NCERT Solutions Class 9 Maths Chapter 13-Surface Areas and Volume

Exercise 13.1

Question 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:
The area of the sheet required for making the box.
The cost of sheet for it, if a sheet measuring 1m^2 costs Rs. 20.

Solution: Given in the question: length (l) of box = 1.5m
Breadth (b) of box = 1.25 m
Depth (h) of box = 0.65m
Box is to be open at top
Area of sheet required.
=\ 2lh+2bh+lb\\ =\ \left[2\times1.5\times0.65+2\times1.25\times0.65+1.5\times1.25\right] m^2\\ =\ (1.95+1.625+1.875)\ m^2\ =\ 5.45\ m^2

(ii) Cost of sheet per \ m^2 area = Rs.20.
Cost of sheet of 5.45\ m^2 area = Rs (5.45×20)
= Rs.109

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

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 white washed.
Total area to be white washed = Area of walls + Area of ceiling of room
=\ 2lh+2bh+lb\\ =\ [2\times5\times3+2\times4\times3+5\times4]\\ =\ (30+24+20)\\ =\ 74\\ Area\ =\ 74\ m^2
Also,
Cost of white wash per m^2 area = Rs.7.50 (Given)
Cost of white washing 74\ m^2 area = Rs. (74×7.50)
= Rs. 555

Question 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 m^2 is Rs.15000, find the height of the hall.
[Hint: Area of the four walls = Lateral surface area.]

Solution:

Let length, breadth, and height of the rectangular hall be l, b, and h respectively.
Area of four walls is given by
=\ 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\ m^2
Cost of painting per square meter area = Rs.10
Cost of painting 250 h square meter area = Rs ( 250h × 10) = Rs.2500h
However, it is given that the cost of painting the walls is Rs. 15000.
15000 = 2500 h
Or h = 6
Therefore, the height of the hall is 6 m.

Question 4. The paint in a certain container is sufficient to paint an area equal to 9.375\ m^2 . 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 is given by
=\ 2(lb\ +bh+lb)\\ =\ \left[2\left(22.5\times10+10\times7.5+22.5\times7.5\right)\right] cm^2\\ =\ 2\left(225+75+168.75\right) cm^2\\ =\ \left(2\times468.75\right) cm^2\\ =\ 937.5\ cm^2
Let n bricks can be painted out by the paint of the container
Area of n bricks = \left(n\times937.5\right)cm^2\ =\ 937.5n\ cm^2
As per given instructions, area that can be painted by the paint of the container = 9.375\ m^2\ =\ 93750\ cm^2
So, we have, 93750 = 937.5n n = 100
Therefore, 100 bricks can be painted out by the paint of the container.

Question 5. A cubical box has each edge 10 cm and another cuboidal box is 12.5cm long, 10 cm wide and 8 cm high
Which box has the greater lateral surface area and by how much?
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
Breadth, b = 10cm
Height, h = 8 cm
Find the lateral surface area for both the figures
Lateral surface area of cubical box is given by
=\ 4\ \left(edge\right)^2\\ =\ 4\left(10\right)^2\\ =\ 400\ cm^2\ \ldots(1)
Lateral surface area of cuboidal box is given by
=\ 2[lh+bh]\\ =\ [2(12.5\times8+10\times8)]\\ =\ (2\times180)\ =\ 360
Therefore, Lateral surface area of cuboidal box is 360 cm^2\.\ \ldots(2)
From (1) and (2), lateral surface area of the cubical box is more than the lateral surface area of the cuboidal box. The difference between both the lateral surfaces is, 40 cm^2\\ .
(Lateral surface area of cubical box – Lateral surface area of cuboidal box
=400 cm^2 –360 cm^2 = 40 cm^2)

Find the total surface area for both the figures
The total surface area of the cubical box is given by
\ =\ 6\left(edge\right)^2\ =\ 6\left(10\ cm\right)^2\ =\ 600 cm^2 ldots(3)
Total surface area of cuboidal box
=\ 2[lh+bh+lb]\\ =\ [2(12.5\times8+10\times8+12.5\times100)]\\ =\ 610
This implies, Total surface area of cuboidal box is 610 cm^2 (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 10 cm^2.
Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm^2

Question 6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including 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 the 12 edges?

Solution:

Length of greenhouse, say l = 30cm
Breadth of greenhouse, say b = 25 cm
Height of greenhouse, say h = 25 cm
Total surface area of greenhouse = Area of the glass is given by
\ =\ 2[lb+lh+bh]\\ =\ [2(30\times25+30\times25+25\times25)]\\ =\ [2(750+750+625)]\\ =\ (2\times2125)\ =\ 4250
i)Total surface area of the glass is 4250\ cm^2

ii) From figure, tape is required along sides AB, BC, CD, DA, EF, FG, GH, HE AH, BE, DG, and CF.
Total length of tape is given by
\ =\ 4(l+b+h)\\ =\ [4(30+25+25)] (after\ substituting\ the\ values)\\ =\ 320
Therefore, 320 cm tape is required for all the 12 edges.

Question 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\ cm^2 , 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 is given by
=\ 2(lb+lh+bh)\\ =\ [2(25\times20+25\times5+20\times5)]\\ =\ [2(500+125+100)]\\ =\ 1450\ cm^2
Extra area required for overlapping is 1450\times\frac{5}{100}\ cm^2\\ =\ 72.5\ cm^2
While considering all over laps, total surface area of bigger box
= \left(1450+72.5\right)cm^2\ =\ 1522.5\ cm^2
Area of cardboard sheet required for 250 such bigger boxes
= \left(1522.5\times250\right)cm^2\ =\ 380625\ cm^2
Smaller Box:
Similarly, total surface area of smaller box is given by
\ =\ \left[2\left(15\times12+15\times5+12\times5\right)\right] cm^2\\ =\ \left[2\left(180+75+60\right)\right] cm^2\\ =\ \left(2\times315\right) cm^2\\ =\ 630\ cm^2
Therefore, extra area required for overlapping \ 630\times\frac{5}{100}\ cm^2=\ 31.5\ cm^2
Total surface area of 1 smaller box while considering all overlaps
=\ \left(630+31.5\right)cm^2\ =\ 661.5\ cm^2
Area of cardboard sheet required for 250 smaller boxes = \left(250\times661.5\right)cm^2\ =\ 165375\ cm^2

In Short:

Now, Total cardboard sheet required
=\ \left(380625+165375\right) cm^2\\ =\ 546000\ cm^2
Given: Cost of 1000 cm2 cardboard sheet = Rs. 4
Therefore, Cost of 546000 cm^2 cardboard sheet = Rs.\frac{546000\times4}{1000}\ =\ Rs.\ 2184
Therefore, the cost of cardboard required for supplying 250 boxes of each kind will be Rs. 2184.

