Class 10 Maths MCQ Questions of Surface Areas and Volumes will help to get the ideas for the world free from Surface Areas and Volumes is given here. The practice of MCQ Questions for Class 10 is the only key of achievement in the exams.

Students will get to know how to solve MCQs dependent on surface regions and volumes of various shapes, like a cone, circle, cylinder and that’s just the beginning. Other concepts discussed in this chapter also include the conversion of a solid from one shape to another shape and then finding the surface area and volume of the new shape formed. In view of the changed exam pattern that includes amounts of MCQ Questions, we have covered all the section shrewd CBSE Class 10 Maths target questions that are required to be remembered for the exam papers.

Sub-Topics Discussed Under the Surface Areas and Volumes

• Basics
• Combination of solids
• Shape conversion of solids

Explore the list of multiple-choice questions below:-

Practice Class 10 MCQ Questions for Maths

1. During conversion of a solid from one shape to another, the volume of the new shape will

(a) increase
(b) decrease
(c) remain unaltered
(d) be doubled

1. A right circular cylinder of radius r cm and height h cm (h>2r) just encloses a sphere of diameter

(a) r cm
(b) 2r cm
(c) h cm
(d) 2h cm

1. If two solid hemispheres of same base radii r, are joined together along their bases, then curved surface area of this new solid is

(a) 4πr2
(b) 6πr2
(c) 3πr2
(d) 8πr2

1. The shape of an ice-cream cone is a combination of:

(a) Sphere + cylinder
(b) Sphere + cone
(c) Hemisphere + cylinder
(d) Hemisphere + cone

1. If a cone is cut parallel to the base of it by a plane in two parts, then the shape of the top of the cone will be a:

(a) Sphere
(b) Cube
(c) Cone itself
(d) Cylinder

1. A cylindrical pencil sharpened at one edge is combination of:

(a) a cone and a cylinder
(b) frustum of a cone and cylinder
(c) a hemisphere and a cylinder
(d) four cylinders

1. A shuttle cock used for playing badminton has the shape of a combination of:

(а) a cylinder and a sphere
(b) a cylinder and a hemisphere
(c) a cone and a sphere
(d) a frustum of a cone and a hemisphere

1. The radius of a wire is decreased to one third. If volume remains the same, the length will become:

(a) 3 times
(b) 6 times
(c) 9 times
(d) 27 times

1. A cylinder and a cone area of same base radius and of same height. The ratio of the volume of cylinder to that of cone is:

(a) 3 : 1
(b) 1 : 3
(c) 2 : 3
(d) 1 : 1

1. If the surface areas of two spheres are in ratio 16 : 9, then their volumes will be in the ratio:

(a) 27 : 64
(b) 64 : 27
(c) 4 : 3
(d) 3 : 4

1. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is

(a) 4πrh + 4πr2
(b) 4πrh − 4πr2
(c) 4πrh + 2πr2
(d) 4πrh − 2πr2

1. The radii of the top and bottom of a bucket of slant height 45cm are 28cm and 7 cm respectively. The curved surface area of the bucket is:

(a) 4950 cm2
(b) 4951 cm2
(c) 4952 cm2
(d) 4953 cm2

1. If we cut a cone in two parts by a plane parallel to the base, then the bottom part left over is the:

(a) Cone
(b) Frustum of cone
(c) Sphere
(d) Cylinder

1. If r is the radius of the sphere, then the surface area of the sphere is given by;

(a) 4 π r2
(b) 2 π r2
(c) π r2
(d) 4/3 π r2

1. If we change the shape of an object from a sphere to a cylinder, then the volume of cylinder will

(a) Increase
(b) Decrease
(c) Remains unchanged
(d) Doubles

1. The radius of the top and bottom of a bucket of slant height 35 cm are 25 cm and 8 cm. The curved surface of the bucket is:

(a) 4000 sq.cm
(b) 3500 sq.cm
(c) 3630 sq.cm
(d) 3750 sq.cm

1. The total surface area of a hemispherical solid having radius 7 cm is

(a) 462 cm2
(b) 294 cm2
(c) 588 cm2
(d) 154 cm2

1. A solid formed on revolving a right angled triangle about its height is

(a) cylinder
(b) sphere
(c) right circular cone
(d) two cones

1. If we join two hemispheres of same radius along their bases, then we get a;

(a) Cone
(b) Cylinder
(c) Sphere
(d) Cuboid

1. Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is

(a) 6πr2
(b) 5πr2
(c) 4πr2
(d) 3πr2

Explanation: During conversion of one solid shape to another, the volume of the new shape will remain unaltered.

Explanation: Because the sphere is enclosed inside the cylinder, therefore the diameter of sphere is equal to the diameter of cylinder which is 2r cm.

Explanation: Because curved surface area of a hemisphere is and here we join two solid hemispheres along their bases of radii r, from which we get a solid sphere.

Hence the curved surface area of new solid = 2πr2 + 2πr2 = 4πr

Explanation: The shape of an ice-cream cone is a combination of a hemisphere and a cone.

Explanation: If we cut a cone into two parts parallel to the base, then the shape of the upper part remains the same.

1. Answer:(a) a cone and a cylinder
2. Answer:(d) a frustum of a cone and a hemisphere

Explanation: Since the total surface area of cylinder of radius r and height h = 2πrh + 2πr2.

When one cylinder is placed over the other cylinder of same height and radius,

Then height of new cylinder = 2h

And radius of the new cylinder = r

Therefore total surface area of new cylinder

= 2πr (2h) + 2πr2

= 4πrh + 2πr

Explanation: Curved Surface area of the bucket = π (R + r) l

⇒ Curved surface area of the bucket = π (28 + 7) X 45

⇒ Curved surface area of the bucket = 4950 cm

2. Answer: (a) 4 π r2

Explanation: If r is the radius of the sphere, then the surface area of the sphere is given by 4 π r2.

Explanation: If we change the shape of a three-dimensional object, the volume of the new shape will be same.

Explanation: Curved surface of bucket = π(R1 + R2) x slant height (l)

Curved Surface = (22/7) x (25 + 8) x 35

CSA = 22 x 33 x 5 = 3630 sq.cm.

Explanation: Reason: Total surface area of hemisphere = 3πr2

= 3 × 22/7 × 7 × 7 = 462 cm2