# NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

## NCERT Solutions for Class 9 Maths Chapter 13 – CBSE Free PDF Download

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes include the accurately designed wide range of solved exercise questions for an excellent understanding. These solutions in Maths for Class 9 are prepared considering the latest CBSE syllabus 2022-23 examination. NCERT for Class 9 Maths Solutions is mainly created as a guide and advantageous reference to help students clear doubts in an effective way. NCERT Solutions for Class 9 Maths are prepared by the teaching faculty having vast teaching experience, along with subject matter experts to serve the purpose.

Class 9 NCERT Solutions are developed keeping in mind the concept-based approach, along with the precise answering method for examinations. Refer to NCERT Solutions for Class 9 to get a better score. They are detailed and well-structured solutions for a good grasp of concept-based knowledge. NCERT for Class 9 Maths Solutions are made available in web and PDF format for ease of access.

## Download the PDF of NCERT Solutions for Class 9 Maths Chapter 13: Surface Areas and Volumes

List of Exercises in Class 9 Maths Chapter 13
Exercise 13.1 solution (9 questions)
Exercise 13.2 solution (8 questions)
Exercise 13.3 solution (9 questions)
Exercise 13.4 solution (5 questions)
Exercise 13.5 solution (5 questions)
Exercise 13.6 solution (8 questions)
Exercise 13.7 solution (9 questions)
Exercise 13.8 solution (10 questions)
Exercise 13.9 solution (3 questions)

## Exercise 13.1 Page No: 213

1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine

(I) The area of the sheet required for making the box.

(ii) The cost of the sheet for it, if a sheet measuring 1m2 costs Rs. 20.

Solution:

Given: length (l) of box = 1.5m

Breadth (b) of box = 1.25 m

Depth (h) of box = 0.65m

(i) Box is to be open at the top.

Area of sheet required.

= 2lh+2bh+lb

= [2×1.5×0.65+2×1.25×0.65+1.5×1.25]m2

= (1.95+1.625+1.875) m2 = 5.45 m2

(ii) Cost of sheet per m2 area = Rs.20

Cost of sheet of 5.45 m2 area = Rs (5.45×20)

= Rs.109.

2. The length, breadth and height of a room are 5 m, 4 m and 3 m, respectively. Find the cost of whitewashing the walls of the room and ceiling at the rate of Rs 7.50 per m2.

Solution:

Length (l) of room = 5m

Breadth (b) of room = 4m

Height (h) of room = 3m

It can be observed that four walls and the ceiling of the room are to be whitewashed.

Total area to be whitewashed = Area of walls + Area of the ceiling of the room

= 2lh+2bh+lb

= [2×5×3+2×4×3+5×4]

= (30+24+20)

= 74

Area = 74 m2

Also,

Cost of whitewash per m2 area = Rs.7.50 (Given)

Cost of whitewashing 74 m2 area = Rs. (74×7.50)

= Rs. 555

3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m2 is Rs.15,000, find the height of the hall.

[Hint: Area of the four walls = Lateral surface area.]

Solution:

Let the length, breadth, and height of the rectangular hall be l, b, and h, respectively.

Area of four walls = 2lh+2bh

= 2(l+b)h

Perimeter of the floor of hall = 2(l+b)

= 250 m

Area of four walls = 2(l+b) h = 250h m2

Cost of painting per square metre area = Rs.10

Cost of painting 250h square metre area = Rs (250h×10) = Rs.2500h

However, it is given that the cost of painting the walls is Rs. 15,000.

15000 = 2500h

Or h = 6

Therefore, the height of the hall is 6 m.

4. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?

Solution:

Total surface area of one brick = 2(lb +bh+lb)

= [2(22.5×10+10×7.5+22.5×7.5)] cm2

= 2(225+75+168.75) cm2

= (2×468.75) cm2

= 937.5 cm2

Let n bricks can be painted out by the paint of the container.

Area of n bricks = (n×937.5) cm2 = 937.5n cm2

As per the given instructions, the area that can be painted by the paint of the container = 9.375 m2 = 93750 cm2

So, we have 93750 = 937.5n

n = 100

Therefore, 100 bricks can be painted out by the paint of the container.

5. A cubical box has each edge 10 cm, and another cuboidal box is 12.5cm long, 10 cm wide, and 8 cm high.

(i) Which box has the greater lateral surface area, and by how much?

(ii) Which box has the smaller total surface area, and by how much?

Solution:

From the question statement, we have

Edge of a cube = 10cm

Length, l = 12.5 cm

Height, h = 8 cm

(i) Find the lateral surface area for both figures.

Lateral surface area of cubical box = 4 (edge)2

= 4(10)2

= 400 cm2 …(1)

Lateral surface area of cuboidal box = 2[lh+bh]

= [2(12.5×8+10×8)]

= (2×180) = 360

Therefore, the lateral surface area of the cuboidal box is 360 cm2. …(2)

From (1) and (2), the lateral surface area of the cubical box is more than the lateral surface area of the cuboidal box. The difference between both lateral surfaces is 40 cm2.

(Lateral surface area of the cubical box – Lateral surface area of cuboidal box=400cm2–360cm2 = 40 cm2)

(ii) Find the total surface area for both figures.

The total surface area of the cubical box = 6(edge)2 = 6(10 cm)2 = 600 cm2…(3)

The total surface area of the cuboidal box

= 2[lh+bh+lb]

= [2(12.5×8+10×8+12.5×100)]

= 610

This implies that the total surface area of the cuboidal box is 610 cm2..(4)

From (3) and (4), the total surface area of the cubical box is smaller than that of the cuboidal box. And their difference is 10cm2.

Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2

6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including the base) held together with tape. It is 30cm long, 25 cm wide, and 25 cm high.

(i) What is the area of the glass?

(ii)How much tape is needed for all 12 edges?

Solution:

Length of the greenhouse, say l = 30cm

Breadth of the greenhouse, say b = 25 cm

Height of greenhouse, say h = 25 cm

(i) Total surface area of greenhouse = Area of the glass = 2[lb+lh+bh]

= [2(30×25+30×25+25×25)]

= [2(750+750+625)]

= (2×2125) = 4250

The total surface area of the glass is 4250 cm2

(ii)

From the figure, the tape is required along sides AB, BC, CD, DA, EF, FG, GH, HE AH, BE, DG, and CF.

Total length of tape = 4(l+b+h)

= [4(30+25+25)] (after substituting the values)

= 320

Therefore, 320 cm tape is required for all 12 edges.

7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm×20cm×5cm, and the smaller of dimension 15cm×12cm×5cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.

Solution:

Let l, b and h be the length, breadth and height of the box.