Question 8. Praveen wanted to make a temporary shelter for her car, by making a box – like structure with tarpaulin that covers all the 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 much tarpaulin 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
Tarpaulin 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
=\ \left[2\left(4\times2.5+3\times2.5\right)+4\times3\right] m^2\\ =\ \left[2\left(10+7.5\right)+12\right] m^2\\ =\ 47 m^2
Therefore, 47\ m^2 tarpaulin will be required

Exercise 13.2

Question 1. The curved surface area of a right circular cylinder of height 14 cm is 88\ cm^2 . 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
Curved surface area of cylinder = 88\ cm^2
We know that, formula to find Curved surface area of cylinder is 2πrh.
So 2\pi rh\ =88\ cm^2 (r is the radius of the base of the cylinder)
2\times\left(\frac{22}{7}\right)\times r\times14\ =\ 88\ cm^2\\ 2r\ =\ 2\ cm\\ d\ =2\ cm
Therefore, the diameter of the base of the cylinder is 2 cm.

Question 2. It is required to make a closed cylindrical tank of height 1m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same? Assume π = \frac{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 = \frac{140}{2}cm = 70cm = 0.7m
Area of sheet required = Total surface are of tank = 2\pi r(r+h)\ unit square
=\ [2\times\left(\frac{22}{7}\right)\times0.7(0.7+1)]\\ =\ 7.48 square meters
Therefore, 7.48 square meters of the sheet are required.

Question 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 inner curved surface area,
outer curved surface area
total surface area
(Assume π=22/7)

Solution:

Let r_1 and r_2 Inner and outer radii of cylindrical pipe
r_1\ =\frac{4}{2}\ cm\ =\ 2\ cm\\ r_2\ =\frac{4.4}{2}\ cm\ =\ 2.2\ cm
Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm
curved surface area of outer surface of pipe = 2\pi r_1h\\ =\ 2\times\left(\frac{22}{7}\right)\times2\times77\ cm^2\\ =\ 968\ cm^2

curved surface area of outer surface of pipe = 2\pi r_2h\\ =\ 2\times\left(\frac{22}{7}\right)\times2.2\times77\ cm^2\\ =\ \left(22\times22\times2.2\right)cm^2 \\ =\ 1064.8\ cm^2

Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe.
=\ 2r_1h+2r_2h+2\pi\left(r_1^2-r_2^2\right)\\ =\ 9668+1064.8+2\times\left(\frac{22}{7}\right)\times(2.22-22)\\ =\ 2031.8+5.28\\ =\ 2038.08\ cm^2
Therefore, the total surface area of the cylindrical pipe is 2038.08\ cm^2.

Question 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 m^2 ? (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 = \frac{84}{2} cm = 42 cm
Now, CSA of roller = 2\pi rh\\ =\ 2\times\left(\frac{22}{7}\right)\times42\times120\\ =\ 31680\ cm^2\
Area of field = 500×CSA of roller
=\ \left(500\times31680\right)cm^2\\ =\ 15840000\ cm^2\\ =\ 1584\ m^2.
Therefore, area of playground is 1584\ m^2.

Question 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\ m^2.
(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 = \frac{diameter}{2} = \frac{50}{2} = 25cm = 0.25m
CSA of pillar is given by
=\ 2\pi rh\\ =\ 2\times\left(\frac{22}{7}\right)\times0.25\times3.5\\ =\ 5.5\ m^2
Cost of painting 1\ m^2 area = Rs. 12.50
Cost of painting 5.5\ m^2 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 m^2 is Rs 68.75.

Question 6. Curved surface area of a right circular cylinder is 4.4\ m^2 . If the radius of the base 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.
Radius of the base of cylinder, r = 0.7m
CSA of cylinder = 2πrh
CSA of cylinder = 4.4m^2
Equating both the equations, we have
2\times\left(\frac{22}{7}\right)\times0.7\times h\ =\ 4.4
Or h = 1
Therefore, the height of the cylinder is 1 m.

Question 7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find
its inner curved surface area,
The cost of plastering this curved surface at the rate of Rs. 40 per m^2.
(Assume π = 22/7)

Solution:

Inner radius of circular well, r = \frac{3.5}{2}m = 1.75m
Depth of circular well, say h = 10m
Inner curved surface area is given by
=\ 2\pi rh\\ =\ (2\times\left(\frac{22}{7}\right)\times1.75\times10)\\ =\ 110\ m^2
Therefore, the inner curved surface area of the circular well is 110\ m^2.

(ii) Cost of plastering 1\ m^2 area = Rs.40
Cost of plastering 110\ m^2 area = Rs (110 × 40)
= Rs.4400
Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

Question 8. In a hot water heating system, there is 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 =\frac{diameter}{2}=\frac{5}{2} 5/2 cm = 2.5cm = 0.025m
Now, CSA of cylindrical pipe = 2\pi rh , where r = radius and h = height of the cylinder
=\ 2\times(22/7)\times0.025\times28\ m^2\\ =\ 4.4m^2
The area of the radiating surface of the system is 4.4m^2.

Question 9. Find The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high.
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 = \frac{4.2}{2} m = 2.1m
The lateral or curved surface area of cylindrical tank is 2\pi rh\\ =\ 2\times(22/7)\times2.1\times4.5\ m^2\\ =\ \left(44\times0.3\times4.5\right)m^2\\ =\ 59.4\ m^2\
Therefore, CSA of tank is 59.4\ m^2.

Total surface area of tank = 2\pi r(r+h)\\ =\ 2\times\left(\frac{22}{7}\right)\times2.1\times(2.1+4.5)\\ =\ 44\times0.3\times6.6\\ =\ 87.12\ m^2
Now, Let S\ m^2 steel sheet be actually used in making the tank.
\ S\left(1\ -\frac{1}{12}\right)\ =\ 87.12\ m^2
This implies, S\ =\ 95.04\ m^2
Therefore, 95.04m^2 steel was used in actual while making such a tank.

Question 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 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 for covering the lampshade. (Assume π = 22/7)

Solution:

Say h = height of the frame of lampshade, looks like cylindrical shape
r = radius
Total height is h = (2.5+30+2.5) cm = 35cm and
r = \frac{20}{2} cm = 10cm
Use curved surface area formula to find the cloth required for covering the lampshade which is 2\pi rh\\ =\ \left(2\times\left(\frac{22}{7}\right)\times10\times35\right)cm^2\\ =\ 2200\ cm^2
Hence, 2200\ cm^2 cloth is required for covering the lampshade.

Question 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:

Radius of the circular end of cylindrical penholder, r = 3cm
Height of penholder, h = 10.5cm
Surface area of a penholder = CSA of pen holder + Area of base of penholder
=\ 2\pi rh+\pi r^2\\ =\ 2\times\left(\frac{22}{7}\right)\times3\times10.5+\left(\frac{22}{7}\right)\times32=\frac{1584}{7}
Therefore, Area of cardboard sheet used by one competitor is \frac{1584}{7}cm^2
So, Area of cardboard sheet used by 35 competitors = 35\times\frac{1584}{7}\ =\ 7920\ cm^2
Therefore, 7920\ cm^2 cardboard sheet will be needed for the competition.

