CBSE Class 10 Surface Areas and Volumes Sure Shot Questions Set 01

CBSE Class 10 Surface Areas and Volumes Sure Shot Questions Set A. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

This CBSE Class 10 Maths practice set covers Chapter 13, Surface Areas and Volumes. It has 85 questions on cones, cylinders, spheres, hemispheres, cubes, cuboids, frustums and combined solids, including surface area and volume, melting and recasting, water flow and tank filling problems, with worked answers, figures and Teacher's Notes.

PRACTICE QUESTIONS

CLASS X: CHAPTER - 13

SURFACE AREAS AND VOLUMES

1. A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the cone and of the remaining solid left out after the cone carved out.

Answer:
1. The largest cone has base radius r = 7 cm and height h = 14 cm.
2. Slant height l = \( \sqrt{r^2 + h^2} \) = \( \sqrt{49 + 196} \) = 7\( \sqrt{5} \) cm.
3. Curved surface area of cone = π r l = \( \frac{22}{7} \) × 7 × 7\( \sqrt{5} \) = 154\( \sqrt{5} \) ≈ 344.35 cm².
4. Surface area of cube = 6 × 14² = 1176 cm².
5. Remaining solid = cube surface − base of cone + curved surface of cone = 1176 − 154 + 154\( \sqrt{5} \) = 1022 + 154\( \sqrt{5} \) ≈ 1366.35 cm².

Teacher's Note:
a) The cone's base circle is removed from the cube face, so subtract π r² only once.
b) The curved surface of the cone is added to the remaining solid.
c) If the total surface of the cone including its base is wanted, it is about 498.35 cm².

2. A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed.

Answer:
1. Volume of sphere = \( \frac{4}{3} \) π (10.5)³ = \( \frac{4}{3} \) × \( \frac{22}{7} \) × 1157.625 = 4851 cm³.
2. Volume of one cone = \( \frac{1}{3} \) π (3.5)² × 3 = 38.5 cm³.
3. Number of cones = 4851 ÷ 38.5 = 126.
4. Hence, 126 cones are formed.

Teacher's Note:
a) Melting and recasting keeps the volume the same.
b) Divide the sphere's volume by one cone's volume to get the count.

3. A canal is 300 cm wide and 120 cm deep. The water in the canal is flowing with a speed of 20 km/h. How much area will it irrigate in 20 minutes if 8 cm of standing water is desired?

Answer:
1. Cross-section of water = 3 m × 1.2 m = 3.6 m².
2. Length of water flowing in 20 minutes = \( \frac{20\,000}{60} \) × 20 = \( \frac{20\,000}{3} \) m.
3. Volume of water = 3.6 × \( \frac{20\,000}{3} \) = 24 000 m³.
4. Standing water depth = 8 cm = 0.08 m.
5. Area irrigated = \( \frac{24\,000}{0.08} \) = 300 000 m².
6. Hence, the area irrigated is 300 000 m² (30 hectares).

Teacher's Note:
a) Convert all units to metres before multiplying.
b) Area equals volume divided by depth, not multiplied.

4. A cone of radius 4 cm is divided into two parts by drawing a plane through the mid point of its axis and parallel to its base. Compare the volumes of the two parts.

Answer:
1. The plane at mid height makes the top cone similar to the whole cone with scale factor \( \frac{1}{2} \).
2. Volume ratio of the top cone to the whole cone = \( \left(\frac{1}{2}\right)^3 \) = \( \frac{1}{8} \).
3. Volume of the frustum = \( \frac{7}{8} \) of the whole cone.
4. Ratio of the two parts = 1 : 7.

Teacher's Note:
a) The ratio does not depend on the height of the cone.
b) Volumes of similar solids scale with the cube of the linear scale factor.

5. Three cubes of a metal whose edges are in the ratio 3:4:5 are melted and converted into a single cube whose diagonal is 12\( \sqrt{3} \) cm. Find the edges of the three cubes.

Answer:
1. Let the edges be 3k, 4k and 5k.
2. Edge of new cube = diagonal ÷ \( \sqrt{3} \) = 12\( \sqrt{3} \) ÷ \( \sqrt{3} \) = 12 cm, so its volume = 1728 cm³.
3. Sum of volumes = (27 + 64 + 125)k³ = 216k³ = 1728, so k³ = 8 and k = 2.
4. Edges of the three cubes = 6 cm, 8 cm and 10 cm.

Teacher's Note:
a) The diagonal of a cube is edge × \( \sqrt{3} \).
b) Use the sum of cubes of the edges, not the sum of the edges.

6. Three metallic solid cubes whose edges are 3 cm, 4 cm and 5 cm are melted and formed into a single cube. Find the edge of the cube so formed.

Answer:
1. Volume of the three cubes = 3³ + 4³ + 5³ = 27 + 64 + 125 = 216 cm³.
2. Edge of the new cube = ∛216 = 6 cm.
3. Hence, the edge of the new cube is 6 cm.

Teacher's Note:
a) Volume is conserved when metal is melted and recast.
b) Check that the answer is a whole number when the cube is perfect.

7. How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9cm × 11cm × 12cm?

Answer:
1. Volume of cuboid = 9 × 11 × 12 = 1188 cm³.
2. Radius of each shot = 1.5 cm, so volume of one shot = \( \frac{4}{3} \) π (1.5)³ = \( \frac{4}{3} \) × \( \frac{22}{7} \) × 3.375 = \( \frac{99}{7} \) cm³.
3. Number of shots = 1188 ÷ \( \frac{99}{7} \) = 84.
4. Hence, 84 shots can be made.

Teacher's Note:
a) Use the radius, which is half the diameter.
b) Check that the total volume of shots does not exceed the cuboid's volume.

8. A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm, respectively. Find the height of the bucket.

Answer:
1. Volume = 28.490 litres = 28 490 cm³.
2. Volume of frustum = \( \frac{1}{3} \) π h (R² + Rr + r²) = \( \frac{1}{3} \) × \( \frac{22}{7} \) × h × (784 + 588 + 441) = \( \frac{5698}{3} \) h.
3. \( \frac{5698}{3} \) h = 28 490, so h = \( \frac{85\,470}{5698} \) = 15 cm.
4. Hence, the height of the bucket is 15 cm.

Teacher's Note:
a) Convert litres to cm³ first, since 1 litre = 1000 cm³.
b) For a frustum, R and r are the two radii, and R² + Rr + r² appears in the formula.

9. A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

Answer:
1. Volume of the whole cone = \( \frac{1}{3} \) π × 64 × 12 = 256π cm³.
2. Top cone has radius 4 cm and height 6 cm, so its volume = \( \frac{1}{3} \) π × 16 × 6 = 32π cm³.
3. Volume of frustum = 256π − 32π = 224π cm³.
4. Ratio = 32π : 224π = 1 : 7.

Teacher's Note:
a) This is the same result as the mid-axis cut in question 4.
b) Subtract the top cone from the whole cone to get the frustum.

10. Two identical cubes each of volume 64 cm3 are joined together end to end. What is the surface area of the resulting cuboid?

Answer:
1. Edge of each cube = ∛64 = 4 cm.
2. The joined cuboid has dimensions 8 cm × 4 cm × 4 cm.
3. Surface area = 2(8 × 4 + 4 × 4 + 8 × 4) = 2(32 + 16 + 32) = 160 cm².

Teacher's Note:
a) Joining two cubes end to end doubles one dimension only.
b) The two faces where the cubes touch are not part of the outside surface.

11. From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid.

Answer:
1. Volume of cube = 7³ = 343 cm³.
2. Volume of cone = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 3² × 7 = 66 cm³.
3. Volume of remaining solid = 343 − 66 = 277 cm³.

Teacher's Note:
a) The cavity volume is removed from the cube.
b) Use the radius 3 cm, not the diameter.

12. Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so formed.

Answer:
1. Slant height of each cone = \( \sqrt{8^2 + 15^2} \) = 17 cm.
2. Curved surface area of one cone = π × 8 × 17 = 136π cm².
3. Surface area of the combined shape = 2 × 136π = 272π cm² = \( \frac{5984}{7} \) ≈ 854.86 cm².

Teacher's Note:
a) The base circles are hidden inside the joined shape, so they are not counted.
b) The slant height is found using the Pythagoras theorem.

13. Two solid cones A and B are placed in a cylindrical tube as shown in the below figure. The ratio of their capacities is 2:1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.

Answer:
1. Both cones have radius r = 3 cm, and their heights add up to the cylinder length: h_A + h_B = 21 cm.
2. Capacity ratio = h_A : h_B = 2 : 1, so h_A = 14 cm and h_B = 7 cm.
3. Capacity of A = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 9 × 14 = 132 cm³.
4. Capacity of B = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 9 × 7 = 66 cm³.
5. Volume of cylinder = \( \frac{22}{7} \) × 9 × 21 = 594 cm³.
6. Volume of remaining portion = 594 − 132 − 66 = 396 cm³.

Teacher's Note:
a) Both cones have the same base radius as the tube, so their heights set their capacities.
b) Heights are in the ratio of capacities because the radius is the same.
c) Check that the three volumes add up to the cylinder's volume.

14. An ice cream cone full of ice cream having radius 5 cm and height 10 cm as shown in the below figure. Calculate the volume of ice cream, provided that 1/6 part is left unfilled with ice cream.

Answer:
1. Volume of cone = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 25 × 10 = \( \frac{5500}{21} \) cm³.
2. Volume of hemisphere = \( \frac{2}{3} \) × \( \frac{22}{7} \) × 125 = \( \frac{5500}{21} \) cm³.
3. Total volume = \( \frac{11000}{21} \) ≈ 523.81 cm³.
4. Ice cream fills \( \frac{5}{6} \) of the total volume = \( \frac{5}{6} \) × 523.81 ≈ 436.51 cm³.

Teacher's Note:
a) The 10 cm height is taken as the height of the conical part, as the question states.
b) Leaving \( \frac{1}{6} \) unfilled means the filled part is \( \frac{5}{6} \) of the whole.
c) The hemisphere and cone have equal volumes here, which makes the check easy.

15. Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.

Answer:
1. Volume of water displaced = π × 3.5² × 5.6 = 68.6π cm³.
2. Radius of one marble = 0.7 cm, so its volume = \( \frac{4}{3} \) π (0.7)³ = \( \frac{4}{3} \) × 0.343π = 0.4573π cm³.
3. Number of marbles = 68.6π ÷ 0.4573π = 150.
4. Hence, 150 marbles should be dropped.

Teacher's Note:
a) The displaced water volume equals the total volume of the marbles.
b) Use the beaker's radius, which is half of 7 cm.

16. How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm.

Answer:
1. Volume of lead piece = 66 × 42 × 21 = 58 212 cm³.
2. Radius of each shot = 2.1 cm, so volume of one shot = \( \frac{4}{3} \) × \( \frac{22}{7} \) × (2.1)³ = 38.808 cm³.
3. Number of shots = 58 212 ÷ 38.808 = 1500.
4. Hence, 1500 shots can be obtained.

Teacher's Note:
a) Use the radius, which is 2.1 cm, in the sphere formula.
b) Multiplying the three dimensions gives the total volume of the lead.