Bigger Box

l = 25cm

b = 20 cm

h = 5 cm

Total surface area of bigger box = 2(lb+lh+bh)

= [2(25×20+25×5+20×5)]

= [2(500+125+100)]

= 1450 cm2

Extra area required for overlapping 1450×5/100 cm2

= 72.5 cm2

While considering all overlaps, the total surface area of the bigger box.

= (1450+72.5) cm2 = 1522.5 cm2

Area of cardboard sheet required for 250 such bigger boxes

= (1522.5×250) cm2 = 380625 cm2

Smaller Box

Similarly, total surface area of the smaller box = [2(15×12+15×5+12×5)] cm2

= [2(180+75+60)] cm2

= (2×315) cm2

= 630 cm2

Therefore, the extra area required for overlapping 630×5/100 cm2 = 31.5 cm2

The total surface area of 1 smaller box while considering all overlaps

= (630+31.5) cm2 = 661.5 cm2

Area of cardboard sheet required for 250 smaller boxes = (250×661.5) cm2 = 165375 cm2

In Short

 Box Dimensions (in cm) Total surface area (in cm2 ) Extra area required for overlapping (in cm2) Total surface area for all overlaps (in cm 2) Area for 250 such boxes (in cm2) Bigger Box l = 25 b = 20 c = 5 1450 1450×5/100 = 72.5 (1450+72.5) = 1522.5 (1522.5×250) = 380625 Smaller Box l = 15 b = 12 h =5 630 630×5/100 = 31.5 (630+31.5) = 661.5 ( 250×661.5) = 165375

Now, total cardboard sheet required = (380625+165375) cm2

= 546000 cm2

Given, cost of 1000 cm2 cardboard sheet = Rs. 4

Therefore, the cost of 546000 cm2 cardboard sheet =Rs. (546000×4)/1000 = Rs. 2184

Therefore, the cost of cardboard required for supplying 250 boxes of each kind will be Rs. 2,184.

8. Praveen wanted to make a temporary shelter for her car by making a box-like structure with a tarpaulin that covers all four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small and therefore negligible, how many tarpaulins would be required to make the shelter of height 2.5m, with base dimensions 4m×3m?

Solution:

Let l, b and h be the length, breadth and height of the shelter.

Given:

l = 4m

b = 3m

h = 2.5m

Tarpaulins will be required for the top and four wall sides of the shelter.

Using formula, area of tarpaulin required = 2(lh+bh)+lb

On putting the values of l, b and h, we get

= [2(4×2.5+3×2.5)+4×3] m2

= [2(10+7.5)+12]m2

= 47m2

Therefore, 47 m2 of tarpaulin will be required.

## Exercise 13.2 Page No: 216

1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder (Assume π =22/7 ).

Solution:

Height of cylinder, h = 14cm

Let the diameter of the cylinder be d.

The curved surface area of cylinder = 88 cm2

We know that the formula to find the Curved surface area of the cylinder is 2πrh.

So 2πrh =88 cm2 (r is the radius of the base of the cylinder)

2×(22/7)×r×14 = 88 cm2

2r = 2 cm

d =2 cm

Therefore, the diameter of the base of the cylinder is 2 cm.

2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square metres of the sheet are required for the same? Assume π = 22/7

Solution:

Let h be the height and r be the radius of a cylindrical tank.

Height of cylindrical tank, h = 1m

Radius = half of diameter = (140/2) cm = 70cm = 0.7m

Area of sheet required = Total surface area of tank = 2πr(r+h) unit square

= [2×(22/7)×0.7(0.7+1)]

= 7.48 square metres

Therefore, 7.48 square metres of the sheet are required.

3. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4cm (see fig. 13.11). Find its

(i) inner curved surface area

(ii) outer curved surface area

(iii) total surface area

(Assume π=22/7)

Solution:

Let r1 and r2 inner and outer radii of the cylindrical pipe.

r1 = 4/2 cm = 2 cm

r2 = 4.4/2 cm = 2.2 cm

Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm

(i) Curved surface area of the outer surface of pipe = 2πr1h

= 2×(22/7)×2×77 cm2

= 968 cm2

(ii) Curved surface area of the outer surface of pipe = 2πr2h

= 2×(22/7)×2.2×77 cm2

= (22×22×2.2) cm2

= 1064.8 cm2

(iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ area of both circular ends of pipe

= 2r1h+2r2h+2π(r12-r22)

= 9668+1064.8+2×(22/7)×(2.22-22)

= 2031.8+5.28

= 2038.08 cm2

Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2.

4. The diameter of a roller is 84 cm, and its length is 120 cm. It takes 500 complete revolutions to

move once over to level a playground. Find the area of the playground in m2 (Assume π = 22/7).

Solution:

A roller is shaped like a cylinder.

Let h be the height of the roller and r be the radius.

h = Length of roller = 120 cm

Radius of the circular end of roller = r = (84/2) cm = 42 cm

Now, CSA of roller = 2πrh

= 2×(22/7)×42×120

= 31680 cm2

Area of field = 500×CSA of roller

= (500×31680) cm2

= 15840000 cm2

= 1584 m2.

Therefore, the area of the playground is 1584 m2.

5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2.

(Assume π = 22/7)

Solution:

Let h be the height of a cylindrical pillar and r be the radius.

Given:

Height cylindrical pillar = h = 3.5 m

Radius of the circular end of pillar = r = diameter/2 = 50/2 = 25cm = 0.25m

CSA of pillar = 2πrh

= 2×(22/7)×0.25×3.5

= 5.5 m2

Cost of painting 1 m2 area = Rs. 12.50

Cost of painting 5.5 m2 area = Rs (5.5×12.50)

= Rs.68.75

Therefore, the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2 is Rs 68.75.

6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height. (Assume π = 22/7)

Solution:

Let h be the height of the circular cylinder and r be the radius.

The radius of the base of the cylinder, r = 0.7m

CSA of cylinder = 2πrh

CSA of cylinder = 4.4m2

Equating both equations, we have

2×(22/7)×0.7×h = 4.4

Or h = 1

Therefore, the height of the cylinder is 1 m.

7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find

(i) its inner curved surface area.

(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2.

(Assume π = 22/7)

Solution:

Inner radius of circular well, r = 3.5/2m = 1.75m

Depth of circular well, say h = 10m

(i) Inner curved surface area = 2πrh

= (2×(22/7 )×1.75×10)

= 110 m2

Therefore, the inner curved surface area of the circular well is 110 m2.

(ii) Cost of plastering 1 m2 area = Rs.40

Cost of plastering 110 m2 area = Rs (110×40)

= Rs.4400

Therefore, the cost of plastering the curved surface of the well is Rs. 4,400.

8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. (Assume π = 22/7)

Solution:

Height of cylindrical pipe = Length of cylindrical pipe = 28m

Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m

Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder

= 2×(22/7)×0.025×28 m2

= 4.4m2

The area of the radiating surface of the system is 4.4m2.

9. Find

(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high.

(ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank? (Assume π = 22/7)

Solution:

Height of cylindrical tank, h = 4.5m

Radius of the circular end , r = (4.2/2)m = 2.1m

(i) The lateral or curved surface area of the cylindrical tank is 2πrh.

= 2×(22/7)×2.1×4.5 m2

= (44×0.3×4.5) m2

= 59.4 m2

Therefore, the CSA of the tank is 59.4 m2.

(ii) Total surface area of tank = 2πr(r+h)

= 2×(22/7)×2.1×(2.1+4.5)

= 44×0.3×6.6

= 87.12 m2

Now, Let S m2 steel sheet be actually used in making the tank.

S(1 -1/12) = 87.12 m2

This implies, S = 95.04 m2

Therefore, 95.04m2 steel was used in actuality while making such a tank.

10. In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth.

The frame has a base diameter of 20 cm and a height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required to cover the lampshade. (Assume π = 22/7)

Solution:

Say h = height of the frame of the lampshade, which looks like a cylindrical shape.

Total height is h = (2.5+30+2.5) cm = 35cm and

r = (20/2) cm = 10cm

Use the curved surface area formula to find the cloth required for covering the lampshade, which is 2πrh.

= (2×(22/7)×10×35) cm2

= 2200 cm2

Hence, 2200 cm2 cloth is required to cover the lampshade.

11. The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π =22/7)

Solution:

The radius of the circular end of the cylindrical penholder, r = 3cm

Height of penholder, h = 10.5cm

Surface area of a penholder = CSA of pen holder + area of base of penholder

= 2πrh+πr2

= 2×(22/7)×3×10.5+(22/7)×32= 1584/7

Therefore, the Area of cardboard sheet used by one competitor is 1584/7 cm2

So, the Area of cardboard sheet used by 35 competitors = 35×1584/7 = 7920 cm2

Therefore, a 7920 cm2 cardboard sheet will be needed for the competition.

## Exercise 13.3 Page No: 221

1. Diameter of the base of a cone is 10.5 cm, and its slant height is 10 cm. Find its curved surface area. (Assume π=22/7)

Solution:

Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm

The slant height of the cone, say l = 10 cm

CSA of the cone is = πrl

= (22/7)×5.25×10 = 165 cm2

Therefore, the curved surface area of the cone is 165 cm2.

2. Find the total surface area of a cone, if its slant height is 21 m and the diameter of its base is 24 m. (Assume π = 22/7)

Solution:

Radius of cone, r = 24/2 m = 12m

Slant height, l = 21 m

Formula: Total Surface area of the cone = πr(l+r)

Total Surface area of the cone = (22/7)×12×(21+12) m2

= 1244.57m2

3. Curved surface area of a cone is 308 cm2, and its slant height is 14 cm. Find

(i) radius of the base and (ii) total surface area of the cone.

(Assume π = 22/7)

Solution:

The slant height of the cone, l = 14 cm

Let the radius of the cone be r.

(i) We know the CSA of cone = πrl

Given: Curved surface area of a cone is 308 cm2

(308 ) = (22/7)×r×14

308 = 44 r

r = 308/44 = 7 cm

The radius of a cone base is 7 cm.

(ii) Total surface area of cone = CSA of cone + Area of base (πr2)

Total surface area of cone = 308+(22/7)×72 = 308+154 = 462 cm2

Therefore, the total surface area of the cone is 462 cm2.

4. A conical tent is 10 m high, and the radius of its base is 24 m. Find

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.

(Assume π=22/7)

Solution:

Let ABC be a conical tent.

Height of conical tent, h = 10 m

Radius of conical tent, r = 24m

Let the slant height of the tent be l.

(i) In the right triangle ABO, we have

AB2 = AO2+BO2(using Pythagoras’ theorem)

l2 = h2+r2

= (10)2+(24)2

= 676

l = 26 m

Therefore, the slant height of the tent is 26 m.

(ii) CSA of tent = πrl

= (22/7)×24×26 m2

Cost of 1 m2 canvas = Rs 70

Cost of (13728/7)m2 canvas is equal to Rs (13728/7)×70 = Rs 137280

Therefore, the cost of the canvas required to make such a tent is Rs 137280.

5. What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. [Use π=3.14]

Solution:

Height of the conical tent, h = 8m

Radius of the base of the tent, r = 6m

Slant height of the tent, l2 = (r2+h2)

l2 = (62+82) = (36+64) = (100)

or l = 10 m

Again, CSA of conical tent = πrl

= (3.14×6×10) m2

= 188.4m2

Let the length of the tarpaulin sheet required be L.

As 20 cm will be wasted,

The effective length will be (L-0.2m).

The breadth of tarpaulin = 3m (given)

Area of sheet = CSA of the tent

[(L–0.2)×3] = 188.4

L-0.2 = 62.8

L = 63 m

Therefore, the length of the required tarpaulin sheet will be 63 m.

6. The slant height and base diameter of the conical tomb are 25m and 14 m, respectively. Find the cost of whitewashing its curved surface at the rate of Rs. 210 per 100 m2. (Assume π = 22/7)

Solution:

Slant height of the conical tomb, l = 25m

Base radius, r = diameter/2 = 14/2 m = 7m

CSA of the conical tomb = πrl

= (22/7)×7×25 = 550

CSA of the conical tomb= 550m2

Cost of whitewashing 550 m2 area, which is Rs (210×550)/100

= Rs. 1155

Therefore, the cost will be Rs. 1155 while whitewashing the tomb.

7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. (Assume π =22/7)

Solution:

Radius of the conical cap, r = 7 cm

Height of the conical cap, h = 24cm

Slant height, l2 = (r2+h2)

= (72+242)

= (49+576)

= (625)

Or l = 25 cm

CSA of 1 conical cap = πrl

= (22/7)×7×25

= 550 cm2

CSA of 10 caps = (10×550) cm2 = 5500 cm2

Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.

8. A bus stop is barricaded from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) =1.02)

Solution:

Given:

Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m

Height of cone, h = 1m

Slant height of cone is l, and l2 = (r2+h2)

Using given values, l2 = (0.22+12)

= (1.04)

Or l = 1.02 m

Slant height of the cone is 1.02 m.

Now,

CSA of each cone = πrl

= (3.14×0.2×1.02)

= 0.64056 m

CSA of 50 such cones = (50×0.64056) = 32.028

CSA of 50 such cones = 32.028 m2

Again,

Cost of painting 1 m2 area = Rs 12 (given)

Cost of painting 32.028 m2 area = Rs (32.028×12)

= Rs.384.336

= Rs.384.34 (approximately)

Therefore, the cost of painting all these cones is Rs. 384.34.

