Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 20 Cylinder Cone And Sphere Surface Area Volume

Exercise 20A

Question 1. The height of a circular cylinder is 20 cm and the radius of its base is 7 cm. Find:
(i) the volume
(ii) the total surface area.
Answer:
Given:
Height = h = 20 cm
Radius of the base = r = 7 cm

(i) Volume of a cylinder = \( \pi r^2 h \)
\( = \frac{22}{7} \times 7 \times 7 \times 20 \text{ cm}^3 \)
\( = 3080 \text{ cm}^3 \)

(ii) Total surface area of a cylinder = \( 2\pi r(h + r) \)
\( = 2 \times \frac{22}{7} \times 7(20 + 7) \text{ cm}^2 \)
\( = 2 \times 22 \times 27 \text{ cm}^2 \)
\( = 1188 \text{ cm}^2 \)
In simple words: We use cylinder formulas to find how much space it takes and how much area we need to cover it. Volume tells us how much water it can hold. Surface area tells us how much paper we need to wrap it.

📝 Teacher's Note: Show students a tin can. Point to the curved side and the two flat circles on top and bottom. This helps them see where the 2πr(h+r) formula comes from.

🎯 Exam Tip: Always write the formulas first, then substitute values. Write units clearly - volume in cm³ and area in cm². You lose marks without units.

Question 2. The inner radius of a pipe is 2.1 cm. How much water can 12 m of this pipe hold?
Answer:
Inner radius of pipe = 2.1 cm
Length of the pipe = 12 m = 1200 cm

Volume = \( \pi r^2 h \)
\( = \frac{22}{7} \times 2.1 \times 2.1 \times 1200 \text{ cm}^3 \)
\( = 16632 \text{ cm}^3 \)
In simple words: We treat the pipe like a cylinder. The water fills up the space inside. So we find the volume using the inside radius and the length of the pipe.

📝 Teacher's Note: Tell students that a pipe is like a hollow cylinder. We only care about the inside space because that is where the water goes. Always convert meters to centimeters when radius is in cm.

🎯 Exam Tip: Remember to convert units - 12 m = 1200 cm. Use the inner radius, not the outer radius, because water goes inside the pipe.

Question 3. A cylinder of circumference 8 cm and length 21 cm rolls without sliding for \( 4\frac{1}{2} \) seconds at the rate of 9 complete rounds per second. Find
(i) distance travelled by the cylinder in \( 4\frac{1}{2} \) seconds, and
(ii) the area covered by the cylinder in \( 4\frac{1}{2} \) seconds
Answer:
Circumference of cylinder = 8 cm
Therefore, radius = \( \frac{C}{2\pi} = \frac{8 \times 7}{2 \times 22} = \frac{14}{11} \) cm
Length of the cylinder (h) = 21 cm

[Diagram: This diagram shows a cylinder rolling along a straight line.]

(i) If distance covered in one revolution is 8 cm, then distance covered in 9 revolutions = 9 × 8 = 72 cm or distance covered in 1 second = 72 cm.

Therefore, distance covered in \( 4\frac{1}{2} \) seconds = 72 × \( \frac{9}{2} \) cm = 324 cm

(ii) Curved surface area = \( 2\pi rh \)
\( = 2 \times \frac{22}{7} \times \frac{14}{11} \times 21 \)
\( = 168 \text{ cm}^2 \)

Area covered in one revolution = 168 cm²
Area covered in 9 revolutions = 168 cm² × 9 = 1512 cm²
Therefore, area covered in 1 second = 1512 cm²

Hence, area covered in \( 4\frac{1}{2} \) seconds = 1512 cm² × \( \frac{9}{2} \) = 6804 cm²
In simple words: When a cylinder rolls, the distance it moves equals how many times it turns × its circumference. The area it covers is the curved surface area × number of turns.

📝 Teacher's Note: Use a toilet paper roll to show rolling. Mark a point on it and show how far it travels in one complete turn. This equals the circumference.

🎯 Exam Tip: First find radius from circumference using C = 2πr. Then use distance = revolutions × circumference. Area covered = curved surface area × number of revolutions.