Exercise 13.3

Question 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= \frac{22}{7})

Solution:

Radius of the base of cone =\frac{diameter}{2} = \frac{10.5}{2} cm = 5.25cm
Slant height of cone, say l = 10 cm
CSA of cone is = \pi rl\\ =\frac{22}{7} \times5.25\times10\ =\ 165\ cm^2
Therefore, the curved surface area of the cone is 165\ cm^2.

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

Solution:

Radius of cone, r = \frac{24}{2}m = 12m
Slant height, l = 21 m
Formula: Total Surface area of the cone = \pi r(l+r)
Total Surface area of the cone = \left(\frac{22}{7}\right)\times12\times\left(21+12\right)m^2\\ =\ 1244.57m^2

Question 3. Curved surface area of a cone is 308\ cm^2 and its slant height is 14 cm. Find
radius of the base and
Total surface area of the cone.
(Assume π = 22/7)

Solution:

Slant height of cone, l = 14 cm
Let the radius of the cone be r.
We know, CSA of cone = πrl
Given: Curved surface area of a cone is 308\ cm^2 \\ (308\ )\ =\ \left(\frac{22}{7}\right)\times r\times14\\ 308\ =\ 44\ r\\ r\ =\frac{308}{44} =\ 7\ cm
Radius of a cone base is 7 cm.

Total surface area of cone = CSA of cone + Area of base (πr2)
Total surface area of cone = 308+\left(\frac{22}{7}\right)\times72\ =\ 308+154\ =\ 462\ cm^2
Therefore, the total surface area of the cone is 462\ cm^2.

Question 4. A conical tent is 10 m high and the radius of its base is 24 m. Find
Slant height of the tent.
Cost of the canvas required to make the tent, if the cost of 1\ m^2 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.
In right triangle ABO, we have
AB^2\ =\ AO^2+BO^2(using\ Pythagoras\ theorem)\\ l^2\ =\ h^2+r^2\\ =\ \left(10\right)^2+\left(24\right)^2\\ =\ 676\\ l\ =\ 26\ m
Therefore, the slant height of the tent is 26 m.

CSA of tent = \pi rl \\ =\frac{22}{7}\times24\times26\ m^2
Cost of 1\ m^2 canvas = Rs 70
Cost of \frac{13728}{7} m^2 canvas is equal to Rs \frac{13728}{7}×70 = Rs 137280
Therefore, the cost of the canvas required to make such a tent is Rs 137280.

Question 5. What length of tarpaulin 3 m wide will be required to make 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 conical tent, h = 8m
Radius of base of tent, r = 6m
Slant height of tent,
l^2\ =\ (r^2+h^2)\\ l^2\ =\ (62+82)\ =\ (36+64)\ =\ (100)\\ or\ l\ =\ 10\ m
Again, CSA of conical tent = \pi rl\\ =\ \left(3.14\times6\times10\right)m^2\\ =\ 188.4\ m^2
Let the length of tarpaulin sheet required be L
As 20 cm will be wasted, therefore,
Effective length will be (L-0.2m).
Breadth of tarpaulin = 3m (given)
Area of sheet = CSA of 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.

Question 6.The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210\ per\ 100\ m^2. (Assume π = 22/7)

Solution:

Slant height of conical tomb, l = 25m
Base radius, r = \frac{diameter}{2} = \frac{14}{2}m = 7m
CSA of conical tomb = \pi rl\\ =\ (22/7)\times7\times25\ =\ 550
CSA of conical tomb= \ 550m^2
Cost of white-washing 550\ m^2 area, which is Rs (210×550)/100
= Rs. 1155
Therefore, cost will be Rs. 1155 while white-washing tomb.

Question 7. A joker’s cap is in the form of 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 conical cap, r = 7 cm
Height of conical cap, h = 24cm
Slant height,
\ l^2\ =\ (r^2+h^2)\\ =\ (72+242)\\ =\ (49+576)\\ =\ (625)
Or l = 25 cm
CSA of 1 conical cap
=\pi rl\\ =\ (22/7)\times7\times25\\ =\ 550\ cm^2
CSA of 10 caps = \left(10\times550\right)cm^2\ =\ 5500\ cm^2
Therefore, the area of the sheet required to make 10 such caps is 5500\ cm^2.

Question 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 m^2 , what will be the cost of painting all these cones? (Use π = 3.14 and take \sqrt{1.04}\ =1.02)
Answer
Given in the question:
Radius of cone, r = \frac{diameter}{2}=\frac{40}{2}cm = 20cm = 0.2 m
Height of cone, h = 1m
Slant height of cone is l, and l^2\ =\ (r^2+h^2)
Using given values, l^2\ =\ (0.22+12)
= (1.04)
Or l = 1.02 m
Slant height of the cone is 1.02 m
Now,
CSA of each cone is given by
=\ \pi rl \\ =\ (3.14\times0.2\times1.02)\\ =\ 0.64056\ m
CSA of 50 such cones = (50×0.64056) = 32.028
CSA of 50 such cones = 32.028\ m^2
Again,
Cost of painting 1\ m^2 area = Rs 12 (given)
Cost of painting 32.028 m^2 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

Question 1. Find the surface area of a sphere of radius:
10.5cm
5.6cm
14cm
(Assume π=22/7)

Solution:

Formula: Surface area of sphere (SA) = 4\pi r^2
Radius of sphere, r = 10.5 cm
SA = 4\times\left(\frac{22}{7}\right)\times10.52\ =\ 1386
Surface area of sphere is 1386\ cm^2

Radius of sphere, r = 5.6cm
Using formula, SA = 4\times\left(\frac{22}{7}\right)\times5.62\ =\ 394.24
Surface area of sphere is 394.24\ cm^2

Radius of sphere, r = 14cm
SA = 4\pi r^2 \\ =\ 4\times\left(\frac{22}{7}\right)\times\left(14\right)^2\\ =\ 2464
Surface area of sphere is 2464\ cm^2

Question 2. Find the surface area of a sphere of diameter:
14cm
21cm
3.5cm
(Assume π = 22/7)

Solution:

Radius of sphere, r = \frac{diameter}{2}= \frac{14}{2} cm = 7 cm
Formula for Surface area of sphere = 4\pi r^2\\ =\ 4\times\left(\frac{22}{7}\right)\times72\ =\ 616
Surface area of a sphere is 616\ cm^2
Radius (r) of sphere =\frac{21}{2} = 10.5 cm
Surface area of sphere = 4\pi r^2\\ =\ 4\times\left(\frac{22}{7}\right)\times10.52\ =\ 1386\
Surface area of a sphere is 1386\ cm^2
Therefore, the surface area of a sphere having diameter 21cm is 1386\ cm^2

Radius(r) of sphere = \frac{3.5}{2} = 1.75 cm
Surface area of sphere = 4\pi r^2\\ =\ 4\times\left(\frac{22}{7}\right)\times1.752\ =\ 38.5\
Surface area of a sphere is 38.5\ cm^2

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

Solution:

Radius of hemisphere, r = 10cm
Formula: Total surface area of hemisphere = 3\pi r^2\\ =\ 3\times3.14\times102\ =\ 942
The total surface area of given hemisphere is 942\ cm^2.