17. How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm.

Answer:
1. Volume of cube = 44³ = 85 184 cm³.
2. Radius of each shot = 2 cm, so volume of one shot = \( \frac{4}{3} \) × \( \frac{22}{7} \) × 8 = \( \frac{704}{21} \) cm³.
3. Number of shots = 85 184 ÷ \( \frac{704}{21} \) = 2541.
4. Hence, 2541 shots can be made.

Teacher's Note:
a) The cube's volume is the edge cubed.
b) Round down only if the division leaves a remainder, since shots must be whole.

18. A wall 24 m long, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm × 16 cm × 10 cm. If the mortar occupies \( \frac{1}{10} \)th of the volume of the wall, then find the number of bricks used in constructing the wall.

Answer:
1. Volume of wall = 24 × 0.4 × 6 = 57.6 m³ = 57 600 000 cm³.
2. Volume of bricks = \( \frac{9}{10} \) × 57 600 000 = 51 840 000 cm³.
3. Volume of one brick = 25 × 16 × 10 = 4000 cm³.
4. Number of bricks = 51 840 000 ÷ 4000 = 12 960.

Teacher's Note:
a) Mortar fills \( \frac{1}{10} \) of the wall, so the bricks fill the rest.
b) Convert metres to centimetres before dividing.

19. Find the number of metallic circular disc with 1.5 cm base diameter and of height 0.2 cm to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

Answer:
1. Volume of one disc = π × 0.75² × 0.2 = 0.1125π cm³.
2. Volume of cylinder = π × 2.25² × 10 = 50.625π cm³.
3. Number of discs = 50.625π ÷ 0.1125π = 450.

Teacher's Note:
a) Use the radius 0.75 cm for the disc and 2.25 cm for the cylinder.
b) π cancels out, so the answer does not depend on its value.

20. A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm, respectively. Find the capacity and surface area of the bucket. Also, find the cost of milk which can completely fill the container, at the rate of Rs 25 per litre (use π = 3.14).

Answer:
1. Capacity = \( \frac{1}{3} \) × 3.14 × 30 × (20² + 20 × 10 + 10²) = 21 980 cm³ = 21.98 litres.
2. Slant height = \( \sqrt{30^2 + 10^2} \) = \( \sqrt{1000} \) ≈ 31.62 cm.
3. Curved surface area = 3.14 × (20 + 10) × 31.62 ≈ 2978.87 cm².
4. Area of bottom = 3.14 × 10² = 314 cm².
5. Surface area of bucket (curved surface and bottom) ≈ 3292.87 cm².
6. Cost of milk = 21.98 × Rs 25 = Rs 549.50.

Teacher's Note:
a) The bucket is open at the top, so the top is not counted in the surface area.
b) Use the slant height for the curved surface, not the vertical height.
c) Always convert cm³ to litres by dividing by 1000 before costing.

21. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. If a cube circumscribes the toy, then find the difference of the volumes of cube and the toy. Also, find the total surface area of the toy.

Answer:
1. Radius of hemisphere = radius of cone = 4 cm.
2. Volume of hemisphere = \( \frac{2}{3} \) π (4)³ = \( \frac{128}{3} \) π cm³.
3. Volume of cone = \( \frac{1}{3} \) π × 16 × 4 = \( \frac{64}{3} \) π cm³.
4. Volume of toy = \( \frac{192}{3} \) π = 64π ≈ 201.14 cm³.
5. Edge of circumscribing cube = 8 cm, so its volume = 512 cm³.
6. Difference = 512 − 201.14 = 310.86 cm³.
7. Slant height of cone = \( \sqrt{16 + 16} \) = 4\( \sqrt{2} \) cm.
8. Total surface area = 2π(4)² + π × 4 × 4\( \sqrt{2} \) = 32π + 16\( \sqrt{2} \)π ≈ 171.69 cm².

Teacher's Note:
a) The base of the hemisphere is not counted in the total surface area.
b) The cube's edge equals the toy's diameter, which is 8 cm.
c) The cone's slant height uses the Pythagoras theorem with radius 4 cm and height 4 cm.

22. A solid metallic hemisphere of radius 8 cm is melted and recasted into a right circular cone of base radius 6 cm. Determine the height of the cone.

Answer:
1. Volume of hemisphere = \( \frac{2}{3} \) π (8)³ = \( \frac{1024}{3} \) π cm³.
2. Volume of cone = \( \frac{1}{3} \) π (6)² h = 12π h cm³.
3. 12π h = \( \frac{1024}{3} \) π, so h = \( \frac{256}{9} \) ≈ 28.44 cm.

Teacher's Note:
a) The volume of a hemisphere is half that of a sphere of the same radius.
b) Melting keeps the volume constant, so equate the two volumes.

23. A rectangular water tank of base 11 m × 6 m contains water upto a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.

Answer:
1. Volume of water = 11 × 6 × 5 = 330 m³.
2. Base area of cylindrical tank = \( \frac{22}{7} \) × 3.5² = 38.5 m².
3. Height of water level = 330 ÷ 38.5 = \( \frac{60}{7} \) ≈ 8.57 m.

Teacher's Note:
a) The volume of water is the same in both tanks.
b) The new level is higher because the cylinder is narrower.

24. A building is in the form of a cylinder surmounted by a hemispherical dome. The base diameter of the dome is equal to \( \frac{2}{3} \) of the total height of the building. Find the height of the building, if it contains 67\( \frac{1}{21} \) m³ of air.

Answer:
1. Let total height = H. Then diameter of dome = \( \frac{2}{3} \) H, so radius r = \( \frac{H}{3} \).
2. Height of cylinder = H − \( \frac{H}{3} \) = \( \frac{2H}{3} \).
3. Volume = π r² × \( \frac{2H}{3} \) + \( \frac{2}{3} \) π r³ = \( \frac{8\pi H^3}{81} \).
4. \( \frac{8\pi H^3}{81} \) = \( \frac{1408}{21} \), which gives H³ = 216 when π = \( \frac{22}{7} \).
5. H = 6 m.
6. Hence, the height of the building is 6 m.

Teacher's Note:
a) Write the dome radius in terms of the total height first.
b) Check by substituting H = 6 m back into the volume formula.

25. How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm provided the thickness of the iron is 1.5 cm. If one cubic cm of iron weighs 7.5 g, find the weight of the box.

Answer:
1. External volume = 36 × 25 × 16.5 = 14 850 cm³.
2. Internal dimensions = 36 − 3 = 33 cm, 25 − 3 = 22 cm, and 16.5 − 1.5 = 15 cm (open top).
3. Internal volume = 33 × 22 × 15 = 10 890 cm³.
4. Iron required = 14 850 − 10 890 = 3960 cm³.
5. Weight = 3960 × 7.5 = 29 700 g = 29.7 kg.

Teacher's Note:
a) Thickness is removed from both sides, so subtract 2 × 1.5 cm from each horizontal side.
b) The box is open at the top, so only 1.5 cm is removed from the height at the bottom.

26. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one fifth of a litre?

Answer:
1. Radius of barrel = 0.25 cm and length = 7 cm.
2. Volume of barrel = \( \frac{22}{7} \) × 0.0625 × 7 = 1.375 cm³.
3. One fifth of a litre = 200 cm³.
4. Number of barrels in bottle = 200 ÷ 1.375 ≈ 145.45.
5. Words written = 3300 × 200 ÷ 1.375 = 480 000 words.

Teacher's Note:
a) Convert 5 mm to 0.5 cm before finding the radius.
b) Words are proportional to the volume of ink used.

27. Water flows at the rate of 10m/minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm?

Answer:
1. Volume of conical vessel = \( \frac{1}{3} \) π (20)² × 24 = 3200π cm³.
2. Water flows out of the pipe at 1000 cm per minute, with radius 0.25 cm.
3. Volume flowing per minute = π × 0.25² × 1000 = 62.5π cm³.
4. Time = 3200π ÷ 62.5π = 51.2 minutes.
5. Hence, it would take about 51.2 minutes to fill the vessel.

Teacher's Note:
a) Convert metres per minute into centimetres per minute before multiplying.
b) Use the radius of the pipe, which is 0.25 cm.

28. A heap of rice is in the form of a cone of diameter 9 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

Answer:
1. Radius = 4.5 m, height = 3.5 m.
2. Volume = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 4.5² × 3.5 = 74.25 m³.
3. Slant height = \( \sqrt{4.5^2 + 3.5^2} \) = \( \sqrt{32.5} \) ≈ 5.70 m.
4. Canvas required = curved surface area = \( \frac{22}{7} \) × 4.5 × 5.70 ≈ 80.63 m².

Teacher's Note:
a) Canvas covers only the curved surface, not the ground.
b) Use the slant height, not the vertical height, for the surface area.

29. A factory manufactures 120000 pencils daily. The pencils are cylindrical in shape each of length 25 cm and circumference of base as 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at Rs 0.05 per dm2.

Answer:
1. Curved surface of one pencil = circumference × length = 1.5 × 25 = 37.5 cm².
2. Total area for 120 000 pencils = 120 000 × 37.5 = 4 500 000 cm².
3. 1 dm² = 100 cm², so total area = 45 000 dm².
4. Cost = 45 000 × Rs 0.05 = Rs 2250.

Teacher's Note:
a) The circumference of the base gives the width of the curved surface directly.
b) Convert cm² to dm² by dividing by 100.

30. Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in pond rise by 21 cm?

Answer:
1. Radius of pipe = 0.07 m, speed = 15 000 m/h.
2. Volume flowing per hour = \( \frac{22}{7} \) × 0.07² × 15 000 = 231 m³.
3. Volume needed to raise the level = 50 × 44 × 0.21 = 462 m³.
4. Time = 462 ÷ 231 = 2 hours.

Teacher's Note:
a) Convert the speed to metres per hour and the diameter to metres.
b) The pond volume is the area of the base multiplied by the rise in level.

31. A solid iron cuboidal block of dimensions 4.4 m × 2.6 m × 1m is recast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

Answer:
1. Volume of block = 4.4 × 2.6 × 1 = 11.44 m³ = 11 440 000 cm³.
2. External radius of pipe = 30 + 5 = 35 cm.
3. Area of cross-section = π(35² − 30²) = 325π cm².
4. Length = 11 440 000 ÷ (325π) = 11 200 cm = 112 m.

Teacher's Note:
a) The thickness is added to the internal radius to get the external radius.
b) Keep all units in centimetres before dividing.

32. 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04m3?

Answer:
1. Volume displaced = 500 × 0.04 = 20 m³.
2. Area of pond = 80 × 50 = 4000 m².
3. Rise in water level = 20 ÷ 4000 = 0.005 m = 0.5 cm.

Teacher's Note:
a) The rise in level equals the displaced volume divided by the pond's area.
b) Convert the answer to centimetres for a clearer result.

33. 16 glass spheres each of radius 2 cm are packed into a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and then the box is filled with water. Find the volume of water filled in the box.

Answer:
1. Volume of box = 16 × 8 × 8 = 1024 cm³.
2. Volume of one sphere = \( \frac{4}{3} \) × \( \frac{22}{7} \) × 8 = \( \frac{704}{21} \) cm³.
3. Volume of 16 spheres = 16 × \( \frac{704}{21} \) ≈ 536.38 cm³.
4. Volume of water = 1024 − 536.38 ≈ 487.62 cm³.