Question 4. How many cubic meters of earth must be dug out to make a well 28 m deep and 2.8 m in diameter? Also, find the cost of plastering its inner surface at Rs 4.50 per sq meter.
Answer:
Radius of the well = \( \frac{2.8}{2} = 1.4 \) m
Depth of the well = 28 m

Therefore, volume of earth dug out = \( \pi r^2 h \)
\( = \frac{22}{7} \times 1.4 \times 1.4 \times 28 \)
\( = \frac{17248}{100} \)
\( = 172.48 \text{ m}^3 \)

Area of curved surface = \( 2\pi rh \)
\( = 2 \times \frac{22}{7} \times 1.4 \times 28 \)
\( = 246.40 \text{ m}^2 \)

Cost of plastering at the rate of Rs 4.50 per sq m
= Rs 246.40 × 4.50
= Rs 1108.80
In simple words: The well is like a cylinder dug into the ground. We find how much earth to remove using volume formula. We plaster only the curved inside wall, not the bottom.

📝 Teacher's Note: Explain that we only plaster the curved surface, not the bottom of the well. The bottom stays as earth. Draw a simple well diagram to show this.

🎯 Exam Tip: Remember diameter ÷ 2 = radius. For cost problems, multiply area by rate per unit area. Always write the final cost with Rs symbol.

Question 5. What length of solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of external diameter 20 cm, 0.25 cm thick and 15 cm long?
Answer:
External diameter of hollow cylinder = 20 cm
Therefore, radius = 10 cm
Thickness = 0.25 cm
Hence, internal radius = (10 - 0.25) = 9.75 cm
Length of cylinder (h) = 15 cm

Volume = \( \pi h(R^2 - r^2) = \pi \times 15(10^2 - 9.75^2) \)
= 15π(100 - 95.0625)cm³
= 15π × 4.9375 cm³

Diameter = 2 cm
Therefore, radius (r) = 1 cm
Let h be the length
then, volume = \( \pi r^2 h = \pi(1 \times 1)h = \pi h \)

Now, according to given condition:
\( \pi h = 15\pi \times 4.9375 \)
\( h = 15 \times 4.9375 \)
\( h = 74.0625 \)

Length of cylinder = 74.0625 cm
In simple words: We melt the solid cylinder and use all that material to make the hollow cylinder. So both volumes must be equal. We find the hollow cylinder volume first, then find what length of solid cylinder gives the same volume.

📝 Teacher's Note: Use play dough to show recasting. Make a solid cylinder, then reshape it into a hollow one. The amount of material stays the same, only the shape changes.

🎯 Exam Tip: For recasting problems, always remember: volume of original = volume of new shape. For hollow cylinder, volume = π(R² - r²)h where R is outer radius, r is inner radius.

Question 6. A cylinder has a diameter of 20 cm. The area of curved surface is 100 sq. cm. Find:
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place.
Answer:
Diameter of the cylinder = 20 cm
Hence, Radius (r) = 10 cm
Height = h cm

(i) Curved surface area = \( 2\pi rh \)
\( 2\pi rh = 100 \text{ cm}^2 \)
\( 2 \times \frac{22}{7} \times 10 \times h = 100 \)
\( h = \frac{100 \times 7}{22 \times 10 \times 2} = \frac{35}{22} \)
\( h = 1.6 \) cm

(ii) Volume of the cylinder = \( \pi r^2 h \)
\( = \frac{22}{7} \times 10 \times 10 \times 1.6 \)
= 502.9 cm³

or
\( = \frac{22}{7} \times 10 \times 10 \times \frac{35}{22} \)
= 500 cm³
In simple words: We know the curved surface area and radius. We use the curved surface formula to find the height first. Then we use height and radius to find volume.

📝 Teacher's Note: Remind students that curved surface area does not include the top and bottom circles. It is only the side surface that curves around.

🎯 Exam Tip: When asked for "correct to one decimal place", round your final answer properly. Show all working steps clearly before rounding.

