**Question 1. Find the volume of a cylinder whose**

**(i) r = 3.5 cm, h = 40 cm**

**(ii) r = 2.8 m, h = 15 m**

**Solution: **

(i)Given that,r = 3.5 cm,

h = 40 cm

As we know that Volume of a cylinder = πr

^{2}h= 22/7 × 3.5 × 3.5 × 40 =

1540 cm^{3}

(ii)Given that,r = 2.8 m,

h =15 m

As we know that Volume of a cylinder = πr

^{2}h= 22/7 × 2.8 × 2.8 × 15 =

369.6 m^{3}

**Question 2. Find the volume of a cylinder, if the diameter (d) of its base and its altitude (h) are :**

**(i) d = 21 cm, h = 10 cm**

**(ii) d = 7 m, h = 24 m**

**Solution:**

(i)Given that,d = 21cm,

r = d/2 = 21/2cm,

h = 10 cm.

As we know that Volume of a cylinder = πr

^{2}h= 22/7 × 21/2 × 21/2 × 10 =

3465 cm^{3}

(ii)Given that,d = 7 m

^{2}r = d/2 = 7/2m

h = 24 m

As we know that Volume of a cylinder = πr

^{2}h= 22/7 × 7/2 × 7/2 × 24 =

924 m^{3}

**Question 3. The area of the base of a right circular cylinder is 616 cm**^{2} and its height is 25 cm. Find the volume of the cylinder.

^{2}and its height is 25 cm. Find the volume of the cylinder.

**Solution: **

Given that,

Area of base of right circular cylinder = 616 cm

^{2}Height of cylinder = 25 cm

Let us assume that radius of cylinder is ‘r’ cm

As we know that Area of base of right circular cylinder = πr

^{2},πr

^{2}= 61622/7 × r

^{2}= 616r

^{2}= 616 × 7/22 = 196r = √196 = 14cm

As we know that, Volume of cylinder = Area of base of right circular cylinder × height

= 616 × 25 =

15400 cm^{3}

**Question 4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find the volume of the cylinder.**

**Solution:**

Given that,

Circumference of base of cylinder = 88 cm

Height of cylinder = 15 cm

Let us assume that ‘r’ be the radius of the cylinder.

As we know that Circumference of base of cylinder = 2πr,

2πr = 88

2 × 22/7 × r = 88

r = 88 × 7 / 2 × 22 = 616/44 = 14cm

Radius of cylinder = 14 cm

We know that Volume of cylinder = πr

^{2}h= 22/7 × 14 × 14 × 15 =

9240 cm^{3}

**Question 5. A hollow cylindrical pipe is 21 dm long. Its outer and inner diameters are 10 cm and 6 cm respectively. Find the volume of the copper used in making** **the pipe.**

**Solution:**

Given that,

Length of cylinder = 21 dm = 210 cm

Outer diameter = 10 cm

Outer radius, R = 10/2 = 5cm

Inner diameter = 6 cm

Inner radius, r = 6/2 = 3cm

As we know that Volume of copper used in making the pipe(hollow) = π (R

^{2 }– r^{2})h^{2}= 22/7 (52 – 32) 210

= 22/7 (25-9) 210 =

10560 cm^{3}

**Question 6. Find the (i) curved surface area (ii) total surface area and (iii) volume of a right circular cylinder whose height is 15 cm and the radius of the base is 7 cm.**

**Solution:**

Given that,

Height of cylinder = 15 cm

Radius of base = 7 cm

(i)We know that formula of Curved surface area = 2πrh= 2 × 22/7 × 7 × 15 = 660 cm

^{2}

(ii)We know that formula of Total surface area = 2πr(h+r)= 2 × 22/7 × 7 (15+7) = 968 cm

^{2}

(iii)We know that formula of Volume of cylinder = πr^{2}h= 22/7 × 7 × 7 × 15 =

2310 cm^{3}

**Question 7. The diameter of the base of a right circular cylinder is 42 cm and its height is 10 cm. Find the volume of the cylinder.**

**Solution: **

Given that,

Diameter of base of cylinder = 42 cm

Radius of base = d/2 = 42/2 = 21cm

Height = 10 cm

We know that Volume of cylinder = πr

^{2}h= 22/7 × 21 × 21 × 10 =

13860 cm^{3}

**Question 8. Find the volume of cylinder, the diameter of whose base is 7 cm and height being 60 cm. Also, find the capacity of the cylinder in litres.**

**Solution: **

Given that,

Diameter of base = 7 cm

Radius of base = d/2 = 7/2 cm

Height of cylinder = 60 cm

As we know that Volume of cylinder = πr

^{2}h= 22/7 × 7/2 × 7/2 × 60 = 2310 cm

^{3}Capacity of cylinder in liters = 2310 / 1000 =

2.31 liters.