## Exercise 13.4 Page No: 225

1. Find the surface area of a sphere of radius

(i) 10.5cm (ii) 5.6cm (iii) 14cm

(Assume π=22/7)

Solution:

Formula: Surface area of a sphere (SA) = 4πr2

(i) Radius of a sphere, r = 10.5 cm

SA = 4×(22/7)×10.52 = 1386

Surface area of a sphere is 1386 cm2

(ii) Radius of a sphere, r = 5.6cm

Using formula, SA = 4×(22/ 7)×5.62 = 394.24

Surface area of a sphere is 394.24 cm2

(iii) Radius of a sphere, r = 14cm

SA = 4πr2

= 4×(22/7)×(14)2

= 2464

Surface area of a sphere is 2464 cm2

2. Find the surface area of a sphere of diameter

(i) 14cm (ii) 21cm (iii) 3.5cm

(Assume π = 22/7)

Solution:

(i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm

Formula for the surface area of sphere = 4πr2

= 4×(22/7)×72 = 616

Surface area of a sphere is 616 cm2

(ii) Radius (r) of sphere = 21/2 = 10.5 cm

Surface area of a sphere = 4πr2

= 4×(22/7)×10.52 = 1386

Surface area of a sphere is 1386 cm2

Therefore, the surface area of a sphere having a diameter 21 cm is 1386 cm2

(iii) Radius(r) of a sphere = 3.5/2 = 1.75 cm

Surface area of a sphere = 4πr2

= 4×(22/7)×1.752 = 38.5

Surface area of a sphere is 38.5 cm2

3. Find the total surface area of a hemisphere of radius 10 cm. [Use π=3.14]

Solution:

Radius of the hemisphere, r = 10cm

Formula: Total surface area of the hemisphere = 3πr2

= 3×3.14×102 = 942

The total surface area of the given hemisphere is 942 cm2.

4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Solution:

Let r1 and r2 be the radii of the spherical balloon and spherical balloon when air is pumped into it, respectively. So,

r1 = 7cm

r2 = 14 cm

Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)

= 4πr12/4πr22

= (r1/r2)2

= (7/14)2 = (1/2)2 = ¼

Therefore, the ratio between the surface areas is 1:4.

5. A hemispherical bowl made of brass has an inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. (Assume π = 22/7)

Solution:

Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm

Formula for the surface area of hemispherical bowl = 2πr2

= 2×(22/7)×(5.25)2 = 173.25

Surface area of the hemispherical bowl is 173.25 cm2

Cost of tin-plating 100 cm2 area = Rs 16

Cost of tin-plating 1 cm2 area = Rs 16 /100

Cost of tin-plating 173.25 cm2 area = Rs. (16×173.25)/100 = Rs 27.72

Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.

6. Find the radius of a sphere whose surface area is 154 cm2. (Assume π = 22/7)

Solution:

Let the radius of the sphere be r.

Surface area of sphere = 154 (given)

Now,

4πr2 = 154

r2 = (154×7)/(4×22) = (49/4)

r = (7/2) = 3.5

The radius of the sphere is 3.5 cm.

7. The diameter of the moon is approximately one-fourth of the diameter of the earth.

Find the ratio of their surface areas.

Solution:

If the diameter of the earth is said d, then the diameter of the moon will be d/4 (as per the given statement).

Radius of moon = ½×d/4 = d/8

Surface area of moon = 4π(d/8)2

Surface area of earth = 4π(d/2)2

The ratio between their surface areas is 1:16.

8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume π =22/7)

Solution:

Given:

Inner radius of the hemispherical bowl = 5cm

Thickness of the bowl = 0.25 cm

Outer radius of the hemispherical bowl = (5+0.25) cm = 5.25 cm

Formula for outer CSA of the hemispherical bowl = 2πr2, where r is the radius of the hemisphere.

= 2×(22/7)×(5.25)2 = 173.25 cm2

Therefore, the outer curved surface area of the bowl is 173.25 cm2.

9. A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in(i) and (ii).

Solution:

(i) Surface area of the sphere = 4πr2, where r is the radius of sphere

(ii) Height of the cylinder, h = r+r =2r

The radius of the cylinder = r

CSA of the cylinder formula = 2πrh = 2πr(2r) (using value of h)

= 4πr2

(iii) Ratio between areas = (Surface area of sphere)/(CSA of Cylinder)

= 4πr2/4πr2 = 1/1

The ratio of the areas obtained in (i) and (ii) is 1:1.

## Exercise 13.5 Page No: 228

1. A matchbox measures 4 cm×2.5cm×1.5cm. What will be the volume of a packet containing 12 such boxes?

Solution:

Dimensions of a matchbox (a cuboid) are l×b×h = 4 cm×2.5 cm×1.5 cm, respectively

Formula to find the volume of matchbox = l×b×h = (4×2.5×1.5) = 15

Volume of matchbox = 15 cm3

Now, volume of 12 such matchboxes = (15×12) cm3 = 180 cm3

Therefore, the volume of 12 matchboxes is 180 cm3.

2. A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold? (1 m3= 1000 l)

Solution:

Dimensions of a cuboidal water tank are l = 6 m and b = 5 m, and h = 4.5 m

Formula to find the volume of the tank, V = l×b×h

Putting the values, we get

V = (6×5×4.5) = 135

The volume of the water tank is 135 m3

Again,

We are given that the amount of water that 1m3 volume can hold = 1000 l

Amount of water, 135 m3volume hold = (135×1000) litres = 135000 litres

Therefore, given cuboidal water tank can hold up to 135000 litres of water.

3. A cuboidal vessel is 10m long and 8m wide. How high must it be made to hold 380 cubic metres of a liquid?

Solution:

Given:

Length of the cuboidal vessel, l = 10 m

Width of the cuboidal vessel, b = 8m

Volume of the cuboidal vessel, V = 380 m3

Let the height of the given vessel be h.

Formula for volume of a cuboid, V = l×b×h

Using the formula, we have

l×b×h = 380

10×8×h= 380

Or h = 4.75

Therefore, the height of the vessels is 4.75 m.

4. Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3.

Solution:

The given pit has its length(l) as 8m, width (b)as 6m and depth (h)as 3 m.

Volume of cuboidal pit = l×b×h = (8×6×3) = 144 (using formula)

The required Volume is 144 m3

Now,

Cost of digging per m3 volume = Rs 30

Cost of digging 144 m3 volume = Rs (144×30) = Rs 4320

5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Solution:

The length (l) and depth (h) of the tank are 2.5 m and 10 m, respectively.

To find: The value of breadth, say b.

Formula to find the volume of a tank = l×b×h = (2.5× b×10) m3= 25b m3

The capacity of tank= 25b m3, which is equal to 25000b litres

Also, the capacity of a cuboidal tank is 50000 litres of water (Given).

Therefore, 25000 b = 50000

This implies that b = 2

Therefore, the breadth of the tank is 2 m.

6. A village, having a population of 4000, requires 150 litres of water per head per day.

It has a tank measuring 20 m×15 m×6 m. For how many days will the water in this tank last?

