CBSE Class 9 Mathematics Surface areas and Volumes Assignment Set 02

Question. The cardboard piece shown below can be folded up along the dotted lines to form a triangular pyramid.Which of the following is a view of the pyramid formed?

Answer : A

Question. The container shown here is filled with water from a uniformly dripping tap.Which of these graphs best represents the water level in the container with respect to time, as it is being filled?

Answer : A

Question. One man takes one day to dig a 4 m long trench. How long would it take 2 men working at the same rate to dig a 16 m long trench?
A. 1 day
B. 2 days
C. 4 days
D. 8 days
Answer : B

Question. A cake is cut into 3 pieces whose weights are in the ratio 2 : 1 : 4.If the third piece weighs 360 g more than the second, how much did the whole cake weigh?

A. 1.44 kg
B. 1.26 kg
C. 840 g
D. 630 g
Answer : C

Question. A solid shape made by joining identical cubes is placed on a table. When viewed from different sides, the solid looks as shown below.Which of these could be the solid?

Answer : D

Question. Sonali has about half a litre of molten wax to make a candle. Which of these candles could she have made using the ENTIRE quantity of wax?

Answer : B

Question. Zubin wants to cover the CURVED SURFACE of an old waste paper basket with colored paper. The dimensions of the basket are shown below. What is the total area that has to be covered with paper? 

A. 220 cm²
B. 440 cm²
C. 880 cm²
D. 1760 cm²
Answer : C

Question. An ant at one corner of an 8 m x 6 m rectangular floor spots a piece of sweet at the opposite corner.What is the MINIMUM distance the ant has to travel to reach the sweet?

A. 7 m
B. 10 m
C. 14 m
D. 28 m
Answer : B

Question. Aftab is checking his weight on a weighing scale. What is the reading on the scale, shown in the Image?

A. 50.3 kg
B. 50.7 kg
C. 52 kg
D. 53.5 kg
Answer : D

Question. Travelling only along the edges and covering a distance of exactly 22 cm, how many different routes can be taken to go from corner P to corner Q of this cuboid?

A. 10
B. 8
C. 6
D. 1
Answer : D

Question. Which of these could be the approximate width of a regular sized basketball court?
A. 5 m
B. 15 m
C. 35 m
D. 50 m
Answer : B

Question. Dalbir has to carve a right circular cylinder of the maximum possible volume from this wooden cuboid shown in the image .What would be the volume (in cm3) of the cylinder?

A. 3000 π
B. 4000 π
C. 4500 π
D. 16000 π
Answer : B

Question. 4 identical triangles are cut off from the 4 corners of a square of side 20 cm as shown in the image.By how much did the perimeter decrease (with respect to the original square)?

A. 4 cm
B. 8 cm
C. 24 cm
D. 28 cm
Answer : B

Question. When an A-4 sized sheet of paper is folded up once through the centre as shown, the 'length to breadth'  ratio of the folded up piece is the same as the 'length to breadth' ratio of the original sheet.What is the 'length to breadth' ratio of an A-4 sized sheet? 

A. √2 : 1
B. 03:02 
C. 02:01
D. (information insufficient) 
Answer : A

Question.  The diamet er of two cones are equal. If their slant heights are in the ratio 5 : 4, find the ratio of their curved surface areas.
(A) 2:3
(B) 4:5
(C) 5:4
(D) 3 :2
Answer : A

Question. The circumference of the base of a 9 m high wooden solid cone is 44 m. Find the slant height of the cone.
(A) √120
(B) √l30m
(C) √150m
(D) 7√5m
Answer : B

Question. A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Take π = 22/7)
(A) 8 cm
(B) 9 cm
(C) 10 cm
(D) 11 cm
Answer : B

Question. Find the volume of a cube whose surface area is 150 cm².
(A) 25J5 cm³
(B) 64 cm³
(C) 125 cm³
(D) 27 cm³
Answer : C

Question. The circumference of the base of 9 m high wooden solid cone is 44 m. Find its volume. (Use π = 22/7)
(A) 235m³
(B) 456m³
(C) 365m³
(D) 462m³
Answer : D

Question. On a particular day, the rain fall recorded on a terrace 6 m long and 5 m broad is 15 cm. Find the quantity of water collected on the terrace.
(A) 300 litres
(B) 450 litres
(C) 3000 litres
(D) 4500 litres
Answer : D

Question. The height of sand in a cylindrical box drops 3 inches when 1 cubic foot of sand is poured out . What is t he diameter, in inches, of the cylinder?
(A) 24/√n
(B) 48/√n
(C) 32/√n
(D) 48/n
Answer : B

Question. The total surface area of a cylinder of height 6.5 em is 220 sq cm. Find its volume.
(A) 25.025 cm³
(B) 2.5025 cm³
(C) 2502.5 cm³
(D) 250.25 cm³
Answer : D

Question. The height of a cylinder is 15 cm. The curved surface area is 660 sq. m. Find its radius.
(Take π as 22/7)
(A) 7 cm
(B) 9 cm
(C) 6 cm
(D) 11 cm
Answer : A

Question. A cylindrical vessel contains 49.896 litres of liquid.Cost of painting its C.S.A.at 2 paise/sq em is ₹ 95.04. What is its total surface area?
(A) 5724 cm²
(B) 7524 cm²
(C) 5742 cm²
(D) 7254 cm²
Answer : B

Case Based MCQs

Case I : Read the following passage and answer the questions from 1 to 5.
Nakul was doing an experiment to find the radius r of a ball. For this he took a cylindrical container with radius R = 7 cm and height 10 cm. He filled the container almost half by water as shown in the figure-1. Now he dropped the ball into the container as in figure-2.
He observed that in figure-2, the water level in the container raised from P to Q i.e, to 3.4 cm.

1. What is the approximate radius of the ball?
(a) 3 cm
(b) 5 cm
(c) 7 cm
(d) 9 cm
Answer : B

2. What is the volume of the cylinder?
(a) 1260 cm²
(b) 540 cm³
(c) 1620 cm³
(d) 1540 cm³
Answer : D

3. What is the volume of the spherical ball?
(a) 620 cm³
(b) 824.26 cm³
(c) 523.81 cm³
(d) 430.1 cm³
Answer : C

4. How many litres of water can be filled in the full container ?
(a) 1.54 litres
(b) 2 litres
(c) 5 litres
(d) 7.5 litres
Answer : B

5. What is the total surface area of the spherical ball?
(a) 441.34 cm²
(b) 314.29 cm²
(c) 620 cm²
(d) 816 cm²
Answer : A

Very Short Answer Type Questions

Question. The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. Find the radius of the sphere.
Answer : 6 cm

Question. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9cm.
Answer : 190.93cm³

Question. A toy is in the form of a cone mounted on a hemi-sphere of same radius. The diameter of the base of the conical part is 7cm and the total height of the toy is 14.5cm. find the volume of the toy.
Answer : 231cm³

Question. The TSA of a solid cylinder is 231cm². If its CSA is 2/3 of its TSA. Find its radius and height.
Answer : 3.5cm, 7cm

Question. How many bags of grain can be stored in a cubic granary 12m x 6m x 5m , if each bag occupies a space of 0.48 m³ ?
Answer : 750

Question. The volume of two cubes are in the ratio 8 : 64 , then find the ratio of their surface areas.
Answer : 4 : 9

Question. A cone and a hemisphere have equal bases and equal volumes. What is the ratio of their heights?
Answer : 2 : 1

Question. The radii of 2 cylinders are in the ratio 3:5 and their heights are in the ratio 2:3. What is the ratio of their curved surface areas.
Answer : 2:5

Question. A cylinder and a cone are of same base radius and of same height. What is the ratio of their volumes?
Answer : 3 : 1

Question. Find the Total Surface Area of a hemispherical solid having radius 7 cm.
Answer : 462

Question. Two cubes each of volume 27cm³ are joined end to end to form a solid. Find the surface area of the solid.
Answer : 90cm²

Question. The length of a hall is 20m and width is 16m. the sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.
Answer : 8.8m

Question. A cone and a cylinder of same radius 3.5cm have same CSA. If height of the cylinder is 14cm then find the slant height of the cone.
Answer : 28cm

Short Answer Type Questions

Question. The radius of the circular part of a hemispherical bowl is 9 cm. Find the total capacity of 21 such bowls.
Answer : Volume of one bowl = Volume of hemisphere
= (2/3)πr³ = (2/3 x 22/7 x 9³) cm³
Capacity of 21 such bowls = 21 × 2/3 x 22/7 × 9³ = 32076 cm³

