# Class 9 RD Sharma Solutions – Chapter 19 Surface Area And Volume of a Right Circular Cylinder – Exercise 19.2 | Set 1

### Question 1. A soft drink is available in two packs:

### (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and

### (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm, Which container has greater capacity and by how much?

**Solution :**

Given that,

The tin can with a rectangular base:

Length= 5 cm,

Breadth = 4 cm,

Height = 15 cm

The plastic cylinder with circular base:

Diameter = 7cm

So the radius of the base = 7/2 cm = 3.5 cm

Height = 10 cm

Now we find the volume of both cans:

Capacity of the tin can = l × b × h = (5 × 4 x 15) cm

^{3}Capacity of plastic cylinder = πR

^{2}H = 22/7 × (3.5)^{2}× 10 cm^{3}= 385 cm^{3}Difference in Capacity = (385 – 300) = 85 cm

^{3}

Hence, the plastic cylinder has greater capacity.

### Question 2. The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10 m, how much concrete mixture would be required to build 14 such pillars?

**Solution: **

Given that,

Radius of the base of a cylindrical pillar= 20 cm

The height of the cylindrical pillar = 10 m

Find how much concrete mixture would be required to build 14 such pillars

So

Volume of the cylindrical pillar = πR

^{2}H= (22/7 × 20

^{2}× 1000)= 8800000/7

= 8.8/7 m

^{3}The volume of 14 pillars = 8.8/7 × 14 = 17.6 m

^{3}

Hence, the volume of the 14 pillars = 17.6 m^{3}

### Question 3. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm^{3} of wood has a mass of 0.6 gm.

**Solution: **

Given that,

The inner diameter of a cylindrical wooden pipe(d

_{1}) = 24 cmSo, the inner radius of a cylindrical pipe(r

_{1}) = 24/2 = 12 cmThe outer diameter of a cylindrical wooden pipe(d

_{2}) = 28 cmSo, the outer radius of a cylindrical pipe(r

_{2}) = 28/2 = 14 cmHeight of cylindrical pipe (h) = 35 cm

Find the mass of the pipe, if 1 cm

^{3}of wood has a mass of 0.6 gmSo,

The Mass of pipe = Volume x density

= π(r

_{2}^{2}– r_{1}^{2})= 22/7 x (14

^{2}− 12^{2}) x 35 = 5720 cm^{3}Mass of 1 cm

^{3}wood = 0.6 gm

Mass of 5720 cm^{3}wood = 5720 × 0.6 = 3432 gm = 3.432 kg

### Question 4. If the lateral surface of a cylinder is 94.2 cm^{2} and its height is 5 cm, find :

### (i) the radius of its base

### (ii) volume of the cylinder [Use pi = 3.141]

**Solution :**

Given that,

The lateral surface of a cylinder = 94.2 cm

^{2}The hight of the cylinder = 5cm

(i)Find the radius of its baseLet’s assume that the radius of cylinder be ‘r’

Curved surface of the cylinder = 2πrh

94.2 = 2 (3.14)r(5)

r = 3 cm

Hence, the radius of the cylinder is 3 cm

(ii)As we know thatThe volume of the cylinder = πr

^{2}h= (3.14 × 32 × 5)

= 141.3 cm

^{3}

Hence, the volume of the cylinder is 141.3 cm^{3}

### Question 5. The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of the metal sheet would be needed to make it?

**Solution: **

Given that,

The height of the cylindrical vessel = 1m

The capacity/volume of the cylinder = 15.4 liters = 0.0154 m

^{3}(As we know 1m^{3}= 1000 liter)Let’s assume that the radius of the circular ends of the cylinders be ‘r’

So the volume of the cylinder is

V = πr

^{2}h0.0154 = (31.4)r

^{2}(1)r = 0.07 m

Now we find the total surface area of a vessel:

TSA = 2πr(r + h)

= 2(3.14 x (0.07) x (0.07 + 1)) = 0.4703 m

^{2}

Hence, we need 0.4703 m^{2 }of the metal sheet

### Question 6. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

**Solution: **

Given that,

The diameter of cylindrical bowl = 7 cm = 3.5 cm

So, the radius = 7/2 cm = 3.5 cm

The bowl is filled with soup to a height =4cm

Now we find the volume soup in 1 bowl

V = πr

^{2}h= 22/7 × 3.5

^{2}× 4 = 154 cm^{3}So the volume soup in 250 bowl

V = (250 × 154) = 38500 cm

^{3}= 38.5 liter

Hence, the soup, hospital has to prepare daily to serve 250 patients is 38.5 liter.