Question 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 r_1 and r_2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So
r_1\ =\ 7cm\\ r_2\ =\ 14\ cm
Now, Required ratio = \frac{initial surface area}{Surface area after pumping air into balloon}
\frac{\ 4\pi r_1^2}{4\pi r_2^2} =\ \left(\frac{r_1}{r_2}\right)^2\\ =\ \left(\frac{7}{14}\right)^2\ =\ \left(\frac{1}{2}\right)^2\ =\frac{1}{4}
Therefore, the ratio between the surface areas is 1:4.

Question 5. A hemispherical bowl made of brass has an inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100\ cm^2 . (Assume π = 22/7)

Solution:
Inner radius of hemispherical bowl, say r = \frac{diameter}{2} = \frac{10.5}{2} cm = 5.25 cm
Formula for Surface area of hemispherical bowl = 2\pi r^2\\ =\ 2\times\left(\frac{22}{7}\right)\times\left(5.25\right)^2\ =\ 173.25
Surface area of hemispherical bowl is 173.25\ cm^2
Cost of tin-plating 100\ cm^2 area = Rs 16
Cost of tin-plating 1\ cm^2 area = Rs \frac{16}{100}
Cost of tin-plating 173.25\ cm^2 area = Rs.\frac{(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\ cm^2 is Rs 27.72.

Question 6. Find the radius of a sphere whose surface area is 154\ cm^2 . (Assume π = 22/7)

Solution:

Let the radius of the sphere be r.
Surface area of sphere = 154 (given)
Now,
4\pi r^2\ =\ 154\\ r^2\ =\ (154\times7)/(4\times22)\ =\ \left(\frac{49}{4}\right)\\ r\ =\ \left(\frac{7}{2}\right)\ =\ 3.5
The radius of the sphere is 3.5 cm

Question 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 diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement)
Radius of earth =\frac{d}{2}
Radius of moon = \frac{1}{2}\times\frac{d}{4}\ =\frac{d}{8}
Surface area of moon = 4\pi\left(\frac{d}{8}\right)^2
Surface area of earth = 4\pi\left(\frac{d}{2}\right)^2
Now,
Ratio of their Surface areas = \frac{4\pi\left(\frac{d}{8}\right)^2}{4\pi\left(\frac{d}{2}\right)^2}=\frac{4}{64}=\frac{1}{16}
Hence, the Ratio of their Surface areas is 1:16.

Question 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 in the question:
Inner radius of hemispherical bowl = 5cm
Thickness of the bowl = 0.25 cm
Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm
Formula for outer CSA of hemispherical bowl = 2\pi r^2 , where r is radius of hemisphere
= 2\times\left(\frac{22}{7}\right)\times\left(5.25\right)^2\ =\ 173.25\ cm^2
Therefore, the outer curved surface area of the bowl is 173.25\ cm^2 .

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

(i)Find surface area of the sphere,
(ii)curved surface area of the cylinder,
(iii)Ratio of the areas obtained in  and .

Solution:

Surface area of sphere = 4\pi r^2 , where r is the radius of sphere
Height of cylinder, h\ =\ r+r\ =2r
Radius of cylinder = r
CSA of cylinder formula = 2\pi rh\ =\ 2\pi r(2r) (using value of h)
= 4\pi r^2
Ratio between areas = \frac{Surface area of sphere}{CSA of Cylinder }=\frac{4\pi r^2}{4\pi r^2\ }=\frac{1}{1}
Now, the Ratio of the areas obtained in (i) and (ii) is 1:1

Exercise 13.5

Question 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\ cm^3
Now, volume of 12 such matchboxes = (15×12)\ cm^3 = 180\ cm^3
Therefore, the volume of 12 matchboxes is 180cm^3.

Question 2. A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold? ( 1\ m^3=\ 1000\ l )

Solution:

Formula to find volume of tank, V = l × b × h
Put the values, we get
V = (6 × 5 × 4.5) = 135
Volume of water tank is 135\ m^3
Again,
We are given that, amount of water that 1m^3 volume can hold = 1000 l
Amount of water, 135\ m^3 volume hold = (135×1000) litres = 135000 litres
Therefore, given cuboidal water tank can hold up to 135000 litres of water.

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

Solution:

Given in the question:
Length of cuboidal vessel, l = 10 m
Width of cuboidal vessel, b = 8m
Volume of cuboidal vessel, V = 380\ m^3
Let the height of the given vessel be h.
Formula for Volume of a cuboid, V\ =\ l\times b\times h
Using formula, we have
l\times b\times h\ =\ 380 10\times8\times h=\ 380\
Or h = 4.75
Therefore, the height of the vessels is 4.75 m.

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

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\times b\times h\ =\ (8\times6\times3)\ =\ 144 (using formula)
Required Volume is 144\ m^3 .
Now,
Cost of digging per m^3 volume = Rs 30
Cost of digging 144\ m^3 volume = Rs (144×30) = Rs 4320

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

Solution:
Length (l) and depth (h) of the tank is 2.5 m and 10 m respectively.
To find: The value of breadth, say b.
Formula to find the volume of a tank = l\times b\times h\ =\ \left(2.5\times\ b\times10\right)m^3=\ 25b\ m^3
Capacity of tank= 25b\ m^3 , which is equal to 25000b liters
Also, capacity of a cuboidal tank is 50000 liters of water (Given)
Therefore, 25000\ b\ =\ 50000
This implies, b = 2
Therefore, the breadth of the tank is 2 m.

Question 6. A village, having a population of 4000, requires 150 liters of water per head per day.
It has a tank measuring 20 m×15 m×6 m. For how many days will the water of 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 the water per head per day = 150 litres
Water consumed by the people in 1 day = (4000\times150)\ litres\ =\ 600000\ litres\ \ldots(1)
Formula to find the capacity of tank, C\ =\ l\times b\times h
Using given data, we have
C\ =\ \left(20\times15\times6\right)m^3=\ 1800\ m^3
Or C = 1800000 litres
Let water in this tank last for d days.
Now, the Water consumed by all people in d days = Capacity of tank (using equation (1))
600000 d =1800000
d = 3
Therefore, the water of this tank will last for 3 days.