Question 7. A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 5 mm thick all round. Find the weight, in kilogram, of 2 metres of the pipe if 1 cm³ of the metal weights 7.7 g.
Answer:
Inner radius of the pipe = r = \( \frac{5}{2} = 2.5 \) cm
External radius of the pipe = R = Inner radius of the pipe + Thickness of the pipe
= 2.5 cm + 0.5 cm
= 3 cm

Length of the pipe = h = 2 m = 200 cm
Volume of the pipe = External Volume - Internal Volume

\( = \pi R^2 h - \pi r^2 h \)
\( = \pi(R^2 - r^2)h \)
\( = \pi(R - r)(R + r)h \)
\( = \frac{22}{7}(3 - 2.5)(3 + 2.5) \times 200 \)
\( = \frac{22}{7} \times 0.5 \times 5.5 \times 200 \)
\( = 1728.6 \text{ cm}^3 \)

Weight of 1 cm³ of metal = 7.7 g
Weight of 1728.6 cm³ of metal = 1728.6 × 7.7 g = 13310.22 g
Weight in kg = \( \frac{13310.22}{1000} = 13.31 \) kg
In simple words: The pipe is hollow, so we find the volume of metal by subtracting the hollow inside volume from the total outside volume. Then we multiply by density to get weight.

📝 Teacher's Note: Show a thick pipe or tube. Explain that thickness = 5 mm = 0.5 cm. The metal is only in the wall of the pipe, not in the hollow center.

🎯 Exam Tip: Convert all units to same system first: 5 mm = 0.5 cm, 2 m = 200 cm. For hollow cylinder, volume = π(R² - r²)h. Don't forget to convert grams to kilograms at the end.

Question 8. A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm × 14 cm × 10.5 cm. Find the rise in level of the water when the solid is submerged.
Answer:
Given:
Diameter of cylindrical container = 42 cm
Therefore, radius (r) = 21 cm
Dimensions of rectangular solid = 22 cm × 14 cm × 10.5 cm

Step 1: Find the volume of the rectangular solid.
Volume of solid = 22 × 14 × 10.5 = 3234 cm³ .....(i)

Step 2: Let height of water rise = h cm
Volume of water displaced = Volume of cylinder with height h
Volume of water displaced = \( \pi r^2 h = \frac{22}{7} \times 21 \times 21 \times h = 22 × 63h \) cm³ .....(ii)

Step 3: Apply the principle of displacement.
Volume displaced = Volume of solid
From (i) and (ii):
22 × 63h = 22 × 14 × 10.5

\( \implies h = \frac{22 \times 14 \times 10.5}{22 \times 63} \)

\( \implies h = \frac{14 \times 10.5}{63} = \frac{147}{63} = \frac{7}{3} \)

\( \implies h = 2\frac{1}{3} \) or 2.33 cm

Rise in water level = 2.33 cm
In simple words: When we put the iron block in water, it pushes water up. The volume of water that rises equals the volume of the iron block. We use this fact to find how much the water level goes up.

📝 Teacher's Note: Show students with a glass of water and a stone. When you drop the stone, water level rises. The volume of risen water equals volume of stone. This is displacement principle.

🎯 Exam Tip: Always write "Volume displaced = Volume of solid". This is the key formula. Show all steps clearly and don't forget units in final answer.

Question 9. A cylindrical container with internal radius of its base 10 cm, contains water up to a height of 7 cm. Find the area of wetted surface of the cylinder.
Answer:
Given:
Internal radius of the cylindrical container = 10 cm
Height of water = 7 cm

Step 1: Identify the wetted surface.
The wetted surface includes:
• Bottom circular base
• Curved side surface up to height 7 cm

Step 2: Calculate the wetted surface area.
Area of wetted surface = Area of base + Curved surface area
= \( \pi r^2 + 2\pi rh \)
= \( \pi r(r + 2h) \)
= \( \pi (r + 2h) \)
= \( \frac{22}{7} \times 10 \times (2 \times 7 + 10) \)
= \( \frac{22}{7} \times 10 \times (14 + 10) \)
= \( \frac{220}{7} \times 24 \)
= \( \frac{220 \times 24}{7} \)
= 754.29 cm²

Area of wetted surface = 754.29 cm²
In simple words: Wetted surface means all parts of the cylinder that touch water. This includes the bottom and the side walls up to water level. We add both areas together.