Teacher's Note:
a) Water fills only the space that the spheres do not occupy.
b) Subtract the total sphere volume from the box volume.

34. A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of Rs. 22 per litre which the container can hold.

Answer:
1. Volume = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 16 × (20² + 20 × 8 + 8²) = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 16 × 624 ≈ 10 458.67 cm³.
2. Capacity = 10.46 litres.
3. Cost of milk = 10.46 × Rs 22 ≈ Rs 230.09.

Teacher's Note:
a) The lower radius is 8 cm and the upper radius is 20 cm, so use both in the frustum formula.
b) Divide the volume in cm³ by 1000 to get litres.

35. A cylindrical bucket of height 32 cm and base radius 18 cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Answer:
1. Volume of sand = π × 18² × 32 = 10 368π cm³.
2. Let the radius of the heap be r. Then \( \frac{1}{3} \) π r² × 24 = 10 368π, so 8r² = 10 368.
3. r² = 1296, so r = 36 cm.
4. Slant height = \( \sqrt{36^2 + 24^2} \) = \( \sqrt{1872} \) = 12\( \sqrt{13} \) ≈ 43.27 cm.

Teacher's Note:
a) The sand volume stays the same when it is poured into the heap.
b) The slant height is the hypotenuse of the right triangle formed by r and h.

36. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket [Use π = 3.14].

Answer:
1. Radius of cylinder and cone = 3 cm, height of cylinder = 12 cm.
2. Height of cone = \( \sqrt{5^2 - 3^2} \) = 4 cm.
3. Curved surface of cylinder = 2π × 3 × 12 = 72π cm².
4. Area of closed lower end = π × 3² = 9π cm².
5. Curved surface of cone = π × 3 × 5 = 15π cm².
6. Total surface area = (72 + 9 + 15)π = 96π = 301.44 cm².
7. Volume = π × 3² × 12 + \( \frac{1}{3} \) π × 3² × 4 = 108π + 12π = 120π = 376.8 cm³.

Teacher's Note:
a) The top of the cylinder is covered by the cone's base, so it is not counted.
b) The slant height is used for the cone's curved surface.
c) Use π = 3.14 throughout as the question requires.

37. A building is in the form a cylinder surmounted by a hemispherical vaulted dome and contains 41\( \frac{19}{21} \) m³ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building?

Answer:
1. Let the total height be H. Then the diameter of the dome is H, so radius = \( \frac{H}{2} \).
2. Height of cylinder = H − \( \frac{H}{2} \) = \( \frac{H}{2} \).
3. Volume = π (\( \frac{H}{2} \))² × \( \frac{H}{2} \) + \( \frac{2}{3} \) π (\( \frac{H}{2} \))³ = \( \frac{5\pi H^3}{24} \).
4. \( \frac{5\pi H^3}{24} \) = \( \frac{880}{21} \), which gives H³ = 64 when π = \( \frac{22}{7} \).
5. H = 4 m.

Teacher's Note:
a) The dome's radius is half of its diameter, which equals half the total height.
b) Check by substituting H = 4 m back into the volume formula.

38. A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl?

Answer:
1. Volume of bowl = \( \frac{2}{3} \) π × 9³ = 486π cm³.
2. Volume of one bottle = π × 1.5² × 4 = 9π cm³.
3. Number of bottles = 486π ÷ 9π = 54.
4. Hence, 54 bottles are needed.

Teacher's Note:
a) A bowl is a hemisphere, so use \( \frac{2}{3} \) π r³.
b) Check whether any liquid remains after the last bottle.

39. A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.

Answer:
1. Volume of cylinder = \( \frac{22}{7} \) × 60² × 180 = 2 036 571.43 cm³.
2. Volume of cone = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 60² × 120 = 452 571.43 cm³.
3. Volume of water left = 2 036 571.43 − 452 571.43 = 1 584 000 cm³ = 504 000π cm³.
4. Hence, the volume of water left is 1 584 000 cm³ (1584 litres).

Teacher's Note:
a) The cone displaces its own volume of water when it is placed in the cylinder.
b) Divide by 1000 to convert cm³ to litres.

40. Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cm/sec in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in tank in half an hour?

Answer:
1. Volume flowing per second = π × 1² × 80 = 80π cm³.
2. Time = 30 minutes = 1800 seconds, so volume = 80π × 1800 = 144 000π cm³.
3. Area of tank base = π × 40² = 1600π cm².
4. Rise in level = 144 000π ÷ 1600π = 90 cm.

Teacher's Note:
a) Convert half an hour into seconds before multiplying.
b) The tank's radius is 40 cm, so its base area uses 40².

41. The rain water from a roof of dimensions 22 m 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.

Answer:
1. Volume of vessel = \( \frac{22}{7} \) × 1² × 3.5 = 11 m³.
2. Area of roof = 22 × 20 = 440 m².
3. Rainfall = 11 ÷ 440 = 0.025 m = 2.5 cm.

Teacher's Note:
a) The vessel's radius is 1 m, which is half its diameter.
b) Rainfall is the depth of water on the roof, found by dividing volume by area.

42. A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.

Answer:
1. Volume of cuboid = 10 × 5 × 4 = 200 cm³.
2. Volume of one conical depression = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 0.5² × 2.1 = 0.55 cm³.
3. Volume of four conical depressions = 2.2 cm³.
4. Volume of cubical depression = 3³ = 27 cm³.
5. Volume of wood = 200 − 2.2 − 27 = 170.8 cm³.

Teacher's Note:
a) Every depression is removed from the cuboid, so subtract each volume.
b) The conical depressions are small, but they must still be counted.

43. A cone of radius 10cm is divided into two parts by drawing a plane through the midpoint of its axis, parallel to its base. Compare the volume of the two parts.

Answer:
1. The plane through the mid point of the axis gives a top cone with scale factor \( \frac{1}{2} \).
2. Volume of the top cone = \( \left(\frac{1}{2}\right)^3 \) = \( \frac{1}{8} \) of the whole cone.
3. Volume of the frustum = \( \frac{7}{8} \) of the whole cone.
4. Ratio of the two parts = 1 : 7.

Teacher's Note:
a) The result is the same for any cone, whatever its height.
b) The answer is the same as in question 4.

44. A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface of the remainder is \( \frac{8}{9} \) of the curved surface of the whole cone. Find the ratio of the line segments into which the cone's altitude is divided by the plane.

Answer:
1. Let the small cone removed be similar to the whole cone with scale factor k.
2. Curved surface of the small cone = k² × curved surface of the whole cone.
3. Curved surface of the remainder = (1 − k²) × curved surface of the whole cone.
4. 1 − k² = \( \frac{8}{9} \), so k² = \( \frac{1}{9} \) and k = \( \frac{1}{3} \).
5. The upper segment of the altitude is \( \frac{1}{3} \) of the whole altitude, and the lower segment is \( \frac{2}{3} \).
6. Ratio of the two segments = 1 : 2.

Teacher's Note:
a) Curved surfaces of similar cones scale with the square of the linear scale factor.
b) The altitude segments scale with the linear factor, not its square.

45. From a solid cylinder of height 24cm and diameter 10cm, two conical cavities of same radius as that of the cylinder are hollowed out. If the height of each conical activity is half the height of cylinder, find the total surface area of the remaining cylinder.

Answer:
1. Radius = 5 cm, cylinder height = 24 cm, and each cone has height 12 cm.
2. Slant height of each cone = \( \sqrt{5^2 + 12^2} \) = 13 cm.
3. Curved surface of cylinder = 2π × 5 × 24 = 240π cm².
4. Curved surface of each cone = π × 5 × 13 = 65π cm².
5. The cones have the same radius as the cylinder, so the two flat ends are completely removed.
6. Total surface area = 240π + 2 × 65π = 370π ≈ 1162.86 cm².

Teacher's Note:
a) When the cone's base equals the cylinder's end, no flat end remains.
b) Add the curved surface of each cone, since it becomes the new inner surface.

46. A farmer connects a pipe of internal diameter 20cm from a canal into a cylindrical tank to her field, which is 10m in diameter and 2m deep. If water flows through the pipe at the rate of 3km/hr, in how much time will the tank be filled?

Answer:
1. Radius of pipe = 0.1 m, and speed = 3000 m/h.
2. Volume flowing per hour = π × 0.1² × 3000 = 30π m³.
3. Volume of tank = π × 5² × 2 = 50π m³.
4. Time = 50π ÷ 30π = \( \frac{5}{3} \) hours = 1 hour 40 minutes.

Teacher's Note:
a) Use the radius of the pipe, which is half of 20 cm.
b) Convert the answer from hours to hours and minutes for a clear result.

47. A toy is in the form of a cone on a hemi-sphere of diameter 7 cm. The toal height of the top is 14.5cm. Find the volume and total surface area of the toy.

Answer:
1. Radius of hemisphere = 3.5 cm, so cone height = 14.5 − 3.5 = 11 cm.
2. Slant height of cone = \( \sqrt{3.5^2 + 11^2} \) = \( \sqrt{133.25} \) ≈ 11.54 cm.
3. Volume = \( \frac{2}{3} \) π (3.5)³ + \( \frac{1}{3} \) π (3.5)² × 11 ≈ 230.74 cm³.
4. Curved surface of hemisphere = 2π(3.5)² = 77 cm².
5. Curved surface of cone = π × 3.5 × 11.54 ≈ 126.98 cm².
6. Total surface area = 77 + 126.98 ≈ 203.98 cm².

Teacher's Note:
a) The total height includes the hemisphere's radius, so subtract it to get the cone's height.
b) The flat base of the hemisphere is not counted in the total surface area.

48. A vessel in the form of hemi-spherical is mounted by a hollow cylinder. The diameter of the bowl is 14cm and the total height of the vessel is 13 cm. Find the capacity of the vessel.

Answer:
1. Radius of bowl = 7 cm, so height of cylinder = 13 − 7 = 6 cm.
2. Volume of hemisphere = \( \frac{2}{3} \) × \( \frac{22}{7} \) × 7³ = \( \frac{2156}{3} \) ≈ 718.67 cm³.
3. Volume of cylinder = \( \frac{22}{7} \) × 7² × 6 = 924 cm³.
4. Capacity = 718.67 + 924 ≈ 1642.67 cm³.

Teacher's Note:
a) The cylinder's height is the total height minus the bowl's radius.
b) The capacity is about 1.64 litres.

49. A cylindrical with radius and height is 4cm and 8cm is filled with Ice-cream and ice-cream is distributed to 10 Children in equal can having hemi-spherical tops. If the height of the conical portion is 4 times the radius of its base, find the radius of the ice-cream cone.

Answer:
1. Volume of ice cream = π × 4² × 8 = 128π cm³.
2. Each child gets \( \frac{1}{10} \) of this, so each cone holds 12.8π cm³.
3. Let the radius of the cone be r. Then its height = 4r, and the volume = \( \frac{1}{3} \) π r² (4r) + \( \frac{2}{3} \) π r³ = 2π r³.
4. 2π r³ = 12.8π, so r³ = 6.4.
5. r ≈ 1.86 cm.

