CBSE Class 9 Mathematics Surface Areas And Volumes Worksheet Set 04

Question 1. Find the area enclosed between two concentric circles of radii 4 cm and 3 cm.
Answer:
Let the radius of the outer circle be \( R = 4 \text{ cm} \) and the radius of the inner circle be \( r = 3 \text{ cm} \).
The area enclosed between the two concentric circles is given by the difference between the areas of the outer and inner circles:
\( \text{Enclosed Area} = \pi R^2 - \pi r^2 \)
\( \implies \text{Enclosed Area} = \pi (R^2 - r^2) \)
\( \implies \text{Enclosed Area} = \frac{22}{7} \times (4^2 - 3^2) \)
\( \implies \text{Enclosed Area} = \frac{22}{7} \times (16 - 9) \)
\( \implies \text{Enclosed Area} = \frac{22}{7} \times 7 \)
\( \implies \text{Enclosed Area} = 22 \text{ cm}^2 \)
Therefore, the area enclosed between the two circles is 22 square centimetres.
In simple words: To find the area between two circular rings, subtract the area of the smaller circle from the larger one.

Exam Tip: Remember to use the formula \( \pi(R^2 - r^2) \) directly to save time and prevent calculation errors.

Question 2. 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. The diameter of the roller is 1.4 m, so its radius is:
\( r = \frac{1.4}{2} = 0.7 \text{ m} \)
The length (height) of the cylinder is \( h = 2 \text{ m} \).
The area covered in one complete revolution is equal to the curved surface area of the cylinder:
\( \text{Curved Surface Area} = 2\pi r h \)
\( \implies \text{Curved Surface Area} = 2 \times \frac{22}{7} \times 0.7 \times 2 \)
\( \implies \text{Curved Surface Area} = 8.8 \text{ m}^2 \)
The area covered by the roller in 5 complete revolutions is:
\( \text{Total Area} = 5 \times 8.8 \)
\( \implies \text{Total Area} = 44 \text{ m}^2 \)
Thus, the total area covered in 5 revolutions is 44 square metres.
In simple words: One turn of the roller covers an area equal to its outer curved side. Multiply this by five to get the total area covered.

Exam Tip: Pay attention to the units - ensure both length and diameter are in metres before starting calculations.

Question 3. A cuboid has total surface area of 40 sq m and its lateral surface area is 26 sq m. Find the area of base.
Answer:
Let the length, breadth, and height of the cuboid be \( l \), \( b \), and \( h \) respectively.
The total surface area (TSA) of a cuboid is given by:
\( \text{TSA} = 2(lb + bh + hl) = 40 \text{ m}^2 \)
The lateral surface area (LSA) of a cuboid is given by:
\( \text{LSA} = 2(l + b)h = 2(lh + bh) = 26 \text{ m}^2 \)
We know that the total surface area is the sum of the lateral surface area and twice the area of the base:
\( \text{TSA} = \text{LSA} + 2 \times (\text{Area of Base}) \)
\( \implies 40 = 26 + 2 \times (\text{Area of Base}) \)
\( \implies 2 \times (\text{Area of Base}) = 40 - 26 \)
\( \implies 2 \times (\text{Area of Base}) = 14 \)
\( \implies \text{Area of Base} = \frac{14}{2} = 7 \text{ m}^2 \)
Hence, the area of the base of the cuboid is 7 square metres.
In simple words: Subtracting the side walls' area from the total area leaves the area of the top and bottom. Halving that gives the base area.

Exam Tip: Understanding the relationship between total surface area, lateral surface area, and base area helps solve such problems without needing individual dimensions.

Question 4. Three metal cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. Find the edge of the new cube. Also find the surface area of the new cube.
Answer:
Let the edges of the three metal cubes be \( a_1 = 3 \text{ cm} \), \( a_2 = 4 \text{ cm} \), and \( a_3 = 5 \text{ cm} \).
The sum of the volumes of these three cubes is:
\( \text{Total Volume} = a_1^3 + a_2^3 + a_3^3 \)
\( \implies \text{Total Volume} = 3^3 + 4^3 + 5^3 \)
\( \implies \text{Total Volume} = 27 + 64 + 125 = 216 \text{ cm}^3 \)
Let the edge of the newly formed single cube be \( a \). Since the total volume remains the same:
\( a^3 = 216 \)
\( \implies a = \sqrt[3]{216} = 6 \text{ cm} \)
Now, the total surface area of the new cube is:
\( \text{Surface Area} = 6a^2 \)
\( \implies \text{Surface Area} = 6 \times 6^2 = 216 \text{ cm}^2 \)
Thus, the edge of the new cube is 6 cm, and its total surface area is 216 square centimetres.
In simple words: Melting shapes means their total space or volume stays the same. Find the combined volume first, then use it to find the new cube's side and surface area.

Exam Tip: Show the intermediate volume sum step clearly to secure full marks for the volume equivalence principle.

Question 5. An iron pipe 20 cm long has exterior diameter equal to 50 cm. If the thickness of the pipe is 1cm, find the whole surface area of the pipe.
Answer:
Let the length of the iron pipe be \( h = 20 \text{ cm} \).
The external diameter is 50 cm, so the external radius is:
\( R = \frac{50}{2} = 25 \text{ cm} \)
Since the thickness of the pipe is 1 cm, the internal radius is:
\( r = 25 - 1 = 24 \text{ cm} \)
The total surface area of a hollow open pipe includes the outer curved surface, the inner curved surface, and the areas of the two circular ring bases:
\( \text{Total Surface Area} = 2\pi R h + 2\pi r h + 2\pi(R^2 - r^2) \)
\( \implies \text{Total Surface Area} = 2\pi [h(R + r) + (R^2 - r^2)] \)
\( \implies \text{Total Surface Area} = 2 \times \frac{22}{7} \times [20(25 + 24) + (25^2 - 24^2)] \)
\( \implies \text{Total Surface Area} = \frac{44}{7} \times [20(49) + (625 - 576)] \)
\( \implies \text{Total Surface Area} = \frac{44}{7} \times [980 + 49] \)
\( \implies \text{Total Surface Area} = \frac{44}{7} \times 1029 \)
\( \implies \text{Total Surface Area} = 44 \times 147 = 6468 \text{ cm}^2 \)
Hence, the whole surface area of the iron pipe is 6468 square centimetres.
In simple words: To find the complete surface area of a hollow pipe, add the outer side, the inner side, and the flat rings on both ends.

Exam Tip: Do not forget to include the area of the two circular rings at the ends when calculating the whole surface area of a pipe.

Question 6. The lateral surface of a cylinder is equal to the curved surface of a cone. If the radius is the same, find the ratio of the height of the cylinder and slant height of the cone.
Answer:
Let the common radius of both the cylinder and the cone be \( r \).
Let the height of the cylinder be \( h \) and the slant height of the cone be \( l \).
The lateral (curved) surface area of the cylinder is \( 2\pi r h \).
The curved surface area of the cone is \( \pi r l \).
According to the given condition:
\( 2\pi r h = \pi r l \)
\( \implies 2h = l \)
\( \implies \frac{h}{l} = \frac{1}{2} \)
Thus, the ratio of the height of the cylinder to the slant height of the cone is \( 1 : 2 \).
In simple words: Since their areas and radii are equal, the cylinder's height must be exactly half of the cone's slant height to match.

