## RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20B.

**Other Exercises**

- RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20A
- RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20B
- RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20C

**Question 1.**

**Solution:**

(i) Radius of the base of the cylinder (r) = 7 cm.

Height (h) = 50 cm.

**Question 2.**

**Solution:**

Radius of cylindrical tank (r) = 1.5 m

and height (h) = 10.5 m

= 74.25 x 1000l

= 74250 l

**Question 3.**

**Solution:**

Radius of the base of pole (r)

= 10 dm

= \(\\ \frac { 10 }{ 100 } \) m

= \(\\ \frac { 1 }{ 10 } \) m

**Question 4.**

**Solution:**

Volume of cylinder = 1.54 m³

= 1540000 cm³

Diameter of its base = 140 cm

Radius (r) = 70 cm

**Question 5.**

**Solution:**

Volume of cylindrical rod = 3850 cm³

Length of rod (h) = 1 m = 100 cm

Let radius of the base of the rod = r

**Question 6.**

**Solution:**

Diameter of closed cylinder = 14 m

Radius = \(\\ \frac { 14 }{ 2 } \)

= 7 m

Height = 5

**Question 7.**

**Solution:**

Circumference of the base of cylinder = 88 cm.

**Question 8.**

**Solution:**

Lateral surface of cylinder = 220 m²

Height (h) = 14 m

Let radius of cylinder = r

**Question 9.**

**Solution:**

Volume of cylinder = 1232 cm³

height (h) = 8cm

Let r be the radius, then

**Question 10.**

**Solution:**

Ratio in radius and height of a cylinder = 7 : 2

Let radius = 7x

then height = 2x

**Question 11.**

**Solution:**

Curved surface area = 4400 cm²

circumference of base = 110 cm

Let r be the radius

**Question 12.**

**Solution:**

In first case,

Side of square base (a) = 5 cm.

and height (h) = 14 cm.

Volume = 5 x 5 x 14 = 350 cm³

In second case,

Radius of the circular base (r) = 3.5 cm.

Height (h) = 12 cm.

Volume = πr²h

= \(\\ \frac { 22 }{ 7 } \) x 3.5 x 3.5 x 12 cm³

= 462 cm²

Hence second type of circular plastic can has greater capacity.

Difference = 462 – 350

= 112 cm³

**Question 13.**

**Solution:**

Diameter of a cylindrical pillar = 48 cm.

Radius (r) = \(\\ \frac { 48 }{ 2 } \) = 24 cm.

\(\\ \frac { 24 }{ 100 } \) m

**Question 14.**

**Solution:**

Length of rectangular vessel (l) = 22 cm.

Breadth (A) = 16 cm.

and height (A) = 14 cm.

**Question 15.**

**Solution:**

Diameter of cylindrical metal = 1 cm.

Radius (r) = \(\\ \frac { 1 }{ 2 } \) cm.

Length. (A) = 11 cm.

Volume = πr²h

**Question 16.**

**Solution:**

Side of a solid cube = 2.2 cm

Volume = (side)³

= (2.2)³

= 10.648 cm³

**Question 17.**

**Solution:**

Diameter of a well = 7 m

Radius (r) = \(\\ \frac { 7 }{ 2 } \) m

**Question 18.**

**Solution:**

Inner diameter of well = 14 m

Inner radius = \(\\ \frac { 14 }{ 2 } \) = 7 m

Depth (h) = 12 m

**Question 19.**

**Solution:**

No. of revolutions = 750

Diameter of road roller = 84 cm

Length (h) = 1 m

**Question 20.**

**Solution:**

Thickness of the metal = 1.5 cm.

External diameter = 12 cm.

**Question 21.**

**Solution:**

Inner diameter of tube = 12 cm.

Inner radius (r) = \(\\ \frac { 12 }{ 2 } \) = 6 cm.

Thickness of metal = 1 m.

Hope given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20B are helpful to complete your math homework.

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