Question. The dome of a building is in the form of a hemisphere. If its radius is 14 cm, find the cost of painting it at the rate of ₹ 3 per sq. cm.
Answer : Since the dome of building is in the shape of hemisphere
⇒ Curved surface area of dome = 2πr²
= 2 × 22/7 × 14 × 14 = 2 × 22 × 2 × 14 = 1232 cm²
Cost of painting the dome at the rate of 1 sq. cm = ₹ 3
∴ Cost of painting of 1232 cm² dome
= ₹ 3 × 1232 = ₹ 3696

Question. A cube of 8 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of its base are 17 cm and 14 cm. Find the rise in water level in the vessel
Answer : Edge of the cube = 8 cm
∴ Volume of the cube = a³ = (8)³ = 512 cm³
If the cube is immersed in the vessel, then the water level rises. Let the rise in water level be x.
Clearly, Volume of the displaced water = Volume of the cube
⇒ Volume of the cube = 17 cm × 14 cm × x cm
⇒ 512 = 17 × 14 × x
⇒ x = 512/17x14 ⇒ x = 512/238 = 2.15 cm

Question. The length of a cold storage is three times its breadth. Its height is 5 m. The area of its four walls (including doors) is 256 m². Find its volume.
Answer : Let length, breadth and height of the cold storage be l, b and h respectively.
Then, l = 3b and h = 5 m.
Now, area of four walls = 256 m²
⇒ 2 (l + b)h = 256 ⇒ 2 (3b + b) × 5 = 256
⇒ 40b = 256 ⇒ b = 6.4 metres
∴ l = 3b = 3 × 6.4 = 19.2 m
Volume of the cold storage = l × b × h
= (19.2 × 6.4 × 5) m3 = 614.4 m³

Question. The curved surface area of a cone is 154 cm². If its radius is x cm and slant height is 7 cm. Find the value of 20x.
Answer : We have, curved surface area = 154 cm²
⇒ πrl = 154 ⇒ r = 154x7/22x7 = 7
Now, r = x cm = 7 cm.
∴ x = 7 ⇒ 20x = 20 × 7 = 140

Question. If the height of a cylinder is 11 cm and area of curved surface is 968 sq. cm. Find the radius of the cylinder.
Answer : Height of cylinder = 11 cm
Curved surface area = 968 cm²
⇒ 2πrh = 968
⇒ 2x22/7 × π × 11 = 968 ⇒ r = 14 cm

Question. The base of a cubical box has a perimeter 250 m. Find the cost of painting its lateral surface area at the rate of ₹ 10 per m².
Answer : Let the side of cubical box be a m.
Perimeter of base of a cubical box = 4a m
⇒ 250 = 4a ⇒ a = 62.5 m
Now, LSA of cubical box = 4a² = 4 × (62.5)² = 15625 m²
Cost of painting 1 m² = ₹ 10
∴ Total cost of painting = 15625 × 10 = ₹ 156250.

Question. A cuboidal oil tin box is 4 m by 2 m by 0.75 m. Find the cost of the tin sheet required for making 20 such tin boxes, if the cost of tin sheet is ₹ 20 per square metre.
Answer : Length, l = 4 m, breadth, b = 2 m and
height, h = 0.75 m
Surface area of one tin box = 2(lb + bh + hl)
= 2 (4 × 2 + 2 × 0.75 + 0.75 × 4)
= 2 (8 + 1.5 + 3) = 2 × 12.5 = 25 m²
∴ Surface area of 20 such tin boxes = (20 × 25) m²
= 500 m²
Now, cost of 1 square metre of tin sheet = ₹ 20
∴ Cost of 500 m² of tin sheet = ₹(20 × 500) = ₹ 10000

Question. Three cubes each of edge 5 cm are joined end to end. Find the surface area of the resulting cuboid
Answer : When three cubes are joined end to end, we get a cuboid such that
Length of the resulting cuboid, l
= 5 cm + 5 cm + 5 cm = 15 cm
Breadth of resulting cuboid, b = 5 cm
Height of the resulting cuboid, h = 5 cm
Surface area of the cuboid = 2 (lb + bh + hl)
= 2 (15 × 5 + 5 × 5 + 5 × 15) cm²
= 2 (75 + 25 + 75) cm² = 350 cm²

Question. Find the diameter of the sphere, whose total surface area is 616 cm²
Answer : Let r be the radius of the sphere.
Total surface area of sphere = 4πr²
⇒ 616 = 4 × 22/7 ×r² ⇒ = r² 616x7/4x22 = 49 ⇒ r = 7 cm
∴ Diameter = 2r = 2 × 7 = 14 cm

Question. Find the volume of a lead pipe 3.5 m long, if the external diameter of the pipe is 2.4 cm, thickness of lead is 3 mm and 1 cm³ of lead weighs 12 g
Answer : External diameter of the pipe = 2.4 cm
External radius of the pipe, (R) = 2 4/2. cm =1.2 cm
Thickness of the pipe = 3 mm = 0.3 cm
Internal radius, (r) = External radius – thickness
= 1.2 cm – 0.3 cm = 0.9 cm
Length of the pipe (h) = 3.5 m = 350 cm
Volume of lead = π(R² – r²)h
= 22/7× [(1.2)² – (0.9)²] × 350 = 22/7× 0.63 × 350 = 693 cm³

Question. Find the total surface area, lateral surface area and the length of diagonal of a cube, each of whose edges measues 20 cm.
(Take √3 = 1.732)
Answer : Here, side (a) = 20 cm
∴ Total surface area of the cube = 6a² = 6(20)²
= 2400 cm²
and, lateral surface area of the cube = 4a² = 4(20)²
= 1600 cm²
Also, length of diagonal of a cube = √3a
= √3 × 20= (1.732 × 20) = 34.64 cm.

Question. A room is 16 m long, 9 m wide and 3 m high. It has two doors, each of dimensions (2 m × 2.5 m)and three windows, each of dimensions (1.6 m × 75 cm). Find the cost of distempering the walls of the room from inside at the rate of ₹ 8 per sq. metre.
Answer : Given, length (l) = 16 m, breadth (b) = 9 m and
height (h) = 3 m
∴ Area of 4 walls of the room = 2(l + b) × h
= 2(16 + 9) × 3 = 150 m².
Area of 2 doors = 2 × (2 × 2.5) = 10 m²
Area of 3 windows = 3 × (16 × 75/100) = 3.6 m².
Area not to be distempered = 10 + 3.6 = 13.6 m²
Area to be distempered = 150 – 13.6 = 136.4 m²
Cost of distempering the walls = ₹ (136.4 × 8) = ₹ 1091.20

Question. The total cost of making a solid spherical ball is ₹ 67914 at the rate of ₹ 14 per cubic metre. Find the radius of this ball.
Answer : Volume of spherical ball = Total cost/Cost of 1m³
⇒ (4/3)πr³ = 67914/14 ⇒ 4/3 x 22/7 x r² = 67914/14
⇒ r³ = 101871/88  ⇒ r³ = 1157.625 ⇒ r = 10.5 m

Question. The external diameter of an iron pipe is 35 cm and its length is 30 cm. If the thickness of the pipe is 2.5 cm, find the curved surface area of the pipe.
Answer : Length of the pipe, h = 30 cm
External radius of the pipe, R = 35/2 cm = 17.5 cm
∴ Thickness of the pipe = 2.5 cm
∴ Internal radius of the pipe, r = (17.5 – 2.5) cm = 15 cm
Now, curved surface area of the pipe = External curved surface area + Internal curved surface area
= 2πRh + 2prh = 2πh (R + r)
= 2 × 22/7 × 30(17.5 + 15) = 2 × 22/7 × 30 × 32.5
= 42900/7 = 6128.57 cm²

Question. A closed cuboidal tank can store 5040 litres of water. The external dimensions of the tank are 2.2 m × 1.7 m × 1.7 m. If the walls of the tank are 5 cm thick, then what is the thickness of the bottom (top) of the tank if they are same?
Answer : Capacity of tank = 5040 litres
= (5040'1000)m³ = 5.040m³
Internal length of tank = (2 2 - 2 x 5/100) m = 2.1m
Internal breadth of tank = (1 7 - 2 x 5/100) m = 1.16 m
Let the thickness of tank in the bottom be x m.
Internal height of tank = (1.7 – 2x) m
Now, Internal volume of tank = Capacity of tank
⇒ 2.1 × 1.6 × (1.7 – 2x) = 5.040
⇒ 1.7 - 2x = 5.040/2.1 x 1.6 = 1.5
⇒ 2x = 1.7 – 1.5 = 0.2
⇒ x = 0.1 m = 0.1 × 100 cm = 10 cm
So, required thickness = 10 cm