Exam Tip: Be careful with the order of the ratio requested: height of cylinder to slant height of cone, which is \( 1 : 2 \).

Question 7. A right circular cylinder just enclosed a sphere of radius r as shown in figure find the surface area of the sphere , curved surface area of the cylinder and also their ratio.
Answer:
Let the radius of the enclosed sphere be \( r \).
1. The surface area of the sphere is:
\( \text{Surface Area of Sphere} = 4\pi r^2 \)

2. Since the cylinder just encloses the sphere:
- The radius of the cylinder's base is equal to the radius of the sphere, \( R = r \).
- The height of the cylinder is equal to the diameter of the sphere, \( h = 2r \).
The curved surface area (CSA) of the cylinder is:
\( \text{CSA of Cylinder} = 2\pi R h \)
\( \implies \text{CSA of Cylinder} = 2\pi (r)(2r) = 4\pi r^2 \)

3. The ratio of their surface areas is:
\( \text{Ratio} = \frac{\text{Surface Area of Sphere}}{\text{CSA of Cylinder}} = \frac{4\pi r^2}{4\pi r^2} = 1 \)
Thus, the ratio is \( 1 : 1 \).

In simple words: When a cylinder perfectly fits around a ball, the height of the cylinder is the ball's diameter. This makes their outer surface areas exactly identical.

Exam Tip: Expressing the height of the cylinder as \( 2r \) is the key substitution step that leads directly to the correct ratio of \( 1 : 1 \).

Question 8. A godown is in the form of a cuboid measuring 60 m x 40 m x 20 m. How many cuboidal boxes can be stored in it if the volume of one box 0.8 m³?
Answer:
The dimensions of the cuboidal godown are \( L = 60 \text{ m} \), \( B = 40 \text{ m} \), and \( H = 20 \text{ m} \).
The volume of the godown is:
\( \text{Volume of Godown} = L \times B \times H \)
\( \implies \text{Volume of Godown} = 60 \times 40 \times 20 = 48000 \text{ m}^3 \)
The volume of one cuboidal box is \( 0.8 \text{ m}^3 \).
The total number of boxes that can be accommodated in the godown is:
\( \text{Number of Boxes} = \frac{\text{Volume of Godown}}{\text{Volume of One Box}} \)
\( \implies \text{Number of Boxes} = \frac{48000}{0.8} = \frac{480000}{8} = 60000 \)
Therefore, 60,000 cuboidal boxes can be stored in the godown.
In simple words: Find the total space inside the warehouse first, then divide it by the space that a single box takes up.

Exam Tip: Ensure that the units of volume for both the godown and the boxes are identical before performing the division.

Question 9. The diameter of a sphere is decreased by 50%. What is the ratio between initial and final curved surface areas?
Answer:
Let the initial radius of the sphere be \( R \), so its initial diameter is \( D = 2R \).
The initial surface area of the sphere is:
\( A_1 = 4\pi R^2 \)
The diameter of the sphere is reduced by 50%, which means the new radius \( R' \) is also reduced by 50%:
\( R' = 0.5R = \frac{R}{2} \)
The final surface area of the sphere is:
\( A_2 = 4\pi (R')^2 \)
\( \implies A_2 = 4\pi \left(\frac{R}{2}\right)^2 = \frac{4\pi R^2}{4} \)
The ratio of the initial surface area to the final surface area is:
\( \text{Ratio} = \frac{A_1}{A_2} = \frac{4\pi R^2}{\frac{4\pi R^2}{4}} = 4 \)
Thus, the ratio of the initial to the final surface area is \( 4 : 1 \).
In simple words: Halving the diameter of a sphere cuts its radius in half. This reduces the surface area to one-fourth of what it originally was.

Exam Tip: Always write down the initial and final states separately with clear variables to avoid arithmetic confusion during ratio calculations.

Question 10. Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14 cm.
Answer:
For the largest cone that can fit inside a cube of edge \( a = 14 \text{ cm} \):
- The base of the cone will touch the bottom face of the cube, so its base diameter is equal to the edge of the cube. Therefore, the radius \( r \) is:
\( r = \frac{14}{2} = 7 \text{ cm} \)
- The vertex of the cone will touch the top face of the cube, so its height \( h \) is equal to the edge of the cube:
\( h = 14 \text{ cm} \)
The volume of the cone is given by:
\( \text{Volume} = \frac{1}{3}\pi r^2 h \)
\( \implies \text{Volume} = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 14 \)
\( \implies \text{Volume} = \frac{1}{3} \times \frac{22}{7} \times 49 \times 14 \)
\( \implies \text{Volume} = \frac{1}{3} \times 22 \times 7 \times 14 \)
\( \implies \text{Volume} = \frac{2156}{3} \approx 718.67 \text{ cm}^3 \)
Thus, the volume of the largest cone is \( 718.67 \text{ cm}^3 \) (or \( \frac{2156}{3} \text{ cm}^3 \)).
In simple words: The biggest cone inside a box has a base as wide as the box and a height as tall as the box. Use these dimensions to find its volume.

Exam Tip: State the relationship between the cone's dimensions and the cube's edge clearly before substituting them into the volume formula.

Question 11. The semi-circular sheet of metal of diameter 28 cm is bent into an open conical cup. Find the depth and the capacity of cup.
Answer:
Let the diameter of the semi-circular sheet be 28 cm, so its radius is \( R = 14 \text{ cm} \).
When this semi-circular sheet is folded to form an open conical cup:
- The radius of the sheet becomes the slant height of the cone, \( l = R = 14 \text{ cm} \).
- The arc length of the semi-circle becomes the circumference of the cone's base. Let \( r \) be the radius of the cone's base:
\( 2\pi r = \pi R \)
\( \implies 2r = R \)
\( \implies 2r = 14 \)
\( \implies r = 7 \text{ cm} \)
Let \( h \) be the depth (height) of the conical cup. Using the Pythagorean relation:
\( h^2 + r^2 = l^2 \)
\( \implies h = \sqrt{l^2 - r^2} \)
\( \implies h = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} = 7\sqrt{3} \approx 12.12 \text{ cm} \)
The capacity (volume) of the cup is:
\( \text{Capacity} = \frac{1}{3}\pi r^2 h \)
\( \implies \text{Capacity} = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 7\sqrt{3} \)
\( \implies \text{Capacity} = \frac{1078\sqrt{3}}{3} \approx 622.37 \text{ cm}^3 \)
Therefore, the depth of the cup is \( 7\sqrt{3} \text{ cm} \) (or approx 12.12 cm) and its capacity is \( \frac{1078\sqrt{3}}{3} \text{ cm}^3 \) (or approx 622.37 cubic centimetres).
In simple words: Folding a semi-circular sheet creates a cone where the sheet's radius is the diagonal edge (slant height) and the curved rim forms the circular base.

Exam Tip: Be sure to establish the correct conversion that the arc length of the sheet equals the base circumference of the cone.