Solution:

Length of the tank = l = 20 m

Breadth of the tank = b = 15 m

Height of the tank = h = 6 m

Total population of a village = 4000

Consumption of water per head per day = 150 litres

Water consumed by the people in 1 day = (4000×150) litres = 600000 litres …(1)

Formula to find the capacity of the tank, C = l×b×h

Using the given data, we have

C = (20×15×6) m3= 1800 m3

Or C = 1800000 litres

Let water in this tank last for d days.

Water consumed by all people in d days = Capacity of the tank (using equation (1))

600000 d =1800000

d = 3

Therefore, the water in this tank will last for 3 days.

7. A godown measures 40 m×25m×15 m. Find the maximum number of wooden crates, each measuring 1.5m×1.25 m×0.5 m, that can be stored in the godown.

Solution:

From the statement, we have

Length of the godown = 40 m

Height = 15 m

Whereas,

Length of the wooden crate = 1.5 m

Height = 0.5 m

The godown and wooden crate are in cuboidal shape. Find the volume of each using the formula V = lbh

Now,

Volume of godown = (40×25×15) m3 = 15000 m3

Volume of a wooden crate = (1.5×1.25×0.5) m3 = 0.9375 m3

Let us consider that, n wooden crates can be stored in the godown, then

Volume of n wooden crates = Volume of godown

0.9375×n =15000

Or n= 15000/0.9375 = 16000

Hence, the number of wooden crates that can be stored in the godown is 16,000.

8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Solution:

Side of a cube = 12cm (Given)

Find the volume of the cube.

Volume of cube = (Side)3 = (12)3cm3= 1728cm3

Surface area of a cube with side 12 cm = 6a2 = 6(12) 2 cm2 …(1)

The cube is cut into eight small cubes of equal volume; say, the side of each cube is p.

Volume of a small cube = p3

Surface area = 6p2 …(2)

Volume of each small cube = (1728/8) cm3 = 216 cm3

Or (p)3 = 216 cm3

Or p = 6 cm

Now, Surface areas of the cubes ratios = (Surface area of the bigger cube)/(Surface area of the smaller cubes)

From equations (1) and (2), we get

Surface areas of the cubes ratios = (6a2)/(6p2) = a2/p2 = 122/62 = 4

Therefore, the required ratio is 4 : 1.

9. A river 3m deep and 40m wide is flowing at the rate of 2km per hour. How much water will fall into the sea in a minute?

Solution:

Given:

Depth of river, h = 3 m

Width of river, b = 40 m

Rate of water flow = 2km per hour = 2000m/60min = 100/3 m/min

Now, volume of water flowed in 1 min = (100/3) × 40 × 3 = 4000m3

Therefore, 4000 m3 water will fall into the sea in a minute.

## Exercise 13.6 Page No: 230

1. The circumference of the base of the cylindrical vessel is 132cm, and its height is 25cm.

How many litres of water can it hold? (1000 cm3= 1L) (Assume π = 22/7)

Solution:

Circumference of the base of cylindrical vessel = 132 cm

Height of vessel, h = 25 cm

Let r be the radius of the cylindrical vessel.

Step 1: Find the radius of the vessel.

We know that the circumference of the base = 2πr, so

2πr = 132 (given)

r = (132/(2 π))

r = 66×7/22 = 21

Step 2: Find the volume of the vessel.

Formula: Volume of cylindrical vessel = πr2h

= (22/7)×212×25

= 34650

Therefore, the volume is 34650 cm3

Since 1000 cm3 = 1L

So, Volume = 34650/1000 L= 34.65L

Therefore, the vessel can hold 34.65 litres of water.

2. The inner diameter of a cylindrical wooden pipe is 24cm, and its outer diameter is 28 cm. The length of the pipe is 35cm. Find the mass of the pipe, if 1cm3 of wood has a mass of 0.6g. (Assume π = 22/7)

Solution:

Inner radius of cylindrical pipe, say r1 = diameter1/ 2 = 24/2 cm = 12cm

Outer radius of cylindrical pipe, say r2 = diameter2/ 2 = 28/2 cm = 14 cm

Height of pipe, h = Length of pipe = 35cm

Now, the Volume of pipe = π(r22-r12)h cm3

Substitute the values.

Volume of pipe = 110×52 cm3 = 5720 cm3

Since Mass of 1 cm3 wood = 0.6 g

Mass of 5720 cm3 wood = (5720×0.6) g = 3432 g or 3.432 kg.

3. A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5cm and width 4cm, having a height of 15 cm and (ii) a plastic cylinder with a circular base of diameter 7cm and height 10cm. Which container has greater capacity, and by how much? (Assume π=22/7)

Solution:

(i) Tin can will be cuboidal in shape.

Dimensions of the tin can are

Length, l = 5 cm

Height, h = 15 cm

Capacity of tin can = l×b×h= (5×4×15) cm3 = 300 cm3

(ii) Plastic cylinder will be cylindrical in shape.

Dimensions of the plastic can are

Radius of the circular end of plastic cylinder, r = 3.5cm

Height , H = 10 cm

Capacity of the plastic cylinder = πr2H

Capacity of the plastic cylinder = (22/7)×(3.5)2×10 = 385

Capacity of the plastic cylinder is 385 cm3

From the results of (i) and (ii), the plastic cylinder has more capacity.

Difference in capacity = (385-300) cm3 = 85cm3

4. If the lateral surface of a cylinder is 94.2cm2 and its height is 5cm, then find

(i) radius of its base (ii) its volume.[Use π= 3.14]

Solution:

CSA of cylinder = 94.2 cm2

Height of cylinder, h = 5cm

(i) Let the radius of the cylinder be r.

Using the CSA of the cylinder, we get

2πrh = 94.2

2×3.14×r×5 = 94.2

r = 3

(ii) Volume of cylinder

The formula for the volume of the cylinder = πr2h

Now, πr2h = (3.14×(3)2×5) (using the value of r from (i))

= 141.3

Volume is 141.3 cm3

5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10m deep. If the cost of painting is at the rate of Rs 20 per m2, find

(i) inner curved surface area of the vessel

(iii) capacity of the vessel

(Assume π = 22/7)

Solution:

(i) Rs 20 is the cost of painting 1 m2 area.

Rs 1 is the cost to paint 1/20 m2 area.

So, Rs 2200 is the cost of painting = (1/20×2200) m2

= 110 m2 area

The inner surface area of the vessel is 110m2.

(ii) Radius of the base of the vessel, let us say r.

Height (h) = 10 m and

Surface area formula = 2πrh

Using the result of (i),

2πrh = 110 m2

2×22/7×r×10 = 110

r = 1.75

(iii) Volume of vessel formula = πr2h

Here r = 1.75 and h = 10

Volume = (22/7)×(1.75)2×10 = 96.25

The volume of vessel is 96.25 m3

Therefore, the capacity of the vessel is 96.25 m3 or 96250 litres.