Teacher's Note:
a) Each child's cone holds exactly one tenth of the total ice cream.
b) The cone and hemisphere together give the factor 2π r³.

50. A tent has cylindrical surmounted by a conical roof. The radius of the cylindrical base is 20m. The total height of tent is 6.3m and height of cylindrical portion is 4.2m, find the volume and surface area of tent.

Answer:
1. Height of cone = 6.3 − 4.2 = 2.1 m.
2. Volume = \( \frac{22}{7} \) × 20² × 4.2 + \( \frac{1}{3} \) × \( \frac{22}{7} \) × 20² × 2.1 = 6160 m³.
3. Slant height of cone = \( \sqrt{20^2 + 2.1^2} \) ≈ 20.11 m.
4. Curved surface of cylinder = 2 × \( \frac{22}{7} \) × 20 × 4.2 = 528 m².
5. Curved surface of cone = \( \frac{22}{7} \) × 20 × 20.11 ≈ 1264.0 m².
6. Surface area of tent ≈ 528 + 1264.0 = 1792.0 m².

Teacher's Note:
a) The conical roof's height is the total height minus the cylinder's height.
b) The surface area counts only the curved parts, since the tent has no floor counted.

51. Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take π = 22/7)

Answer:
1. Radius = 1.75 cm, height of cone = 5 − 1.75 = 3.25 cm.
2. Slant height of cone = \( \sqrt{1.75^2 + 3.25^2} \) ≈ 3.69 cm.
3. Curved surface of cone = \( \frac{22}{7} \) × 1.75 × 3.69 ≈ 20.31 cm².
4. Curved surface of hemisphere = 2 × \( \frac{22}{7} \) × 1.75² = 19.25 cm².
5. Area to colour = 20.31 + 19.25 ≈ 39.56 cm².

Teacher's Note:
a) The flat base of the hemisphere is joined to the cone and is not coloured.
b) Subtract the radius from the total height to get the cone's height.

52. A wooden toy rocket is in the shape of a cone mounted on a cylinder. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take π = 3.14)

Answer:
1. Height of cylinder = 26 − 6 = 20 cm.
2. Slant height of cone = \( \sqrt{2.5^2 + 6^2} \) = 6.5 cm.
3. Area painted orange = curved surface of cone = 3.14 × 2.5 × 6.5 ≈ 51.03 cm².
4. Area painted yellow = curved surface of cylinder = 2 × 3.14 × 1.5 × 20 = 188.4 cm².

Teacher's Note:
a) Use the radius of each part, which is half its diameter.
b) The exposed flat ring of the cone's base is not counted here.

53. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Answer:
1. Cone height = 15.5 − 3.5 = 12 cm.
2. Slant height = \( \sqrt{3.5^2 + 12^2} \) = 12.5 cm.
3. Curved surface of cone = \( \frac{22}{7} \) × 3.5 × 12.5 = 137.5 cm².
4. Curved surface of hemisphere = 2 × \( \frac{22}{7} \) × 3.5² = 77 cm².
5. Total surface area = 137.5 + 77 = 214.5 cm².

Teacher's Note:
a) The total height includes the hemisphere's radius, so subtract it first.
b) The base of the hemisphere is not part of the outside surface.

54. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2

Answer:
1. Radius = 2 m, cylinder height = 2.1 m, slant height of cone = 2.8 m.
2. Curved surface of cylinder = \( \frac{22}{7} \) × 2 × 2 × 2.1 = 26.4 m².
3. Curved surface of cone = \( \frac{22}{7} \) × 2 × 2.8 = 17.6 m².
4. Canvas required = 26.4 + 17.6 = 44 m².
5. Cost = 44 × Rs 500 = Rs 22 000.

Teacher's Note:
a) Canvas covers only the curved surfaces, not the floor.
b) The slant height is given directly, so no Pythagoras step is needed.

55. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

Answer:
1. Radius = 0.7 cm, height = 2.4 cm, slant height of cavity = \( \sqrt{0.7^2 + 2.4^2} \) = 2.5 cm.
2. Curved surface of cylinder = 2π × 0.7 × 2.4 = 3.36π cm².
3. Area of the bottom base = π × 0.7² = 0.49π cm².
4. Curved surface of cone cavity = π × 0.7 × 2.5 = 1.75π cm².
5. The top circle is fully open because the cone's base equals it, so it is not counted.
6. Total surface area = (3.36 + 0.49 + 1.75)π = 5.6π ≈ 17.59 cm² ≈ 18 cm².

Teacher's Note:
a) The cavity's curved surface becomes part of the outside surface.
b) Round to the nearest whole square centimetre as asked.

56. A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Take π = 3.14)

Answer:
1. Radius = 2.5 cm, height = 10 cm.
2. Apparent capacity = 3.14 × 2.5² × 10 = 196.25 cm³.
3. Volume of hemispherical raised portion = \( \frac{2}{3} \) × 3.14 × 2.5³ ≈ 32.71 cm³.
4. Actual capacity = 196.25 − 32.71 ≈ 163.54 cm³.

Teacher's Note:
a) The apparent capacity ignores the raised bottom.
b) The actual capacity is the apparent capacity minus the hemisphere's volume.

57. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take π = 3.14)

Answer:
1. Radius = 2 cm, cone height = 2 cm.
2. Volume of toy = \( \frac{2}{3} \) × 3.14 × 2³ + \( \frac{1}{3} \) × 3.14 × 2² × 2 = 8π ≈ 25.12 cm³.
3. Cylinder radius = 2 cm, height = 2 + 2 = 4 cm.
4. Volume of cylinder = 3.14 × 2² × 4 = 50.24 cm³.
5. Difference = 50.24 − 25.12 = 25.12 cm³.

Teacher's Note:
a) The cylinder's height equals the hemisphere's radius plus the cone's height.
b) The difference here equals the toy's volume, which is a neat check.

58. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Answer:
1. Radius = 1.4 cm, so the cylindrical middle part has height = 5 − 2 × 1.4 = 2.2 cm.
2. Volume of cylinder = \( \frac{22}{7} \) × 1.4² × 2.2 ≈ 13.55 cm³.
3. Volume of two hemispheres = \( \frac{4}{3} \) × \( \frac{22}{7} \) × 1.4³ ≈ 11.50 cm³.
4. Volume of one gulab jamun ≈ 25.05 cm³.
5. Syrup in one gulab jamun = 30% of 25.05 ≈ 7.52 cm³.
6. Syrup in 45 gulab jamuns ≈ 45 × 7.52 ≈ 338.2 cm³.

Teacher's Note:
a) The length of 5 cm includes both hemispherical ends.
b) Two hemispheres together make one full sphere.

59. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Answer:
1. Volume of cuboid = 15 × 10 × 3.5 = 525 cm³.
2. Volume of one conical depression = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 0.5² × 1.4 ≈ 0.3667 cm³.
3. Volume of four depressions = 1.4667 cm³.
4. Volume of wood = 525 − 1.4667 ≈ 523.53 cm³.

Teacher's Note:
a) Only the four conical depressions are removed in this stand.
b) The depressions are small, but each one must still be subtracted.

60. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer:
1. Volume of water in vessel = \( \frac{1}{3} \) π × 5² × 8 = \( \frac{200}{3} \) π cm³.
2. Water that flows out = \( \frac{1}{4} \) × \( \frac{200}{3} \) π = \( \frac{50}{3} \) π cm³.
3. Volume of one lead shot = \( \frac{4}{3} \) π (0.5)³ = \( \frac{\pi}{6} \) cm³.
4. Number of shots = \( \frac{50}{3} \) π ÷ \( \frac{\pi}{6} \) = 100.

Teacher's Note:
a) The volume that overflows equals the volume of the shots that displace it.
b) Take π out of both sides before dividing.

61. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14)

Answer:
1. Volume of first cylinder = 3.14 × 12² × 220 = 99 475.2 cm³.
2. Volume of second cylinder = 3.14 × 8² × 60 = 12 057.6 cm³.
3. Total volume = 111 532.8 cm³.
4. Mass = 111 532.8 × 8 = 892 262.4 g ≈ 892.26 kg.

Teacher's Note:
a) Use the radius 12 cm for the base diameter of 24 cm.
b) Divide grams by 1000 to get kilograms.

62. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Answer:
1. Volume of cone = \( \frac{1}{3} \) π × 60² × 120 = 144 000π cm³.
2. Volume of hemisphere = \( \frac{2}{3} \) π × 60³ = 144 000π cm³.
3. Volume of solid = 288 000π cm³.
4. Volume of cylinder = π × 60² × 180 = 648 000π cm³.
5. Water left = 648 000π − 288 000π = 360 000π ≈ 1 131 428.57 cm³ (using π = \( \frac{22}{7} \)).

Teacher's Note:
a) The solid displaces its own volume of water when it is placed in the cylinder.
b) This is about 1131 litres of water left in the cylinder.

63. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Answer:
1. Radius of sphere = 4.25 cm, so volume of sphere = \( \frac{4}{3} \) × 3.14 × 4.25³ ≈ 321.4 cm³.
2. Radius of neck = 1 cm and length = 8 cm, so volume of neck = 3.14 × 1² × 8 = 25.12 cm³.
3. Total volume = 321.4 + 25.12 ≈ 346.5 cm³.
4. The child's figure of 345 cm³ is not exact, since the volume is about 346.5 cm³.
5. Hence, she is not correct in her exact measurement.

Teacher's Note:
a) The neck is a cylinder, so add its volume to the sphere's volume.
b) A small difference in reading can come from measuring error.

64. A cone of height 24 cm and radius of base 6 cm is made up of modeling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.

Answer:
1. Volume of cone = \( \frac{1}{3} \) π × 6² × 24 = 288π cm³.
2. Let the radius of the sphere be R. Then \( \frac{4}{3} \) π R³ = 288π, so R³ = 216.
3. R = 6 cm.

Teacher's Note:
a) Reshaping keeps the volume of clay the same.
b) Cube root of 216 is 6, so check by cubing 6.

65. Selvi's house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44 m × 95cm. The overhead tank has its radius 60 cm and height 95 cm. Find the height of the water left in the sump after the overhead tank has been completely filled with water from the sump which had been full. Compare the capacity of the tank with that of the sump. (Use π = 3.14)

Answer:
1. Volume of sump = 1.57 × 1.44 × 0.95 ≈ 2.148 m³.
2. Volume of overhead tank = 3.14 × 0.6² × 0.95 ≈ 1.074 m³.
3. Height of water removed from sump = 1.074 ÷ (1.57 × 1.44) ≈ 0.475 m.
4. Height of water left in sump = 0.95 − 0.475 = 0.475 m = 47.5 cm.
5. Ratio of tank capacity to sump capacity = 1.074 : 2.148 = 1 : 2.

Teacher's Note:
a) The sump's water level drops by the same height that the tank's volume represents.
b) Convert all lengths to metres before calculating.

66. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.

Answer:
1. Volume of rod = π × 0.5² × 8 = 2π cm³.
2. Length of wire = 18 m = 1800 cm.
3. Cross-section of wire = 2π ÷ 1800 = \( \frac{\pi}{900} \) cm².
4. Radius of wire = \( \frac{1}{30} \) cm, so the thickness (diameter) = \( \frac{1}{15} \) cm ≈ 0.67 mm.