Question 12. 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 embankment.
Answer:
The well is cylindrical. Its inside diameter is 10 m, so the inner radius \( r = 5 \text{ m} \).
The depth of the well is \( h = 14 \text{ m} \).
The volume of earth dug out of the well is:
\( \text{Volume of Earth} = \pi r^2 h \)
\( \implies \text{Volume of Earth} = \pi \times 5^2 \times 14 = 350\pi \text{ m}^3 \)
The earth is spread around the well to form a circular ring-shaped embankment of width 5 m.
- Inner radius of the embankment \( r_1 = 5 \text{ m} \)
- Outer radius of the embankment \( r_2 = 5 + 5 = 10 \text{ m} \)
The base area of this embankment ring is:
\( \text{Base Area} = \pi(r_2^2 - r_1^2) = \pi(10^2 - 5^2) = 75\pi \text{ m}^2 \)
Let the height of the embankment be \( H \). Since the volume of the embankment equals the volume of the dug-out earth:
\( \text{Base Area} \times H = \text{Volume of Earth} \)
\( \implies 75\pi \times H = 350\pi \)
\( \implies H = \frac{350}{75} = \frac{14}{3} \approx 4.67 \text{ m} \)
Hence, the height of the embankment is approximately 4.67 metres.
In simple words: The dirt pulled from the well is used to make a ring wall around the opening. Divide the volume of dirt by the flat area of the ring to find the wall's height.

Exam Tip: Keep the calculations in terms of \( \pi \) and cancel it out at the end to make calculations simple and highly accurate.

Question 13. The radius and slant height of a cone are in the ratio 4 : 7. If its curved surface area is 792 sq cm, find its radius.
Answer:
Let the radius \( r \) and the slant height \( l \) of the cone be \( 4x \) and \( 7x \), where \( x \) is a common constant ratio multiplier.
The curved surface area of a cone is given by:
\( \text{Curved Surface Area} = \pi r l = 792 \text{ cm}^2 \)
\( \implies \frac{22}{7} \times 4x \times 7x = 792 \)
\( \implies 22 \times 4x^2 = 792 \)
\( \implies 88x^2 = 792 \)
\( \implies x^2 = \frac{792}{88} = 9 \)
\( \implies x = 3 \)
Therefore, the radius of the cone is:
\( r = 4 \times 3 = 12 \text{ cm} \)
The radius of the cone is 12 cm.
In simple words: Use the ratio to write the radius and slant height in terms of a variable, plug them into the area formula, solve for that variable, and calculate the radius.

Exam Tip: When dealing with ratios in surface area, remember that the variable will be squared (e.g., \( x^2 \)), so make sure to take the square root at the end.

Question 14. The diameter of 0.84 m long roller is 1.5 m. If it takes 100 complete revolutions to level a playground, find the cost of levelling it at the rate of 50 paise per square metre.
Answer:
The cylindrical roller has:
- Length (height) \( h = 0.84 \text{ m} \)
- Diameter = 1.5 m, so the radius is \( r = 0.75 \text{ m} \).
The area covered in one single revolution is equal to its curved surface area:
\( \text{Area per Revolution} = 2\pi r h \)
\( \implies \text{Area per Revolution} = 2 \times \frac{22}{7} \times 0.75 \times 0.84 \)
\( \implies \text{Area per Revolution} = 2 \times 22 \times 0.75 \times 0.12 = 3.96 \text{ m}^2 \)
The total area leveled in 100 revolutions is:
\( \text{Total Area} = 100 \times 3.96 = 396 \text{ m}^2 \)
The cost of levelling is 50 paise (Rs. 0.50) per square metre. Hence, the total cost is:
\( \text{Total Cost} = 396 \times 0.50 = \text{Rs. } 198 \)
Thus, the cost of levelling the playground is Rs. 198.
In simple words: Find the outer side area of the roller, multiply it by 100 to get the total ground space, then multiply by the cost per square metre.

Exam Tip: Converting 50 paise to Rs. 0.50 early helps avoid unit mismatch errors in the final cost calculation.

Question 15. Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of sum of the surface areas of the three cubes.
Answer:
Let the side length of each of the three identical cubes be \( a \).
The total surface area of one cube is \( 6a^2 \).
The sum of the surface areas of the three individual cubes is:
\( \text{Sum of Areas} = 3 \times 6a^2 = 18a^2 \)
When these three cubes are placed side-by-side in a row, they form a cuboid with dimensions:
- Length \( l = 3a \)
- Breadth \( b = a \)
- Height \( h = a \)
The total surface area of this new cuboid is:
\( \text{TSA of Cuboid} = 2(lb + bh + hl) \)
\( \implies \text{TSA of Cuboid} = 2(3a \times a + a \times a + a \times 3a) \)
\( \implies \text{TSA of Cuboid} = 2(3a^2 + a^2 + 3a^2) = 2(7a^2) = 14a^2 \)
The ratio of the total surface area of the new cuboid to the sum of the surface areas of the three cubes is:
\( \text{Ratio} = \frac{14a^2}{18a^2} = \frac{7}{9} \)
Thus, the ratio is \( 7 : 9 \).
In simple words: When cubes are joined, some of their outer faces touch and get hidden inside. This makes the surface area of the combined cuboid smaller than the sum of separate cubes.

Exam Tip: Be sure to write down the dimensions of the new cuboid in terms of the cube's edge before applying the surface area formula.

Question 16. The cost of papering four walls of a room at 90 paise per square metre is 157.50. The height of the room is 5 metres. Find the length and the breadth of the room if they are in the ratio 4:1.
Answer:
Let the length \( l \) and breadth \( b \) of the room be in the ratio \( 4:1 \), so \( l = 4x \) and \( b = x \).
The total cost of papering the four walls is Rs. 157.50 at a rate of 90 paise (Rs. 0.90) per square metre.
The area of the four walls (which is the lateral surface area of the room) is:
\( \text{Area of Four Walls} = \frac{\text{Total Cost}}{\text{Rate per sq m}} \)
\( \implies \text{Area of Four Walls} = \frac{157.50}{0.90} = 175 \text{ m}^2 \)
The height of the room is \( h = 5 \text{ m} \). The formula for the area of four walls is:
\( \text{Area of Four Walls} = 2(l + b)h \)
\( \implies 175 = 2(4x + x) \times 5 \)
\( \implies 175 = 10(5x) \)
\( \implies 50x = 175 \)
\( \implies x = \frac{175}{50} = 3.5 \text{ m} \)
Now, we can find the dimensions of the room:
- Length \( l = 4 \times 3.5 = 14 \text{ m} \)
- Breadth \( b = 3.5 \text{ m} \)
Thus, the length of the room is 14 m and the breadth is 3.5 m.
In simple words: Find the wall area by dividing total spending by the price per square metre. Use the wall area and height to calculate the room's length and width.

Exam Tip: Remember that the area of four walls of a room is given by the lateral surface area formula, which is \( 2(l + b)h \).

