Class 10 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Q1: The radii for the top as well as the bottom of a bucket for the slant height of 45 cm are 28 cm and 7 cm, respectively. The curved surface area of the bucket is?

Ans: Slant height of the bucket = 45 cm

Top radius is = r1 = 28cm 

Bottom radius is = r2 = 7cm 

Curved surface area of the bucket is = πl(r1+r2)  

=22/7 * 45 * (28+7)

= 22/7 *45 *35

=4950cm²      

Q2. If two identical solid cubes of side ‘x’ are joined end to end, then the total surface area of the resulting cuboid is 12x2. Is it true?  

Ans: ∵ The total surface area of a cube of side x is 6x²
When they are joined end to end, the length becomes 2x
∴Total surface area
= 2[lh + bh + hl]

= 2 [(2x · x) + (x · x) + (2x · x)]

= 2 [2x² + x² + 2x²]

= 2 [5x²] = 10x² ≠ 12x²
∴ False   

Q3. If a solid cone of base radius ‘r’ and height ‘h’ is placed over a solid cylinder having same base radius ‘r’ and height ‘h’ as that of the cone, then the curved surface area of the shape is Is it true?

Ans: ∵ Curved surface area of a cone  

And curved surface area of the cylinder = 2πrh
∴ The curved surface area of the combination

∴ True.

Q4. A cylinder and a cone are of the same base radius and same height. Find the ratio of the volumes of the cylinder of that of the cone.

Ans: Let the base radius = r and height = h

⇒ The required ratio = 3: 1

Q5:  If two solid hemispheres for the same base radius r are joined together along with their bases, what is the curved surface area of this new solid?

Ans: The radius of the hemisphere = r

We know curved surface area = 2πr2  

The curved surface area of two solid hemisphere

= 2 * 2πr²

= 4πr²  

Q6: Vol. and surface area of a solid hemisphere are numerically equal. What is diameter of hemisphere?

Ans:

Vol. of hemisphere = S. A. of solid hemisphere 2 ,

⇒ 23πr2 = 3πr2

⇒ r = 92

∴ diameter = 2r = 9 units

Q7: Volumes for two spheres are in the ratio 64: 27. The ratio for their surface areas is?

Ans: Assume two-sphere having radius r1 and r2

As per the question, 

volume of the first sphere / volume of the second sphere = 64/27

= (4/3 *πr1³)/ (4/3 *πr2³) = 64/27

(r1/r2)³ = 64/27

r1/r2 = 3√(64/27) =4/3

Ratio for their surface area is = (4 *πr1²)/ (4 *πr2²) = r1²/r2² = (r1/r2)² = (4 /3)2 = 16/9

 So, the required ratio is 16:9

Q8: Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1. What is the ratio of their volumes?

Ans: Given,

Ratio of heights of two cones = 1 : 3

Ratio of radii = 3 : 1

Let h and 3h be the height of two cones.

Also, 3r and r be the corresponding radii of cones.

So, r1 = 3r, h1 = h, r2 = r, h2 = 3h.

Ratio of volumes = [(1/3)πr1²h1]/ [(1/3)πr2²h2]

= [(3r)² h]/[r² (3h)] = (9r²h)/(3r²h)

= 3/1

Hence, the ratio of volumes = 3:1