 # Class 10 Maths Chapter 13 Surface Areas and Volumes MCQs

Class 10 Maths Chapter 13 (Surface areas and volumes) MCQs are provided here online, along with their solutions, for students who are preparing for the board exams. These objective questions are presented here, chapter-wise, with detailed explanations, as per CBSE and NCERT curriculum, by which students can easily score good marks.

## 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

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

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

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

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

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

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 8cm. The curved surface of the bucket is:

(a)4000 sq.cm

(b)3500 sq.cm

(c)3630 sq..cm

(d)3750 sq.cm

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

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

C.S = 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

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 30cm. The total surface area of the tank is:

(a)30m

(b)3.3m

(c)30.3m

(d)3300m

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