Question 17. A hollow cylinder is made of iron of height 1 m. Its inner diameter is 54 cm and thickness of iron sheet of cylinder is 9 cm. Find the weight of the hollow cylinder if 1 c.c. of iron weighs 8 gm.
Answer:
Let the height of the hollow cylinder be \( h = 1 \text{ m} = 100 \text{ cm} \).
The inner diameter of the cylinder is 54 cm, so the inner radius is:
\( r = \frac{54}{2} = 27 \text{ cm} \)
The thickness of the iron sheet is 9 cm, so the outer radius of the cylinder is:
\( R = r + \text{thickness} = 27 + 9 = 36 \text{ cm} \)
The volume of iron used in the hollow cylinder is given by:
\( \text{Volume of Iron} = \pi(R^2 - r^2)h \)
\( \implies \text{Volume of Iron} = \frac{22}{7} \times (36^2 - 27^2) \times 100 \)
\( \implies \text{Volume of Iron} = \frac{22}{7} \times (36 - 27)(36 + 27) \times 100 \)
\( \implies \text{Volume of Iron} = \frac{22}{7} \times 9 \times 63 \times 100 \)
\( \implies \text{Volume of Iron} = 22 \times 9 \times 9 \times 100 = 178200 \text{ cm}^3 \)
Since 1 c.c. (cubic centimetre) of iron weighs 8 gm, the total weight of the cylinder is:
\( \text{Weight} = 178200 \times 8 = 1425600 \text{ gm} \)
\( \implies \text{Weight} = \frac{1425600}{1000} = 1425.6 \text{ kg} \)
Therefore, the weight of the hollow cylinder is 1425.6 kg.
In simple words: Find the volume of the iron by subtracting the inner empty space from the outer cylinder shape, then multiply by the weight factor.

Exam Tip: Be sure to convert the height of the cylinder from metres to centimetres so all units are consistent before calculating the volume.

Question 18. A well with 8 m inner diameter is dug 21 m deep. Earth taken out of it is spread all around to a width of 3 m to form an embankment. Find the height of embankment.
Answer:
The cylindrical well has:
- Inner diameter = 8 m, so the inner radius \( r = 4 \text{ m} \).
- Depth \( h = 21 \text{ m} \).
The volume of earth taken out of the well is:
\( \text{Volume of Earth} = \pi r^2 h \)
\( \implies \text{Volume of Earth} = \pi \times 4^2 \times 21 = 336\pi \text{ m}^3 \)
This earth is spread around the well to form an embankment of width 3 m.
- Inner radius of the embankment \( r_1 = 4 \text{ m} \)
- Outer radius of the embankment \( r_2 = r_1 + \text{width} = 4 + 3 = 7 \text{ m} \)
The base area of the embankment ring is:
\( \text{Base Area} = \pi(r_2^2 - r_1^2) = \pi(7^2 - 4^2) = \pi(49 - 16) = 33\pi \text{ m}^2 \)
Let the height of the embankment be \( H \). The volume of the embankment is equal to the volume of the dug-out earth:
\( \text{Base Area} \times H = \text{Volume of Earth} \)
\( \implies 33\pi \times H = 336\pi \)
\( \implies H = \frac{336}{33} = \frac{112}{11} \approx 10.18 \text{ m} \)
Hence, the height of the embankment is approximately 10.18 metres.
In simple words: Calculate the volume of dirt removed from the well. Divide this volume by the surface area of the circular embankment path to get the embankment's height.

Exam Tip: Embankment problems require finding the area of the ring shape (outer area minus inner area) which acts as the base of the embankment.

Question 19. A plastic box 1.25 m long, 1.05 m wide and 75 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 and also find the cost of sheet for it, if a sheet measuring 1 sq m cost 20.
Answer:
The dimensions of the open plastic box are:
- Length \( l = 1.25 \text{ m} \)
- Breadth \( b = 1.05 \text{ m} \)
- Depth (height) \( h = 75 \text{ cm} = 0.75 \text{ m} \)
Since the box is open at the top, the surface area of the sheet required will exclude the top face:
\( \text{Area of Sheet} = \text{Lateral Surface Area} + \text{Area of Base} \)
\( \text{Area of Sheet} = 2(l + b)h + lb \)
\( \implies \text{Area of Sheet} = 2(1.25 + 1.05) \times 0.75 + 1.25 \times 1.05 \)
\( \implies \text{Area of Sheet} = 2(2.30) \times 0.75 + 1.3125 \)
\( \implies \text{Area of Sheet} = 4.60 \times 0.75 + 1.3125 \)
\( \implies \text{Area of Sheet} = 3.45 + 1.3125 = 4.7625 \text{ m}^2 \)
The cost of the sheet is Rs. 20 per square metre. Hence, the total cost of the sheet is:
\( \text{Total Cost} = 4.7625 \times 20 = \text{Rs. } 95.25 \)
Thus, the area of sheet required is \( 4.7625 \text{ m}^2 \) and the cost is Rs. 95.25.
In simple words: An open box has 5 faces (4 walls and 1 base). Find the total area of these 5 faces and multiply it by the rate to find the cost.

Exam Tip: Be sure to convert the depth from centimetres to metres before starting the area calculation to keep the units consistent.

Question 20. A reservoir is in the form of rectangular parallelepiped. Its length is 20 m. If 18 kl of water is removed from the reservoir, the water level goes down by 15 cm. Find the width of the reservoir.
Answer:
Let the length, width, and height of the reservoir be \( l \), \( w \), and \( h \).
The volume of water removed is 18 kl. Since \( 1 \text{ kl} = 1000 \text{ litres} = 1 \text{ m}^3 \), the volume of water removed is:
\( V = 18 \text{ m}^3 \)
The removal of this water causes the water level to drop by \( 15 \text{ cm} = 0.15 \text{ m} \).
The volume of water removed is equal to the volume of the cuboid formed by the length, width, and the change in height:
\( V = l \times w \times \text{decrease in level} \)
\( \implies 18 = 20 \times w \times 0.15 \)
\( \implies 18 = 3 \times w \)
\( \implies w = \frac{18}{3} = 6 \text{ m} \)
Hence, the width of the reservoir is 6 metres.
In simple words: The volume of water removed matches a flat box shape with the reservoir's base and a height of 15 cm. Use this to find the missing width.

Exam Tip: Use the direct conversion \( 1 \text{ kilolitre (kl)} = 1 \text{ cubic metre (m}^3) \) to convert volume units immediately.

Question 21. If v is the volume of a cuboid of dimension a, b, c and s is its surface area, then prove that \( \frac{1}{v} = \frac{2}{s} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \)
Answer:
The volume of a cuboid with dimensions \( a, b, c \) is:
\( v = abc \)
The total surface area of this cuboid is:
\( s = 2(ab + bc + ca) \)
Let us simplify the right-hand side (RHS) of the given equation:
\( \text{RHS} = \frac{2}{s} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \)
Taking the common denominator for the terms inside the parentheses:
\( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ca + ab}{abc} \)
Substituting this back into the expression for RHS:
\( \text{RHS} = \frac{2}{s} \left( \frac{ab + bc + ca}{abc} \right) \)
Since \( s = 2(ab + bc + ca) \), we can substitute \( ab + bc + ca = \frac{s}{2} \):
\( \text{RHS} = \frac{2}{s} \times \left( \frac{\frac{s}{2}}{abc} \right) \)
\( \implies \text{RHS} = \frac{2}{s} \times \frac{s}{2abc} \)
\( \implies \text{RHS} = \frac{1}{abc} \)
Since \( v = abc \):
\( \text{RHS} = \frac{1}{v} = \text{LHS} \)
Hence proved.
In simple words: Put the formulas for total area and volume into the equation's right side, find a common denominator, and simplify to show it matches the left side.

Exam Tip: To solve proof problems easily, start working on the more complex side (RHS) and reduce it step-by-step to the simpler side (LHS).