Question 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 statement, we have
Length of the godown = 40 m
Breadth = 25 m
Height = 15 m
Whereas,
Length of the wooden crate = 1.5 m
Breadth = 1.25 m
Height = 0.5 m
Since godown and wooden crate are in cuboidal shape. Find the volume of each using formula, \ V\ =\ lbh .
Now,
Volume of godown = \left(40\times25\times15\right)m^3\ =\ 15000\ m^3
Volume of a wooden crate = \left(1.5\times1.25\times0.5\right)m^3\ =\ 0.9375\ m^3
Let us consider, n wooden crates can be stored in the godown, then
Volume of n wooden crates = Volume of godown
\ 0.9375\times n\ =15000
Or n=\frac{15000}{0.9375}\ =\ 16000
Hence, the number of wooden crates that can be stored in the godown is 16,000.

Question 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 cube:
Volume of cube = \left(Side\right)^3\ =\ \left(12\right)^3cm^3=\ 1728cm^3
Surface area of a cube with side 12 cm = 6a^2\ =\ 6\left(12\right)^2\ cm^2\ \ldots(1)
Cube is cut into eight small cubes of equal volume, say side of each cube is p.
Volume of a small cube = p^3
Surface area = 6p^2\ \ldots(2)
Volume of each small cube = \left(\frac{1728}{8}\right)cm^3\ =\ 216\ cm^3
Or \left(p\right)^3\ =\ 216\ cm^3
Or p = 6 cm
Now, Surface areas of the cubes ratios = \frac{Surface area of bigger cube}{Surface area of smaller cubes}
From equation (1) and (2), we get
Surface areas of the cubes ratios = \frac{6a^2}{6p^2}\ =\frac{a^2}{p^2}\ =\frac{122}{62}\ =\ 4
Therefore, the required ratio is 4: 1.

Question 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 in the question:
Depth of river, h = 3 m
Width of river, b = 40 m
Rate of water flow = 2km per hour = 2000\ m/min\ \ \ =\frac{100}{3}\ m/min\
Now, Volume of water flowed in 1 min = \left(\frac{100}{3}\right)\ \times\ 40\ \times\ 3\ =\ 4000m^3
Therefore, 4000\ m^3 water will fall into the sea in a minute.

Exercise 13.6

Question 1. The circumference of the base of cylindrical vessel is 132cm and its height is 25cm. How many liters of water can it hold? ( 1000\ cm^3=\ 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 vessel
We know that, circumference of base = 2\pi r , so
Given, in the question,
2\pi r\ =\ 132\ \\ r\ =\ \left(\frac{132}{2\ \pi}\right)\\ r\ =\ 66\times\frac{7}{22}\ =\ 21\
Radius is 21 cm
Step 2: Find the volume of vessel
Formula: Volume of cylindrical vessel = \pi r^2h \\ =\ \left(\frac{22}{7}\right)\times{21}^2\times25\\ =\ 34650
Therefore, volume is 34650\ cm^3
Since, 1000\ cm^3\ =\ 1L
So, Volume = \frac{34650}{1000}\ L=\ 34.65L
Therefore, vessel can hold 34.65 liters of water.

Question 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 1cm^3 of wood has a mass of 0.6g. (Assume π = 22/7)

Solution:

Inner radius of cylindrical pipe, say \ r_1\ =\frac{diameter_1}{2}\ =\frac{24}{2}\ cm\ =\ 12cm
Outer radius of cylindrical pipe, say r_2\ =\frac{diameter_2}{2}\ =\frac{28}{2}\ cm\ =\ 14\ cm
Height of pipe, h = Length of pipe = 35cm
Now, the Volume of pipe = \pi(r_2^2-r_1^2)h\ cm^3
Substitute the values.
Volume of pipe = 110\times52\ cm^3\ =\ 5720\ cm^3
Since, Mass of 1\ cm^3 wood = 0.6 g
Mass of 5720\ cm^3 wood = (5720\times0.6)\ g\ =\ 3432\ g\ or\ 3.432\ kg

Question 3. A soft drink is available in two packs a tin can with a rectangular base of length 5cm and width 4cm, having a height of 15 cm and a plastic cylinder with 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 tin can are
Length, l = 5 cm
Breadth, b = 4 cm
Height, h = 15 cm
Capacity of tin can = \ l\times b\times h=\ \left(5\times4\times15\right)cm^3\ =\ 300\ cm^3

(ii)Plastic cylinder will be cylindrical in shape. (iii) Dimensions of plastic can are:
Radius of circular end of plastic cylinder, r = 3.5cm
Height, H = 10 cm
Capacity of plastic cylinder = \pi r^2H
Capacity of plastic cylinder = \left(\frac{22}{7}\right)\times\left(3.5\right)^2\times10\ =\ 385
Capacity of plastic cylinder is 385\ cm^3
From results of (i) and (ii), plastic cylinder has more capacity.
Difference in capacity = \left(385-300\right)cm^3\ =\ 85cm^3

Question 4. If the lateral surface of a cylinder is 94.2\ cm^2 and its height is 5cm, then find
radius of its base
its volume.[Use π= 3.14]

Solution:

CSA of cylinder = \ 94.2\ cm^2
Height of cylinder, h = 5cm

Let the radius of the cylinder be r.
Using CSA of cylinder, we get
2\pi rh\ =\ 94.2 \\ 2\times3.14\times r\times5\ =\ 94.2\\ r\ =\ 3
Radius is 3 cm [/katex]

Volume of cylinder
Formula for volume of cylinder = \pi r^2h
Now, \pi r^2h\ =\ (3.14\times\left(3\right)^2\times5) (using value of r from (i))
= 141.3
Volume is 141.3\ cm^3

Question 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 m^2 , find
inner curved surface area of the vessel
radius of the base
capacity of the vessel
(Assume π = 22/7)

Solution:

Rs 20 is the cost of painting \ 1\ m^2 area.
Rs 1 is the cost to paint \frac{1}{20}m^2 area
So, Rs 2200 is the cost of painting = \left(\frac{1}{20}\times2200\right)m^2
= 110\ m^2\ area
The inner surface area of the vessel is 110m^2 .

Radius of the base of the vessel, let us say r.
Height (h) = 10 m and
Surface area formula = 2\pi rh
Using result of (i)
2\pi rh\ =\ 110\ m^2 \\ 2\times22/7\times r\times10\ =\ 110 \\ r\ =\ 1.75
Radius is 1.75 m.

Volume of vessel formula = \pi r^2h
Here r = 1.75 and h = 10
Volume = \ (22/7)\times\left(1.75\right)^2\times10\ =\ 96.25
Volume of vessel is 96.25\ m^3
Therefore, the capacity of the vessel is 96.25\ m^3 or 96250 liters.

Question 6. The capacity of a closed cylindrical vessel of height 1m is 15.4 liters. How many square meters of 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 liters = 0.0154\ m^3
Let r be the radius of the circular end.
Now,
Capacity of cylindrical vessel = \ \left(\frac{22}{7}\right)\times r^2\times1\ =\ 0.0154
After simplifying, we get, r = 0.07 m
Again, total surface area of vessel = 2\pi r(r+h) \\ =\ 2\times\frac{22}{7}\times0.07(0.07+1)\\ =\ 0.44\times1.07\\ =\ 0.4708
Total surface area of vessel is \ 0.4708\ m^2\
Therefore, 0.4708\ m^2 of the metal sheet would be required to make the cylindrical vessel.

