and $$\displaystyle S_{2} $$ be the surface area of a sphere and the curved surface of the circumscribed cylinder then $$ \displaystyle S_{1} $$ is equal to

A
$$ \displaystyle S_{2} $$

B
$$ \displaystyle 2S_{2} $$

C
$$ \displaystyle \frac{1}{2}S_{2} $$

D
$$ \displaystyle \frac{2}{3}S_{2} $$

Solution

Let the radius of the sphere be $$r.$$ Then the radius of the circumscribing cylinder will be $$r$$ as well.

Height$$(H)$$ of the cylinder will be equal to the diameter of the sphere that is equal to $$2r$$.

So, the surface area of the sphere will be $$S_1= 4\pi r^2\quad\quad\dots(i)$$

And, curved surface area of the cylinder,
$$S_2= 2 \pi rH$$
$$ = (2\pi r)2r $$
$$= 4\pi r^2\quad\quad\dots(ii)$$

From $$(i)$$ and $$(ii)$$, we can conclude that $$S_1=S_2$$.

Hence, option $$A$$ is correct.
Question 10
1 / -0

A conical flask of base radius r and height h is full of milk. The milk is now poured into a cylindrical flask of radius 2r. What is the height to which the milk will rise in the flask?

A
$$\displaystyle\frac{h}{3}$$

B
$$\displaystyle\frac{h}{6}$$

C
$$\displaystyle\frac{h}{9}$$

D
$$\displaystyle\frac{h}{12}$$

Solution

If H be the height of the milk through which it rises in the cylinder, then

$$\displaystyle\pi(2r)^2H=\frac{1}{3}\pi r^2h\implies H=\dfrac{h}{12}$$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 13

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Surface Areas and Volumes Test - 13

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Question 1

$$21.56$$ litres

$$215.6$$ litres

$$21.560$$ litres

$$2156$$ litres

Solution

Question 2

a cylinder and a sphere

a cylinder and a hemisphere

a sphere and a cone

frustum of a cone and a hemisphere

Solution

Question 3

a cone

frustum of a cone

a cylinder

a sphere

Solution

Question 4

a frustum of a cone

cone

cylinder

sphere

Solution

Question 5

increase

decrease

remain unaltered

be doubled

Solution

Question 6

Sphere

Cylinder

Cone

Cuboid

Solution

Question 7

True

False

Insufficient Data

None of the above

Solution

Question 8

$$V=\frac {\pi r^2}{3}(h+H+2r)$$

$$V=\frac {\pi r^2}{3}(h+2H+2r)$$

$$V=\frac {\pi r^2}{3}(h+3H+4r)$$

$$V=\frac {\pi r^2}{3}(h+3H+2r)$$

Solution

Question 9

$$ \displaystyle S_{2} $$

$$ \displaystyle 2S_{2} $$

$$ \displaystyle \frac{1}{2}S_{2} $$

$$ \displaystyle \frac{2}{3}S_{2} $$

Solution

Question 10

$$\displaystyle\frac{h}{3}$$

$$\displaystyle\frac{h}{6}$$

$$\displaystyle\frac{h}{9}$$

$$\displaystyle\frac{h}{12}$$

Solution

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