# Free Surface Areas and Volumes 03 Practice Test - 10th Grade

### Question 1

A bucket is in the form of a frustum of a cone, its depth is 15 cm and the diameters of the top and the bottom are 56 cm and 42 cm respectively. How many litres of water can the bucket hold ? (Take π = 227)

#### SOLUTION

Solution :R = 28 cm

r = 21 cm

h = 15 cm

Capacity of the bucket = 13πh(R2+r2+Rr)

= 13×227×15×[(28)2+(21)2+(28)(21)]cm3

= 227×5×[784+441+588]cm3

= 227×5×1813cm3

= 22 × 5 × 259 cm3

= 28490 cm3

= 284901000 liters

= 28.49 liters

### Question 2

There is a coin. Ram attaches a conical attachment to one flat end of coin. The conical attachment has same radius as coin. What is the surface area of the combined solid?

Coin base area + Coin C.S.A + Hemisphere C.S.A

Coin base area + Coin C.S.A + cone C.S.A

Total surface area of coin + total surface area of cone

Total surface area of cone

#### SOLUTION

Solution :B

This leaves only the following surfaces for the combined solid-

1. One flat surface of coin

2. Curved surface of coin

3. Curved surface area of cone

These 3 areas add up together to give total surface area of combined solid.

### Question 3

A tent is in the form of a cylinder of diameter 4.2 m and height 4 m, is surmounted by a cone of equal base and height 2.8 m. Find the cost of canvas required for making the tent at ₹100 per sq.m.

₹ 6590

₹ 7590

₹ 8590

₹ 9876

#### SOLUTION

Solution :B

The total surface area of the tent:

T=2πrh1+πrlπ=227;r=2.1 mh1=height of the cylinder=4 mh2=height of the cone=2.8 ml= slant height of the conel=√2.82+2.12=3.5 m

So,

T=(2×227×2.1×4+227×2.1×3.5)T=75.9 m2

Therefore, the total cost of canvas at ₹100 per sq meter = 75.9×100=₹ 7590.

### Question 4

If Hemakshi reshaped a cone of height h cm and radius of base r cm into a sphere, then, which of the following options is always correct?

Volume of cone = Volume of sphere

Surface area of cone = Surface area of sphere

Radius of cone = Radius of sphere

#### SOLUTION

Solution :A

If any solid is reshaped into a new solid then volume always remains constant.

### Question 5

If the radii of the ends of a bucket, 45 cm high, are 28 cm and 7 cm respectively, then the total surface area of the bucket is:

#### SOLUTION

Solution :C

Total surface area of the bucket

=TSA=π×l×(r1+r2)+π×(r12+r22)Where π=227,

l= Slant height of the bucket

=√(h)2+(r1−r2)2

r1=28cm and r2 = 7 cm

l= Slant height of the bucket

=√(45)2+(21)2 cm=49.6 cm

Therefore,

TSA=[227×49.6×(28+7)+ 227×(282+72)]=8074 cm2

### Question 6

A cylindrical pipe has inner diameter of 7 cm and water flows through it at the rate of 192.5 litres per minute. Find the rate of flow in kilometres per hour.

1

3

5

6

#### SOLUTION

Solution :B

In one minute 192.5 litres of water flows.

So, The Volume of water that flows in one hour = (192.50 × 60) liters. [∵1 hour=60 mins ]

Volume in cm3 = (192.5 × 60 × 1000) cm3 [∵1 litre=1000 cm3 ]

Inner radius of the pipe = 3.5 cm.Let the length of column of water that flows in 1 hour be h cm.

Then,227 × 3.5 × 3.5 × h = 192.5 × 60 × 1000

h= 300000 cm = 3 km

Hence, the rate of flow = 3 km per hour.

### Question 7

A piece of cloth is required to completely cover a solid object. The solid object is composed of a hemisphere and a cone surmounted on it. If the common radius is 7 m and height of the cone is 1 m, what is the area of cloth required?

262.39 m2

463.39 m2

662.39 m2

563 m2

#### SOLUTION

Solution :B

Given

r = 7m

h = 1m

Surface area of hemisphere = 2πr2

=2×227×(7)2 = 308 m2.For calculating the surface area of a cone we need to calculate its slant height,

l = √r2+h2l = √49+1=√50 m.

Surface area of cone = πrl= 227×7×√50 = 155.39 m2.

So, area of cloth required = (308 + 155.39)m2 = 463.39 m2

### Question 8

A road roller was used for levelling a road of width 2 m. It was observed that, the road roller required 25 complete revolutions to level the entire road. If the radius and the length of the roller is 7 m and 2 m respectively, then length of the road is ____. [Take π = 3.14].

1200 m

2100 m

1100 m

1234 m

#### SOLUTION

Solution :C

Given, Radius of Cylinder, r=7 m

length of roller, h=2 m

Width of road, B=2 m

Length of road be ′L′

Number of revolution taken = 25

Roller rolls on its curved surface area.

∴ Surface area of = CSA of cylinder

one revolution

=2πrh=2×227×7×2=88 m2Total area covered in 25 revolution =25×88=2200 m2

This area will be equal to the area of the road levelled

Area of road = L×B=2200⇒L×2=2200⇒L=1100 m

∴ Length of the level road is 1100 m.

### Question 9

Ram has a semicircular disc. He rotates it about its diameter by 360 degrees. When he rotates the disc, a volume of air in his room gets swept.The name of the object/shape that exactly occupies this volume is

#### SOLUTION

Solution :The shape obtained when a semi circular disc is rotated by 360 degrees is a sphere. If it was rotated by 180 degrees instead of 360 degrees, then we obtain a hemisphere. The line segment AB will be diameter of the sphere formed.

### Question 10

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. The number of lead shots dropped in the vessel is ______.

100

50

#### SOLUTION

Solution :A

Let number of lead shots = n

Height of cone =h=8 cmRadius of cone =r1=5 cm

Volume of water present in the cone = Volume of cone =13.π.(r1)2.h=(13×227×52×8 cm2)

Volume of water flown out =(14×13×227×52×8) cm3Radius of lead shot =r=0.5 cm

Volume of each lead shot ... =43.π.r3=(43×227×(0.5)3) cm3

According to given situation, n lead shots are thrown into cone such that 1/4th of water present in cone flows out.

It means volume of n lead shots = volume of water flown out

⇒ (n×43×227×(0.5)3)=(14×13×227×52×8)

⇒ n=14×13×52×8×1(0.53)×34

⇒ n=100