Question. The slant height of a cone is 25 cm and  the vertical height is 24 cm. Find the radius and the total surface area of the cone.
Answer : Here, h = 24 cm and slant height (l) = 25 cm
Let r be the radius of cone. Then,
l² = h² + r²
⇒ r² = 252 – 242 = 49 ⇒ r = 7
Total surface area of the cone = πr(l + r)
= [22/7 x 7 x (25+7)] cm² = 704 cm²

Question. The length and breadth of a rectangular solid are respectively 35 cm and 20 cm. If its volume is 7000 cm³, then find its height (in cm).
Answer : Let h be the height of the solid.
∴ Volume of cuboid = l × b × h
⇒ 7000 = 35 × 20 × h ⇒ h = 7000 = 700/10 cm

Question. If volume and surface area of a sphere is numerically equal then find its radius (in units).
Answer : Let r be the radius of the sphere.
∴ Volume of sphere = Surface area of sphere
∴ (4/3)πr³ = 4πr² ⇒ r³/r² = 3 ⇒ r = 3 units

Question. Coins of same size (say 10 rupee coin) are placed one above the other and a cylindrical block is obtained. The volume of this block is 67.76 cm³. Find the number of coins arranged in the block, if thickness of each coin is 2 mm and radius of each coin is 1.4 cm.
Answer : Let h be the height of cylindrical block and n be the
number of coins used to obtain it.
Volume of block = πr2h ⇒ 67.76 = 22/7 × 1.4 × 1.4 × h
⇒ h = 67.76x7/22x1.4x1.4 = 11cm = 110mm       [∴ 1 cm = 10 mm]
Now, n × thickness of a coin = height of block
⇒ n × 2 = 110 ⇒ n = 5

Question 1. A small indoor greenhouse is made of entirely glass planes (including) bases together with tape. It is 30 cm long, 25 cm wide and 25 cm high. What is the area of the glass? How much tape is needed for all the 12 edges?
Answer:
To find the area of the glass, we calculate the total surface area of the cuboid-shaped greenhouse using the formula \( 2(lw + wh + hl) \), where \( l = 30 \) cm, \( w = 25 \) cm, and \( h = 25 \) cm.
Total Surface Area = \( 2(30 \times 25 + 25 \times 25 + 25 \times 30) \)
\( = 2(750 + 625 + 750) \)
\( = 2(2125) = 4250 \text{ cm}^2 \).
Hence, the area of the glass is \( 4250 \text{ cm}^2 \).

To find the length of the tape needed for all 12 edges of the cuboid, we use the formula \( 4(l + w + h) \).
Total length of tape = \( 4(30 + 25 + 25) \)
\( = 4(80) = 320 \text{ cm} \).
Thus, \( 320 \text{ cm} \) of tape is required.
In simple words: To find the glass area, calculate the total surface area of the box. To find the tape needed, add up the lengths of all twelve outer borders.

Exam Tip: Remember that a cuboid has 4 lengths, 4 widths, and 4 heights, which is why the edge length formula is multiplied by 4.

Question 2. Find the surface area of a cube whose edge is 11 cm.
Answer:
The total surface area of a cube is calculated using the formula \( 6a^2 \), where \( a \) represents the length of one edge.
Given that the edge \( a = 11 \) cm:
Surface Area = \( 6 \times 11^2 \)
\( = 6 \times 121 = 726 \text{ cm}^2 \).
Therefore, the surface area of the cube is \( 726 \text{ cm}^2 \).
In simple words: To get the surface area of a cube, find the area of one square face and multiply it by six since a cube has six identical faces.

Exam Tip: Always write the unit as square centimeters (\( \text{cm}^2 \)) for area calculations to avoid losing minor marks.

Question 3. The dimensions of a cuboid are in the ratio 1:2:3 and its total surface area is 88 m2. Find the dimensions of the cuboid.
Answer:
Let the dimensions of the cuboid be \( x \), \( 2x \), and \( 3x \).
The formula for the total surface area of a cuboid is \( 2(lw + wh + hl) \).
Given that the total surface area is \( 88 \text{ m}^2 \):
\( 2(x(2x) + 2x(3x) + 3x(x)) = 88 \)
\( 2(2x^2 + 6x^2 + 3x^2) = 88 \)
\( 2(11x^2) = 88 \)
\( 22x^2 = 88 \)

\( \implies x^2 = 4 \)
\( \implies x = 2 \text{ m} \) (since length cannot be negative).

Substituting \( x = 2 \) back into the ratio, we get the dimensions:
Length = \( x = 2 \text{ m} \)
Breadth = \( 2x = 4 \text{ m} \)
Height = \( 3x = 6 \text{ m} \).
In simple words: Use a variable like \( x \) to write the dimensions as parts of a ratio, plug them into the surface area formula, solve for \( x \), and then multiply to get the final measurements.

Exam Tip: Be careful not to confuse the units — the surface area is in square meters, so the final dimensions must be in meters.

Question 4. Three cubes each of side 5 cm are joined end to end. Find the surface area of the resulting cuboid.
Answer:
When three identical cubes of side \( 5 \text{ cm} \) are joined end to end, the resulting shape is a cuboid with the following dimensions:
Length (\( l \)) = \( 5 + 5 + 5 = 15 \text{ cm} \)
Breadth (\( w \)) = \( 5 \text{ cm} \)
Height (\( h \)) = \( 5 \text{ cm} \)

The total surface area of this new cuboid is:
Surface Area = \( 2(lw + wh + hl) \)
\( = 2(15 \times 5 + 5 \times 5 + 5 \times 15) \)
\( = 2(75 + 25 + 75) \)
\( = 2(175) = 350 \text{ cm}^2 \).
Therefore, the surface area of the resulting cuboid is \( 350 \text{ cm}^2 \).
In simple words: Joining three cubes end to end only increases the total length. The width and height stay the same as a single cube's side.

Exam Tip: A common mistake is simply multiplying the surface area of one cube by three. Remember that the faces where the cubes touch are hidden inside, which reduces the overall external surface area.

Question 5. cuboidal oil tin is 30 cm by 40 cm by 50 cm. Find the cost of the tin required for making 20 such tins if the cost of tin sheet is Rs 20 per square meter.
Answer:
First, we find the total surface area of one cuboidal tin with dimensions \( l = 30 \text{ cm} = 0.3 \text{ m} \), \( w = 40 \text{ cm} = 0.4 \text{ m} \), and \( h = 50 \text{ cm} = 0.5 \text{ m} \).
Surface area of one tin = \( 2(lw + wh + hl) \)
\( = 2(0.3 \times 0.4 + 0.4 \times 0.5 + 0.5 \times 0.3) \)
\( = 2(0.12 + 0.20 + 0.15) \)
\( = 2(0.47) = 0.94 \text{ m}^2 \).

The total surface area required for 20 such tins is:
Total Area = \( 20 \times 0.94 = 18.8 \text{ m}^2 \).

Now, we calculate the cost at the rate of Rs. 20 per square meter:
Total Cost = \( 18.8 \times 20 = \text{Rs. } 376 \).
In simple words: Convert the dimensions to meters first, calculate the surface area of one container, multiply by twenty to find the total metal needed, and then multiply by the price.

Exam Tip: Converting centimeters to meters at the very beginning avoids complicated unit conversions of area later on (where \( 1 \text{ m}^2 = 10000 \text{ cm}^2 \)).

Question 6. The floor of a rectangular hall has a perimeter of 250 m. Its height is 6 m. Find the cost of painting its four walls at the rate of Rs. 6 per square meter.
Answer:
The area of the four walls of a rectangular room is equal to its lateral surface area, given by the formula:
Area of four walls = \( 2(l + w)h \)

Since the perimeter of the floor is \( 2(l + w) = 250 \text{ m} \) and the height \( h = 6 \text{ m} \):
Area of four walls = \( 250 \times 6 = 1500 \text{ m}^2 \).

Now, the cost of painting at Rs. 6 per square meter is:
Total Cost = \( 1500 \times 6 = \text{Rs. } 9000 \).
In simple words: The perimeter of the floor multiplied by the height of the room gives the total area of the four walls. Multiply this area by the cost per square meter to get the final bill.

Exam Tip: Notice that you do not need to find the individual length and width of the hall because the product \( 2(l+w) \) is already given directly as the perimeter.

Question 7. Agrawal sweets was placing an order for making cardboard boxes for packing their sweets. Two sizes of the boxes were required. The bigger of dimension 25 cm x 20 cm x 5 cm and the smaller of dimension 15 cm x 12 cm x 5 cm. 5 % of the total surface is required extra, for all the overlaps. If the cost of cardboard is Rs. 4 for 1000 cm3, find the cost of cardboard required for supplying 250 boxes of each kind.
Answer:
Note: Cardboard is measured by its area, so "1000 cm3" in the question is a typographic error in the source and refers to \( 1000 \text{ cm}^2 \).