6. The capacity of a closed cylindrical vessel of height 1m is 15.4 litres. How many square metres of the metal sheet would be needed to make it? (Assume π = 22/7)

Solution:

Height of cylindrical vessel, h = 1 m

Capacity of cylindrical vessel = 15.4 litres = 0.0154 m3

Let r be the radius of the circular end.

Now,

Capacity of cylindrical vessel = (22/7)×r2×1 = 0.0154

After simplifying, we get r = 0.07 m

Again, the total surface area of the vessel = 2πr(r+h)

= 2×22/7×0.07(0.07+1)

= 0.44×1.07

= 0.4708

Total surface area of the vessel is 0.4708 m2

Therefore, 0.4708 m2 of the metal sheet would be required to make the cylindrical vessel.

7. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm, and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite. (Assume π = 22/7)

Solution:

Radius of pencil, r1 = 7/2 mm = 0.7/2 cm = 0.35 cm

Radius of graphite, r2 = 1/2 mm = 0.1/2 cm = 0.05 cm

Height of pencil, h = 14 cm

Formula to find the volume of wood in pencil = (r12-r22)h cubic units

Substituting values, we have,

= [(22/7)×(0.352-0.052)×14]

= 44×0.12

= 5.28

This implies that the volume of wood in pencil = 5.28 cm3

Again,

Volume of graphite = r22h cubic units

Substituting the values, we have,

= (22/7)×0.052×14

= 44×0.0025

= 0.11

So, the volume of graphite is 0.11 cm3.

8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7cm. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients? (Assume π = 22/7)

Solution:

Diameter of the cylindrical bowl = 7 cm

Radius of the cylindrical bowl, r = 7/2 cm = 3.5 cm

Bowl is filled with soup to a height of4cm, so h = 4 cm

Volume of the soup in one bowl= πr2h

(22/7)×3.52×4 = 154

Volume of the soup in one bowl is 154 cm3

Therefore,

Volume of the soup given to 250 patients = (250×154) cm3= 38500 cm3

= 38.5litres.

## Exercise 13.7 Page No: 233

1. Find the volume of the right circular cone with

(i) radius 6cm, height 7 cm (ii) radius 3.5 cm, height 12 cm (Assume π = 22/7)

Solution:

Volume of cone = (1/3) πr2h cube units

Where r be radius and h be the height of the cone

(i) Radius of the cone, r = 6 cm

Height of the cone, h = 7cm

Ley V be the volume of the cone, so we have

V = (1/3)×(22/7)×36×7

= (12×22)

= 264

The volume of the cone is 264 cm3.

(ii) Radius of the cone, r = 3.5cm

Height of the cone, h = 12cm

Volume of the cone = (1/3)×(22/7)×3.52×7 = 154

Hence,

The volume of the cone is 154 cm3.

2. Find the capacity in litres of a conical vessel with

(i) radius 7cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm

(Assume π = 22/7)

Solution:

(i) Radius of the cone, r =7 cm

Slant height of the cone, l = 25 cm

or h = 24

Height of the cone is 24 cm

Now,

Volume of the cone, V = (1/3) πr2h (formula)

V = (1/3)×(22/7) ×72×24

= (154×8)

= 1232

So, the volume of the vessel is 1232 cm3

Therefore, the capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm3)

= 1.232 Liters.

(ii) Height of the cone, h = 12 cm

Slant height of the cone, l = 13 cm

r = 5

Hence, the radius of the cone is 5 cm.

Now, Volume of the cone, V = (1/3)πr2h

V = (1/3)×(22/7)×52×12 cm3

= 2200/7

Volume of the cone is 2200/7 cm3

Now, Capacity of the conical vessel= 2200/7000 litres (1L = 1000 cm3)

= 11/35 litres

3. The height of a cone is 15cm. If its volume is 1570cm3, find the diameter of its base. (Use π = 3.14)

Solution:

Height of the cone, h = 15 cm

Volume of cone =1570 cm3

Let r be the radius of the cone

As we know, volume of the cone, V = (1/3) πr2h

So, (1/3) πr2h = 1570

(1/3)×3.14×r2 ×15 = 1570

r2 = 100

r = 10

Radius of the base of the cone 10 cm.

4. If the volume of a right circular cone of height 9cm is 48πcm3, find the diameter of its base.

Solution:

Height of cone, h = 9cm

Volume of cone =48π cm3

Let r be the radius of the cone.

As we know, volume of the cone, V = (1/3) πr2h

So, 1/3 π r2(9) = 48 π

r2 = 16

r = 4

Radius of the cone is 4 cm.

So, diameter = 2×Radius = 8

Thus, diameter of the base is 8cm.

5. A conical pit of a top diameter 3.5m is 12m deep. What is its capacity in kilolitres?

(Assume π = 22/7)

Solution:

Diameter of conical pit = 3.5 m

Radius of conical pit, r = diameter/ 2 = (3.5/2)m = 1.75m

Height of pit, h = Depth of pit = 12m

Volume of cone, V = (1/3) πr2h

V = (1/3)×(22/7) ×(1.75)2×12 = 38.5

Volume of the cone is 38.5 m3

Hence, capacity of the pit = (38.5×1) kiloliters = 38.5 kiloliters.

6. The volume of a right circular cone is 9856cm3. If the diameter of the base is 28cm, find

(i) height of the cone

(ii) slant height of the cone

(iii) curved surface area of the cone

(Assume π = 22/7)

Solution:

Volume of a right circular cone = 9856 cm3

Diameter of the base = 28 cm

(i) Radius of cone, r = (28/2) cm = 14 cm

Let the height of the cone be h

Volume of cone, V = (1/3) πr2h

(1/3) πr2h = 9856

(1/3)×(22/7) ×14×14×h = 9856

h = 48

The height of the cone is 48 cm.

Slant height of the cone is 50 cm.

(iii) curved surface area of cone = πrl

= (22/7)×14×50

= 2200

Curved surface area of the cone is 2200 cm2.

7. A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Solution:

Height (h)= 12 cm

Radius (r) = 5 cm, and

Slant height (l) = 13 cm

Volume of cone, V = (1/3) πr2h

V = (1/3)×π×52×12

= 100π

Volume of the cone so formed is 100π cm3.

8. If the triangle ABC in Question 7 is revolved about the side 5cm, then find the volume of the solids so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Solution:

A right-angled ΔABC is revolved about its side 5cm, a cone will be formed of radius as 12 cm, height as 5 cm, and slant height as 13 cm.

Volume of cone = (1/3) πr2h, where r is the radius and h is the height of the cone.

= (1/3)×π×12×12×5

= 240 π

The volume of the cones formed is 240π cm3.