Teacher's Note:
a) The volume of the rod stays the same when it is drawn into wire.
b) The thickness of a wire is taken as its diameter.

67. A hemispherical tank full of water is emptied by a pipe at the rate of 3\( \frac{4}{7} \) litres per second. How much time will it take to empty half the tank, if it is 3m in diameter? (Take π = 22/7)

Answer:
1. Radius of tank = 1.5 m.
2. Volume of full hemisphere = \( \frac{2}{3} \) × \( \frac{22}{7} \) × 1.5³ ≈ 7.07 m³.
3. Half the tank = 3.54 m³ = 3535.71 litres.
4. Rate = 3\( \frac{4}{7} \) = \( \frac{25}{7} \) litres per second.
5. Time = 3535.71 ÷ \( \frac{25}{7} \) = 990 seconds = 16 minutes 30 seconds.

Teacher's Note:
a) Half the tank means half of its water, not half of its height.
b) Convert litres per second and cubic metres into the same unit before dividing.

68. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Answer:
1. Volume of earth = \( \frac{22}{7} \) × 3.5² × 20 = 770 m³.
2. Area of platform = 22 × 14 = 308 m².
3. Height of platform = 770 ÷ 308 = 2.5 m.

Teacher's Note:
a) The earth dug out from the well equals the volume of the well.
b) Use the radius, which is 3.5 m, not the diameter.

69. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Answer:
1. Volume of earth = \( \frac{22}{7} \) × 1.5² × 14 = 99 m³.
2. Outer radius of embankment = 1.5 + 4 = 5.5 m, inner radius = 1.5 m.
3. Area of the ring = \( \frac{22}{7} \) (5.5² − 1.5²) = 88 m².
4. Height of embankment = 99 ÷ 88 = 1.125 m.

Teacher's Note:
a) The embankment is a ring, so its area is outer minus inner.
b) The earth dug out forms the ring's volume.

70. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Answer:
1. Volume of container = π × 6² × 15 = 540π cm³.
2. Volume of one cone = \( \frac{1}{3} \) π × 3² × 12 = 36π cm³.
3. Volume of hemispherical top = \( \frac{2}{3} \) π × 3³ = 18π cm³.
4. Volume of one ice cream cone = 36π + 18π = 54π cm³.
5. Number of cones = 540π ÷ 54π = 10.

Teacher's Note:
a) Each cone includes its hemispherical top in the filling.
b) Use the radius 3 cm for the cone, which is half of 6 cm.

71. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Answer:
1. Volume of cuboid = 5.5 × 10 × 3.5 = 192.5 cm³.
2. Radius of coin = 0.875 cm, thickness = 0.2 cm.
3. Volume of one coin = \( \frac{22}{7} \) × 0.875² × 0.2 = 0.48125 cm³.
4. Number of coins = 192.5 ÷ 0.48125 = 400.

Teacher's Note:
a) Convert the thickness from mm to cm before multiplying.
b) Use the radius, which is half of 1.75 cm.

72. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Answer:
1. This is the same as question 35. Volume of sand = π × 18² × 32 = 10 368π cm³.
2. Let the radius of the heap be r. Then \( \frac{1}{3} \) π r² × 24 = 10 368π, so r² = 1296 and r = 36 cm.
3. Slant height = \( \sqrt{36^2 + 24^2} \) = 12\( \sqrt{13} \) ≈ 43.27 cm.

Teacher's Note:
a) The volume of sand stays the same when it is poured into the heap.
b) This question repeats question 35 with the same answer.

73. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Answer:
1. Length of water flowing in 30 minutes = 10 000 ÷ 2 = 5000 m.
2. Volume of water = 6 × 1.5 × 5000 = 45 000 m³.
3. Standing water depth = 0.08 m.
4. Area irrigated = 45 000 ÷ 0.08 = 562 500 m² = 56.25 hectares.

Teacher's Note:
a) Convert 30 minutes into hours before multiplying by the speed.
b) Area irrigated equals the volume of water divided by the depth needed.

74. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Answer:
1. This is the same as question 46.
2. Radius of pipe = 0.1 m, speed = 3000 m/h, so volume flowing per hour = 30π m³.
3. Volume of tank = 50π m³.
4. Time = 50π ÷ 30π = \( \frac{5}{3} \) hours = 1 hour 40 minutes.

Teacher's Note:
a) Use the radius of the pipe, which is half of 20 cm.
b) This question repeats question 46 with the same answer.

75. Hanumappa and his wife Gangamma are busy making jaggery out of sugarcane juice. They have processed the sugarcane juice to make the molasses, which is poured into moulds in the shape of a frustum of a cone having the diameters of its two circular faces as 30 cm and 35 cm and the vertical height of the mould is 14 cm. If each cm3 of molasses has mass about 1.2 g, find the mass of the molasses that can be poured into each mould. (Take π = 22/7)

Answer:
1. Radii = 15 cm and 17.5 cm, height = 14 cm.
2. Volume = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 14 × (17.5² + 17.5 × 15 + 15²) = 11 641.67 cm³.
3. Mass = 11 641.67 × 1.2 = 13 970 g ≈ 13.97 kg.

Teacher's Note:
a) Halve the diameters to get the two radii before using the frustum formula.
b) Convert grams to kilograms by dividing by 1000.

76. An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold.

Answer:
1. Radii = 22.5 cm and 12.5 cm. Height of frustum = 40 − 6 = 34 cm.
2. Slant height = \( \sqrt{34^2 + 10^2} \) ≈ 35.44 cm.
3. Curved surface of frustum = \( \frac{22}{7} \) × 35 × 35.44 ≈ 3898.41 cm².
4. Curved surface of cylindrical base = 2 × \( \frac{22}{7} \) × 12.5 × 6 ≈ 471.43 cm².
5. Area of the bottom circle = \( \frac{22}{7} \) × 12.5² ≈ 491.07 cm².
6. Area of metal sheet ≈ 3898.41 + 471.43 + 491.07 ≈ 4860.91 cm².
7. Volume of water = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 34 × (22.5² + 22.5 × 12.5 + 12.5²) ≈ 33 615.48 cm³.

Teacher's Note:
a) The bottom circle separates the bucket from the hollow base, so only the frustum holds water.
b) The cylindrical base is hollow and is not part of the water capacity.
c) The water volume is about 33.6 litres.

77. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2. (Take π = 3.14)

Answer:
1. Slant height = \( \sqrt{16^2 + 12^2} \) = 20 cm.
2. Volume = \( \frac{1}{3} \) × 3.14 × 16 × (20² + 20 × 8 + 8²) = 10 449.92 cm³ ≈ 10.45 litres.
3. Cost of milk = 10.45 × Rs 20 ≈ Rs 209.00.
4. Curved surface = 3.14 × (20 + 8) × 20 = 1758.4 cm².
5. Area of base = 3.14 × 8² = 200.96 cm².
6. Area of metal sheet = 1758.4 + 200.96 = 1959.36 cm².
7. Cost of sheet = 1959.36 × Rs 8 ÷ 100 ≈ Rs 156.75.

Teacher's Note:
a) The container is open at the top, so only the base is added to the curved surface.
b) Sheet cost is charged per 100 cm², so divide the area by 100 before multiplying.

78. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \( \frac{1}{16} \) cm, find the length of the wire.

Answer:
1. Half the vertical angle is 30°, so the base radius R = 20 tan 30° = \( \frac{20}{\sqrt{3}} \) cm.
2. Radius of the top of the frustum r = \( \frac{R}{2} \) = \( \frac{10}{\sqrt{3}} \) cm, and the frustum height is 10 cm.
3. Volume of frustum = \( \frac{1}{3} \) π × 10 × (R² + Rr + r²) = \( \frac{7000\pi}{9} \) cm³.
4. Radius of wire = \( \frac{1}{32} \) cm, so its cross-section = \( \frac{\pi}{1024} \) cm².
5. Length of wire = \( \frac{7000\pi}{9} \) ÷ \( \frac{\pi}{1024} \) = \( \frac{7\,168\,000}{9} \) ≈ 796 444 cm ≈ 7964.4 m.

Teacher's Note:
a) The vertical angle is 60°, so each half-angle at the apex is 30°.
b) The wire's diameter is given, so use half of it as the radius.

79. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.

Answer:
1. Hypotenuse = 5 cm. Altitude from the right angle to the hypotenuse = \( \frac{3 \times 4}{5} \) = 2.4 cm, which is the common radius.
2. Heights of the two cones along the hypotenuse: \( \sqrt{3^2 - 2.4^2} \) = 1.8 cm and \( \sqrt{4^2 - 2.4^2} \) = 3.2 cm.
3. Volume = \( \frac{1}{3} \) π × 2.4² × (1.8 + 3.2) = 9.6π ≈ 30.17 cm³.
4. Surface area = π × 2.4 × (3 + 4) = 16.8π ≈ 52.8 cm².

Teacher's Note:
a) The common base radius is the altitude to the hypotenuse, not a leg.
b) The two cones together share one base circle, which is inside the double cone.

80. The decorative block shown in Fig. is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Take π = 22/7).

Answer:
1. Surface area of cube = 6 × 5² = 150 cm².
2. Area of the hemisphere's flat base covered by the hemisphere = π × 2.1² = 13.86 cm² is removed from the top face and added back in the hemisphere.
3. Curved surface of hemisphere = 2π × 2.1² = 27.72 cm².
4. Total surface area = 150 − 13.86 + 27.72 = 163.86 cm².

Teacher's Note:
a) The hemisphere's flat base covers part of the cube top, so it is not counted twice.
b) Only the curved part of the hemisphere is exposed.

81. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in above Fig.. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Answer:
1. Curved surface of the cylinder = 2π × 3.5 × 10 = 70π cm².
2. Each hemispherical cavity has curved surface = 2π × 3.5² = 24.5π cm².
3. Two cavities give 49π cm².
4. Total surface area = 70π + 49π = 119π = 374 cm².

Teacher's Note:
a) The flat ends of the cylinder are completely removed by the hemispheres.
b) The cavities add their curved surfaces to the outside of the article.

82. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water level in the cylindrical vessel rises by 3\( \frac{5}{9} \) cm. find the diameter of the cylindrical vessel.

Answer:
1. Radius of sphere = 6 cm, so volume = \( \frac{4}{3} \) π × 216 = 288π cm³.
2. Rise in water level = 3\( \frac{5}{9} \) = \( \frac{32}{9} \) cm.
3. Let the radius of the vessel be R. Then π R² × \( \frac{32}{9} \) = 288π, so R² = 81.
4. R = 9 cm, so the diameter of the vessel = 18 cm.

Teacher's Note:
a) The volume of water displaced equals the sphere's volume.
b) Convert the mixed fraction to an improper fraction before dividing.

83. An iron pillar has lower part in the form of a right circular cylinder and the upper part is in the form of a right circular cone. The radius of the base of the cone and cylinder is 8cm. The cylindrical part is 240cm high and the conical part is 36cm high. Find the weight of the pillar if 1 cm3 of iron weighs 7.5 grams.