Question 22. The area of three adjacent faces of a cuboid are x, y and z. If the volume is V, prove that \( V^2 = xyz \).
Answer:
Let the dimensions of the cuboid be length \( a \), breadth \( b \), and height \( c \).
The volume \( V \) of the cuboid is:
\( V = abc \)
The areas of three adjacent faces of the cuboid are given by:
\( x = ab \)
\( y = bc \)
\( z = ca \)
Multiplying these three face areas together:
\( xyz = (ab) \times (bc) \times (ca) \)
\( \implies xyz = a^2 b^2 c^2 \)
\( \implies xyz = (abc)^2 \)
Since \( V = abc \), substituting \( V \) into the equation:
\( xyz = V^2 \)
Thus, \( V^2 = xyz \). Hence proved.
In simple words: The product of the three face areas is the square of the product of length, width, and height, which is exactly the square of the volume.

Exam Tip: Define the three side variables explicitly first to make showing the multiplication property straightforward.

Question 23. A wall of the length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm x 12 cm x 8 cm, how many bricks would be required?
Answer:
First, let us find the volume of the wall. Converting all dimensions of the wall into centimetres:
- Length \( L = 10 \text{ m} = 1000 \text{ cm} \)
- Height \( H = 4 \text{ m} = 400 \text{ cm} \)
- Thickness \( T = 24 \text{ cm} \)
The volume of the wall is:
\( \text{Volume of Wall} = L \times H \times T \)
\( \implies \text{Volume of Wall} = 1000 \times 400 \times 24 = 9,600,000 \text{ cm}^3 \)
The dimensions of each brick are \( l = 24 \text{ cm} \), \( b = 12 \text{ cm} \), and \( h = 8 \text{ cm} \).
The volume of one brick is:
\( \text{Volume of One Brick} = l \times b \times h \)
\( \implies \text{Volume of One Brick} = 24 \times 12 \times 8 = 2304 \text{ cm}^3 \)
The number of bricks required is:
\( \text{Number of Bricks} = \frac{\text{Volume of Wall}}{\text{Volume of One Brick}} \)
\( \implies \text{Number of Bricks} = \frac{9600000}{2304} \approx 4166.67 \)
Thus, approximately 4167 bricks (or 4166.67 bricks) would be required to build the wall.
In simple words: Divide the total wall volume by the volume of a single brick to find the total number of bricks needed.

Exam Tip: Be sure to keep all units in centimetres to avoid extremely large metric conversion discrepancies.

Question 24. How many litres of water flow out of a pipe having an area of cross-section of 5 sq cm. in one minute if the speed of the water in the pipe is 30 cm/sec?
Answer:
The cross-sectional area of the pipe is \( 5 \text{ cm}^2 \).
The speed of water inside the pipe is \( 30 \text{ cm/sec} \).
The volume of water flowing out per second is:
\( \text{Volume per second} = \text{Area} \times \text{Speed} \)
\( \implies \text{Volume per second} = 5 \times 30 = 150 \text{ cm}^3 \)
In one minute (which contains 60 seconds), the volume of water that flows out is:
\( \text{Total Volume in 1 minute} = 150 \times 60 = 9000 \text{ cm}^3 \)
We know that \( 1000 \text{ cm}^3 = 1 \text{ litre} \). Converting the volume into litres:
\( \text{Volume in Litres} = \frac{9000}{1000} = 9 \text{ litres} \)
Thus, 9 litres of water flow out of the pipe in one minute.
In simple words: The water flowing from a pipe forms a cylinder every second. Find the volume of this cylinder for 60 seconds, and then convert it to litres.

Exam Tip: Do not miss the final step of dividing the cubic centimetre volume by 1000 to convert the answer into litres.

Question 25. A circular tent is cylindrical to a height of 3 metres 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 canvas 5 m wide to make the required tent.
Answer:
The tent consists of a cylindrical base and a conical top. The diameter is 105 m, so the radius of both parts is:
\( r = \frac{105}{2} = 52.5 \text{ m} \)
For the cylindrical portion:
- Height \( h = 3 \text{ m} \)
- Curved Surface Area (CSA) of the cylinder is:
\( \text{CSA of Cylinder} = 2\pi r h \)
\( \implies \text{CSA of Cylinder} = 2 \times \frac{22}{7} \times 52.5 \times 3 = 990 \text{ m}^2 \)
For the conical portion:
- Slant height \( l = 53 \text{ m} \)
- Curved Surface Area (CSA) of the cone is:
\( \text{CSA of Cone} = \pi r l \)
\( \implies \text{CSA of Cone} = \frac{22}{7} \times 52.5 \times 53 = 8745 \text{ m}^2 \)
The total surface area of the canvas required is the sum of both curved surface areas:
\( \text{Total Area} = 990 + 8745 = 9735 \text{ m}^2 \)
The width of the canvas is 5 m. The length of the canvas is:
\( \text{Length} = \frac{\text{Total Area}}{\text{Width}} \)
\( \implies \text{Length} = \frac{9735}{5} = 1947 \text{ m} \)
Therefore, the required length of the canvas is 1947 metres.
In simple words: Add the wall area of the cylinder to the roof area of the cone to find total fabric area, then divide by the fabric's width.

Exam Tip: The base of the tent does not require canvas, so calculate only the curved surface areas for both the cylinder and cone.

Question 26. The radius and slant height of a cone are in the ratio 4:7. If its curved surface area is 792 sq cm, find its radius.
Answer:
Let the radius of the cone be \( r = 4x \) and its slant height be \( l = 7x \), where \( x \) is a scaling constant.
The curved surface area of the cone is:
\( \text{Curved Surface Area} = \pi r l = 792 \text{ cm}^2 \)
\( \implies \frac{22}{7} \times (4x) \times (7x) = 792 \)
\( \implies 22 \times 4x^2 = 792 \)
\( \implies 88x^2 = 792 \)
\( \implies x^2 = 9 \)
\( \implies x = 3 \)
Now, we calculate the radius:
\( r = 4 \times 3 = 12 \text{ cm} \)
Thus, the radius of the cone is 12 cm.
In simple words: Express radius and slant height with a multiplier. Use the area equation to find this multiplier, then multiply it back to find the radius.

Exam Tip: Double-check your final multiplication step to make sure you found the radius (\( 4x \)) and not the slant height (\( 7x \)).

Question 27. A powder tin is in cylindrical shape, whose base has a diameter of 14 cm and height 20 cm. A label is wrapped around the surface of the container. If the label is pasted leaving 2 cm from the top and the bottom. What is the area of the label?
Answer:
The powder tin is cylindrical. Its base diameter is 14 cm, so its radius is:
\( r = \frac{14}{2} = 7 \text{ cm} \)
The height of the cylinder is 20 cm. The label is pasted leaving a gap of 2 cm from both the top and the bottom, so the height of the label is:
\( h_{\text{label}} = 20 - 2 - 2 = 16 \text{ cm} \)
The area of the label is equal to the curved surface area of this cylindrical portion:
\( \text{Area of Label} = 2\pi r h_{\text{label}} \)
\( \implies \text{Area of Label} = 2 \times \frac{22}{7} \times 7 \times 16 \)
\( \implies \text{Area of Label} = 2 \times 22 \times 16 = 704 \text{ cm}^2 \)
Therefore, the area of the label is 704 square centimetres.
In simple words: Subtract 2 cm from the top and 2 cm from the bottom of the height to find the label's height, then use the cylinder's curved area formula.