So, the required ratio = (the result of question 7) / (the result of question 8) = (100π)/(240π) = 5/12 = 5:12.

9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas.

(Assume π = 22/7)

Solution:

Radius (r) of heap = (10.5/2) m = 5.25

Height (h) of heap = 3m

Volume of heap = (1/3)πr2h

= (1/3)×(22/7)×5.25×5.25×3

= 86.625

The volume of the heap of wheat is 86.625 m3.

Again,

= (22/7)×5.25×6.05

= 99.825

Therefore, the area of the canvas is 99.825 m2.

## Exercise 13.8 Page No: 236

1. Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m

(Assume π =22/7)

Solution:

(i) Radius of the sphere, r = 7 cm

Using, Volume of the sphere = (4/3) πr3

= (4/3)×(22/7)×73

= 4312/3

Hence, volume of the sphere is 4312/3 cm3

(ii) Radius of the sphere, r = 0.63 m

Using, volume of sphere = (4/3) πr3

= (4/3)×(22/7)×0.633

= 1.0478

Hence, volume of the sphere is 1.05 m3 (approx).

2. Find the amount of water displaced by a solid spherical ball of diameter

(i) 28 cm (ii) 0.21 m

(Assume π =22/7)

Solution:

(i) Diameter = 28 cm

Radius, r = 28/2 cm = 14cm

Volume of the solid spherical ball = (4/3) πr3

Volume of the ball = (4/3)×(22/7)×143 = 34496/3

Hence, volume of the ball is 34496/3 cm3

(ii) Diameter = 0.21 m

Radius of the ball =0.21/2 m= 0.105 m

Volume of the ball = (4/3 )πr3

Volume of the ball = (4/3)× (22/7)×0.1053 m3

Hence, volume of the ball = 0.004851 m3

3. The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3? (Assume π=22/7)

Solution:

Given,

Diameter of a metallic ball = 4.2 cm

Radius(r) of the metallic ball, r = 4.2/2 cm = 2.1 cm

Volume formula = 4/3 πr3

Volume of the metallic ball = (4/3)×(22/7)×2.1 cm3

Volume of the metallic ball = 38.808 cm3

Now, using the relationship between density, mass and volume,

Density = Mass/Volume

Mass = Density × volume

= (8.9×38.808) g

= 345.3912 g

Mass of the ball is 345.39 g (approx).

4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Solution:

Let the diameter of the earth be “d”. Therefore, the radius of the earth will be d/2.

Diameter of the moon will be d/4, and the radius of the moon will be d/8.

Find the volume of the moon.

Volume of the moon = (4/3) πr3 = (4/3) π (d/8)3 = 4/3π(d3/512)

Find the volume of the earth

Volume of the earth = (4/3) πr3= (4/3) π (d/2)3 = 4/3π(d3/8)

Fraction of the volume of the earth is the volume of the moon

Answer: Volume of the moon is of the 1/64 volume of the earth.

5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22/7)

Solution:

Diameter of the hemispherical bowl = 10.5 cm

Radius of the hemispherical bowl, r = 10.5/2 cm = 5.25 cm

Formula for volume of the hemispherical bowl = (2/3) πr3

Volume of the hemispherical bowl = (2/3)×(22/7)×5.253 = 303.1875

Volume of the hemispherical bowl is 303.1875 cm3

Capacity of the bowl = (303.1875)/1000 L = 0.303 litres(approx.)

Therefore, the hemispherical bowl can hold 0.303 litres of milk.

6. A hemispherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume π = 22/7)

Solution:

Inner Radius of the tank, (r ) = 1m

Outer Radius (R ) = 1.01m

Volume of the iron used in the tank = (2/3) π(R3– r3)

Put values,

Volume of the iron used in the hemispherical tank = (2/3)×(22/7)×(1.013– 13) = 0.06348

So, volume of the iron used in the hemispherical tank is 0.06348 m3.

7. Find the volume of a sphere whose surface area is 154 cm2. (Assume π = 22/7)

Solution:

Let r be the radius of a sphere.

Surface area of the sphere = 4πr2

4πr2 = 154 cm2 (given)

r2 = (154×7)/(4 ×22)

r = 7/2

Now,

Volume of the sphere = (4/3) πr3

8. A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of Rs. 4989.60. If the cost of white-washing is 20 per square metre, find the

(i) inside surface area of the dome (ii) volume of the air inside the dome

(Assume π = 22/7)

Solution:

(i) Cost of whitewashing the dome from inside = Rs 4989.60

Cost of whitewashing 1m2 area = Rs 20

CSA of the inner side of dome = 498.96/2 m2  = 249.48 m2

(ii) Let the inner radius of the hemispherical dome be r.

CSA of the inner side of dome = 249.48 m2 (from (i))

Formula to find CSA of a hemisphere = 2πr2

2πr2 = 249.48

2×(22/7)×r2 = 249.48

r2 = (249.48×7)/(2×22)

r2 = 39.69

r = 6.3

So, the radius is 6.3 m.

Volume of air inside the dome = Volume of hemispherical dome

Using the formula, the volume of the hemisphere = 2/3 πr3

= (2/3)×(22/7)×6.3×6.3×6.3

= 523.908

= 523.9(approx.)

Answer: The volume of air inside the dome is 523.9 m3.

9. Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the

(i) radius r’ of the new sphere,

(ii) ratio of Sand S’.

Solution:

Volume of the solid sphere = (4/3)πr3

Volume of twenty seven solid sphere = 27×(4/3)πr3 = 36 π r3

(i) New solid iron sphere radius = r’

Volume of this new sphere = (4/3)π(r’)3

(4/3)π(r’)3 = 36 π r3

(r’)3 = 27r3

r’= 3r

Radius of the new sphere will be 3r (thrice the radius of the original sphere)

(ii) Surface area of the iron sphere of radius r, S =4πr2

Surface area of the iron sphere of radius r’= 4π (r’)2

Now

S/S’ = (4πr2)/( 4π (r’)2)

S/S’ = r2/(3r’)2 = 1/9

The ratio of S and S’ is 1: 9.

10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm3) is needed to fill this capsule? (Assume π = 22/7)

Solution:

Diameter of the capsule = 3.5 mm

Radius of the capsule, say r = diameter/ 2 = (3.5/2) mm = 1.75mm

Volume of the spherical capsule = 4/3 πr3

Volume of the spherical capsule = (4/3)×(22/7)×(1.75)3 = 22.458

Answer: The volume of the spherical capsule is 22.46 mm3.

## Exercise 13.9 Page No: 236

1. A wooden bookshelf has external dimensions as follows: Height = 110cm, Depth = 25cm,

Breadth = 85cm (see fig. 13.31). The thickness of the plank is 5cm everywhere. The external faces are to be polished, and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.