Question 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, r_1\ =\frac{7}{2}\ mm\ =\frac{0.7}{2}\ cm\ =\ 0.35\ cm
Radius of graphite, r_2\ =\frac{1}{2}\ mm\ =\frac{0.1}{2}\ cm\ =\ 0.05\ cm
Height of pencil, h = 14 cm
Formula to find, volume of wood in pencil = (r_1^2-r_2^2)\ h cubic units
Substitute the values, we have
=\ [\left(\frac{22}{7}\right)\times({0.35}^2-{0.05}^2)\times14] \\ =\ 44\times0.12\\ =\ 5.28\
This implies, volume of wood in pencil = 5.28\ cm^3
Again,
Volume of graphite = r_2^2 h cubic units Substitute the values, we have =\ \left(\frac{22}{7}\right)\times{0.05}^2\times14 \\ =\ 44\times0.0025\\ =\ 0.11
So, the volume of graphite is 0.11\ cm^3 .

Question 8. 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? (Assume π = 22/7)

Solution:

Diameter of cylindrical bowl = 7 cm
Radius of cylindrical bowl, r = 7/2 cm = 3.5 cm
Bowl is filled with soup to a height of4cm, so h = 4 cm Volume of soup in one bowl= \pi r^2h \\ \left(\frac{22}{7}\right)\times3.52\times4\ =\ 154
Volume of soup in one bowl is 154\ cm^3
Therefore,
Volume of soup given to 250 patients = \left(250\times154\right)cm^3=\ 38500\ cm^3
= 38.5litres.

Exercise 13.7

Question 1.
Find the volume of the right circular cone with
Radius 6cm, height 7 cm
Radius 3.5 cm, height 12 cm (Assume π = 22/7)
Solution:
Volume of cone = \ \left(\frac{1}{3}\right)\ \pi r^2h cube units
Where r be radius and h be the height of the cone
Radius of cone, r = 6 cm
Height of cone, h = 7cm
Say, V be the volume of the cone, we have
V\ =\ \left(\frac{1}{3}\right)\times\left(\frac{22}{7}\right)\times36\times7 \\ =\ (12\times22)\\ =\ 264
The volume of the cone is 264\ cm^3 .

Radius of cone, r = 3.5cm
Height of cone, h = 12cm
Volume of cone = \left(\frac{1}{3}\right)\times\left(\frac{22}{7}\right)\times3.52\times7\ =\ 154
Hence,
The volume of the cone is 154\ cm^3 .

Question 2. Find the capacity in liters of a conical vessel with
radius 7cm, slant height 25 cm
height 12 cm, slant height 13 cm
(Assume π = 22/7)

Solution:

Radius of cone, r =7 cm
Slant height of cone, l = 25 cm
Height of cone, h=\sqrt{l^2-r^2} \\ h=\sqrt{{25}^2-7^2} \\ h=\sqrt{625-49}
Or h = 24
Height of the cone is 24 cm
Now,
Volume of cone, V\ =\ \left(\frac{1}{3}\right)\pi r^2h\ \\ V\ =\ \left(\frac{1}{3}\right)\times\left(\frac{22}{7}\right)\ \times72\times24 \\ =\ (154\times8) \\ =\ 1232\
So, the volume of the vessel is 1232\ cm^3
Therefore, capacity of the conical vessel = (1232/1000) liters (because 1L\ =\ 1000\ cm^3 )
= 1.232 Liters.

Height of cone, h = 12 cm
Slant height of cone, l = 13 cm
Radius of cone,
r=\sqrt{l^2-h^2} \\ r=\sqrt{{13}^2-{12}^2} \\ r=\sqrt{169-144} \\ r=5
Hence, the radius of cone is 5 cm.
Now, Volume of cone, V\ =\ \left(\frac{1}{3}\right)\pi r^2h \\ V\ =\ \left(\frac{1}{3}\right)\times\left(\frac{22}{7}\right)\times52\times12\ cm^3 \\ =\ 2200/7
Volume of cone is \frac{2200}{7}cm^3
Now, Capacity of the conical vessel= 2200/7000 liters (1L\ =\ 1000\ cm^3 )
= 11/35 liters

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

Solution:

Height of the cone, h = 15 cm
Volume of cone = 1570\ cm^3
Let r be the radius of the cone
As we know: Volume of cone, V\ =\ \left(\frac{1}{3}\right)\pi r^2h
So, \left(\frac{1}{3}\right)\pi r^2h\ =\ 1570 \\ \left(\frac{1}{3}\right)\times3.14\times r^2\ \times15\ =\ 1570 \\ r^2\ =\ 100 \\ r\ =\ 10\
Radius of the base of the cone is 10 cm.

Question 4. If the volume of a right circular cone of height 9cm is 48\pi\ cm^3 , find the diameter of its base.

Solution:

Height of cone, h = 9cm
Volume of cone = 48\pi\ cm^3
Let r be the radius of the cone.
As we know: Volume of cone, V\ =\ \left(\frac{1}{3}\right)\pi r^2h
So, \ 1/3\ \pi\ r^2(9)\ =\ 48\ \pi \\ r^2\ =\ 16 \\ r\ =\ 4\
Radius of the cone is 4 cm.
So, diameter = 2 × Radius = 8
Thus, the diameter of the base is 8cm.

Question 5. A conical pit of top diameter 3.5m is 12m deep. What is its capacity in kiloliters?
(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\ =\ \left(\frac{1}{3}\right)\pi r^2h \\ V\ =\ (\frac{1}{3})\times(\frac{22}{7})\ \times\left(1.75\right)^2\times12\ =\ 38.5
Volume of cone is 38.5\ m^3
Hence, capacity of the pit = (38.5×1) kiloliters = 38.5 kiloliters.

Question 6. The volume of a right circular cone is 9856\ cm^3 . If the diameter of the base is 28cm, find
height of the cone
slant height of the cone
curved surface area of the cone
(Assume π = 22/7)

Solution:

Volume of a right circular cone = 9856\ cm^3
Diameter of the base = 28 cm
Radius of cone, r = (28/2) cm = 14 cm
Let the height of the cone be h
Volume of cone, V\ =\ \left(\frac{1}{3}\right)\pi r^2h \\ \left(\frac{1}{3}\right)\pi r^2h\ =\ 9856 \\ \left(\frac{1}{3}\right)\times\left(\frac{22}{7}\right)\ \times14\times14\times h\ =\ 9856 \\ h\ =\ 48
The height of the cone is 48 cm.