Answer:
1. Volume of cylinder = \( \frac{22}{7} \) × 8² × 240 = 48 384 cm³.
2. Volume of cone = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 8² × 36 = 2304 cm³.
3. Total volume = 50 688 cm³.
4. Weight = 50 688 × 7.5 = 380 160 g = 380.16 kg.

Teacher's Note:
a) The radius is the same for both parts, so only the heights differ.
b) Convert grams to kilograms at the end.

84. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see below figure)

Answer:
1. Height of frustum = 22 − 10 = 12 cm, with radii R = 9 cm and r = 4 cm.
2. Slant height of frustum = \( \sqrt{12^2 + 5^2} \) = 13 cm.
3. Curved surface of frustum = \( \frac{22}{7} \) × (9 + 4) × 13 ≈ 531.43 cm².
4. Curved surface of cylinder = 2 × \( \frac{22}{7} \) × 4 × 10 ≈ 251.43 cm².
5. Area of tin sheet = 531.43 + 251.43 ≈ 782.86 cm².

Teacher's Note:
a) The funnel is open at both ends, so no base is added.
b) The slant height of the frustum comes from the difference of the radii, 5 cm, and its height.

85. The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm (see above sided Fig). Find its volume, the curved surface area and the total surface area. (Take π = 22/7)

Answer:
1. Slant height = \( \sqrt{45^2 + (28 - 7)^2} \) = \( \sqrt{2466} \) ≈ 49.66 cm.
2. Volume = \( \frac{1}{3} \) × \( \frac{22}{7} \) × 45 × (28² + 28 × 7 + 7²) = 48 510 cm³.
3. Curved surface area = \( \frac{22}{7} \) × (28 + 7) × 49.66 ≈ 5462.47 cm².
4. Total surface area = 5462.47 + \( \frac{22}{7} \) × 28² + \( \frac{22}{7} \) × 7² = 5462.47 + 2464 + 154 ≈ 8080.47 cm².

Teacher's Note:
a) The slant height uses the difference of the two radii, which is 21 cm.
b) The total surface area adds both circular ends to the curved surface.
c) The volume is in cm³, which is about 48.51 litres.

[Take \( \pi = \frac{22}{7} \), (Unless stated otherwise)]

Question. A metallic sphere of radius \( 4.2 \text{ cm} \) is melted and recast into the shape of a cylinder of radius \( 6 \text{ cm} \). Find the height of the cylinder.
Answer: Radius of the sphere \( (r_1) = 4.2 \text{ cm} \)
∴ Volume of the sphere \( \left( \frac{4}{3} \pi r_1^3 \right) = \frac{4}{3} \times \frac{22}{7} \times \frac{42}{10} \times \frac{42}{10} \times \frac{42}{10} \text{ cm}^3 \)
Radius of the cylinder \( (r_2) = 6 \text{ cm} \)
Let ‘h’ be the height of the cylinder
∴ Volume of the cylinder \( = \pi r_2^2 h = \frac{22}{7} \times 6 \times 6 \times h \text{ cm}^3 \)
Since, Volume of the metallic sphere = Volume of the cylinder.
\( \Rightarrow \frac{4}{3} \times \frac{22}{7} \times \frac{42}{10} \times \frac{42}{10} \times \frac{42}{10} = \frac{22}{7} \times 6 \times 6 \times h \)
\( \Rightarrow h = \frac{4}{3} \times \frac{22}{7} \times \frac{42}{10} \times \frac{42}{10} \times \frac{42}{10} \times \frac{7}{22} \times \frac{1}{6} \times \frac{1}{6} \text{ cm} \)
\( = \frac{4 \times 7 \times 7 \times 4}{10 \times 10 \times 10} \text{ cm} = \frac{2744}{1000} \text{ cm} = 2.744 \text{ cm} \).

Question. Metallic spheres of radii \( 6 \text{ cm} \), \( 8 \text{ cm} \) and \( 10 \text{ cm} \), respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Answer: Radii of the given spheres are:
\( r_1 = 6 \text{ cm} \)
\( r_2 = 8 \text{ cm} \)
\( r_3 = 10 \text{ cm} \)
⇒ Volume of the given spheres are:
\( V_1 = \frac{4}{3} \pi r_1^3 \), \( V_2 = \frac{4}{3} \pi r_2^3 \) and \( V_3 = \frac{4}{3} \pi r_3^3 \)
∴ Total volume of the given spheres
\( = V_1 + V_2 + V_3 \)
\( = \frac{4}{3} \pi r_1^3 + \frac{4}{3} \pi r_2^3 + \frac{4}{3} \pi r_3^3 = \frac{4}{3} \pi [r_1^3 + r_2^3 + r_3^3] \)
\( = \frac{4}{3} \times \frac{22}{7} \times [6^3 + 8^3 + 10^3] \text{ cm}^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times [216 + 512 + 1000] \text{ cm}^3 = \frac{4}{3} \times \frac{22}{7} \times 1728 \text{ cm}^3 \)
Let the radius of the new big sphere be R.
∴ Volume of the new sphere
\( = \frac{4}{3} \times \pi \times R^3 = \frac{4}{3} \times \frac{22}{7} \times R^3 \)
Since, the two volume must be equal.
∴ \( \frac{4}{3} \times \frac{22}{7} \times R^3 = \frac{4}{3} \times \frac{22}{7} \times 1728 \text{ cm}^3 \)
\( \Rightarrow R^3 = 1728 \)
\( \Rightarrow R^3 = 2^3 \times 2^3 \times 3^3 \)
\( \Rightarrow R^3 = (2 \times 2 \times 3)^3 \)
\( \Rightarrow R = 2 \times 2 \times 3 \)
\( \Rightarrow R = 12 \text{ cm} \)
Thus, the required radius of the resulting sphere \( = 12 \text{ cm} \).

Question. A \( 20 \text{ m} \) deep well with diameter \( 7 \text{ m} \) is dug and the earth from digging is evenly spread out to form a platform \( 22 \text{ m} \) by \( 14 \text{ m} \). Find the height of the platform.
Answer: Diameter of the cylindrical well \( = 7 \text{ m} \)
\( \Rightarrow \) Radius of the cylinder \( (r) = \frac{7}{2} \text{ m} \)
Depth of the well \( (h) = 20 \text{ m} \)
∴ Volume \( = \pi r^2 h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 \text{ m}^3 = 22 \times 7 \times 5 \text{ m}^3 \)
⇒ Volume of the earth taken out \( = 22 \times 7 \times 5 \text{ m}^3 \)
Now this earth is spread out to form a cuboidal platform having
length \( = 22 \text{ m} \)
breadth \( = 14 \text{ m} \)
Let ‘h’ be the height of the platform.
∴ Volume of the platform \( = 22 \times 14 \times h \text{ m}^3 \)
∴ \( 22 \times 14 \times h = 22 \times 7 \times 5 \)
\( \Rightarrow h = \frac{22 \times 7 \times 5}{22 \times 14} = \frac{5}{2} \text{ m} = 2.5 \text{ m} \).
Thus, the required height of the platform is \( 2.5 \text{ m} \).

Question. A well of diameter \( 3 \text{ m} \) is dug \( 14 \text{ m} \) deep. The earth taken out of it has been spread evenly all around it in shape of a circular ring of width \( 4 \text{ m} \) to form an embankment. Find the height of the embankment.  
Answer: Diameter of cylindrical well \( (d) = 3 \text{ m} \)
\( \Rightarrow \) Radius of the cylindrical well \( (r) = \frac{3}{2} \text{ m} = 1.5 \text{ m} \)
Depth of the well \( (h) = 14 \text{ m} \)
∴ Volume \( = \pi r^2 h = \frac{22}{7} \times \left( \frac{15}{10} \right)^2 \times 14 \text{ m}^3 \)
\( = \frac{22 \times 15 \times 15 \times 14}{7 \times 10 \times 10} \text{ m}^3 = 11 \times 3 \times 3 \text{ m}^3 = 99 \text{ m}^3 \)
Let the height of the embankment \( = \text{‘H’ metre} \).
Internal radius of the embankment \( (r) = 1.5 \text{ m} \).
External radius of the embankment \( R = (4 + 1.5) \text{ m} = 5.5 \text{ m} \).
∴ Volume of the embankment
\( = \pi R^2 H - \pi r^2 H = \pi H [R^2 - r^2] = \pi H (R + r) (R - r) \)
\( = \frac{22}{7} \times H (5.5 + 1.5) (5.5 - 1.5) = \frac{22}{7} \times H \times 7 \times 4 \text{ m}^3 \)
Since, Volume of the embankment = Volume of the cylindrical well
∴ \( \frac{22}{7} \times H \times 7 \times 4 = 99 \)
\( \Rightarrow H = \frac{99 \times 7}{22 \times 7 \times 4} = \frac{9}{8} \text{ m} = 1.125 \text{ m} \)
Thus, the required height of the embankment \( = 1.125 \text{ m} \).

Question. How many silver coins, \( 1.75 \text{ cm} \) in diameter and of thickness \( 2 \text{ mm} \), must be melted to form a cuboid of dimensions \( 5.5 \text{ cm} \times 10 \text{ cm} \times 3.5 \text{ cm} \)?
Answer: For a circular coin:
Diameter \( = 1.75 \text{ cm} \)
\( \Rightarrow \) Radius \( (r) = \frac{175}{200} \text{ cm} \)
Thickness \( (h) = 2 \text{ mm} = \frac{2}{10} \text{ cm} \)
∴ Volume \( = \pi r^2 h = \frac{22}{7} \times \left( \frac{175}{200} \right)^2 \times \frac{2}{10} \text{ cm}^3 \) [∵ A coin is like a cylinder]
For a cuboid:
Length \( (l) = 10 \text{ cm} \), Breadth \( (b) = 5.5 \text{ cm} \)
and Height \( (h) = 3.5 \text{ cm} \)
∴ Volume \( = 10 \times \frac{55}{10} \times \frac{35}{10} \text{ cm}^3 \)
Number of coins
Let the number of coins need to melt be ‘n’
∴ \( n = \left( 10 \times \frac{55}{10} \times \frac{35}{10} \right) \div \left( \frac{22}{7} \times \frac{175}{200} \times \frac{175}{200} \times \frac{2}{10} \right) \)
\( = \left( 10 \times \frac{55}{10} \times \frac{35}{10} \right) \times \frac{7}{22} \times \frac{200}{175} \times \frac{200}{175} \times \frac{10}{2} = 16 \times 25 = 400 \)
Thus, the required number of coins \( = 400 \).

