# Free Surface Areas and Volumes 03 Practice Test - 9th Grade

### Question 1

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

#### SOLUTION

Solution :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)

=2h(l+b)

=2×5 ×(10+20)

=10 ×30

=300 cm2

### Question 2

A vessel is in the shape of a cube of side 30 m. How much water can it hold?

#### SOLUTION

Solution :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 of cubic vessel = (side)3

= (30)3

= 27000 m3.

### Question 3

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?

1:2

2:3

3:1

3:5

#### SOLUTION

Solution :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 .

⇒h=2r

Ratio of their surface areas

=Surface area of ballSurface area of cylindrical tin

=4 π r22 π r(r+h)

=2rr+h

=2rr+2r

=23

=2:3

Hence, required ratio is 2:3

### Question 4

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

25%

50%

75%

40%

#### SOLUTION

Solution :C

Let the radius of the old cylinder be R and that of the new cylinder be r.

Then, r=R2

Volume of old cylinder=πR2h

Volume of new cylinder=πr2h

=π×(R2)2×h

=π×R2×h4

Hence, reduction in volume

=Volume of old cylinder - volume of new cylinder

=πR2h−π×R2×h4=34πR2h

∴ Percentage reduction

=Reduction in volumeVolume of old cylinder×100

=34πR2hπR2h×100

=75%

### Question 5

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

#### SOLUTION

Solution :C

Volume of the big iron ball will be equal to the volume of 27 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.⇒ 43×πR3=27×(43×πr3)

⇒R3=27r3R=3r=3×5=15 cm.

### Question 6

Raghu needs to make a cylindrical aluminum tube. What will be the area of the aluminum sheet required to make the tube, if the length and radius of the tube should be 1 m and 3.5 cm respectively [Top and bottom of the tube are of another material]?

(use π=227)

2200 m2

220 m2

2200 cm2

220 cm2

#### SOLUTION

Solution :C

Given that,

Radius of the tube = 3.5 cm

Length of the tube = 1 m = 100 cm

Curved surface area of cylindrical tube

=2πrh

=2×227×3.5×100

=2×22×0.5×100

=2200 cm2

Hence, the required area of the aluminium sheet is 2200 cm2.

### Question 7

Pavan built a conical flask using 550 m2 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

#### SOLUTION

Solution :B

The area of aluminium sheet = Curved surface area of flask

⇒550 m2=πrl

⇒550=227×7×l

⇒550=22l

⇒l=25 mNow, height of flask

⇒h=√(l2−r2)

⇒h=√252−72

⇒h=24 m

Hence, the volume of conical flask

=13 πr2h

=13×227×72× 24

=1232 m3We know that 1 m 3=1000 litres

So, volume in litres =1232000 litres

### Question 8

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.

#### SOLUTION

Solution :Water that will flow out will be equal to the volume of spherical ball dropped into the vessel.

So,43 π r3 = π R2 H

43 × 227 × (0.7)3 = 227 × (7)2 ×H

⇒H=0.933333 cm

1 m=10−6μm∴H=9333.33 μm

### Question 9

The radius and height of a cone are in the ratio 4: 3. The area of the base is 154cm^{2}. The area of the curved surface in cm2

#### SOLUTION

Solution :rh = 43

Base area = πr2=154

⇒r=7cm so h=34×7=214

⇒l=√(h2+r2)=354⇒ So, curved surface area = πrl

⇒ 227×(7)×354=192.5 cm2

### Question 10

The dimensions of a cuboid are in the ratio of 1 : 2 : 3 and its total surface area is 88 m2. The volume of cuboid is

#### SOLUTION

Solution :The sides will be x,2x,3x

Total surface area =2(x.2x+2x.3x+x.3x)

⇒2(2x2+6x2+3x2)=88⇒22x2=88

⇒x=2

So sides are 2 cm, 4 cm and 6 cm

So volume =2×4×6=48m3