# CBSE Class 10 Mathematics Surface Area And Volume Worksheet Set B

Read and download free pdf of CBSE Class 10 Mathematics Surface Area And Volume Worksheet Set B. Students and teachers of Class 10 Mathematics can get free printable Worksheets for Class 10 Mathematics Chapter 13 Surface Area and Volume in PDF format prepared as per the latest syllabus and examination pattern in your schools. Class 10 students should practice questions and answers given here for Mathematics in Class 10 which will help them to improve your knowledge of all important chapters and its topics. Students should also download free pdf of Class 10 Mathematics Worksheets prepared by school teachers as per the latest NCERT, CBSE, KVS books and syllabus issued this academic year and solve important problems with solutions on daily basis to get more score in school exams and tests

## Worksheet for Class 10 Mathematics Chapter 13 Surface Area and Volume

Class 10 Mathematics students should refer to the following printable worksheet in Pdf for Chapter 13 Surface Area and Volume in Class 10. This test paper with questions and answers for Class 10 will be very useful for exams and help you to score good marks

### Class 10 Mathematics Worksheet for Chapter 13 Surface Area and Volume

Surface Areas and Volumes

Q.- A bird bath for garden in the shape of a cylinder with a hemispherical depression at one end (see figure). The height of the cylinder is 1.45 m and its radius is 30 cm.Find the total surface area of the bird-bath.
(Take π = 22/7 ) Sol. Let h be height of the cylinder and r the common radius of the cylinder and hemisphere. Then, the total surface area of the bird-bath
= CSA of cylinder + CSA of hemisphere
= 2πrh + 2πr2 = 2πr2 = 2πr(h + r)
= 2 × 22/7 × 30 (145 + 30) cm2
= 33000 cm2 = 3.3 m2

Q.- A juice seller was serving his customers using glasses as shown in figure. The inner diameter of the cylindrical glass was 5 cm,but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm. Find the apparent capacity of the glass and its actual capacity. (Use π= 3.14) Sol. Since the inner diameter of the glass = 5 cm and height = 10 cm, the apparent capacity of the glass = πr2h
= 3.14 × 2.5 × 2.5 × 10 cm3 = 196.25 cm3
But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.
i.e. it is less by 2/3 πr3

= 2/3 × 3.14 × 2.5 × 2.5 × 2.5 cm3 = 32.71 cm3
So, the actual capacity of the glass = apparent capacity of glass – volume of the hemisphere
= (196.25 – 32.71) cm3
= 163.54 cm2

Q.- A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volume of the cylinder and the toy. (Take π = 3.14) Sol. Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere (see figure). The radius BO of the hemisphere
(as well as of the cone) = 1/2 × 4 cm = 2 cm
So, volume of the toy = 2/3 πr3 + 1/3 πr2h = 25.12 cm3
Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular cylinder
= HP = BO = 2 cm, and its height is
EH = AO + OP = (2 + 2) cm = 4 cm
So, the volume required
= Volume of the right circular cylinder – volume of the toy
= (3.14 × 22 × 4 – 25.12) cm3
= 25.12 cm3
= 25.12 cm3
Hence, the required difference of the two
volumes = 25.12 cm3.

Q.- A cone of height 24 cm and radius of base 6 cm is made up of modeling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.
Sol. Volume of cone = 1/3 × π × 6 × 6 × 24 cm3
If r is the radius of the sphere, then its volume is 4/3 πr3.
Since the volume of clay in the form of the cone and the sphere remains the same, we have.

4/3 × π × r3 = 1/3 × π × 6 × 6 × 24
r3 = 3 × 3 × 24 = 33 × 23
r = 3 × 2 = 6
Therefore, the radius of the sphere is 6 cm

Q.-  Selvi's house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44 m × 95 cm.The overhead tank has its radius 60 cm and height 95 cm. Find the height of the water left in the sump after the overhead tank has been completely filled with water from the sump which had been full. Compare the capacity of the tank with that of the sump. (Use π = 3.14)

Sol. The volume of water in the overhead tank equals the volume of the water removed from the shump.
Now the volume of water in the overhead
tank (cylinder) = πr2h
= 3.14 × 0.6 × 0.6 × 0.95 m3
The volume of water in the sump when full
= l × b × h = 1.57 × 1.44 × 0.95 m3
The volume of water left in the sump after filling the tank
= (1.57×1.44×0.95)–(3.14 × 0.6 × 0.6 × 0.95)] m3
= (1.57 × 0.6 × 0.6 × 0.95 × 2) m3
So, the height of the water left in the sump
= volume of water left in the sump/λ×b Therefore, the capacity of the tank is half the capacity of the sump.

Q.- A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.
Sol. The volume of the rod = π ×(1/2)× 8 cm3
= 2π cm3.
The length of the new wire of the same volume = 18 m = 1800 cm
If r is the radius (in cm) of cross section of the wire, its volume = π × r2 × 1800 cm3
Therefore, π × r2 × 1800 = 2π
i.e. r2 =1/900
i.e. r =1/30
So, The diameter of the cross section i.e. the thickness of the wire is1/15 cm, i.e. 0.67 mm (approx.)

More question-

1.The lateral surface area of right circular cylinder with base radius 7cm and height 10 cm is:

2.The lateral surface area of cylinder is 176cm2 & base area 38.5cm2. Then its volume is

(A) 803cm3

(B) 380cm3

(C) 308cm3

(D) 830cm3

3.Ratio of curved surface areas of two cylinders with equal radii is:

(A) H2 : h2

(B) 2H : h

(C) H : h

(D) None

4.Two cubes of 12cm edge are joined end to end. Find the surface area of the resulting cuboid.

5.Three cubes of sides 6 cm edge are joined end to end. Find the surface area of the resulting cuboid.

6.A solid sphere of radius 6cm is melted and recast into small spherical balls each of diameter 0.6cm.Find the number of balls thus obtained.

7.How many spherical bullets can be made out of a solid cube of lead whose edge measures 55cm, each bullet being 10 cm in diameter?

8.The area of the base of a cone is 616 sq. cm. If its height is 48 cm then its total surface area is:

(A) 2681cm2 (B) 2861cm2 (C) 2816cm2 (D) None

9.Ratio of lateral surface areas of two cylinders with equal heights is .

(A) R : r (B) H : h (C) R2 : r2 (D) None

Please click the below link to access CBSE Class 10 Mathematics Surface Area And Volume Worksheet Set B

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### CBSE Class 10 Mathematics Chapter 13 Surface Area and Volume Worksheet

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### Worksheet for Mathematics CBSE Class 10 Chapter 13 Surface Area and Volume

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#### Chapter 13 Surface Area and Volume worksheet Mathematics CBSE Class 10

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#### Chapter 13 Surface Area and Volume CBSE Class 10 Mathematics Worksheet

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#### Worksheet for CBSE Mathematics Class 10 Chapter 13 Surface Area and Volume

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