Exam Tip: Be careful to subtract 2 cm twice (once for the top, once for the bottom) to find the correct height of the label.

Question 28. A sphere of diameter 7 cm is dropped in a right circular cylinder vessel partly filled with water. The diameter of the cylindrical vessel is 14 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Answer:
For the dropped sphere:
- Diameter = 7 cm, so the radius \( r_{\text{sphere}} = \frac{7}{2} = 3.5 \text{ cm} \).
The volume of the sphere is:
\( V_{\text{sphere}} = \frac{4}{3}\pi r_{\text{sphere}}^3 = \frac{4}{3}\pi \left(\frac{7}{2}\right)^3 = \frac{343}{6}\pi \text{ cm}^3 \)
For the cylindrical vessel:
- Diameter = 14 cm, so the radius \( r_{\text{cylinder}} = 7 \text{ cm} \).
Let the rise in the water level be \( h \).
The volume of the displaced water (which is cylindrical in shape with height \( h \)) is equal to the volume of the submerged sphere:
\( V_{\text{displaced}} = \pi r_{\text{cylinder}}^2 h \)
\( \implies \pi \times 7^2 \times h = \frac{343}{6}\pi \)
\( \implies 49 \times h = \frac{343}{6} \)
\( \implies h = \frac{343}{6 \times 49} \)
\( \implies h = \frac{7}{6} \approx 1.17 \text{ cm} \)
Thus, the water level will rise by \( \frac{7}{6} \text{ cm} \) or approximately 1.17 centimetres.
In simple words: Dropping a ball into a glass of water pushes the water level up. The volume of this extra water cylinder equals the volume of the ball.

Exam Tip: Use fractions like \( 7/2 \) instead of decimals to simplify terms and avoid complex division errors.

Question 29. The diameter of a sphere is decreased by 50%. By what percent will its curved surface area decrease?
Answer:
Let the initial diameter of the sphere be \( D \), and the initial surface area be \( A_1 \):
\( A_1 = \pi D^2 \)
The diameter is decreased by 50%, so the final diameter \( D' \) is:
\( D' = 0.5D \)
The final surface area \( A_2 \) is:
\( A_2 = \pi (D')^2 = \pi (0.5D)^2 = 0.25 \pi D^2 = 0.25 A_1 \)
The decrease in surface area is:
\( \text{Decrease in Area} = A_1 - A_2 = A_1 - 0.25 A_1 = 0.75 A_1 \)
The percentage decrease in surface area is:
\( \text{Percentage Decrease} = \frac{\text{Decrease in Area}}{\text{Initial Area}} \times 100\% \)
\( \implies \text{Percentage Decrease} = \frac{0.75 A_1}{A_1} \times 100\% = 75\% \)
Thus, the curved surface area of the sphere decreases by 75%.
In simple words: Since surface area depends on the square of the size, cutting the diameter in half reduces the area to a quarter, a 75% drop.

Exam Tip: Percentage change can be calculated directly by looking at the square of the scale factor (e.g., \( 0.5^2 = 0.25 \), showing a 75% reduction).

Question 30. Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of sum of the surface areas of the three cubes.
Answer:
Let the side of each cube be \( a \).
The sum of the surface areas of the three individual cubes is:
\( \text{Sum of Areas} = 3 \times 6a^2 = 18a^2 \)
When placed in a row, the new cuboid has dimensions \( l = 3a \), \( b = a \), and \( h = a \).
The total surface area of this new cuboid is:
\( \text{TSA of Cuboid} = 2(lb + bh + hl) = 2(3a^2 + a^2 + 3a^2) = 14a^2 \)
The ratio of the total surface area of the new cuboid to the sum of surface areas of the three cubes is:
\( \text{Ratio} = \frac{14a^2}{18a^2} = \frac{7}{9} \)
Therefore, the ratio is \( 7 : 9 \).
In simple words: Putting cubes together covers up some sides, reducing the overall exposed surface area.

Exam Tip: Be prepared for this question to appear under different marks; writing out each step clearly is crucial for full points.

Question 31. How many bricks will be required for a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures 25 cm x 11.25 cm x 6 cm?
Answer:
First, let us convert all the dimensions of the wall into centimetres:
- Length \( L = 8 \text{ m} = 800 \text{ cm} \)
- Height \( H = 6 \text{ m} = 600 \text{ cm} \)
- Thickness \( T = 22.5 \text{ cm} \)
The volume of the wall is:
\( \text{Volume of Wall} = L \times H \times T \)
\( \implies \text{Volume of Wall} = 800 \times 600 \times 22.5 = 10,800,000 \text{ cm}^3 \)
The dimensions of each brick are \( l = 25 \text{ cm} \), \( b = 11.25 \text{ cm} \), and \( h = 6 \text{ cm} \).
The volume of one brick is:
\( \text{Volume of One Brick} = 25 \times 11.25 \times 6 = 1687.5 \text{ cm}^3 \)
The number of bricks required is:
\( \text{Number of Bricks} = \frac{\text{Volume of Wall}}{\text{Volume of One Brick}} \)
\( \implies \text{Number of Bricks} = \frac{10800000}{1687.5} = 6400 \)
Thus, exactly 6400 bricks are required to build the wall.
In simple words: Find the total space of the wall and divide it by the space of one brick to find the total bricks.

Exam Tip: Double-check your decimal division when dividing by 11.25 to avoid common calculation errors.

Question 32. The internal measurements of a cuboidal room are 10 m x 4 m x 6 m. Find the cost of white washing of the walls at the rate of Rs. 5 per square metre.
Answer:
The internal dimensions of the room are length \( l = 10 \text{ m} \), breadth \( b = 4 \text{ m} \), and height \( h = 6 \text{ m} \).
The area of the four walls to be white-washed is equal to the lateral surface area (LSA) of the room:
\( \text{Area of Walls} = 2(l + b)h \)
\( \implies \text{Area of Walls} = 2(10 + 4) \times 6 \)
\( \implies \text{Area of Walls} = 2 \times 14 \times 6 = 168 \text{ m}^2 \)
The rate of white-washing is Rs. 5 per square metre. The total cost is:
\( \text{Total Cost} = 168 \times 5 = \text{Rs. } 840 \)
Thus, the total cost of white-washing the walls is Rs. 840.
In simple words: Find the total surface area of the four walls using the height and base perimeter, then multiply by the cost per square metre.

Exam Tip: Be careful to only include the four walls (lateral surface area) and not the floor or ceiling, unless explicitly stated.

