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 3 of 6 pages

Question 21. A vessel is in the shape of a cube of side 30

m

. How much water can it hold?

29000 m3
27000 m3
26000 m3
28000 m3

Discuss Question

Answer: Option B. -> 27000 m3
:
B
In order to find the quantity of water that the vessel can hold, we have to find the volume of the vessel.
Side length of the cubic vessel = 30

m

.
Volume ofcubic vessel =

(s i d e)^{3}

(30)^{3}

= 27000

m^{3}

Question 22. A cricket ball of radius

r

fits exactly into a cylindrical tin as shown in the figure below. What will be the ratio between the surface areas of the cricket ball and the tin?

A Cricket Ball Of Radius R Fits Exactly Into A Cylindrical T...

Discuss Question

Answer: Option B. -> 2:3
:
B
Given that, the radius of the sphere is

r

.
Since the cricket ball fits exactly inside the cylinder tin, the height of the cylinder (

h

) will be equal to the diameter of the ball .

\Rightarrow h = 2 r

Ratio of their surface areas

= \frac{S u r f a c e a r e a o f b a l l}{S u r f a c e a r e a o f c y l i n d r i c a l t i n}

= \frac{4 π r^{2}}{2 π r (r + h)}

= \frac{2 r}{r + h}

= \frac{2 r}{r + 2 r}

= \frac{2}{3}

= 2 : 3

Hence, required ratio is 2:3

Question 23. Pavan filled water in a cylindrical vessel of radius 7cm and height 14cm. Then he gently dropped a spherical ball of radius 0.7cm. By how much should the height be increased if he doesn't want water to overflow when the spherical ball is dropped into it? Give the answer in micrometres and correct up to 2 decimal places.
___

Discuss Question

:
Water that will flowout will be equal to the volume of spherical ball dropped into the vessel.
So,

\frac{4}{3}

π

r^{3}

π

R^{2}

\frac{4}{3}

\times

\frac{22}{7}

\times

{(0.7)}^{3}

\frac{22}{7}

\times

{(7)}^{2}

\times H

\Rightarrow H = 0.933333 c m

1 m = 10^{- 6} μ m

∴ H = 9333.33 μ m

Question 24. The radius and height of a cone are in the ratio 4: 3. The area of the base is 154cm². The area of the curved surface in

c m^{2}

Discuss Question

\frac{r}{h}

\frac{4}{3}

Base area =

π

r^{2} = 154

\Rightarrow r = 7 c m

h = \frac{3}{4} \times 7 = \frac{21}{4}

\Rightarrow l = \sqrt{(h^{2} + r^{2})} = \frac{35}{4}

\Rightarrow

So, curved surface area =

π

r l

\Rightarrow

\frac{22}{7} \times (7) \times \frac{35}{4} = 192.5 c m^{2}

Question 25. Pavan built a conical flask using 550

m^{2}

of aluminium sheet. If the radius of the flask is 7 m, then how much water can be filled in the flask(in litres)? [The bottom of the flask is of another material.]

1232 litres
1232000 litres
12300 litres
123 litres

Discuss Question

Answer: Option B. -> 1232000 litres
:
B
Thearea of aluminium sheet = Curved surface area of flask

\Rightarrow 550 m^{2} = π r l

\Rightarrow 550 = \frac{22}{7} \times 7 \times l

\Rightarrow 550 = 22 l

\Rightarrow l = 25 m

Now, height of flask

\Rightarrow h = \sqrt{(l^{2} - r^{2})}

\Rightarrow h = \sqrt{25^{2} - 7^{2}}

\Rightarrow h = 24 m

Hence, the volume of conical flask

= \frac{1}{3}

π

r^{2} h

= \frac{1}{3} \times \frac{22}{7} \times 7^{2} \times 24

= 1232 m^{3}

We know that 1 m

^{3} = 1000 l i t r e s

So, volume in litres

= 1232000 l i t r e s

Question 26. If the radius of a cylinder is reduced by 50%, then the volume of the cylinder will be reduced by_____.

Discuss Question

Answer: Option C. -> 75%
:
C
Let the radius of the old cylinder be

R

and that of the new cylinder be

r

Then, r = \frac{R}{2}

Volume of old cylinder = π R^{2} h

Volume of new cylinde r = π r^{2} h

= π \times (\frac{R}{2})^{2} \times h

= \frac{π \times R^{2} \times h}{4}

Hence, reduction in volume

= Volumeof old cylinder - volume ofnew cylinder

= π R^{2} h - \frac{π \times R^{2} \times h}{4} = \frac{3}{4} π R^{2} h

∴

Percentage reduction

= \frac{Reduction in volume}{Volume of old cylinder} \times 100

= \frac{\frac{3}{4} π R^{2} h}{π R^{2} h} \times 100

= 75 %

Question 27. The dimensions of a cuboid are in the ratio of 1 : 2 : 3 and its total surface area is 88

m^{2}

. The volume of cuboid is___

m^{3}

Discuss Question

:
The sides will be

x, 2 x, 3 x

Total surface area

= 2 (x .2 x + 2 x .3 x + x .3 x)

\Rightarrow 2 (2 x^{2} + 6 x^{2} + 3 x^{2}) = 88

\Rightarrow 22 x^{2} = 88

\Rightarrow x = 2

So sides are 2 cm, 4 cm and 6 cm
So volume

= 2 \times 4 \times 6 = 48 m^{3}

Question 28. A box has length, breadth and height of 10 cm, 20 cm and 5 cm respectively. The lateral surface area of the box is ______.

100 cm2
200 cm2
300 cm2
400 cm2

Discuss Question

Answer: Option C. -> 300 cm2
:
C
Given,
length

(l)

= 10cm
breadth (b) = 20cm
height (h) = 5cm
Since each side's measurementis different, we can assume itas cuboid.
Lateral surface area of the box (cuboid)

= 2 h (l + b)

= 2 \times 5 \times (10 + 20)

= 10 \times 30

= 300 c m^{2}

Question 29.

A box has length, breadth and height of 10 cm, 20 cm and 5 cm respectively. The lateral surface area of the box is ______.

100 $c m^{2}$
200 $c m^{2}$
300 $c m^{2}$
400 $c m^{2}$

Discuss Question

Answer: Option C. -> 300

c m^{2}

:
C

Given,
length $(l)$ = 10 cm
breadth (b) = 20 cm
height (h) = 5 cm
Since each side's measurement is different, we can assume it as cuboid.
Lateral surface area of the box (cuboid)
$= 2 h (l + b)$
$= 2 \times 5 \times (10 + 20)$
$= 10 \times 30$
$= 300 c m^{2}$