**Question 9. A rectangular strip 25 cm× 7 cm is rotated about the longer side. Find the volume of the solid, thus generated.**

**Solution: **

Given that,

Dimensions of Rectangular Strip = 25 cm × 7 cm

When it rotated about its longer side then it will become,

Radius of base = 7 cm

Height of cylinder = 25 cm

As we know that Volume of cylinder = πr

^{2}h= 22/7 × 7 × 7 × 25 =

3850 cm^{3}

**Question 10. A rectangular sheet of paper, 44 cm × 20 cm, is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.**

**Solution:**

Given that,

Dimensions of rectangular sheet = 44cm × 20cm

When it rolled along its length it will become,

Radius of base = length/2π

= 44×7 / 2×22 = 7cm

Height of cylinder = 20 cm

We know that Volume of cylinder = πr

^{2}h= 22/7 × 7 × 7 × 20 =

3080 cm^{3}

**Question 11. The volume and the curved surface area of cylinder are 1650 cm**^{3} and 660 cm^{2} respectively. Find the radius and height of the cylinder.

^{3}and 660 cm

^{2}respectively. Find the radius and height of the cylinder.

**Solution:**

Given that,

Volume of cylinder = 1650 cm

^{3}Curved surface area = 660 cm

^{2}Volume of Cylinder/Curved Surface Area = 1650/660

πr

^{2}h/ 2πrh = 1650/660r/ 2 = 5/2

r = 5cm

Surface area = 660 cm

^{2 }(Given)As we know that Surface Area of Cylinder = 2πrh

2πrh = 660

2 × 22/7 × 5 × h = 660

h = 660×7 / 2×22×5

= 4620/220 =

21cm

Hence, Radius is 5cm and height is 21cm.

**Question 12. The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3. Calculate the ratio of their volumes.**

**Solution: **

Given that,

Ratio of radii of two cylinder = 2:3

Radius of 1st Cylinder = r1

Radius of 2nd Cylinder = r2

r1/r2 = 2/3

Ratio of their heights = 5:3

Height of 1st Cylinder = h1

Height of 2nd Cylinder = h2

h1/h2 = 5/3

Volume of 1st Cylinder = v1

Volume of 2nd Cylinder = v2

v1 / v2 = π(r1)

^{2}h1 / π(r2)^{2}h2= 22 × 5 / 32 × 3

= 4×5 / 9×3 =

20/27

Hence, Ratio of Volumes of two Cylinder’s is 20 : 27.

**Question 13. The ratio between the curved surface area and the total surface area of a right circular cylinder is 1:2. Find the volume of the cylinder, if its total surface area is 616 cm**^{2}.

^{2}.

**Solution: **

Given that,

Total surface area of cylinder = 616 cm

^{2}Ratio between Curved Surface Area and Total Surface Area of Cylinder = 1:2

2πrh / 2πr (r+h) = 1/2

h / (r+h) = 1/2

2h =r+h

Hence, r = h

As we know, 2πr (h+r) = 616

2πr (r+r) = 616

2πr (2r) = 616

4πr

^{2}= 616r

^{2}= 616/4π= 616×7 / 4×22 = 49

r = √49 = 7

Hence, Radius = 7cm and Height = 7cm

As we know that Volume of cylinder = πr

^{2}h= 22/7 × 7 × 7 × 7 =

1078 cm^{3}

**Question 14. The curved surface area of a cylinder is 1320 cm**^{2} and its base has diameter 21 cm. Find the volume of the cylinder.

^{2}and its base has diameter 21 cm. Find the volume of the cylinder.