Question 33. A cylinder is within the cube touching all the vertical faces. A cone is inside the cylinder. If the height and base of the cone is same as cylinder, find the ratio of their volumes.
Answer:
Let the edge of the cube be \( a \).
1. The volume of the cube is:
\( V_{\text{cube}} = a^3 \)
2. The cylinder is placed inside the cube touching all vertical faces, which means:
- The diameter of the circular base of the cylinder is equal to the edge of the cube, so the radius \( r = \frac{a}{2} \).
- The height of the cylinder is \( h = a \).
The volume of the cylinder is:
\( V_{\text{cylinder}} = \pi r^2 h = \pi \left(\frac{a}{2}\right)^2 a = \frac{\pi a^3}{4} \)
3. The cone is inside the cylinder with the same height and base as the cylinder:
The volume of the cone is:
\( V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{\pi a^3}{4} = \frac{\pi a^3}{12} \)
Now, let us find the ratio of their volumes (Cube : Cylinder : Cone):
\( \text{Ratio} = a^3 : \frac{\pi a^3}{4} : \frac{\pi a^3}{12} \)
\( \implies \text{Ratio} = 1 : \frac{\pi}{4} : \frac{\pi}{12} \)
Multiplying by 12:
\( \text{Ratio} = 12 : 3\pi : \pi \)
(If the ratio of only the Cylinder to the Cone is requested, it is simply \( 3 : 1 \)).
In simple words: The cylinder has a circle base fitting inside the square box, and the cone inside is exactly one-third of the cylinder's volume.

Exam Tip: Providing both the triple ratio (Cube : Cylinder : Cone) and the double ratio (Cylinder : Cone) ensures you meet any specific interpretation of the question.

Question 34. The external length, breadth and height of a closed rectangular wooden box are 18 cm, 10 cm and 6 cm respectively and thickness of wood is ½ cm. When the box is empty, it weighs 15 kg and when filled with sand it weighs 100 kg. Find the weight of one cubic cm of wood and one cubic cm of sand.
Answer:
The external dimensions of the closed box are:
- Length \( L = 18 \text{ cm} \)
- Breadth \( B = 10 \text{ cm} \)
- Height \( H = 6 \text{ cm} \)
The external volume of the box is:
\( V_{\text{external}} = 18 \times 10 \times 6 = 1080 \text{ cm}^3 \)
Since the thickness of the wood is \( 0.5 \text{ cm} \), the internal dimensions are:
- Internal Length \( l = 18 - 2(0.5) = 17 \text{ cm} \)
- Internal Breadth \( b = 10 - 2(0.5) = 9 \text{ cm} \)
- Internal Height \( h = 6 - 2(0.5) = 5 \text{ cm} \)
The internal volume of the box (where sand is filled) is:
\( V_{\text{internal}} = 17 \times 9 \times 5 = 765 \text{ cm}^3 \)
The volume of the wood is:
\( V_{\text{wood}} = V_{\text{external}} - V_{\text{internal}} = 1080 - 765 = 315 \text{ cm}^3 \)

1. Weight of one cubic cm of wood:
The empty box (which is made entirely of wood) weighs 15 kg.
\( \text{Weight of } 1 \text{ cm}^3 \text{ of wood} = \frac{15 \text{ kg}}{315 \text{ cm}^3} = \frac{1}{21} \text{ kg} \approx 0.0476 \text{ kg} \text{ (or } 47.62 \text{ gm)} \)

2. Weight of one cubic cm of sand:
The weight of the sand is the difference between the filled weight and the empty weight:
\( \text{Weight of Sand} = 100 \text{ kg} - 15 \text{ kg} = 85 \text{ kg} \)
\( \text{Weight of } 1 \text{ cm}^3 \text{ of sand} = \frac{85 \text{ kg}}{765 \text{ cm}^3} = \frac{1}{9} \text{ kg} \approx 0.111 \text{ kg} \text{ (or } 111.11 \text{ gm)} \)
Thus, the weight of 1 cubic cm of wood is approximately 0.0476 kg, and that of sand is approximately 0.111 kg.
In simple words: Find the space the wood occupies by subtracting the inner space from the outer space. Divide the empty weight by this wood space. Then, divide the sand weight by the inner space.

Exam Tip: Because the box is closed, you must subtract twice the thickness (\( 2 \times 0.5 = 1 \text{ cm} \)) from every external dimension to find the internal dimensions.

Question 35. How many litres of water flows out of a pipe having an area of cross-section of 5 sq. cm in one minute, if the speed of water in the pipe is 30 cm/sec?
Answer:
The pipe's cross-sectional area is \( 5 \text{ cm}^2 \).
The velocity of the water stream is \( 30 \text{ cm/sec} \).
The volume of water exiting the pipe per second is:
\( \text{Volume per second} = 5 \times 30 = 150 \text{ cm}^3 \)
In one minute (60 seconds), the volume of water is:
\( \text{Total Volume in 1 minute} = 150 \times 60 = 9000 \text{ cm}^3 \)
Since \( 1000 \text{ cm}^3 \) is equivalent to \( 1 \text{ litre} \):
\( \text{Volume in Litres} = \frac{9000}{1000} = 9 \text{ litres} \)
Therefore, 9 litres of water flows out in one minute.
In simple words: Find the volume of water coming out each second and multiply it by 60 to get the volume for a full minute, then convert to litres.

Exam Tip: Be sure to write the formula "Volume = Area of Cross-section × Speed × Time" to show your conceptual understanding.

Question 36. A hemispherical bowl of internal diameter 40 cm contains a liquid. This liquid is to be filled in cylindrical bottles of radius 2 cm and height 8 cm. How many bottles are required to empty the bowl?
Answer:
For the hemispherical bowl:
- Internal diameter = 40 cm, so the internal radius is \( R = 20 \text{ cm} \).
The volume of liquid inside the bowl is:
\( V_{\text{bowl}} = \frac{2}{3}\pi R^3 = \frac{2}{3}\pi \times 20^3 = \frac{16000}{3}\pi \text{ cm}^3 \)
For each cylindrical bottle:
- Radius \( r = 2 \text{ cm} \)
- Height \( h = 8 \text{ cm} \)
The volume of one bottle is:
\( V_{\text{bottle}} = \pi r^2 h = \pi \times 2^2 \times 8 = 32\pi \text{ cm}^3 \)
The number of bottles required to empty the bowl is:
\( \text{Number of Bottles} = \frac{V_{\text{bowl}}}{V_{\text{bottle}}} \)
\( \implies \text{Number of Bottles} = \frac{\frac{16000}{3}\pi}{32\pi} \)
\( \implies \text{Number of Bottles} = \frac{16000}{3 \times 32} \)
\( \implies \text{Number of Bottles} = \frac{500}{3} \approx 166.67 \)
Since we need a whole number of bottles to completely hold all the liquid, we require 167 bottles.
In simple words: Divide the total volume of liquid in the hemispherical bowl by the volume of a single bottle to find the total bottles needed.

Exam Tip: If the division results in a fraction, round up to the next integer because a partial bottle still represents an extra physical bottle needed.

