Surface Areas And Volumes(9th Grade > Mathematics ) Questions and answers for exam Preparation

9th Grade > Mathematics

SURFACE AREAS AND VOLUMES MCQs

Total Questions : 58 | Page 1 of 6 pages

Question 1. There is a conical tent whose slant height is 14

m

. If the curved surface area of cone is 308

m^{2}

,then its base area is _______.

There Is A Conical Tent Whose slant Height Is 14 M. If The ...

132 m2
224 m2
254 m2
154 m2

Discuss Question

Answer: Option D. -> 154 m2
:
D
The curved surface area of a cone =

π r l

where,

r

is radius and

l

is slant height.

\Rightarrow 308 = π r l

\Rightarrow \frac{22}{7} \times r \times 14 = 308

\Rightarrow r = \frac{308 \times 7}{22 \times 14}

\Rightarrow r = 7 m

Base area

= π r^{2}

= \frac{22}{7} \times 7^{2}

= 154 m^{2}

Question 2. 27 spherical iron balls of radius 5 cm each are melted and recasted into a big sphere.The radius of the big sphere is _____.

3 cm
9 cm
15 cm
5 cm

Discuss Question

Answer: Option C. -> 15 cm
:
C
Volumeof the big iron ballwill be equal to the volume of27 small iron balls.
Let the radius of the big sphere be R cm.
Let the radius of the small sphere be
r cm = 5 cm.

\Rightarrow \frac{4}{3} \times π R^{3} = 27 \times (\frac{4}{3} \times π r^{3})

\Rightarrow R^{3} = 27 r^{3}

R = 3 r = 3 \times 5 = 15

cm.

Question 3. An iron spherical ball of radius 21cm is melted and it is re-casted in hemispheres of radius 7cm. The number of balls formed will be
___

Discuss Question

:
The volume will remain constant.
Let's assume

x

such hemispherical balls will be formed.
So

\frac{4}{3} π R^{3} = x \frac{2}{3} π r^{3}

(Where R is the radius of spherical ball and r is the radius of thehemispheres)

\Rightarrow \frac{4}{3} π {(21)}^{3} = x \frac{2}{3} π {(7)}^{3}

\Rightarrow x = \frac{2 \times 21 \times 21 \times 21}{7 \times 7 \times 7}

\Rightarrow x = 54

Question 4. The radius of a hemisphere having total surface area of 1848 sq. cm is 7 cm.

True
False

Discuss Question

Answer: Option B. -> False
:
B
Total surface area

= 3 π r^{2}

\Rightarrow 3 \times \frac{22}{7} \times r^{2} = 1848

\Rightarrow r^{2} = \frac{1848 \times 7}{3 \times 22}

\Rightarrow r^{2} = 196

\Rightarrow r = 14

cm
Thus, above statement is false.

Question 5. A conical dome of a palace is supported by a cylindrical pillar, both having the same radius. If the ratio of the height of the pillar to that of dome is 4:3, then what is the ratio of their volumes?

Discuss Question

Answer: Option C. -> 4:1
:
C
Given that, the ratio of the heightof the pillar to that of dome is 4:3 .

\Rightarrow \frac{h_{c y l i n d e r}}{h_{c o n e}} = \frac{4}{3}

We know that,

v o l u m e o f c y l i n d e r = π r^{2} h

v o l u m e o f c o n e = \frac{1}{3} π r^{2} h

.
Now, ratio of volume of cylinder to the volume of cone

= \frac{π r^{2} h_{c y l i n d e r}}{\frac{1}{3} π r^{2} h_{c o n e}}

= \frac{π r^{2} \times 4}{\frac{1}{3} π r^{2} \times 3}

= \frac{4}{1}

= 4 : 1

Question 6. The total surface area in

m^{2}

of a cuboid with dimensions of 26m, 14m and 6.5m respectively is ___ m

^{2}

Discuss Question

:
Total Surface Area = 2( l x b + b x h + h x l)
= 2(26 x14 + 14 x6.5 + 6.5 x26)
= 2(364 + 91 + 169

= 1248 m^{2}

Question 7. The ratio of the radius to the height of a right cone is 3:4. Then the ratio of total surface area to curved surface of the cone is____.

Discuss Question

Answer: Option D. -> 8:5
:
D
Given that,
r:h = 3:4
Letr =

3 x

and h =

4 x

From the relation

l^{2} = h^{2} + r^{2}

,
The Slant height:

l = \sqrt{(3 x)^{2} + (4 x)^{2}}

\Rightarrow l = 5 x

We know that,
Total surface area of cone

= π r (r + l)

Curved surface area of cone

= π r l

Hence, the required ratio

= \frac{T o t a l s u r f a c e a r e a o f c o n e}{C u r v e d s u r f a c e a r e a}

= \frac{π r (r + l)}{π r l}

= \frac{π \times 3 x (3 x + 5 x)}{π \times 3 x \times 5 x}

= 8 : 5

Question 8. The water in a cubical tank of side 10m is transferred to completely fill a cuboidal tank of length 5m and breadth 10 and height h. Then the height of the cuboidal tank (in metres) is ___

Discuss Question

:
Volume of cube = (side)

^{3}

Volume of cuboid = (length)

\times

(breadth)

\times

(height)
So height of cuboidal tank can be found by equating the two

∴ height = \frac{{side}^{3}}{length \times breadth}

= \frac{10^{3}}{5 \times 10}

∴

Height = 20m

Question 9. Meena needs to serve mango juice to her guests in cylindrical tumblers of radius 7 cm up to a height of 10 cm. If she wants to serve 25 guests, how much juice should she prepare?

35 L
42.5 L
38.5 L
50 L

Discuss Question

Answer: Option C. -> 38.5 L
:
C
We know that,

V o l u m e o f a c y l i n d r i c a l c o n t a i n e r

π

r^{2} h

1000 c m^{3} = 1 L

Hence, the volume of each cylindrical tumbler

= \frac{22}{7} \times 7^{2} \times 10

= 1540 c m^{3}

So, the total volume of juice required

= 1540 \times 25

= 38500 {c m}^{3}

= 38.5 L

Question 10. The cost of painting the curved surface area of a cylindrical pillar of height 10 m and radius 3.5 m at ₹ 10/- per sq. meter is ₹ 2200/-.
(