First, find the total surface area of one bigger box:
\( A_{\text{bigger}} = 2(25 \times 20 + 20 \times 5 + 5 \times 25) \)
\( = 2(500 + 100 + 125) = 2(725) = 1450 \text{ cm}^2 \).

Next, find the total surface area of one smaller box:
\( A_{\text{smaller}} = 2(15 \times 12 + 12 \times 5 + 5 \times 15) \)
\( = 2(180 + 60 + 75) = 2(315) = 630 \text{ cm}^2 \).

Total surface area for one box of each type = \( 1450 + 630 = 2080 \text{ cm}^2 \).
Total surface area for 250 boxes of each type = \( 250 \times 2080 = 520,000 \text{ cm}^2 \).

Add \( 5\% \) extra area for overlaps:
Extra cardboard needed = \( 5\% \text{ of } 520,000 = 26,000 \text{ cm}^2 \).
Total cardboard required = \( 520,000 + 26,000 = 546,000 \text{ cm}^2 \).

Calculate the cost at Rs. 4 per \( 1000 \text{ cm}^2 \):
Total Cost = \( \frac{546,000}{1000} \times 4 = 546 \times 4 = \text{Rs. } 2184 \).
In simple words: Find the outer area of both boxes, add them together, multiply by 250, and then add 5% extra for the overlapping cardboard folds. Finally, calculate the total price based on the rate per 1000 square centimeters.

Exam Tip: Be sure to sum up both areas before applying the 5% extra and multiplying by 250 to keep the calculations simple and reduce the chance of errors.

Question 8. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2 find the diameter of the base.
Answer:
The curved surface area (CSA) of a cylinder is given by the formula \( 2\pi r h \).
Given that \( h = 14 \text{ cm} \) and CSA = \( 88 \text{ cm}^2 \):
\( 2 \times \frac{22}{7} \times r \times 14 = 88 \)
\( 88 \times r = 88 \)

\( \implies r = 1 \text{ cm} \).

The diameter of the base is \( 2 \times r = 2 \times 1 = 2 \text{ cm} \).
In simple words: Set up the curved surface area equation using the given height, solve for the radius, and then double it to find the diameter.

Exam Tip: Always check if the question asks for the radius or the diameter. It is a very common mistake to stop after finding \( r \).

Question 9. The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it cover in 5 revolutions?
Answer:
The garden roller is cylindrical in shape.
Diameter = \( 1.4 \text{ m} \implies \) Radius (\( r \)) = \( 0.7 \text{ m} \).
Length (\( h \)) = \( 2 \text{ m} \).

The area covered in one full revolution is equal to the curved surface area of the cylinder:
Area of one revolution = \( 2\pi r h \)
\( = 2 \times \frac{22}{7} \times 0.7 \times 2 \)
\( = 2 \times 22 \times 0.1 \times 2 = 8.8 \text{ m}^2 \).

Area covered in 5 revolutions is:
Total Area = \( 5 \times 8.8 = 44 \text{ m}^2 \).
In simple words: Find the curved side area of the roller since only that part rolls on the ground. Multiply this single-turn area by five to find the total ground covered.

Exam Tip: The length of the roller acts as the height (\( h \)) of the cylinder when using the formula.

Question 10. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm , the outer diameter is 4.4 cm .Find
a) Inner curved surface area
b) Outer curved surface area
c) Total surface area
Answer:
Given:
Length of the pipe (\( h \)) = \( 77 \text{ cm} \)
Inner radius (\( r \)) = \( \frac{4}{2} = 2 \text{ cm} \)
Outer radius (\( R \)) = \( \frac{4.4}{2} = 2.2 \text{ cm} \)

a) Inner curved surface area:
\( \text{Inner CSA} = 2\pi r h = 2 \times \frac{22}{7} \times 2 \times 77 \)
\( = 2 \times 22 \times 2 \times 11 = 968 \text{ cm}^2 \).

b) Outer curved surface area:
\( \text{Outer CSA} = 2\pi R h = 2 \times \frac{22}{7} \times 2.2 \times 77 \)
\( = 2 \times 22 \times 2.2 \times 11 = 1064.8 \text{ cm}^2 \).

c) Total surface area:
The total surface area includes the inner curved surface, the outer curved surface, and the areas of the two ring-shaped circular ends at the top and bottom.
Area of one circular end = \( \pi(R^2 - r^2) \)
\( = \frac{22}{7} \times (2.2^2 - 2^2) \)
\( = \frac{22}{7} \times (4.84 - 4) = \frac{22}{7} \times 0.84 = 22 \times 0.12 = 2.64 \text{ cm}^2 \).
Area of both circular ends = \( 2 \times 2.64 = 5.28 \text{ cm}^2 \).

Total Surface Area = Inner CSA + Outer CSA + Area of both ends
\( = 968 + 1064.8 + 5.28 = 2038.08 \text{ cm}^2 \).
In simple words: Work out the inner and outer side areas separately. To get the total surface area, add these two side areas together with the thin ring-shaped flat edges at both open ends.

Exam Tip: Do not forget to add the areas of the two circular ring ends when finding the total surface area of a hollow pipe.

Question 11. A cylindrical vessel without lid, has to be tin coated on both the sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin- coating at the rate of Rs. 3.50 per 1000 cm2.
Answer:
Given:
Radius (\( r \)) = \( 70 \text{ cm} \)
Height (\( h \)) = \( 1.4 \text{ m} = 140 \text{ cm} \)

Since the cylindrical vessel has no lid, we only coat the curved surface and the circular base on both the inner and outer sides.
Area of one side (inner or outer) = Curved Surface Area + Base Area
\( = 2\pi r h + \pi r^2 \)
\( = \left(2 \times \frac{22}{7} \times 70 \times 140\right) + \left(\frac{22}{7} \times 70 \times 70\right) \)
\( = (2 \times 22 \times 10 \times 140) + (22 \times 10 \times 70) \)
\( = 61,600 + 15,400 = 77,000 \text{ cm}^2 \).

Since both the inside and outside are coated, the total surface area is:
Total Area = \( 2 \times 77,000 = 154,000 \text{ cm}^2 \).

The cost of coating is Rs. 3.50 per \( 1000 \text{ cm}^2 \):
Total Cost = \( \frac{154,000}{1000} \times 3.50 = 154 \times 3.50 = \text{Rs. } 539 \).
In simple words: Find the area of the curved walls plus the bottom base. Since you are painting both the inside and the outside, double this total area, and then calculate the price based on the given rate.

Exam Tip: Ensure all units are identical before starting. Convert the height from meters to centimeters first since the rate is given in square centimeters.

Question 12. A lampshade is cylindrical in shape, it has to be covered with a decorative cloth, the frame has the base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to given to it for folding it over the top and the bottom of the frame. Find out how much cloth is needed for covering the lamp- shade.
Answer:
Given:
Base diameter = \( 20 \text{ cm} \implies \) radius (\( r \)) = \( 10 \text{ cm} \)
Height of the frame = \( 30 \text{ cm} \)

A margin of \( 2.5 \text{ cm} \) is required at both the top and the bottom for folding. Therefore, the total height (\( H \)) of the cloth required is:
\( H = 30 + 2.5 + 2.5 = 35 \text{ cm} \).

The area of the cloth needed is equal to the curved surface area of the cylinder with this new height:
Area of cloth = \( 2\pi r H \)
\( = 2 \times \frac{22}{7} \times 10 \times 35 \)
\( = 2 \times 22 \times 10 \times 5 = 2200 \text{ cm}^2 \).
In simple words: Add the top and bottom margin heights to the frame height. Use this new combined height to calculate the curved side area of the cylinder.

Exam Tip: Do not forget to add the margin twice (once for the top fold and once for the bottom fold) to the height before calculating the area.

Question 13. The diameter of a cone is 14 cm and its slant height is 9 cm. Find the area of its curved surface.
Answer:
Given:
Diameter = \( 14 \text{ cm} \implies \) radius (\( r \)) = \( 7 \text{ cm} \)
Slant height (\( l \)) = \( 9 \text{ cm} \)

The curved surface area (CSA) of a cone is calculated using the formula:
\( \text{CSA} = \pi r l \)
\( = \frac{22}{7} \times 7 \times 9 \)
\( = 22 \times 9 = 198 \text{ cm}^2 \).
Thus, the curved surface area is \( 198 \text{ cm}^2 \).
In simple words: Halve the diameter to find the base radius, and then multiply it by the slant height and pi to get the curved area of the cone.