Question 37. An open box is made of wood 3 cm thick. Its external length, breadth and height are 1.48 m, 1.16 m and 8.3 dm. Find the cost of painting the inner surface of Rs. 1000 per 10 sq metres.
Answer:
First, let us convert all external dimensions into centimetres:
- External Length \( L = 1.48 \text{ m} = 148 \text{ cm} \)
- External Breadth \( B = 1.16 \text{ m} = 116 \text{ cm} \)
- External Height \( H = 8.3 \text{ dm} = 83 \text{ cm} \)
Since the box is open at the top, the thickness of 3 cm is subtracted twice from the length and breadth, but only once from the height (since there is no top lid):
- Internal Length \( l = 148 - 2(3) = 142 \text{ cm} = 1.42 \text{ m} \)
- Internal Breadth \( b = 116 - 2(3) = 110 \text{ cm} = 1.10 \text{ m} \)
- Internal Height \( h = 83 - 3 = 80 \text{ cm} = 0.8 \text{ m} \)
The internal surface area to be painted includes the four walls and the base of the open box:
\( \text{Inner Surface Area} = 2(l + b)h + lb \)
\( \implies \text{Inner Surface Area} = 2(1.42 + 1.10) \times 0.8 + (1.42 \times 1.10) \)
\( \implies \text{Inner Surface Area} = 2(2.52) \times 0.8 + 1.562 \)
\( \implies \text{Inner Surface Area} = 4.032 + 1.562 = 5.594 \text{ m}^2 \)
The cost of painting is Rs. 1000 per 10 square metres. Thus, the rate is Rs. 100 per square metre. The total cost is:
\( \text{Total Cost} = 5.594 \times 100 = \text{Rs. } 559.40 \)
Thus, the total cost of painting the inner surface is Rs. 559.40.
In simple words: Subtract wood thickness to find inner dimensions (subtract only once for the height because the box has no lid). Find the total area of the bottom and four sides, then calculate the price.

Exam Tip: For open boxes, always remember to subtract the wall thickness only once from the height when determining the internal height.

Question 38. The ratio of radii of two right circular cylinders is 2:5 and that of their heights is 1:4, find the ratio of their Volumes.
Answer:
Let the radii of the two cylinders be \( r_1 \) and \( r_2 \), so \( \frac{r_1}{r_2} = \frac{2}{5} \).
Let the heights of the two cylinders be \( h_1 \) and \( h_2 \), so \( \frac{h_1}{h_2} = \frac{1}{4} \).
The volume \( V \) of a right circular cylinder is \( \pi r^2 h \). The ratio of their volumes is:
\( \frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} \)
\( \implies \frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right) \)
\( \implies \frac{V_1}{V_2} = \left(\frac{2}{5}\right)^2 \times \left(\frac{1}{4}\right) \)
\( \implies \frac{V_1}{V_2} = \frac{4}{25} \times \frac{1}{4} = \frac{1}{25} \)
Thus, the ratio of their volumes is \( 1 : 25 \).
In simple words: The volume of a cylinder depends on the square of its radius and its height. Multiply the squared radius ratio by the height ratio.

Exam Tip: Show the separation of the radius and height ratios clearly in your expression to demonstrate a systematic approach.

Question 39. A wooden toy is in the form of a cone. The diameter of the base of the cone is 6 cm and the height of the cone is 4 cm. Find the cost of painting the toy at the rate of Rs. 5 per 100 cm².
Answer:
For the conical toy:
- Diameter of the base = 6 cm, so its radius is \( r = 3 \text{ cm} \).
- Height of the cone is \( h = 4 \text{ cm} \).
First, find the slant height \( l \) of the cone:
\( l = \sqrt{r^2 + h^2} \)
\( \implies l = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm} \)
The total surface area of the toy to be painted is:
\( \text{Total Surface Area} = \pi r (r + l) \)
\( \implies \text{Total Surface Area} = \frac{22}{7} \times 3 \times (3 + 5) \)
\( \implies \text{Total Surface Area} = \frac{22}{7} \times 24 = \frac{528}{7} \approx 75.43 \text{ cm}^2 \)
The rate of painting is Rs. 5 per 100 \( \text{cm}^2 \). The total cost is:
\( \text{Total Cost} = \text{Total Surface Area} \times \frac{5}{100} \)
\( \implies \text{Total Cost} = \frac{528}{7} \times \frac{5}{100} = \frac{2640}{700} \approx \text{Rs. } 3.77 \)
Thus, the cost of painting the wooden toy is approximately Rs. 3.77.
In simple words: Find the slant height of the cone, calculate the total surface area (including the bottom circular base), and then apply the painting rate.

Exam Tip: A toy cone is a solid object, so you must calculate the *total* surface area (which includes the base) rather than just the curved surface area.

Question 40. An open cuboidal box is made of 3 cm thick wood. Its external length, breadth and height are 1.48 m, 1.16 m and 8.3 dm respectively. Find the cost of painting the inner surface at the rate of Rs. 50 per square metre.
Answer:
Converting all external dimensions into centimetres:
- External Length \( L = 1.48 \text{ m} = 148 \text{ cm} \)
- External Breadth \( B = 1.16 \text{ m} = 116 \text{ cm} \)
- External Height \( H = 8.3 \text{ dm} = 83 \text{ cm} \)
Since the box is open at the top, the wood thickness of 3 cm is subtracted twice from the length and breadth, but only once from the height:
- Internal Length \( l = 148 - 2(3) = 142 \text{ cm} = 1.42 \text{ m} \)
- Internal Breadth \( b = 116 - 2(3) = 110 \text{ cm} = 1.10 \text{ m} \)
- Internal Height \( h = 83 - 3 = 80 \text{ cm} = 0.8 \text{ m} \)
The internal surface area to be painted (comprising the four walls and the base) is:
\( \text{Inner Surface Area} = 2(l + b)h + lb \)
\( \implies \text{Inner Surface Area} = 2(1.42 + 1.10) \times 0.8 + (1.42 \times 1.10) \)
\( \implies \text{Inner Surface Area} = 2(2.52) \times 0.8 + 1.562 \)
\( \implies \text{Inner Surface Area} = 4.032 + 1.562 = 5.594 \text{ m}^2 \)
The rate of painting is Rs. 50 per square metre. The total cost is:
\( \text{Total Cost} = 5.594 \times 50 = \text{Rs. } 279.70 \)
Hence, the total cost of painting the inner surface of the box is Rs. 279.70.
In simple words: Find the inner dimensions, calculate the total area of the five inside faces, and then multiply by the cost per square metre.

Exam Tip: Writing down step-by-step conversions of dimensions to standard metric units like metres is crucial for checking your work and securing partial marks.

Question 41. A conical vessel having internal radius 3 cm and height 25 cm is full of water. The water is emptied into a cylindrical vessel with internal radius of 10 cm. Find the height to which the water rises.
Answer:
For the conical vessel:
- Internal radius \( r_1 = 3 \text{ cm} \)
- Height \( h_1 = 25 \text{ cm} \)
The volume of water inside the conical vessel is:
\( V = \frac{1}{3}\pi r_1^2 h_1 \)
\( \implies V = \frac{1}{3}\pi \times 3^2 \times 25 = 75\pi \text{ cm}^3 \)
Let the height of water rise in the cylindrical vessel be \( h_2 \).
For the cylindrical vessel:
- Internal radius \( r_2 = 10 \text{ cm} \)
The volume of water transferred is the same, which forms a cylinder of radius \( r_2 \) and height \( h_2 \):
\( V = \pi r_2^2 h_2 \)
\( \implies 75\pi = \pi \times 10^2 \times h_2 \)
\( \implies 100 \times h_2 = 75 \)
\( \implies h_2 = \frac{75}{100} = 0.75 \text{ cm} \)
Therefore, the water will rise to a height of 0.75 centimetres in the cylindrical vessel.
In simple words: The water's total volume does not change when poured from the cone to the cylinder. Set their volume equations equal to solve for the height.

Exam Tip: Keeping \( \pi \) in the volume equations allows it to cancel out neatly, reducing complex decimal multiplications.