**Solution: **

Given that,

Diameter of base = 21 cm

Radius of base = d/2 = 21/2 cm

Curved surface area = 1320 cm

^{2}As we know that Curved surface Area of Cylinder = 2πrh

2πrh = 1320

2 × 22/7 × 21/2 × h = 1320

h = 1320×7×2 / 2×22×21

= 18480/924 = 20cm

As we know that Volume of Cylinder = πr

^{2}h= 22/7 × 21/2 × 21/2 × 20 =

6930 cm^{3}

**Question 15. The ratio between the radius of the base and the height of a cylinder is 2:3. Find the total surface area of the cylinder, if its volume is 1617 cm3.**

**Solution: **

Given that,

Ratio between radius and height of a cylinder = 2:3

r/h = 2/3

h = 3/2 r

Volume of cylinder = 1617 cm

^{3}As we know that Volume of Cylinder = πr

^{2}hπr

^{2}h = 161722/7 × r

^{2}× 3/2r = 1617r

^{3}= 1617×7×2 / 22×3 = 343r =

^{3}√343= 7cm

Hence, Radius = 7 cm

Height = 3/2r = 3/2 × 7 = 21/2 = 10.5cm

As we know that Total Surface Area of Cylinder = 2πr (r+h)

= 2 × 22/7 × 7 (10.5+7) =

770 cm^{2}

**Question 16. The curved surface area of a cylindrical pillar is 264 m**^{2} and its volume is 924 m^{3}. Find the diameter and the height of the pillar.

^{2}and its volume is 924 m

^{3}. Find the diameter and the height of the pillar.

**Solution:**

Given that,

Curved surface area of cylinder = 264 m

^{2},Volume = 924 m

^{3},Volume of Cylinder / Curved Surface Area of Cylinder

πr

^{2}h / 2πrh = 924 / 264r/2 = 924 / 264

r = 924×2 / 264 = 7m

Hence, Radius = 7 m

Diameter of cylinder = 2 × radius = 2×7 =

14mCurved surface area = 264 m

^{2}(Given)2πrh = 264

2 × 22/7 × 7 × h = 264

h = 264×7 / 2×22×7 =

6m

Hence, Height of cylinder is 6m & Diameter of cylinder is 14m.

**Question 17. Two circular cylinders of equal volumes have their heights in the ratio 1:2. Find the ratio of their radii.**

**Solution:**

Given that,

Ratio of their height = 1:2,

Height of 1st Cylinder = h1,

Height of 2nd Cylinder = h2

h1 / h2 = 1/2

Volume of 1st Cylinder, V1 = Volume of 2nd Cylinder, V2

V1 = V2

π(r1

^{2})h1 = π(r2^{2 })h2r1

^{2}/ r2^{2}= 2/1r1 / r2 = √(2/1) =

√2 / 1

Hence, the Ratio of their radii is √2:1

**Question 18. The height of a right circular cylinder is 10/5 m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find** **the volume of the cylinder.**

**Solution: **

Given that,

Height of cylinder = 10.5 m

3(A+A) = 2 Curved Surface Area (where, A = Circular Area of Box)

3×2A = 2(2πrh)

6A = 4πrh

6πr

^{2}= 4πrhr

^{2}/r = 4πh/6πr = 2/3 h = 2×10.5 / 3 = 7m

As we know that Volume of Cylinder = πr

^{2}h= 22/7 × 7 × 7 × 10.5 =

1617 m^{3}

**Question 19. How many cubic meters of earth must be dug-out to sink a well 21 m deep and 6m diameter?**

**Solution:**

Given that,

Height of cylinder = 21m,

Diameter of well = 6m,

Radius of well = d/2 = 6/2 = 3m

We know that Volume of Earth that must be dug out from this well is = πr

^{2}h= 22/7 × 3 × 3 × 21 =

594 m^{3}

**Question 20. The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the trunk is 3 m, find the volume of the timber that can be obtained** **from the trunk.**

**Solution: **

Given that,

Length of the trunk = 3m = 300 cm,

Circumference of trunk of tree = 176 cm

As we know that Circumference of Trunk of tree = 2πr

2πr = 176

2 × 22/7 × r = 176

r = 176×7 / 2×22 = 28cm

Hence, Radius = 28cm

We know that Volume of Timber can be obtained from trunk of tree = πr

^{2}h= 22/7 × 28 × 28 × 300 = 7392 cm

^{3}= 0.74 m^{3}

### Chapter 22 Mensuration III (Surface Area And Volume Of Right Circular Cylinder) – Exercise 22.2 | Set 2

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