Exam Tip: Verify that the value provided is indeed the slant height (\( l \)) and not the vertical height (\( h \)) before using the formula.

Question 14. The radius and the slant height are in the ratio 4:7. If its curved surface area is 792 cm2, find its radius.
Answer:
Let the radius \( r = 4x \) and slant height \( l = 7x \).
The curved surface area of a cone is \( \pi r l = 792 \text{ cm}^2 \).
\( \frac{22}{7} \times 4x \times 7x = 792 \)
\( 22 \times 4 \times x^2 = 792 \)
\( 88x^2 = 792 \)

\( \implies x^2 = 9 \)
\( \implies x = 3 \text{ cm} \).

Now, find the radius:
Radius = \( 4x = 4 \times 3 = 12 \text{ cm} \).
In simple words: Represent the radius and slant height with a multiplier variable, plug them into the curved surface area equation, find the multiplier, and then calculate the radius.

Exam Tip: Always double check your final answer by substituting the values back into the ratio to see if they yield the given surface area.

Question 15. The radius of the cone is 7 cm and its curved surface area is 176 cm2.
Answer:
The question is cut off in the source, but it asks to find the slant height (\( l \)) of the cone.
The formula for curved surface area is \( \pi r l \).
Given that radius \( r = 7 \text{ cm} \) and CSA = \( 176 \text{ cm}^2 \):
\( \frac{22}{7} \times 7 \times l = 176 \)
\( 22 \times l = 176 \)

\( \implies l = 8 \text{ cm} \).

Thus, the slant height of the cone is \( 8 \text{ cm} \).
In simple words: Use the curved surface area formula with the given radius to solve for the missing slant height.

Exam Tip: If you encounter a cut-off question on an exam, solve for the most logical missing dimension (in this case, the slant height \( l \)).

Question 16. A jokers cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet needed to make 10 such caps.
Answer:
Given:
Radius (\( r \)) = \( 7 \text{ cm} \)
Height (\( h \)) = \( 24 \text{ cm} \)

First, find the slant height (\( l \)) of the cone using the Pythagorean relation:
\( l = \sqrt{r^2 + h^2} \)
\( = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ cm} \).

The area of the sheet required for one cap is its curved surface area:
\( \text{Area of 1 cap} = \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \text{ cm}^2 \).

For 10 caps, the total area of the sheet needed is:
Total Area = \( 10 \times 550 = 5500 \text{ cm}^2 \).
In simple words: Use the height and base radius to calculate the slant length of the cone. Then find the curved surface area of one cap and multiply it by ten.

Exam Tip: Since a cap is open at the bottom, we only need to calculate the curved surface area, not the total surface area.

Page 92

Question 17. A circus tent is cylindrical to a height of 3 m and conical above it, if its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide required to make the tent.
Answer:
Given:
Diameter = \( 105 \text{ m} \implies \) radius (\( r \)) = \( 52.5 \text{ m} \)
Height of cylindrical portion (\( h \)) = \( 3 \text{ m} \)
Slant height of conical portion (\( l \)) = \( 53 \text{ m} \)

Total surface area of the tent canvas = Curved surface area of cylinder + Curved surface area of cone
\( = 2\pi r h + \pi r l = \pi r (2h + l) \)
\( = \frac{22}{7} \times 52.5 \times (2(3) + 53) \)
\( = 22 \times 7.5 \times (6 + 53) \)
\( = 165 \times 59 = 9735 \text{ m}^2 \).

Since the width of the canvas is \( 5 \text{ m} \):
Length of canvas required = \( \frac{\text{Total Area}}{\text{Width}} = \frac{9735}{5} = 1947 \text{ m} \).
In simple words: Find the curved side areas of both the cylinder and the cone, add them together to get the total fabric area, and then divide by the width of the canvas roll to find the required length.

Exam Tip: Double check that you do not include the base of the cylinder or the base of the cone in your area calculations, as a tent is open inside.

Question 18. There are two cones, the surface area of one cone is twice the surface area of the other cone. The slant height of the latter is twice that of the former. Find the ratio of their radii.
Answer:
Let the first cone have radius \( r_1 \), slant height \( l_1 \), and curved surface area \( C_1 \).
Let the second cone (latter) have radius \( r_2 \), slant height \( l_2 \), and curved surface area \( C_2 \).

According to the problem:
\( C_1 = 2 C_2 \)
\( l_2 = 2 l_1 \)

Using the formula for curved surface area of a cone (\( C = \pi r l \)):
\( \pi r_1 l_1 = 2 (\pi r_2 l_2) \)
Substitute \( l_2 = 2 l_1 \) into the equation:
\( r_1 l_1 = 2 \times r_2 \times (2l_1) \)
\( r_1 l_1 = 4 r_2 l_1 \)
\( r_1 = 4 r_2 \)

Therefore, the ratio of their radii is:
\( \frac{r_1}{r_2} = \frac{4}{1} \implies 4:1 \).
In simple words: Set up the surface area ratio equation using the variables, substitute the given relationship for the slant heights, and solve to find the ratio between the two radii.

Exam Tip: Be precise about which cone is which (former vs. latter) to make sure you do not invert the final ratio.

Question 19. Find the surface area of a sphere of radius 7 cm.
Answer:
The total surface area of a sphere is given by the formula \( 4\pi r^2 \).
Given that the radius \( r = 7 \text{ cm} \):
Surface Area = \( 4 \times \frac{22}{7} \times 7^2 \)
\( = 4 \times \frac{22}{7} \times 49 \)
\( = 4 \times 22 \times 7 = 616 \text{ cm}^2 \).
Thus, the surface area of the sphere is \( 616 \text{ cm}^2 \).
In simple words: Plug the given radius into the sphere surface area formula and calculate the result.

Exam Tip: Spheres only have one surface area formula because they do not have separate curved and flat surfaces.

Question 20. The radius of a hemi-spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of the surface areas of the balloon in the two cases.
Answer:
Let the initial radius be \( r_1 = 7 \text{ cm} \) and the final radius be \( r_2 = 14 \text{ cm} \).
The surface area of a hemisphere (curved or total) is proportional to the square of its radius (\( r^2 \)).
Ratio of surface areas = \( \frac{2\pi r_1^2}{2\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 \)
\( = \left(\frac{7}{14}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).

Thus, the ratio of the surface areas in the two cases is \( 1:4 \).
In simple words: Since the area depends on the square of the radius, when the radius doubles, the surface area becomes four times larger.

Exam Tip: You can skip calculating the actual numerical areas when asked for a ratio, as the constant terms (\( 2\pi \) or \( 3\pi \)) cancel out completely.

Question 21. The internal and the external diameters of a hollow hemi-spherical vessel are 24 cm and 25 cm respectively. The cost of painting one square meter of the surface is 7 paise. Find the total cost of painting the vessel all over.
Answer:
Note: While the question paper lists the unit as "one square meter" due to a typographical error, the standard physical problem uses "one square centimeter" because the vessel's dimensions are in centimeters. We will solve using "7 paise per square centimeter".

Given:
Inner radius (\( r \)) = \( \frac{24}{2} = 12 \text{ cm} \)
Outer radius (\( R \)) = \( \frac{25}{2} = 12.5 \text{ cm} \)

Total surface area to be painted includes the inner curved surface, the outer curved surface, and the flat circular rim at the top:
\( \text{Total Surface Area} = 2\pi r^2 + 2\pi R^2 + \pi(R^2 - r^2) \)
\( = \pi(2r^2 + 2R^2 + R^2 - r^2) \)
\( = \pi(3R^2 + r^2) \)
\( = \frac{22}{7} \times (3 \times 12.5^2 + 12^2) \)
\( = \frac{22}{7} \times (3 \times 156.25 + 144) \)
\( = \frac{22}{7} \times (468.75 + 144) \)
\( = \frac{22}{7} \times 612.75 \text{ cm}^2 \).

Now, calculate the cost of painting at the rate of 7 paise per \( \text{cm}^2 \):
Total Cost = \( \text{Area} \times \text{Rate} \)
\( = \left( \frac{22}{7} \times 612.75 \right) \times 7 \text{ paise} \)
\( = 22 \times 612.75 = 13,480.5 \text{ paise} = \text{Rs. } 134.81 \).
In simple words: Find the inner surface area, the outer surface area, and the area of the ring on top. Sum these areas up and multiply by the cost of painting.

Exam Tip: Notice how the 7 in the denominator of \( \pi = \frac{22}{7} \) cancels out perfectly with the painting rate of 7 paise, leaving an easy calculation.