Question. A cylindrical bucket, \( 32 \text{ cm} \) high and with radius of base \( 18 \text{ cm} \), is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is \( 24 \text{ cm} \), find the radius and slant height of the heap.  
Answer: For the cylindrical bucket:
Radius \( (r) = 18 \text{ cm} \)
Height \( (h) = 32 \text{ cm} \)
Volume \( = \pi r^2 h = \frac{22}{7} \times (18)^2 \times 32 \text{ cm}^3 \)
\( \Rightarrow \) Volume of the sand \( = \frac{22}{7} \times 18 \times 18 \times 32 \text{ cm}^3 \)
For the conical heap:
Height \( (H) = 24 \text{ cm} \)
Let radius of the base be \( (R) \).
∴ Volume of conical heap \( = \frac{1}{3} \pi R^2 H = \frac{1}{3} \times \frac{22}{7} \times R^2 \times 24 \text{ cm}^3 \)
Radius of the conical heap of sand:
∵ Volume of the conical heap of sand = Volume of the sand
∴ \( \frac{1}{3} \times \frac{22}{7} \times R^2 \times 24 = \frac{22}{7} \times 18 \times 18 \times 32 \)
\( \Rightarrow R^2 = \frac{22}{7} \times 18 \times 18 \times 32 \times \frac{3 \times 7}{22 \times 24} = 18 \times 18 \times 4 = 18^2 \times 2^2 \)
\( \Rightarrow R = 18 \times 2 \text{ cm} = 36 \text{ cm} \)
Slant Height
Let ‘l’ be the slant height of the conical heap of the sand.
∴ \( l^2 = R^2 + H^2 \)
\( \Rightarrow l^2 = 36^2 + 24^2 \)
\( \Rightarrow l^2 = (12 \times 3)^2 + (12 \times 2)^2 \)
\( \Rightarrow l^2 = 12^2 [3^2 + 2^2] = 12^2 \times 13 \)
\( \Rightarrow l = \sqrt{12^2 \times 13} = 12\sqrt{13} \text{ cm} \)
Thus, the required radius \( = 36 \text{ cm} \) and slant height \( = 12\sqrt{13} \text{ cm} \).

Question. Water in a canal, \( 6 \text{ m} \) wide and \( 1.5 \text{ m} \) deep, is flowing with a speed of \( 10 \text{ km/h} \). How much area will it irrigate in \( 30 \text{ minutes} \), if \( 8 \text{ cm} \) of standing water is needed? 
Answer: Width of the canal \( = 6 \text{ m} \)
Depth of the canal \( = 1.5 \text{ m} \)
Length of the water column in \( 1 \text{ hr} = 10 \text{ km} \)
∴ Length of the water column in \( 30 \text{ minutes} \) (i.e., \( \frac{1}{2} \text{ hr} \)) \( = \frac{10}{2} \text{ km} = 5000 \text{ m} \)
∴ Volume of water flown in \( \frac{1}{2} \text{ hr} \)
\( = 6 \times 1.5 \times 5000 \text{ m}^3 = 6 \times \frac{15}{10} \times 5000 \text{ m}^3 = 45000 \text{ m}^3 \)
Since the above amount (volume) of water is spread in the form of a cuboid of height as \( 8 \text{ cm} = \frac{8}{100} \text{ m} \).
Let the area of the cuboid \( = a \)
∴ Volume of the cuboid = Area \( \times \) Height \( = a \times \frac{8}{100} \text{ m}^3 \)
Thus, \( a \times \frac{8}{100} = 45000 \)
\( \Rightarrow a = \frac{45000 \times 100}{8} = \frac{4500000}{8} \text{ m}^2 \)
\( \Rightarrow a = 562500 \text{ m}^2 = \frac{562500}{10000} \text{ hectares} = 56.25 \text{ hectares} \)
Thus, the required area \( = 56.25 \text{ hectares} \).

Question. A farmer connects a pipe of internal diameter \( 20 \text{ cm} \) from a canal into a cylindrical tank in her field, which is \( 10 \text{ m} \) in diameter and \( 2 \text{ m} \) deep. If water flows through the pipe at the rate of \( 3 \text{ km/h} \), in how much time will the tank be filled?
Answer: Diameter of the pipe \( = 20 \text{ cm} \)
\( \Rightarrow \) Radius of the pipe \( (r) = \frac{20}{2} \text{ cm} = 10 \text{ cm} \)
Since, the water flows through the pipe at \( 3 \text{ km/hr} \).
∴ Length of water column per hour \( (h) = 3 \text{ km} = 3 \times 1000 \text{ m} = 3000 \times 100 \text{ cm} = 300000 \text{ cm} \).
∴ Volume of water per hour \( = \pi r^2 h = \pi \times 10^2 \times 300000 \text{ cm}^3 = \pi \times 30000000 \text{ cm}^3 \)
Now, for the cylindrical tank,
Diameter \( = 10 \text{ m} \Rightarrow \) Radius \( (R) = \frac{10}{2} \text{ m} = 5 \times 100 \text{ cm} = 500 \text{ cm} \)
Height \( (H) = 2 \text{ m} = 2 \times 100 \text{ cm} = 200 \text{ cm} \)
∴ Volume of the cylindrical tank \( = \pi R^2 H = \pi \times (500)^2 \times 200 \text{ cm}^3 \)
Now, time required to fill the tank
\( = \frac{\text{Volume of the tank}}{\text{Volume of water flown in 1 hour}} = \frac{\pi \times 500 \times 500 \times 200}{\pi \times 30000000} \text{ hrs} \)
\( = \frac{5 \times 5 \times 2}{30} \text{ hrs} = \frac{5}{3} \text{ hrs} = \frac{5}{3} \times 60 \text{ minutes} = 100 \text{ minutes} \).

Question. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per \( \text{cm}^3 \).
Answer: Since, diameter of the cylinder = 10 cm
\(\therefore\) Radius of the cylinder \((r) = \frac{10}{2} \text{ cm} = 5 \text{ cm}\)
\(\Rightarrow\) Length of wire in completely one round
\(2\pi r = 2 \times 3.14 \times 5 \text{ cm} = 31.4 \text{ cm}\)
\(\because\) Diameter of wire = 3 mm = \( \frac{3}{10} \text{ cm} \)
\(\therefore\) The thickness of cylinder covered in one round = \( \frac{3}{10} \text{ cm} \)
\(\Rightarrow\) Number of rounds (turns) of the wire to cover 12 cm
\( = \frac{12}{3/10} = 12 \times \frac{10}{3} = 40 \)
\(\therefore\) Length of wire required to cover the whole surface = Length of wire required to complete 40 rounds
\( = 40 \times 31.4 \text{ cm} = 1256 \text{ cm} \)
Now, radius of the wire = \( \frac{3}{2} \text{ mm} = \frac{3}{20} \text{ cm} \)
\(\therefore\) Volume of wire = \( \pi r^2 l \)
\( = \frac{22}{7} \times \frac{3}{20} \times \frac{3}{20} \times 1256 \text{ cm}^3 \)
\(\because\) Density of wire = 8.88 \( \text{gm/cm}^3 \)
\(\therefore\) Weight of the wire = [Volume of the wire] \(\times\) Density
\( = \left[ \frac{22}{7} \times \frac{3}{20} \times \frac{3}{20} \times 1256 \right] \times 8.88 \text{ gm} = \frac{22}{7} \times \frac{3}{20} \times \frac{3}{20} \times 1256 \times \frac{888}{100} \text{ gm} \)
\( = 788 \text{ g (approx.)} \)

Question. A cistern, internally measuring 150 cm \(\times\) 120 cm \(\times\) 110 cm, has 129600 \( \text{cm}^3 \) of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each being 22.5 cm \(\times\) 7.5 cm \(\times\) 6.5 cm?
Answer: \(\because\) Dimensions of the cistern are 150 cm, 120 cm and 110 cm.
\(\therefore\) Volume of the cistern = \( 150 \times 120 \times 110 \text{ cm}^3 = 1980000 \text{ cm}^3 \)
Volume of water contained in the cistern = 129600 \( \text{cm}^3 \)
\(\therefore\) Free space (volume) which is not filled with water
\( = 1980000 - 129600 \text{ cm}^3 = 1850400 \text{ cm}^3 \)
Now, Volume of one brick = \( 22.5 \times 7.5 \times 6.5 \text{ cm}^3 \)
\( = \frac{225}{10} \times \frac{75}{10} \times \frac{65}{10} \text{ cm}^3 = 1096.875 \text{ cm}^3 \)
\(\therefore\) Volume of water absorbed by one brick = \( \frac{1}{17} (1096.875) \text{ cm}^3 \)
Let ‘n’ bricks can be put in the cistern.
\(\therefore\) Volume of water absorbed by ‘n’ bricks = \( \frac{n}{17} (1096.875) \)
\(\therefore\) [Volume occupied by ‘n’ bricks] = [(free space in the cistern) + (volume of water absorbed by n-bricks)]
\( \Rightarrow [n \times (1096.875)] = [1850400 + \frac{n}{17} (1096.875)] \)
\( \Rightarrow 1096.875 n - \frac{n}{17} (1096.875) = 1850400 \)
\( \Rightarrow n \left( 1 - \frac{1}{17} \right) \times 1096.875 = 1850400 \)
\( \Rightarrow \frac{16}{17} n = \frac{1850400}{1096.875} \)
\( \Rightarrow n = \frac{1850400 \times 1000}{1096875} \times \frac{17}{16} = 1792.4102 \approx 1792 \)
Thus, 1792 bricks can be put in the cistern.

Question. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is \( 97280 \text{ km}^2 \), show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Answer: Volume of three rivers = \( 3 \{(Surface area of a river) \times Depth\} \)
\( = 3 \left\{ (1072 \text{ km} \times \frac{75}{1000} \text{ km}) \times \frac{3}{1000} \text{ km} \right\} = 3 \left\{ \frac{241200}{1000000} \text{ km}^3 \right\} \)
\( = \frac{723600}{1000000} \text{ km}^3 = 0.7236 \text{ km}^3 \)
Volume of rainfall = (surface area) \(\times\) (height of rainfall)
\( = 97280 \times \frac{10}{100 \times 1000} \text{ km}^3 \) [\(\because 10 \text{ cm} = \frac{10}{100 \times 1000} \text{ km}\)]
\( = \frac{9728}{1000} \text{ km}^3 = 9.728 \text{ km}^3 \)
Since, \( 0.7236 \text{ km}^3 \neq 9.728 \text{ km}^3 \)
\(\therefore\) The additional water in the three rivers is not equivalent to the rainfall.

