Class 10 Maths Chapter 13 Surface Areas and Volumes MCQs
Class 10 Maths Chapter 13 MCQs (Surface areas and volumes) are provided here online, along with their solutions. Students who are preparing for the board exams can refer to these multiple-choice questions to score good marks in the final examination. These objective questions are framed as per the CBSE syllabus (2021-2022) and NCERT curriculum. To get all class 10 Maths chapter-wise MCQs, Click here.
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Class 10 Maths MCQs for Surface Areas and Volumes
1. The shape of an ice-cream cone is a combination of:
(a) Sphere + cylinder
(b) Sphere + cone
(c) Hemisphere + cylinder
(d) Hemisphere + cone
Answer: (d) Hemisphere + cone
The shape of an ice-cream cone is a combination of a hemisphere and a cone.
2. 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
Answer: (c) Cone itself
Explanation: If we cut a cone into two parts parallel to the base, then the shape of the upper part remains the same.
3. 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
Answer: (b) Frustum of cone
Explanation: See the figure below
4. 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
Answer: (a) 4 π r2
If r is the radius of the sphere, then the surface area of the sphere is given by 4 π r2.
5. 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
Answer: (c) Remains unchanged
Explanation: If we change the shape of a three-dimensional object, the volume of the new shape will be same.
6. Fifteen solid spheres are made by melting a solid metallic cone of base diameter 2cm and height 15cm. The radius of each sphere is:
(a) ½
(b) ¼
(c) 1/3√2
(d) 1/3√4
Answer: (d) 1/3√4
Explanation: Volume of 15 spheres = Volume of a cone
15 x (4/3) π r3 = ⅓ πr2h
5×4 π r3 = ⅓ π 12(15)
20r3 = 5
r3 = 5/20 = ¼
r = 1/3√4
7. 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
Answer: (c) 3630 sq.cm
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.
8. If a cylinder is covered by two hemispheres shaped lid of equal shape, then the total curved surface area of the new object will be
(a) 4πrh + 2πr2
(b) 4πrh – 2πr2
(c) 2πrh + 4πr2
(d) 2πrh + 4πr
Answer: (c) 2πrh + 4πr2
Explanation: Curved surface area of cylinder = 2πrh
The curved surface area of hemisphere = 2πr2
Here, we have two hemispheres.
So, total curved surface area = 2πrh + 2(2πr2) = 2πrh + 4πr2
9. A tank is made of the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and radius is 30 cm. The total surface area of the tank is:
(a) 30 m
(b) 3.3 m
(c) 30.3 m
(d) 3300 m
Answer: (b) 3.3 m
Explanation: Total surface area of tank = CSA of cylinder + CSA of hemisphere
= 2πrh + 2πr2= 2π r(h + r)
= 2 x 22/7 x 30(145 + 30) cm2
=33000 cm2
= 3.3 m2
10. If we join two hemispheres of same radius along their bases, then we get a;
(a) Cone
(b) Cylinder
(c) Sphere
(d) Cuboid
Answer: (c) Sphere
If we join two hemispheres of same radius along their bases, then we get a Sphere.
11. A cylindrical pencil sharpened at one edge is the combination of
(a) a cone and a cylinder
(b) frustum of a cone and a cylinder
(c) a hemisphere and a cylinder
(d) two cylinders
Answer: (a) a cone and a cylinder
A cylindrical pencil sharpened at one edge is the combination of a cone and a cylinder.
12. A shuttle cock used for playing badminton has the shape of the combination of
(a) a cylinder and a sphere
(b) a cylinder and a hemisphere
(c) a sphere and a cone
(d) frustum of a cone and a hemisphere
Answer: (d) frustum of a cone and a hemisphere
A shuttle cock used for playing badminton has the shape of the combination of frustum of a cone and a hemisphere.