Question 22. A storage tank consists of a circular cylinder, with a hemisphere adjoined at either ends. If the external diameter of the cylinder be 1.4 cm and its length be 5 m, what will be the cost of painting it on the outside at the rate of Rs. 10 per square meter?
Answer:
Note: The printed diameter "1.4 cm" is a typographic error in the source and should be \( 1.4 \text{ m} \) to align with the rest of the tank's dimensions.

Given:
Radius (\( r \)) = \( \frac{1.4}{2} = 0.7 \text{ m} \)
Length of the cylinder (\( h \)) = \( 5 \text{ m} \)

The total outside surface area to be painted is the curved surface of the cylinder plus the surface areas of the two adjoined hemispheres (which together form a single complete sphere):
Total Surface Area = Curved surface area of cylinder + Surface area of sphere
\( = 2\pi r h + 4\pi r^2 = 2\pi r(h + 2r) \)
\( = 2 \times \frac{22}{7} \times 0.7 \times (5 + 2(0.7)) \)
\( = 4.4 \times (5 + 1.4) \)
\( = 4.4 \times 6.4 = 28.16 \text{ m}^2 \).

Calculate the cost at Rs. 10 per square meter:
Total Cost = \( 28.16 \times 10 = \text{Rs. } 281.60 \).
In simple words: The tank is a cylinder with rounded dome ends. Find the side area of the cylinder and the area of the two ends combined, add them, and multiply by the cost of paint.

Exam Tip: The two hemispherical ends combine to make one sphere of the same radius, so you can save time by using the single sphere formula \( 4\pi r^2 \).

Question 23. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 6 cm and its slant height is 5 cm. Find the cost of painting the toy at the rate of Rs 5 per 1000 cm2.
Answer:
Given:
Diameter = \( 6 \text{ cm} \implies \) radius (\( r \)) = \( 3 \text{ cm} \)
Slant height (\( l \)) = \( 5 \text{ cm} \)

The total surface area of the toy is the curved surface area of the cone plus the curved surface area of the hemispherical base:
Total Area = \( \pi r l + 2\pi r^2 = \pi r(l + 2r) \)
\( = \frac{22}{7} \times 3 \times (5 + 2(3)) \)
\( = \frac{66}{7} \times (5 + 6) \)
\( = \frac{66}{7} \times 11 = \frac{726}{7} \approx 103.71 \text{ cm}^2 \).

Calculate the cost at Rs. 5 per \( 1000 \text{ cm}^2 \):
Total Cost = \( \frac{726}{7} \times \frac{5}{1000} \approx \text{Rs. } 0.52 \).
In simple words: Find the side area of the cone and the curved bottom of the hemisphere. Add them together to get the total area, and then calculate the price based on the rate.

Exam Tip: Since the cone sits directly on top of the hemisphere, the circular base is hidden inside and should not be included in the surface area.

Question 24. A matchbox measures 4 cm x 2.5 cm x 1.5 cm. What will be the volume of a packet containing 12 such boxes?
Answer:
The volume of a single cuboidal matchbox is calculated as:
\( V = \text{length} \times \text{width} \times \text{height} \)
\( V = 4 \times 2.5 \times 1.5 = 15 \text{ cm}^3 \).

The volume of a packet containing 12 such boxes is:
Total Volume = \( 12 \times 15 = 180 \text{ cm}^3 \).
In simple words: Find the volume of one matchbox by multiplying its length, width, and height, and then multiply that number by twelve to find the packet's total volume.

Exam Tip: Volume is expressed in cubic units (\( \text{cm}^3 \)), so make sure to use the correct exponent.

Question 25. The capacity of a tank, which is cuboidal in shape, is 50,000 l. Find the breadth of the tank if its length and depth are respectively 2.5 m and 10 m.
Answer:
First, convert the capacity from liters to cubic meters. Since \( 1000 \text{ liters} = 1 \text{ m}^3 \):
Volume (\( V \)) = \( \frac{50,000}{1000} = 50 \text{ m}^3 \).

Now, use the cuboid volume formula:
\( V = \text{length} \times \text{breadth} \times \text{depth} \)
\( 50 = 2.5 \times b \times 10 \)
\( 50 = 25b \)

\( \implies b = 2 \text{ m} \).

Thus, the breadth of the tank is \( 2 \text{ m} \).
In simple words: Convert the liters to cubic meters, then divide this volume by the product of the length and height to find the missing width.

Exam Tip: Remember the metric conversion factor: \( 1 \text{ m}^3 = 1000 \text{ liters} \). This is a fundamental conversion in volume problems.

Question 26. A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm. Find the rise in water level in the vessel.
Answer:
When the cube is fully submerged, the volume of water displaced is equal to the volume of the cube.
Volume of cube = \( \text{edge}^3 = 9^3 = 729 \text{ cm}^3 \).

The water in the rectangular vessel rises in a cuboidal shape with a base area of \( 15 \text{ cm} \times 12 \text{ cm} \). Let the rise in height be \( h \):
Volume of displaced water = \( \text{Base Area} \times \text{Rise in height} \)
\( 729 = 15 \times 12 \times h \)
\( 729 = 180h \)

\( \implies h = \frac{729}{180} = 4.05 \text{ cm} \).

Thus, the rise in water level is \( 4.05 \text{ cm} \).
In simple words: Calculate the volume of the cube first. This volume represents the exact amount of water that is pushed upward. Divide it by the floor area of the container to find how high the water rises.

Exam Tip: Use the principle of displacement: Volume of submerged solid = Volume of raised liquid.

Question 27. A solid cube of side 12 cm is cut into 8 cubes of equal volume. What will be the side of the new cube? Also find the ratio between their surface areas.
Answer:
Volume of the large cube = \( 12^3 = 1728 \text{ cm}^3 \).
Since it is divided into 8 cubes of equal volume:
Volume of each small cube = \( \frac{1728}{8} = 216 \text{ cm}^3 \).

Let the side of the new cube be \( a \):
\( a^3 = 216 \)

\( \implies a = 6 \text{ cm} \).

Now, find the ratio of their surface areas:
Surface Area of large cube (\( S_1 \)) = \( 6 \times 12^2 \)
Surface Area of small cube (\( S_2 \)) = \( 6 \times 6^2 \)
Ratio = \( \frac{S_1}{S_2} = \frac{6 \times 12^2}{6 \times 6^2} = \frac{144}{36} = \frac{4}{1} \implies 4:1 \).
In simple words: Divide the volume of the large cube by eight to get the volume of a small cube, then take the cube root to find its side. The surface area ratio is the squared ratio of their side lengths.

Exam Tip: The ratio of surface areas of two similar solids is always the square of the ratio of their corresponding linear dimensions: \( (12/6)^2 = 2^2 = 4 \).

Question 28. How many cubic centimeters of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If one cubic cm of iron weighs 15 g, find the weight of the empty box in kg.
Answer:
Given:
External dimensions: Length (\( L \)) = \( 36 \text{ cm} \), Breadth (\( B \)) = \( 25 \text{ cm} \), Height (\( H \)) = \( 16.5 \text{ cm} \)
Thickness (\( t \)) = \( 1.5 \text{ cm} \)

Since the box is open at the top, the thickness is subtracted twice from length and breadth, but only once from height to get the internal dimensions:
Internal Length (\( l \)) = \( 36 - 2(1.5) = 33 \text{ cm} \)
Internal Breadth (\( b \)) = \( 25 - 2(1.5) = 22 \text{ cm} \)
Internal Height (\( h \)) = \( 16.5 - 1.5 = 15 \text{ cm} \)

External Volume = \( 36 \times 25 \times 16.5 = 14,850 \text{ cm}^3 \)
Internal Volume = \( 33 \times 22 \times 15 = 10,890 \text{ cm}^3 \)

Volume of iron used = External Volume - Internal Volume
\( = 14,850 - 10,890 = 3960 \text{ cm}^3 \).

Now, find the weight of the box:
Weight = \( 3960 \times 15 \text{ g} = 59,400 \text{ g} \)
Weight in kg = \( \frac{59,400}{1000} = 59.4 \text{ kg} \).
In simple words: Subtract the wall thickness from the outer dimensions to find the inner space size. The volume of the iron is the outer box volume minus the inner empty volume. Multiply this iron volume by the density to get the final weight.

Exam Tip: Be very careful when calculating internal dimensions for an open box. Do not subtract the thickness twice from the height because there is no lid at the top.

Question 29. A well with 10 m inside diameter is dug 14 m deep. Earth taken out of it is spread all around to a width of 5 m to form an embankment. Find the height of the embankment.
Answer:
Given:
Diameter of well = \( 10 \text{ m} \implies \) radius (\( r \)) = \( 5 \text{ m} \)
Depth of well (\( h \)) = \( 14 \text{ m} \)

Volume of earth dug out = Volume of cylinder
\( = \pi r^2 h = \pi \times 5^2 \times 14 = 350\pi \text{ m}^3 \).