VERY SHORT ANSWER TYPE QUESTIONS

Question. A cylinder, a cone and a hemisphere have the same values for r and h. Find the ratio of their volumes.
Answer: Let radius be ‘r’ and height ‘h’.
\(\therefore r = h\)
[Volume of a cylinder] : [Volume of a cone] : [Volume of a hemisphere]
\( \Rightarrow \pi r^2 h : \frac{1}{3} \pi r^2 h : \frac{2}{3} \pi r^3 \)
\( \Rightarrow h : \frac{1}{3} h : \frac{2}{3} r \) [Dividing by \(\pi r^2\)]
\( \Rightarrow h : \frac{1}{3} h : \frac{2}{3} h \) [\(\because r = h\)]
\( \Rightarrow 1 : \frac{1}{3} : \frac{2}{3} \) [Dividing by \(h\)]
\( \Rightarrow 3 : 1 : 2 \) [multiplying by 3]
\(\therefore\) The required ratio = 3 : 1 : 2

Question. If two identical solid cubes of side ‘x’ are joined end to end, then the total surface area of the resulting cuboid is \( 12x^2 \). Is it true?
Answer: \(\because\) The total surface area of a cube of side x is \( 6x^2 \)
When they are joined end to end, the length becomes 2x
\(\therefore\) Total surface area = \( 2[lh + bh + hl] = 2 [(2x \cdot x) + (x \cdot x) + (2x \cdot x)] \)
\( = 2 [2x^2 + x^2 + 2x^2] = 2 [5x^2] = 10x^2 \neq 12x^2 \)
\(\therefore\) False

Question. A spherical ball is melted to make eight new identical balls. Then the radius of each new ball is \( \frac{1}{8} \) th of the radius of the original ball. Is it true?
Answer: \(\because\) Radius ‘R’ of original ball \( \Rightarrow \) volume = \( \frac{4}{3} \pi R^3 \)
And Radius ‘r’ of the new ball \( \Rightarrow \) volume = \( \frac{4}{3} \pi r^3 \)
\(\therefore \frac{4}{3} \pi R^3 = 8 \left[ \frac{4}{3} \pi r^3 \right] \)
\( \Rightarrow R^3 = (2r)^3 \Rightarrow R = 2r \) or \( r = \frac{1}{2} R \)
\( \Rightarrow \) Radius of the new ball = Half the radius of original ball.
\(\therefore\) False

Question. If a solid cone of base radius ‘r’ and height ‘h’ is placed over a solid cylinder having same base radius ‘r’ and height - ‘h’ as that of the cone, then the curved surface area of the shape is \( \pi r (\sqrt{h^2 + r^2} + 2h) \). Is it true?
Answer: \(\because\) Curved surface area of a cone = \(\pi rl = \pi r \sqrt{h^2 + r^2} \) [\(\because l = \sqrt{h^2 + r^2}\)]
And curved surface area of the cylinder = \( 2\pi rh \)
\(\therefore\) The curved surface area of the combination = \( \pi r \sqrt{h^2 + r^2} + 2\pi rh \)
\(\therefore\) True.

SHORT ANSWER TYPE QUESTIONS

Question. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Answer: Here, radius (r) = 4.2 cm
\(\therefore\) Volume of the sphere = \( \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left( \frac{42}{10} \right)^3 \text{ cm}^3 \)
Let the height of the cylinder = h
\(\because\) Radius of the cylinder (R) = 6 cm
\(\therefore\) Volume of the cylinder = \( \pi R^2 h = \pi \times (6)^2 \times h \)
Since, [Volume of the cylinder] = [Volume of the sphere]
\(\therefore \pi \times 6^2 \times h = \frac{4}{3} \pi \times \frac{42}{10} \times \frac{42}{10} \times \frac{42}{10} \)
\( \Rightarrow h = \frac{4}{3} \times \frac{42 \times 42 \times 42}{1000} \times \frac{1}{36} = \frac{2744}{1000} \text{ cm} = 2.744 \text{ cm} \)

LONG ANSWER TYPE QUESTIONS

Question. A bucket is in the form of a frustum of a cone. Its depth is 15 cm and the diameters of the top and the bottom are 56 cm and 42 cm respectively. Find how much water can the bucket hold? [use \( \pi = \frac{22}{7} \)]
Answer: Here \( r_1 = \frac{56}{2} = 28 \text{ cm} \)
\( r_2 = \frac{42}{2} = 21 \text{ cm} \)
Height \( h = 15 \text{ cm} \)
Since the volume of a frustum of a cone = \( \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1r_2) \)
\(\therefore\) Volume of the bucket
\( = \frac{1}{3} \times \frac{22}{7} \times 15 [28^2 + 21^2 + 28 \times 21] \text{ cm}^3 \)
\( = \frac{110}{7} [784 + 441 + 588] \text{ cm}^3 = \frac{110}{7} \times 1813 \text{ cm}^3 \)
\( = 110 \times 259 \text{ cm}^3 = 28490 \text{ cm}^3 \)
\( = \frac{28490}{1000} \text{ litres} = 28.49 \text{ litres} \)

Question. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be \( \frac{1}{27} \text{th} \) of the volume of the given cone, at what height above the base is the section made?
Answer: In the figure, we have
\(\Delta ABC \sim \Delta ADE\) [By AA similarity]
\(\Rightarrow \frac{BC}{DE} = \frac{AB}{AD}\)
[\(\because\) corresponding sides of similar \(\Delta\)s are proportional.]
\(\Rightarrow \frac{r}{R} = \frac{h}{30}\) ...(1)
Here, volume of the small cone = \( \frac{1}{3} \pi r^2 h \)
Volume of the given cone = \( \frac{1}{3} \pi R^2 H \)
Since, [Volume of the small cone] = \( \frac{1}{27} \) [Volume of the given cone]
\(\Rightarrow \frac{1}{3} \pi r^2 h = \frac{1}{27} \left( \frac{1}{3} \pi R^2 H \right) \)
\(\Rightarrow \frac{r^2}{R^2} \times \frac{h}{H} = \frac{1}{27} \)
\(\Rightarrow \left( \frac{h}{30} \right)^2 \times \frac{h}{30} = \frac{1}{27} \)
\(\Rightarrow \left( \frac{h}{30} \right)^3 = \frac{1}{27} = \left( \frac{1}{3} \right)^3 \)
\(\Rightarrow \frac{h}{30} = \frac{1}{3} \Rightarrow h = 10 \text{ cm} \)
\(\therefore\) The required height \( BD = AD - AB = (30 - 10) \text{ cm} = 20 \text{ cm} \)

Question. A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and slant height of the conical portion is 53 m, find the area of the canvas needed to make the tent. [Use \( \pi = \frac{22}{7} \)].
Answer: For cylindrical part:
We have, radius (\(r\)) = 52.5 m
Height (\(h\)) = 3 m
Curved surface area = \( 2\pi rh \)
For the conical part
Slant height (\(l\)) = 53 m
Radius (\(r\)) = 52.5 m
\(\therefore\) Curved surface area = \(\pi rl\)
Area of the canvas = \( 2\pi rh + \pi rl \)
\( = \pi r (2h + l) \)
\( = \frac{22}{7} \times 52.5 \times (2 \times 3 + 53) \text{ m}^2 = \frac{22}{7} \times \frac{525}{10} \times 59 \text{ m}^2 \)
\( = \frac{22 \times 75 \times 59}{10} \text{ m}^2 = 11 \times 15 \times 59 \text{ m}^2 = 9735 \text{ m}^2 \)

Question. An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part in 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.
Answer: Here, height of the cylindrical part \( h = 240 \text{ cm} \)
Height of the conical part, \( H = 36 \text{ cm} \)
Radius \( r = 8 \text{ cm} \)
Now, the total volume of the pillar
= [volume of the cylindrical part] + [volume of the conical part]
\( = [\pi r^2 h] + \left[ \frac{1}{3} \pi r^2 H \right] = \pi r^2 \left( h + \frac{1}{3} H \right) \)
\( = \frac{22}{7} \times 8^2 \left( 240 + \frac{1}{3} \times 36 \right) \text{ cm}^3 = \frac{22 \times 64}{7} [240 + 12] \text{ cm}^3 \)
\( = \frac{22 \times 64}{7} \times 252 \text{ cm}^3 = 22 \times 64 \times 36 \text{ cm}^3 = 50688 \text{ cm}^3 \)
Weight of the pillar \( = 7.8 \times 50688 \text{ g} = \frac{78}{10} \times \frac{50688}{1000} \text{ kg} \)
\( = \frac{3953664}{10000} \text{ kg} = 395.3664 \text{ kg} \approx 395.36 \text{ kg} \)

Question. The diameter of a sphere is 28 cm. Find the cost of painting it all around at ₹ 0.10 per square cm.
Answer: ₹ 246.40

Question. The perimeter of one face of a wooden cube is 20 cm. Find its weight if 1 cm³ of wood weighs 8.25 g.
Answer: 1031.25 g

Question. The radii of two cylinders are in the ratio of \( 1 : \sqrt{3} \). If the volumes of two cylinders be same, find the ratio of their respective heights.
Answer: 3 : 1

Question. A solid composed of a cylinder with hemispherical ends. If the whole length of the solid is 108 cm and the diameter of the hemispherical end is 36 cm, find the cost of polishing its surface at the rate of 70 paise per square cm.
Answer: ₹ 8553.60

Question. A 5 m wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of ₹ 25 per metre [Use \( \pi = \frac{22}{7} \)] 
Answer: Hint: Slant height \( = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = 25 \text{ m} \)
\(\therefore\) Curved S.A \( = \pi rl = \frac{22}{7} \times 25 \times 7 \text{ m}^2 = 550 \text{ m}^2 \)
Area of the cloth required = length \(\times\) breadth
\( = l \times 5 = 550 \Rightarrow l = \frac{550}{5} = 110 \text{ m} \)
Cost of the cloth \( = Rs 25 \times 110 = Rs 2750 \text{ per metre} \).

Question. A girl empties a cylindrical bucket, full of sand, of base radius 18 cm and height 32 cm, on the floor to form a conical heap of sand. If the height of this conical heap is 24 cm, then find its slant height correct up to one place of decimal. 
Answer: Hint: Let radius and height of cylindrical bucket = \( R \) and \( H \) respectively
Let \( r \) and ‘\( h \)’ be radius and height of conical heap.
\(\therefore \pi R^2 H = \frac{1}{3} \pi r^2 h \Rightarrow r = \sqrt{\frac{3 \times H \times R^2}{h}} = \sqrt{\frac{3 \times 18 \times 18 \times 32}{24}} = 36 \)
\(\therefore l = \sqrt{r^2 + h^2} = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872} = 43.3 \text{ cm} \).

Question. The largest possible sphere is carved out of a wooden solid cube of side 7 cm. Find the volume of the wood left. [Use \( \pi = \frac{22}{7} \)] 
Answer: Hint: Volume of the wooden cube \( = 7 \times 7 \times 7 \text{ cm}^3 = 343 \text{ cm}^3 \)
Radius of the sphere \( = \frac{1}{2} \text{ side of the cube} = \frac{1}{2} (7 \text{ cm}) = 3.5 \text{ cm} \)
\(\therefore\) Volume of the sphere \( = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times \left( \frac{7}{2} \right)^3 \text{ cm}^3 = 179.66 \text{ cm}^3 \)
\(\Rightarrow\) Volume of the wood left \( = [ \text{volume of cube} ] - [ \text{volume of sphere} ] = (343 - 179.66) \text{ cm}^3 = 163.34 \text{ cm}^3 \).

Question. 150 spherical marbles, each of radius 1.4 cm, are dropped in a cylindrical vessel of radius 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel. 
Answer: Hint: Volume of 150 spherical marbles \( = 150 \times \text{[volume of a spherical marble]} \)
\( = 150 \times \frac{4}{3} \times \frac{22}{7} \times (1.4)^3 = 548.8 \text{ cm}^3 \)
\(\therefore\) Let the level of water rises \( h \) cm.
\(\therefore\) Volume of ‘\( h \)’ high water in the vessel \( = \pi r^2 h \text{ cm}^3 = \frac{22}{7} \times 7 \times 7 \times h = 154 h \text{ cm}^3 \)
\(\Rightarrow 154 h = 548.8 \Rightarrow h = \frac{5488}{10} \times \frac{1}{154} = 3.56 \text{ cm} \).