Slant Height of Cone,
l=\sqrt{r^2+h^2} \\ l=\sqrt{{14}^2+{48}^2} \\ l=\sqrt{196+2304} \\ l=50
Hence, the Slant height of the Cone is 50 Cm.

curved surface area of cone = \pi rl \\ =\ \left(\frac{22}{7}\right)\times14\times50 \\ =\ 2200
Curved surface area of the cone is 2200\ cm^2 .

Question 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\ =\ \left(\frac{1}{3}\right)\pi r^2h \\ V\ =\ \left(\frac{1}{3}\right)\times\pi\times5^2\times12 \\ =\ 100\pi\
Volume of the cone so formed is 100\pi\ cm^3 .

Question 8. If the triangle ABC in the 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 = \left(\frac{1}{3}\right)\pi r^2h ; where r is the radius and h be the height of cone
= (1/3)\ \times\pi\times12\times12\times5 \\ =\ 240\ \pi
The volume of the cones of formed is 240\pi\ cm^3 .
So, required ratio = (result of question 7) / (result of question 8) = \frac{100\pi}{240\pi}\ =\frac{5}{12}\ =\ 5:12.

Question 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 = \ \left(\frac{1}{3}\right)\pi r^2h \\ =\ \left(\frac{1}{3}\right)\times\left(\frac{22}{7}\right)\times5.25\times5.25\times3 \\ =\ 86.625
The volume of the heap of wheat is 86.625\ m^3 .
Again,
Area of Canvas required= CSA of Cone= \pi rl, Where l=\sqrt{r^2+h^2}
After Substituting the Values, we have
CSA of Cone= [\frac{22}{7}\times5.25\times\sqrt{\left(5.25\right)^2+3^2}] \\ =(\frac{22}{7})\times5.25\times6.05 \\ =99.825
Therefore, the area of the Canvas is 99.825\ m^2 .

Exercise 13.8

Question 1. Find the volume of a sphere whose radius is
7 cm
0.63 m
(Assume π =22/7)

Solution:

Radius of sphere, r = 7 cm
Using, Volume of sphere = \left(\frac{4}{3}\right)\pi r^3 \\ =\ \left(\frac{4}{3}\right)\times\left(\frac{22}{7}\right)\times73 \\ =\frac{4312}{3}
Hence, volume of the sphere is \frac{4312}{3}cm^3
Radius of sphere, r = 0.63 m
Using, volume of sphere = \left(\frac{4}{3}\right)\pi r^3 \\ =\ \left(\frac{4}{3}\right)\times\left(\frac{22}{7}\right)\times0.633 \\ =\ 1.0478
Hence, volume of the sphere is 1.05\ m^3 (approx).

Question 2. Find the amount of water displaced by a solid spherical ball of diameter
28 cm
0.21 m
(Assume π =22/7)

Solution:

Diameter = 28 cm
Radius, r = 28/2 cm = 14cm
Volume of the solid spherical ball = \left(\frac{4}{3}\right)\ \pi r^3
Volume of the ball = \left(\frac{4}{3}\right)\times\left(\frac{22}{7}\right)\times143\ =\frac{34496}{3}
Hence, volume of the ball is \frac{34496}{3}cm^3
Diameter = 0.21 m
Radius of the ball =0.21/2 m= 0.105 m
Volume of the ball = \left(\frac{4}{3}\right)\pi r^3
Volume of the ball = \left(\frac{4}{3}\right)\times\ \left(\frac{22}{7}\right)\times0.1053\ m^3
Hence, volume of the ball = 0.004851\ m^3

Question 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 cm^3 ? (Assume π=22/7)

Solution:

Given in the question,
Diameter of a metallic ball = 4.2 cm
Radius(r) of the metallic ball, r = 4.2/2 cm = 2.1 cm
Volume formula = \frac{4}{3}\pi r^3
Volume of the metallic ball = \left(\frac{4}{3}\right)\times\left(\frac{22}{7}\right)\times2.1\ cm^3
Volume of the metallic ball = 38.808\ cm^3
Now, using relationship between, density, mass and volume,
Density = Mass/Volume
Mass = Density × volume
=\ (8.9\times38.808)\ g \\ =\ 345.3912\ g
Mass of the ball is 345.39 g (approx).

Question 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 earth be “d”. Therefore, the radius of earth will be will be d/2
Diameter of moon will be d/4 and the radius of moon will be d/8
Find the volume of the moon:
Volume of the moon = \left(\frac{4}{3}\right)\pi r^3\ =\ \left(\frac{4}{3}\right)\pi\ \left(\frac{d}{8}\right)^3\ =\frac{4}{3\pi}\ \left(\frac{d^3}{512}\right)
Find the volume of the earth:
Volume of the earth = \left(\frac{4}{3}\right)\pi r^3=\ \left(\frac{4}{3}\right)\ \pi\ \left(\frac{d}{2}\right)^3\ =\ 4/3\pi\left(\frac{d^3}{8}\right)
Fraction of the volume of the earth is the volume of the moon
Volume of the moon/ Volume of the earth = \frac{\frac{4}{3}\pi\left(\frac{d^3}{512}\right)}{\frac{4}{3}\pi\left(\frac{d^3}{8}\right)}=\frac{8}{512}=\frac{1}{64}
Hence, the volume of the moon is of the 1/64 volume of earth.

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

Solution:

Diameter of hemispherical bowl = 10.5 cm
Radius of hemispherical bowl, r = 10.5/2 cm = 5.25 cm
Formula for volume of the hemispherical bowl = \left(\frac{2}{3}\right)\ \pi r^3
Volume of the hemispherical bowl = \left(\frac{2}{3}\right)\times\left(\frac{22}{7}\right)\times5.253\ =\ 303.1875
Volume of the hemispherical bowl is 303.1875\ cm^3
Capacity of the bowl = (303.1875)/1000 L = 0.303 liters (approx.)
Therefore, a hemispherical bowl can hold 0.303 liters of milk.

Question 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 = \left(\frac{2}{3}\right)\ \pi(R^3– r^3)
Put values,
Volume of the iron used in the hemispherical tank = \left(\frac{2}{3}\right)\times\left(\frac{22}{7}\right)\times(1.013– 13) = 0.06348
So, volume of the iron used in the hemispherical tank is 0.06348\ m^3 .

Question 7. Find the volume of a sphere whose surface area is 154\ cm^2 . (Assume π = 22/7)

Solution:

Let r be the radius of a sphere.
Surface area of sphere = 4\pi r^2
Given, in the question,
4\pi r^2\ =\ 154\ cm^2\ \\ r^2\ =\ (154\times7)/(4\ \times22) \\ r\ =\frac{7}{2}
Radius is 7/2 cm
Now,
Volume of the sphere = \left(\frac{4}{3}\right)\pi r^3
Volume of the Sphere= \left(\frac{4}{3}\right)\times\left(\frac{22}{7}\right)\times\left(\frac{7}{2}\right)^3=179\frac{2}{3}
Now, Volume of Sphere is 179\frac{2}{3}\ cm^3 .