Solution:

External dimensions of book self:

Length, l = 85cm

Height, h = 110 cm

External surface area of the shelf while leaving out the front face of the shelf.

= lh+2(lb+bh)

= [85×110+2(85×25+25×110)] = (9350+9750) = 19100

External surface area of the shelf is 19100 cm2

Area of front face = [85×110-75×100+2(75×5)] = 1850+750

So, the area is 2600 cm2

Area to be polished = (19100+2600) cm2 = 21700 cm2.

Cost of polishing 1 cm2 area = Rs 0.20

Cost of polishing 21700 cm2 area Rs. (21700×0.20) = Rs 4340

Dimensions of the row of the bookshelf

Length(l) = 75 cm

Breadth (b) = 20 cm and

Height(h) = 30 cm

Area to be painted in one row= 2(l+h)b+lh = [2(75+30)× 20+75×30] = (4200+2250) = 6450

So, the area is 6450 cm2.

Area to be painted in 3 rows = (3×6450)cm2 = 19350 cm2.

Cost of painting 1 cm2 area = Rs. 0.10

Cost of painting 19350 cm2 area = Rs (19350 x 0.1) = Rs 1935

Total expense required for polishing and painting = Rs. (4340+1935) = Rs. 6275

Answer: The cost for polishing and painting the surface of the bookshelf is Rs. 6275.

2. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in fig. 13.32. Eight such spheres are used forth is the purpose and are to be painted silver. Each support is a cylinder of radius 1.5cm and height 7cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2, and black paint costs 5 paise per cm2.

Solution:

Diameter of the wooden sphere = 21 cm

Radius of the wooden sphere, r = diameter/ 2 = (21/2) cm = 10.5 cm

Formula: Surface area of the wooden sphere = 4πr2

= 4×(22/7)×(10.5)2 = 1386

So, the surface area is 1386 cm3

Radius of the circular end of cylindrical support = 1.5 cm

Height of the cylindrical support = 7 cm

Curved surface area = 2πrh

= 2×(22/7)×1.5×7 = 66

So, CSA is 66 cm2

Now,

Area of the circular end of cylindrical support = πr2

= (22/7)×1.52

= 7.07

Area of the circular end is 7.07 cm2

Again,

Area to be painted silver = [8 ×(1386-7.07)] = 8×1378.93 = 11031.44

Area to be painted is 11031.44 cm2

Cost for painting with silver colour = Rs(11031.44×0.25) =Rs 2757.86

Area to be painted black = (8×66) cm2 = 528 cm2

Cost for painting with black colour =Rs (528×0.05) = Rs26.40

Therefore, the total painting cost is

= Rs(2757.86 +26.40)

= Rs 2784.26

3. The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?

Solution:

Let the diameter of the sphere be “d”.

Radius of the sphere, r1 = d/2

New radius of the sphere, say r2 = (d/2)×(1-25/100) = 3d/8

Curved surface area of the sphere, (CSA)1 = 4πr12 = 4π×(d/2)2 = πd2 …(1)

Curved surface area of the sphere when the radius is decreased (CSA)2 = 4πr22 = 4π×(3d/8)2 = (9/16)πd2 …(2)

From equations (1) and (2), we have

Decrease in surface area of sphere = (CSA)1 – (CSA)2

= πd2 – (9/16)πd2

= (7/16)πd2

= (7d2/16d2)×100 = 700/16 = 43.75% .

Therefore, the percentage decrease in the surface area of the sphere is 43.75%.

 Also Access NCERT Exemplar for Class 9 Maths Chapter 13 CBSE Notes for Class 9 Maths Chapter 13

### NCERT Solutions for Class 9 Maths Chapter 13

NCERT Solutions Class 9 Maths Chapter 13 help students find the surface areas and volumes of cuboids and cylinders, cones, and spheres. This chapter explains how the area is found by multiplying the length and breadth of various objects. NCERT Solutions Class 9 Maths Chapter 13 include all types of exercise problems from basic to advance level questions to prepare students for competitive examinations. NCERT Solutions for Class 9 Maths Chapter 13 also include the activities explained in a better way for the understanding of the concepts before solving the questions. NCERT Solutions for Class 9 Maths Chapter 13 include the following topics:

• Surface area of a cuboid and a cube
• Surface area of a right circular cylinder
• Surface area of a right circular cone
• Surface area of a sphere
• Volume of a cuboid
• Volume of a cylinder
• Volume of a right circular cone
• Volume of a sphere
• NCERT Solutions for Class 9 Chapter 13 explain the formation of various geometrical objects.
• They signify the importance of different formulas in finding the surface area and volume of various objects.
• Details of the cuboid, cube, right circular cone, cylinder, hemisphere, and sphere are given.

### Key Features of NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes

• List the important formula to find surface areas and volumes of the cube, cuboid, cylinder, cone, and sphere.
• They are designed for the students to remember the formulas and apply them relevantly.
• These solutions will be useful for CBSE exams, competitive exams, and even for Maths Olympiads.
• Answers are aimed at providing an effortless solution to finding the surface area and volumes.
• Provide completely solved solutions to all the questions present in the respective NCERT textbooks.

Apart from the questions from the NCERT Textbook, students are also advised to solve questions from other study materials. It will not only provide a strong conceptual knowledge but also improves their abilities to solve difficult questions effortlessly.

Disclaimer:

Dropped Topics – 13.1 Introduction, 13.2 Surface area of a cuboid and cube, 13.3 Surface area of right circular cylinder, 13.6 Volume of cuboid and 13.7 Volume of cylinder.

## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 13

### Why should we follow NCERT Solutions for Class 9 Maths Chapter 13?

NCERT Solutions for Class 9 Maths Chapter 13 are the best reference materials that offer complete and quality information about different Maths concepts. The questions that have been given in the solutions have been solved in an easy-to-remember format, which further helps students to understand and remember the answers clearly. To score good marks, practising these solutions for Class 9 Maths can help to a great extent.

### How BYJU’S NCERT Solutions for Class 9 Maths Chapter 13 help the students in preparing for CBSE exams?

Our solution module uses various examples and diagrams to explain the questions wherever necessary. Students aiming to secure an excellent score in CBSE  examinations must solve the NCERT Solutions for Class 9. These NCERT Solutions for Class 9 Maths Chapter 13 help the students in gaining a better knowledge of the topics covered. Solving the questions from each exercise will ensure that the students score high marks in the CBSE exams.

### What are the topics covered under NCERT Solutions for Class 9 Maths Chapter 13?

NCERT Solutions for Class 9 Maths Chapter 13 includes the following topics:
1. Surface area of a cuboid and a cube
2. Surface area of a right circular cylinder
3. Surface area of a right circular cone
4. Surface area of a sphere
5. Volume of a cuboid
6. Volume of a cylinder
7. Volume of a right circular cone
8. Volume of a sphere