The embankment is spread around the well to a width of \( 5 \text{ m} \).
Inner radius of embankment (\( r_1 \)) = \( 5 \text{ m} \)
Outer radius of embankment (\( r_2 \)) = \( 5 + 5 = 10 \text{ m} \)

Area of the embankment ring = \( \pi(r_2^2 - r_1^2) \)
\( = \pi(10^2 - 5^2) = 75\pi \text{ m}^2 \).

Let the height of the embankment be \( H \).
Volume of embankment = Volume of earth dug out
\( 75\pi \times H = 350\pi \)
\( 75H = 350 \)

\( \implies H = \frac{350}{75} = \frac{14}{3} \approx 4.67 \text{ m} \).
In simple words: Find the volume of the dug-out earth. The earth is spread in a circular ring around the well. Divide the earth's volume by the flat surface area of this ring to get its height.

Exam Tip: Keep the calculations in terms of \( \pi \) as long as possible so they cancel out on both sides of the equation at the end.

Question 30. How many liters of water can flow out of a pipe having an area of cross-section of 5 cm2 in one minute, if the speed of water in the pipe is 30 cm/sec?
Answer:
Given:
Area of cross-section (\( A \)) = \( 5 \text{ cm}^2 \)
Speed of water (\( v \)) = \( 30 \text{ cm/sec} \)

Volume of water flowing out in 1 second:
\( V_{\text{per sec}} = A \times v = 5 \times 30 = 150 \text{ cm}^3 \).

Volume of water flowing out in 1 minute (60 seconds):
\( V_{\text{per min}} = 150 \times 60 = 9000 \text{ cm}^3 \).

Convert the volume to liters (where \( 1000 \text{ cm}^3 = 1 \text{ liter} \)):
Volume in liters = \( \frac{9000}{1000} = 9 \text{ liters} \).
In simple words: Multiply the cross-section area by the flow speed to find the water volume released per second. Multiply that by sixty seconds to get the volume per minute, and then convert cubic centimeters to liters.

Exam Tip: Make sure to match the time units — if the speed is per second, multiply by 60 to find the flow rate per minute.

Question 31. The ratio between the radii of the base and the height of the cylinder is 2 : 3 what is the Total surface area if the volume of the cylinder is 1617 cm3.
Answer:
Let the radius of the base \( r = 2x \) and the height \( h = 3x \).
The volume of a cylinder is given by \( \pi r^2 h \).
Given that the volume is \( 1617 \text{ cm}^3 \):
\( \frac{22}{7} \times (2x)^2 \times (3x) = 1617 \)
\( \frac{22}{7} \times 4x^2 \times 3x = 1617 \)
\( \frac{264}{7} x^3 = 1617 \)
\( x^3 = \frac{1617 \times 7}{264} \)
\( x^3 = \frac{147 \times 7}{24} = \frac{49 \times 7}{8} = \frac{343}{8} \)

\( \implies x = \frac{7}{2} = 3.5 \text{ cm} \).

Now, find the dimensions:
Radius (\( r \)) = \( 2 \times 3.5 = 7 \text{ cm} \)
Height (\( h \)) = \( 3 \times 3.5 = 10.5 \text{ cm} \)

Total Surface Area = \( 2\pi r(r + h) \)
\( = 2 \times \frac{22}{7} \times 7 \times (7 + 10.5) \)
\( = 44 \times 17.5 = 770 \text{ cm}^2 \).
In simple words: Set up the volume equation with the ratio variable to solve for \( x \). Use it to find the radius and height, and then calculate the total surface area of the cylinder.

Exam Tip: Simplify the fraction step-by-step by dividing both numerator and denominator by common factors to find the cube root easily.

Question 32. The trunk of a tree is cylindrical in shape and its circumference is 176 c m. If the length of the trunk is 3 m. Find the volume of timber that can be obtained from the trunk.
Answer:
Given:
Circumference of the cylinder (\( 2\pi r \)) = \( 176 \text{ cm} \)
Length of the trunk (\( h \)) = \( 3 \text{ m} = 300 \text{ cm} \)

First, find the radius \( r \):
\( 2 \times \frac{22}{7} \times r = 176 \)
\( \frac{44}{7} r = 176 \)

\( \implies r = \frac{176 \times 7}{44} = 28 \text{ cm} \).

Now, find the volume of the cylindrical trunk:
\( \text{Volume} = \pi r^2 h = \frac{22}{7} \times 28 \times 28 \times 300 \)
\( = 22 \times 4 \times 28 \times 300 = 739,200 \text{ cm}^3 \).
In cubic meters:
Volume = \( \frac{739,200}{1,000,000} = 0.7392 \text{ m}^3 \).
In simple words: Find the radius from the circumference, then plug the radius and the length of the tree trunk into the cylinder volume formula.

Exam Tip: Be sure to keep your units consistent: either convert all measurements to centimeters or all measurements to meters before starting.

Question 33. Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.
Answer:
Given:
Total weight of wire = \( 13.2 \text{ kg} = 13,200 \text{ g} \)
Density of copper = \( 8.4 \text{ g/cm}^3 \)
Diameter of wire = \( 4 \text{ mm} = 0.4 \text{ cm} \implies \) radius (\( r \)) = \( 0.2 \text{ cm} \)

First, find the volume of the copper wire:
\( \text{Volume} = \frac{\text{Total Weight}}{\text{Density}} = \frac{13,200}{8.4} = \frac{11,000}{7} \text{ cm}^3 \).

Since the wire is cylindrical, use the cylinder volume formula to find its length (\( L \)):
\( \text{Volume} = \pi r^2 L \)
\( \frac{11,000}{7} = \frac{22}{7} \times (0.2)^2 \times L \)
\( 11,000 = 22 \times 0.04 \times L \)
\( 11,000 = 0.88 L \)

\( \implies L = \frac{11,000}{0.88} = 12,500 \text{ cm} = 125 \text{ m} \).
In simple words: Divide the total weight of the wire by the weight of one cubic centimeter to get the total volume of copper. Then set this equal to the cylinder volume formula and solve for the wire's length.

Exam Tip: Remember to convert the diameter from millimeters to centimeters to match the density units (\( \text{g/cm}^3 \)).

Question 34. The diameter of a right circular cone is 8 cm and its volume is 48 \(\pi\) cm3. What is the height of the cone?
Answer:
Given:
Diameter = \( 8 \text{ cm} \implies \) radius (\( r \)) = \( 4 \text{ cm} \)
Volume (\( V \)) = \( 48\pi \text{ cm}^3 \)

Using the cone volume formula:
\( V = \frac{1}{3} \pi r^2 h \)
\( 48\pi = \frac{1}{3} \pi \times 4^2 \times h \)
\( 48 = \frac{16}{3} h \)
\( h = \frac{48 \times 3}{16} \)

\( \implies h = 9 \text{ cm} \).

Thus, the height of the cone is \( 9 \text{ cm} \).
In simple words: Plug the radius and the volume containing \(\pi\) into the volume formula. Cancel \(\pi\) on both sides and solve for the vertical height of the cone.

Exam Tip: Keeping \( \pi \) as a symbol rather than substituting its value allows for quick cancellation, saving valuable exam time.

Question 35. The volume of a cone is 18480 cm3. If the height of the cone is 40 cm. Find the radius of the base.
Answer:
Using the cone volume formula:
\( V = \frac{1}{3} \pi r^2 h \)
Given that Volume = \( 18,480 \text{ cm}^3 \) and height \( h = 40 \text{ cm} \):
\( 18,480 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 40 \)
\( 18,480 = \frac{880}{21} r^2 \)
\( r^2 = \frac{18,480 \times 21}{880} \)
\( r^2 = 21 \times 21 = 441 \)

\( \implies r = \sqrt{441} = 21 \text{ cm} \).

Thus, the radius of the base is \( 21 \text{ cm} \).
In simple words: Plug the height and total volume into the cone volume equation and solve to find the squared radius, then take the square root.

Exam Tip: When dividing large numbers, look for simple factors like 10 or 11 first to reduce the expression quickly.

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Question 36. A right triangle ABC with its sides 5 cm, 12 cm and 13 cm is revolved about its side of 12 cm. Find the volume of the right circular cone so formed.
Answer:
When the right triangle is revolved around its \( 12 \text{ cm} \) side, that side acts as the vertical axis (height) of the cone, while the perpendicular \( 5 \text{ cm} \) side sweeps out the circular base. The hypotenuse of \( 13 \text{ cm} \) forms the slant height.
So, height (\( h \)) = \( 12 \text{ cm} \), radius (\( r \)) = \( 5 \text{ cm} \).