### Question 7. A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.

**Solution: **

Given that,

Garden roller height = 63 cm,

Garden roller outer circumference = 440 cm,

Garden roller thickness = 4 cm

Find the volume of iron.

So, let’s assume that the R be the external radius and the inner radius be ‘r’

2πR = 440

2 x 22/7 x R = 440

R = 70

Now we find the value of inner radius:

r = R – 4

70 – 4 = 66cm

Now we find the volume of the iron:

V = π (R

^{2}− r^{2}) x h= 22/7 x (70

^{2}− 66^{2}) x 63= 22/7 x 4 x 136 x 63 =

107712 cm^{3}

Hence, the volume of the iron is 107712 cm^{3}

### Question 8. A solid cylinder has a total surface area of 231cm^{2}. Its curved surface area is 2/3 of the total surface area. Find the volume of the cylinder.

**Solution: **

Given that,

Total surface area = 231cm

^{2},Curved surface area = 2/3 x (Total Surface Area)

So,

Curved surface area = 2/3 x 231 = 154

As we know that,

the total surface area of cylinder = 2πrh + 2πr

^{2}2πrh + 2πr

^{2 }= 231 —————-(i)Where, 2πrh is the curved surface area, So

154 + 2πr

^{2}= 2312πr

^{2}= 231 – 1542πr

^{2}= 772 x 22/7 x r

^{2}= 77r

^{2}= (7×7) / (2×2)r = 7/2

The radius of cylinder = 7/2

Now we find the height of the cylinder

So, as we know that

Curved surface area = 2πrh

2πrh = 154

2 x 22/7 x 7/2 x h = 154

h = 154/22 = 7

So, the height of cylinder = 7

Now we find the volume of the cylinder:

Volume = πr

^{2}h= 22/7 x 7/2 x 7/2 x 7 = 269.5 cm

^{3}

So, the volume of the cylinder is 269.5 cm^{3}

### Question 9. The cost of painting the total outside surface of a closed cylindrical oil tank at 50 paise per square decimetre is Rs 198. The height of the tank is 6 times the radius of the base of the tank. Find the volume corrected to 2 decimal places.

**Solution: **

Let’s assume that the radius of the tank = r dm

So, the height of the tank(h) = 6r dm

It is given that the cost of painting = 50 paisa per dm

^{2 }So, the total cost of painting = Rs 198

= 2πr(r + h) = 198

= 2 × 22/7 × r(r + 6r) × 1/2 = 198

r = 3 dm

Hence the radius of the tank is 3 dm

Therefore, h = (6 × 3) dm = 18 dm

As we know that,

Volume of the tank = πr

^{2}h= 22/7 × 9 × 18 = 509.14 dm

^{3}

### Question 10. 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 and the ratio of their curved surfaces.

**Solution: **

Given that the ratio of the radii of two cylinders = 2:3

The ratio of the heights two cylinders = 5:3

So, let’s assume that the radius of the two cylinders are 2x and 3x

The height of the two cylinders is 5y and 3y

Find: The ratio of their volumes and the ratio of their curved surfaces

So, for the ratio of their volumes:

We have

Volume of cylinder A/ Volume of cylinder B = π (r)

^{2 }h/π (R)^{2 }H= π (2x)

^{2 }5y/π (3x)^{2 }3y = 20/27Hence, the ratio of the volumes of two cylinders are 20:27.

So, for the ratio of their surface area:

We have

Surface area of cylinder A / Surface area of cylinder B = 2πrh/2πRH

= (2π × 2x × 5y) / (2π × 3x × 3y) = 10 / 9

Hence, the ratio of the surface area of two cylinders are 10:9.

### Question 11. 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}.

**Solution: **

Given that

Total surface area (TSA) = 616 cm

^{2}The ratio between the curved surface area and the total surface area of a right circular cylinder = 1 : 2

Find: the volume of the cylinder

According to the question

Curved Surface Area / Total Surface Area = 1/2

CSA = 1/2 x TSA

CSA = 1/2 x 616

CSA = 308 cm

^{2}Now, we find the total surface area

TSA = 2πrh + 2πr

^{2}616 = CSA + 2πr2

616 = 308 + 2πr

^{2}2πr

^{2}= 616 – 3082πr

^{2}= 308πr

^{2}= 308/2r

^{2}= 308/2πr = 7 cm

Since, CSA = 308 cm

^{2}2πrh = 308

2 x 22/7 x 7 x h = 308

h = 7cm

Now we find the volume of cylinder

V = πr

^{2}x h= 22/7 x 7 x 7 x 7

= 22 x 49

= 1078 cm

^{3}

Hence, the volume of cylinder is1078 cm^{3}

### Question 12. The curved surface area of a cylinder is 1320 cm^{2 }and its base had diameter 21 cm. Find the height and volume of the cylinder.