📝 Teacher's Note: Fill a glass with water and ask students which parts get wet. Only the bottom and sides up to water level get wet, not the full height of glass.

🎯 Exam Tip: Remember wetted surface = base area + curved surface area (only up to water height). Don't include the full cylinder height, only water height.

Question 10. Find the total surface area of an open pipe of length 50 cm, external diameter 20 cm and internal diameter 6 cm.
Answer:
Given:
Length of open pipe = 50 cm
External diameter = 20 cm ⇒ External radius (R) = 10 cm
Internal diameter = 6 cm ⇒ Internal radius (r) = 3 cm

Step 1: Calculate curved surface area of pipe (both outer and inner).
Surface area of pipe open from both sides =
= 2πRh + 2πrh
= 2πh(R + r)
= \( 2 \times \frac{22}{7} \times 50 \times (10 + 3) \)
= 4085.71 cm²

Step 2: Calculate area of upper and lower circular parts.
Area of upper and lower part =
= \( 2\pi(R^2 - r^2) \)
= \( 2 \times \frac{22}{7} \times (10^2 - 3^2) \)
= \( 2 \times \frac{22}{7} \times (100 - 9) \)
= \( 2 \times \frac{22}{7} \times 91 \)
= 572 cm²

Step 3: Find total surface area.
Total surface area = 4085.71 + 572 = 4657.71 cm²

Total surface area = 4657.71 cm²
In simple words: An open pipe has outer curved surface, inner curved surface, and two ring-shaped ends. We add all these areas to get total surface area.

📝 Teacher's Note: Show students a toilet paper roll. It has outer surface, inner surface, and two ends. An open pipe is similar but the ends are rings, not full circles.

🎯 Exam Tip: For open pipe, curved surface area = outer + inner = 2πRh + 2πrh. End area = 2π(R² - r²). Add both to get total. Don't forget the factor of 2.

Question 11. The height and radius of base of a cylinder are in the ratio 3:1. If its volume is 1029π cm³, find its total surface area.
Answer:
Given:
Ratio between height and radius of a cylinder = 3:1
Volume = 1029π cm³ .....(i)
Let radius of the base = r
Then height = 3r

Step 1: Find the radius using volume formula.
Volume = \( \pi r^2 h = \pi \times r^2 \times 3r = 3\pi r^3 \) .....(ii)

From (i) and (ii):
\( 3\pi r^3 = 1029\pi \)
\( r^3 = \frac{1029\pi}{3\pi} = \frac{1029}{3} = 343 \)
\( r = 7 \)

Therefore, radius = 7 cm and height = 3 × 7 = 21 cm

Step 2: Calculate total surface area.
Total surface area = \( 2\pi r(h + r) \)
= \( 2 \times \frac{22}{7} \times 7 \times (21 + 7) \)
= \( 2 \times \frac{22}{7} \times 7 \times 28 \)
= 1232 cm²

Total surface area = 1232 cm²
In simple words: We used the ratio to write height in terms of radius. Then we used the volume formula to find the actual radius. Finally we calculated total surface area using the formula.

📝 Teacher's Note: When ratio is given, always express both values in terms of one variable. This makes calculations easier. Students often forget this step.

🎯 Exam Tip: Write "Let radius = r, then height = 3r" clearly. Use volume formula to find r. Then substitute in surface area formula. Show each step with proper working.

Question 12. The radius of a solid right circular cylinder increases by 20% and its height decreases by 20%. Find the percentage change in its volume.
Answer:
Given:
Let the radius of original cylinder = 100 cm
And height of original cylinder = 100 cm

Step 1: Calculate original volume.
Original volume = \( \pi r^2 h \)
= π × (100)² × 100
= 1000000π cm³

Step 2: Find new dimensions after changes.
New radius = r' = 100 + 20% of 100 = 120 cm
New height = h' = 100 - 20% of 100 = 80 cm

Step 3: Calculate new volume.
New volume = \( \pi r'^2 h' \)
= π × (120)² × 80
= 1152000π cm³

Step 4: Find percentage change.
Increase in volume = New volume - Original volume
= 1152000π - 1000000π
= 152000π cm³

Percentage change = \( \frac{\text{Increase in volume}}{\text{Original volume}} \times 100\% \)
= \( \frac{152000π}{1000000π} \times 100\% \)
= 15.2%

Volume increases by 15.2%
In simple words: Even though height decreased, radius increased more (because radius is squared in volume formula). So overall volume increased by 15.2%.