Volume of the cone is:
\( V = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \pi \times 5^2 \times 12 \)
\( = 4 \times 25\pi = 100\pi \text{ cm}^3 \approx 314.29 \text{ cm}^3 \).
In simple words: Revolving the triangle forms a cone. The side it rotates around becomes the height, and the other short side becomes the base radius. Plug these into the volume formula to find the answer.

Exam Tip: The side about which the triangle is revolved always becomes the height of the resulting cone.

Question 37. A cone of radius 5 cm is filled with water. If the water is poured in a cylinder of radius 10 cm, the height of the water rises by 2 cm , find the height of the cone.
Answer:
The volume of water poured from the cone equals the volume of the water rising in the cylinder.
Let the height of the cone be \( H_c \).
Volume of water in cone = \( \frac{1}{3} \pi \times r_{\text{cone}}^2 \times H_c = \frac{1}{3} \pi \times 5^2 \times H_c = \frac{25}{3}\pi H_c \).

Volume of water raised in cylinder = \( \pi \times r_{\text{cyl}}^2 \times h_{\text{rise}} = \pi \times 10^2 \times 2 = 200\pi \text{ cm}^3 \).

Equating the two volumes:
\( \frac{25}{3}\pi H_c = 200\pi \)
\( 25 H_c = 600 \)

\( \implies H_c = \frac{600}{25} = 24 \text{ cm} \).

Thus, the height of the cone is \( 24 \text{ cm} \).
In simple words: Calculate the volume of water in the cylinder based on how high it rose. This volume is identical to the cone's volume, which lets you solve for the cone's height.

Exam Tip: Since water is transferred without loss, equate the volume formulas of the two shapes to solve for the unknown parameter.

Question 38. A solid cube of side 7 cm is melted to make a cone of height 5 cm, find the radius of the base of the cone.
Answer:
When a solid is melted and recast into another shape, its total volume remains the same.
Volume of the solid cube = \( \text{side}^3 = 7^3 = 343 \text{ cm}^3 \).

Using the volume formula for the new cone with height \( h = 5 \text{ cm} \):
\( \text{Volume of cone} = \frac{1}{3} \pi r^2 h \)
\( 343 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 5 \)
\( 343 = \frac{110}{21} r^2 \)
\( r^2 = \frac{343 \times 21}{110} \)
\( r^2 = \frac{7203}{110} \approx 65.48 \)

\( \implies r = \sqrt{65.48} \approx 8.09 \text{ cm} \).

Thus, the base radius of the cone is approximately \( 8.09 \text{ cm} \).
In simple words: Set the volume of the cube equal to the volume formula of the cone. Solve for the radius, taking the square root at the end.

Exam Tip: Write down your rounding steps clearly when dealing with non-perfect squares to ensure you receive full step-marks.

Question 39. Find the volume of the largest right circular cone that can be fitted in a cube of edge 14 cm.
Answer:
The largest cone that can fit inside a cube of edge \( a = 14 \text{ cm} \) will have:
1. A base diameter equal to the edge of the cube: \( \text{Diameter} = 14 \text{ cm} \implies \) radius (\( r \)) = \( 7 \text{ cm} \).
2. A height equal to the edge of the cube: \( h = 14 \text{ cm} \).

Using the cone volume formula:
\( V = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 14 \)
\( = \frac{1}{3} \times 22 \times 7 \times 14 \)
\( = \frac{2156}{3} \approx 718.67 \text{ cm}^3 \).
In simple words: The widest part of the cone matches the cube's width, and the cone's height matches the cube's height. Use these measurements to calculate the cone's volume.

Exam Tip: Visualizing the cone inside the cube makes it clear that the cone's diameter and height are both equal to the cube's side length.

Question 40. A solid lead ball of radius 7 cm was melted and then drawn into a wire of diameter 0.2 cm. Find the length of the wire.
Answer:
The volume of the sphere (lead ball) equals the volume of the cylindrical wire.
Volume of lead ball = \( \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \times 7^3 = \frac{1372}{3}\pi \text{ cm}^3 \).

Radius of the wire (\( r \)) = \( \frac{0.2}{2} = 0.1 \text{ cm} \).
Let the length of the wire be \( H \).
Volume of wire = \( \pi r^2 H = \pi \times (0.1)^2 \times H = 0.01\pi H \).

Equating the volumes:
\( 0.01\pi H = \frac{1372}{3}\pi \)
\( 0.01 H = \frac{1372}{3} \)
\( H = \frac{1372}{3 \times 0.01} = \frac{137,200}{3} \approx 45,733.33 \text{ cm} \).

In meters:
Length of wire = \( \frac{45,733.33}{100} = 457.33 \text{ m} \).
In simple words: Calculate the volume of the sphere. Set this equal to the volume formula of the thin cylinder (wire) and solve to find its length.

Exam Tip: Be sure to keep track of the units, converting the final length from centimeters to meters to make the final answer more readable.

Question 41. How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter.
Answer:
Volume of lead cube = \( \text{edge}^3 = 44^3 = 85,184 \text{ cm}^3 \).

Radius of each spherical bullet (\( r \)) = \( \frac{4}{2} = 2 \text{ cm} \).
Volume of one spherical bullet = \( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 2^3 = \frac{704}{21} \text{ cm}^3 \).

Number of bullets = \( \frac{\text{Volume of cube}}{\text{Volume of one bullet}} \)
\( = \frac{85,184}{\frac{704}{21}} = \frac{85,184 \times 21}{704} = 121 \times 21 = 2541 \).
Thus, \( 2541 \) bullets can be made.
In simple words: Find the total volume of the cube and the volume of a single tiny bullet. Divide the cube's volume by the bullet's volume to find the total quantity.

Exam Tip: Instead of calculating decimals for intermediate steps, keep your numbers as fractions to let terms simplify nicely at the end.

Question 42. 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 S and S'
Answer:
Volume of 27 small spheres = \( 27 \times \left(\frac{4}{3}\pi r^3\right) \).
Let the radius of the large sphere be \( r' \).
Volume of the new large sphere = \( \frac{4}{3}\pi (r')^3 \).

Since the volumes are equal:
\( \frac{4}{3}\pi (r')^3 = 27 \times \left(\frac{4}{3}\pi r^3\right) \)
\( (r')^3 = 27 r^3 \)

\( \implies r' = 3r \).

Now, find the ratio of their surface areas \( S \) and \( S' \):
\( S = 4\pi r^2 \)
\( S' = 4\pi (r')^2 = 4\pi (3r)^2 = 36\pi r^2 \)

Ratio of \( S \) and \( S' \) is:
\( \frac{S}{S'} = \frac{4\pi r^2}{36\pi r^2} = \frac{1}{9} \implies 1:9 \).
In simple words: Combine the volumes of the twenty-seven spheres to find the size of the new sphere, showing its radius is three times larger. Then square this scale factor to find the surface area ratio.

Exam Tip: When the radius is scaled by a factor \( k \), the volume scales by \( k^3 \) and the surface area scales by \( k^2 \).

Question 43. A dome of a building is in the form of hemisphere. From inside, it was white washed at the cost of Rs.498.96. If the cost of whitewashing it, is Rs.2.00 per square meter, find the inside surface area of the dome and the volume of air inside the dome.
Answer:
First, find the inner curved surface area of the dome:
\( \text{Inside Surface Area} = \frac{\text{Total Cost of Whitewashing}}{\text{Rate per square meter}} = \frac{498.96}{2.00} = 249.48 \text{ m}^2 \).

Since the dome is a hemisphere, its inside surface area is \( 2\pi r^2 \):
\( 2\pi r^2 = 249.48 \)
\( 2 \times \frac{22}{7} \times r^2 = 249.48 \)
\( r^2 = \frac{249.48 \times 7}{44} = 39.69 \)

\( \implies r = \sqrt{39.69} = 6.3 \text{ m} \).

Now, find the volume of air inside the dome (volume of hemisphere):
\( \text{Volume} = \frac{2}{3} \pi r^3 \)
\( = \frac{2}{3} \times \frac{22}{7} \times 6.3 \times 6.3 \times 6.3 \)
\( = \frac{2}{3} \times 22 \times 0.9 \times 39.69 \)
\( = 44 \times 0.3 \times 39.69 = 13.2 \times 39.69 = 523.908 \text{ m}^3 \).
In simple words: Divide the total whitewashing bill by the rate to find the inside area of the dome. Use this area to find the radius, and then calculate the volume of air inside using the hemisphere volume formula.

Exam Tip: Be careful to use the hemisphere volume formula (\( \frac{2}{3}\pi r^3 \)) rather than the full sphere volume formula.