13. A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that 1/8 space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is
(a) 142296
(b) 142396
(c) 142496
(d) 142596
Answer: (a) 142296
Explanation:
Volume of cube = 223 = 10648 cm3
Volume of cube that remains unfilled = (1/8) × 10648 = 1331 cm3
volume occupied by spherical marbles = 10648 − 1331 = 9317 cm3
Radius of the spherical marble = 0.5/2 = 0.25 cm = 1/4 cm
Volume of 1 spherical marble = (4/3) × (22/7) × (1/4)3 = 11/168 cm3
Numbers of spherical marbles = n = 9317 × (11/168) = 142296
14. A solid piece of iron in the form of a cuboid of dimensions 49 cm × 33 cm × 24 cm, is moulded to form a solid sphere. The radius of the sphere is
(a) 21 cm
(b) 23 cm
(c) 25 cm
(d) 19 cm
Answer: (a) 21 cm
Explanation:
For the given cuboid,
Length, l = 49 cm
Breadth, b = 33 cm
Height, h = 24 cm
Volume of cube = 49 × 33 × 24 cm3
Let r be the radius of the sphere.
Volume of sphere = 4/3 πr3
Volume of cuboid = volume of sphere molded
49 × 33 × 24 = 4/3 πr3
⇒ πr3 = 29106
⇒ r3 = 29106 × (22/7)
⇒ r3 = 9261
⇒ r3 = (21)3
⇒ r = 21 cm
Hence, the radius of sphere is 21 cm
15. 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
Answer: (b) 2r cm
Explanation:
Radius of cylinder = r
Height = h > 2r
Since the sphere is enclosed by a cylinder, the radius of the cylinder will be the radius of the sphere.
Therefore, diameter of sphere = 2r cm
16. The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is
(a) 32.7 litres
(b) 33.7 litres
(c) 34.7 litres
(d) 31.7 litres
Answer: (a) 32.7 litres
Explanation:
Given,
The height of the bucket = h = 35 cm
Diameter of one circular end of bucket = 44 cm
Then the radius R = 22 cm
Diameter of another end = 24 cm
Then the radius r = 12 cm
We know that Volume of the bucket = (1/3)πh[R2 + r2 + Rr]
= (1/3) × (22/7) × 35 × [(22)2 + (12)2 + 22 × 12]
= (35/3) × (22/7) × (484 + 144 + 264)
= (5 × 22 × 892)/3
= 32706.6 cm3
= 32.7 litres.
17. Two identical solid cubes of side a are joined end to end. Then the total surface area of the resulting cuboid is
(a) 12a2
(b) 10a2
(c) 8a2
(d) 11a2
Answer: (b) 10a2
Explanation:
The total surface area of a cube having side a = 6a2
If two identical faces of side a are joined together, then the total surface area of the cuboid so formed is 10a2.
18. A solid cylinder of radius r and height h is placed over another cylinder of same height and radius. The total surface area of the shape so formed is
(a) 4πrh + 4πr2
(b) 2πrh + 4πr2
(c) 2πrh + 2πr2
(d) 4πrh + 2πr2
Answer: (d) 4πrh + 2πr2
Explanation:
We know that,
The total surface area of cylinder = 2πrh + 2πr2
When one cylinder is placed over the other cylinder of same height and radius, then height of the new cylinder will be 2h and radius will be r.
Thus, the total surface area of the shape so formed = 2πr(2h) + 2πr2 = 4πrh + 2πr2
19. The number of shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm is approximately equal to
(a) 84
(b) 90
(c) 92
(d) 80
Answer: (a) 84
Explanation:
Volume of cuboidal lead solid = 9 cm × 11 cm × 12 cm = 1188 cm3
Radius of lead shot = 3/2 cm = 1.5 cm
Volume of each shot = (4/3)πr3
= (4/3) × (22/7) × 1.5 × 1.5 × 1.5
= 14.143 cm3
Number of lead shots can be made = 1188/14.143 = 84 (approx.)
20. 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
Answer: (c) 4πr2
Explanation:
When two hemispheres are joined together along their bases, a sphere of the same base radius is formed.
Curved Surface Area of a sphere = 4πr2
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