📝 Teacher's Note: Use easy numbers like 100 for original values. This makes percentage calculations simple. Emphasize that radius has more effect because it is squared.

🎯 Exam Tip: Always assume convenient values like 100 for percentage problems. Calculate original and new values separately, then find percentage change. Don't forget the final answer with % sign.

Question 13. The radius of a solid right circular cylinder decreases by 20% and its height increases by 10%. Find the percentage change in its: (i) volume (ii) curved surface area
Answer:
Given:
Let radius of original cylinder = 100 cm
And height of original cylinder = 100 cm

Step 1: Calculate original values.
Original volume = \( \pi r^2 h = \pi \times (100)^2 \times 100 = 1000000\pi \) cm³

New radius = r' = 100 - 20% = 80 cm
New height = h' = 100 + 10% = 110 cm

Step 2: Calculate new volume.
New volume = \( \pi r'^2 h' = \pi \times (80)^2 \times 110 = 704000\pi \) cm³

Step 3: Find percentage change in volume.
Decrease in volume = Original volume - New volume
= 1000000π - 704000π = 296000π cm³

(i) Percentage change in volume = \( \frac{\text{Decrease in volume}}{\text{Original volume}} \times 100\% \)
= \( \frac{296000\pi}{1000000\pi} \times 100\% \) = 29.6%

Volume decreases by 29.6%

Step 4: Calculate curved surface areas.
(ii) Original curved surface area = 2πrh = 2π × 100 × 100 = 20000π cm²
New curved surface area = 2πr'h' = 2π × 80 × 110 = 17600π cm²

Decrease in curved surface area = 20000π - 17600π = 2400π cm²

Percentage change in curved surface area = \( \frac{2400\pi}{20000\pi} \times 100\% \) = 12%

Curved surface area decreases by 12%
In simple words: When radius decreases, both volume and curved surface area decrease. Volume decreases more because radius is squared in volume formula.

📝 Teacher's Note: Show students that radius has bigger effect on volume (r²) than on curved surface area (r). That's why volume change is bigger.

🎯 Exam Tip: Do volume and curved surface area separately. Use formulas V = πr²h and CSA = 2πrh. Show decrease/increase clearly and calculate percentage change properly.

Question 14. Find the minimum length in cm and correct to nearest whole number of the thin metal sheet required to make a hollow and closed cylindrical box of diameter 20 cm and
Answer:
Note: The question appears to be incomplete in the provided content as the height of the cylinder is not specified.

In simple words: To solve this problem, we need the height of the cylindrical box which is missing from the question.

📝 Teacher's Note: This question is incomplete. We need both diameter and height to find the area of metal sheet required for a closed cylindrical box.

🎯 Exam Tip: For complete cylindrical box, total surface area = 2πr(r + h). We need both radius and height to solve such problems.

Question 14. Find the area of metal sheet required, if 10% of it is wasted in cutting, overlapping, etc.
Answer:
Given:
Height of the cylindrical box = h = 35 cm
Base radius of the cylindrical box = r = 10 cm
Width of metal sheet = 1 m = 100 cm

Step 1: Find the area of metal sheet required.
Area of metal sheet required = Total surface area of the box
\( \Rightarrow \) Length × Width = \( 2\pi r(r + h) \)

Step 2: Calculate the length.
\( \Rightarrow \) Length × 100 = \( 2 \times \frac{22}{7} \times 10(10 + 35) \)
\( \Rightarrow \) Length × 100 = \( 2 \times \frac{22}{7} \times 10 \times 45 \)
\( \Rightarrow \) Length = \( \frac{2 \times 22 \times 10 \times 45}{100 \times 7} = 28.28 \text{ cm} = 28 \text{ cm} \)