Question 8. A dome of a building is in the form of a hemi sphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing is Rs20 per square meter, find the
inside surface area of the dome
volume of the air inside the dome
(Assume π = 22/7)

Solution:

Cost of white-washing the dome from inside = Rs 4989.60
Cost of white-washing 1m^2 area = Rs 20
CSA of the inner side of dome = \frac{498.96}{2}m^2\ =\ 249.48\ m^2

Let the inner radius of the hemispherical dome be r.
CSA of inner side of dome = 249.48\ m^2 (from (i))
Formula to find CSA of a hemisphere = 2\pi r^2 \\ 2\pi r^2\ =\ 249.48 \\ 2\times\left(\frac{22}{7}\right)\times r^2\ =\ 249.48 \\ r^2\ =\frac{249.48\times7}{2\times22}\\ r^2\ =\ 39.69 \\ r\ =\ 6.3
So, radius is 6.3 m
Volume of air inside the dome = Volume of hemispherical dome
Using formula, volume of the hemisphere = \frac{2}{3}\pi r^3 \\ =\ \left(\frac{2}{3}\right)\times\left(\frac{22}{7}\right)\times6.3\times6.3\times6.3 \\ =\ 523.908 \\ =\ 523.9 (approx.)
Hence, the Volume of air inside the dome is 523.9\ m^3 .

Question 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
radius r’ of the new sphere,
Ratio of Sand S’.

Solution:

Volume of the solid sphere = \left(\frac{4}{3}\right)\pi r^3
Volume of twenty seven solid sphere = 27\times\left(\frac{4}{3}\right)\pi r^3\ =\ 36\ \pi\ r^3
(i) New solid iron sphere radius = r’
Volume of this new sphere = \left(\frac{4}{3}\right)\pi(r’)^3 \\ \left(\frac{4}{3}\right)\pi(r’)^3 = 36 π r^3 \\ (r’)^3 = 27r^3 \\ r’= 3r
Radius of new sphere will be 3r (thrice the radius of original sphere)

(ii) Surface area of iron sphere of radius r, S\ =4\pi r^2
Surface area of iron sphere of radius r’= 4π (r’)^2
Now
S/S’ = (4πr^2)/( 4π (r’)^2) \\ S/S’ = r^2/(3r’)^2 = 1/9
The ratio of S and S’ is 1: 9.

Question 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 capsule = 3.5 mm
Radius of capsule, say r = diameter/ 2 = (3.5/2) mm = 1.75mm
Volume of spherical capsule = \frac{4}{3}\pi r^3
Volume of spherical capsule = \left(\frac{4}{3}\right)\times\left(\frac{22}{7}\right)\times\left(1.75\right)^3\ =\ 22.458
The volume of the spherical capsule is 22.46\ mm^3 .

Exercise 13.9

Question 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 cm^2 and the rate of painting is 10 paise per cm^2 , find the total expenses required for polishing and painting the surface of the bookshelf.

Solution:

External dimensions of book self,
Length, l = 85cm
Breadth, b = 25 cm
Height, h = 110 cm
External surface area of shelf while leaving out the front face of the shelf
=\ lh+2(lb+bh) \\ =\ [85\times110+2(85\times25+25\times110)] = (9350+9750) = 19100
External surface area of shelf is 19100\ cm^2
Area of front face = [85\times110-75\times100+2(75\times5)] = 1850+750
So, area is 2600\ cm^2
Area to be polished = \left(19100+2600\right)cm^2\ =\ 21700\ cm^2 .
Cost of polishing 1\ cm^2 area = Rs 0.20
Cost of polishing 21700\ cm^2 area Rs. (21700×0.20) = Rs 4340
Dimensions of row of the book shelf
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)\times\ 20+75\times30] = (4200+2250) = 6450
So, area is 6450\ cm^2 .
Area to be painted in 3 rows = \left(3\times6450\right)cm^2\ =\ 19350\ cm^2.
Cost of painting 1\ cm^2 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
Hence, the cost for polishing and painting the surface of the book shelf is Rs. 6275.

Question 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 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 cm^2 and black paint costs 5 paise per cm^2 .

Solution:

Diameter of wooden sphere = 21 cm
Radius of wooden sphere, r = diameter/ 2 = (21/2) cm = 10.5 cm
Formula: Surface area of wooden sphere = 4\pi r^2 \\ =\ 4\times\left(\frac{22}{7}\right)\times\left(10.5\right)^2\ =\ 1386

So, surface area is 1386\ cm^3
Radius of the circular end of cylindrical support = 1.5 cm
Height of cylindrical support = 7 cm
Curved surface area = 2\pi rh \\ =\ 2\times(22/7)\times1.5\times7\ =\ 66
So, CSA is 66\ cm^2
Now,
Area of the circular end of cylindrical support = \pi r^2 \\ =\ \left(\frac{22}{7}\right)\times1.52 \\ =\ 7.07
Area of the circular end is 7.07\ cm^2
Again,
Area to be painted silver = [8\ \times\ (1386-7.07)] = 8×1378.93 = 11031.44
Area to be painted is 11031.44\ cm^2
Cost for painting with silver color = Rs (11031.44×0.25) =Rs 2757.86
Area to be painted black = \left(8\times66\right)cm^2\ =\ 528\ cm^2
Cost for painting with black color =Rs (528×0.05) = Rs26.40
Therefore, the total painting cost is:
= Rs (2757.86 +26.40)
= Rs 2784.26.

Question 3. The diameter of a sphere is decreased by 25%. By what percent does it’s curved surface area decrease?

Solution:

Let the diameter of the sphere be “d”.
Radius of sphere, r_1\ =\ d/2
New radius of sphere, say r_2\ =\ \left(\frac{d}{2}\right)\times\left(1-\frac{25}{100}\right)\ =\frac{3d}{8}
Curved surface area of sphere, \ \left(CSA\right)_1\ =\ 4\pi r_1^2\ =\ 4\pi\times\left(\frac{d}{2}\right)^2\ =\ \pi d^2\ \ldots(1)
Curved surface area of sphere when radius is decreased \left(CSA\right)_2\ =\ 4\pi r_2^2\ =\ 4\pi\times\left(\frac{3d}{8}\right)^2=\ \left(\frac{9}{16}\right)\pi d^2\ \ldots(2)
From equation (1) and (2), we have
Decrease in surface area of sphere = \left(CSA\right)_1\ – (CSA)_2 \\ =\ \pi d^2\ – (9/16)πd^2 \\ =\ \left(\frac{7}{16}\right)\pi d^2
Percentage decrease in Surface area of Sphere= \frac{\left(CSA\right)_1-\left(CSA\right)_2}{\left(CSA\right)_1}\times100 \\ =\left(\frac{7d^2}{16d^2}\right)\times100 \\ =\frac{700}{16} \\ =43.75%
Hence, the Percentage decrease in the surface area of the sphere is 43.75% .