**Solution: **

Given that

The curved surface area of a cylinder = 1320 cm

^{2 }Diameter of its base = 21 cm

So, radius = 21/2 = 10.5 cm

r = 21/2 = 10.5 cm

Find: the height and volume of the cylinder.

So, the curved surface area of a cylinder is

CSA = 2πrh

2 x 22/7 x 10.5 x h = 1320

h = 1320/66 = 20 cm

So the height of the cylinder is 20 cm

Now we find the volume of cylinder

V = πr

^{2}h= 22/7 x 10.5 x 10.5 x 20

= 22 x 1.5 x 10.5 x 20 = 6930 cm

^{3}

Hence, the volume of cylinder is 6930 cm^{3}

### Question 13. 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 cm^{3}.

**Solution: **

Given that,

The volume of the cylinder = 1617 cm

^{3}The ratio between the radius of the base and the height of a cylinder = 2:3

r/h = 2/3

r = 2/3 x h ——————–(i)

Find: The total surface area of the cylinder

So, we find the volume of cylinder

V = πr

^{2}h1617 = 22/7 x (2/3 x h)

^{2}x h1617 = 22/7 x (2/3 x h)

^{3}h

^{3}= (1617 x 7 x 3) / 22 x 4h = 10.5 cm

From, eqn. (i), we get

r = 2/3 x 10.5 = 7 cm

Now we find the total surface area of cylinder

TSA = 2πr (h + r)

= 2 x 22/7 x 7(10.5 + 7)

= 44 x 17.5

= 770 cm

^{2}

Hence, the total surface area of cylinder is 770 cm^{2}

### Question 14. A rectangular sheet of paper, 44 cm x 20 cm, is rolled along its length of form cylinder. Find the volume of the cylinder so formed.

**Solution:**

Given that,

The dimensions of the rectangular sheet of paper = 44 cm x 20 cm

So,

Length = 44 cm,

Height = 20 cm

Find: The volume of the cylinder

Curved Surface Area = 2πr

2πr = 44

r = 44/2π

r = 44/2π = 7 cm

Hence, the radius of the cylinder is 7 cm

Now, we find the volume of cylinder

V = πr

^{2}h= 22/7 x 7 x 7 x 20

= 154 x 20 = 3080 cm

^{3}

Volume of cylinder is 3080 cm^{3}

### Question 15. The curved surface area of the cylindrical pillar is 264 m^{2} and its volume is 924 m^{3}. Find the diameter and the height of the pillar.

**Solution: **

Given that,

The curved surface area of the cylindrical pillar = 264 m

^{2}The volume of the cylindrical pillar = 924 m

^{3}We have to find the diameter and the height of the pillar

So,

Volume of the cylinder

V = πr

^{2}hπ x r

^{2}x h = 924πrh(r) = 924

πrh = 924/r

As we know that the curved surface area of the cylinder

CSA = 2πrh

264 = 2πrh …(1)

Substitute πrh in this eq and we get,

2 x 924/r = 264

r = 1848/264 = 7 m

Substitute r value in eq (i) and we get,

2 x 22/7 x 7 x h = 264

h = 264/44 = 6 m

Hence, the diameter = 2r = 2(7) = 14 m and height = 6 m

### Question 16. Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Find the ratio of two radii.

**Solution:**

Let’s assume that we have two cylinders,

So, the radius of the cylinders = r

_{1}, r_{2}The height of the cylinders = h

_{1}, h_{2}The volume of the cylinders = v

_{1}, v_{2}According to the question

It is given that the h

_{1}/h_{2}= 1/2 and v_{1}= v_{2}We have to find the ratio of two radii

So,

v

_{1}/v_{2}= (r_{1}/r_{2})^{2}x (h_{1}/h_{2})As v

_{1}= v_{2}v

_{1}/v_{1}= (r_{1}/r_{2})^{2}x (1/2)1 = (r

_{1}/r_{2})^{2}x (1/2)(r

_{1}/r_{2})^{2}= (2/1)(r

_{1}/r_{2}) = √2 / 1

Hence, the ratio of the radii are √2:1

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the ** Demo Class for First Step to Coding Course, **specifically **designed for students of class 8 to 12. **

The students will get to learn more about the world of programming in these **free classes** which will definitely help them in making a wise career choice in the future.