Step 3: Find area and cost.
Area of metal sheet = Length × Width = 28 × 100 = 2800 cm² = 0.28 m²
Cost of the sheet at the rate of Rs.56 per m² = Rs. (56 × 0.28) = Rs. 15.68

Step 4: Find total sheet required with 10% wastage.
Let the total sheet required be x.
Then, x - 10% of x = 2800 cm²
\( \Rightarrow x - \frac{10}{100} \times x = 2800 \)
\( \Rightarrow 9x = 2800 \)
\( \Rightarrow x = 3111 \text{ cm}^2 \)

Total metal sheet required = 3111 cm²
In simple words: We first find how much metal we need for the box. Then we add 10% extra because some metal gets wasted when cutting and joining pieces.

📝 Teacher's Note: Show students that wastage means we need more material than the exact calculation. If 10% is wasted, then 90% is useful. So total needed = useful amount ÷ 0.9.

🎯 Exam Tip: Always set up the equation correctly: total - wastage = useful amount. Show all steps clearly to get full marks.

Question 15. 3080 cm³ of water is required to fill a cylindrical vessel completely and 2310 cm³ of water is required to fill it upto 5 cm below the top. Find :
(i) radius of the vessel.
(ii) height of the vessel.
(iii) wetted surface area of the vessel when it is half-filled with water.
Answer:
Given:
Volume when completely filled = 3080 cm³
Volume when 5 cm below top = 2310 cm³

Let r be the radius of the cylindrical vessel and h be its height.

Step 1: Set up equations.
Volume of cylindrical vessel = Volume of water filled in it
\( \Rightarrow \pi r^2 h = 3080 \)
\( \Rightarrow \frac{22}{7} \times r^2 \times h = 3080 \)
\( \Rightarrow r^2 \times h = 980 \) ....(i)

Step 2: Find radius using the 5 cm difference.
Volume of cylindrical vessel of height 5 cm = (3080 - 2310) cm³
\( \Rightarrow \pi r^2 \times 5 = 770 \)
\( \Rightarrow \frac{22}{7} \times r^2 \times 5 = 770 \)
\( \Rightarrow r^2 = 49 \)
\( \Rightarrow r = 7 \text{ cm} \)

Step 3: Find height.
Substituting r² = 49 in (i), we get
49 × h = 980
\( \Rightarrow h = 20 \text{ cm} \)

Step 4: Find wetted surface area when half-filled.
Wetted surface area of the vessel when it is half-filled with water
= 2πrh + πr²
= πr(2h + r)
= \( \frac{22}{7} \times 7(2 \times 10 + 7) \) [Half-filled ⇒ Height = \( \frac{20}{2} = 10 \text{ cm} \)]
= 22 × 27
= 594 cm²

(i) Radius = 7 cm
(ii) Height = 20 cm
(iii) Wetted surface area = 594 cm²
In simple words: We used the difference in water volumes to find the radius first. Then we found the height. Wetted area means the area that touches water.

📝 Teacher's Note: Explain that "5 cm below top" means the height of water is (h-5). The difference in volumes gives us a 5 cm high cylinder with the same radius.

🎯 Exam Tip: Always find radius first from the volume difference. Then use the full volume to get height. Show each step with clear working.

Question 16. Find the volume of the largest cylinder formed when a rectangular piece of paper 44 cm by 33 cm is rolled along it :
(i) shorter side.
(ii) longer side.
Answer:
Given:
Length of rectangular paper = 44 cm
Breadth of rectangular paper = 33 cm

(i) When the paper is rolled along its shorter side:
Height of cylinder = h = 44 cm
Circumference of cross-section = 2πr = 33 cm
\( \Rightarrow 2 \times \frac{22}{7} \times r = 33 \)
\( \Rightarrow r = \frac{33 \times 7}{2 \times 22} = 5.25 \text{ cm} \)

Volume of cylinder = \( \pi r^2 h = \frac{22}{7} \times 5.25 \times 5.25 \times 44 = 3811.5 \text{ cm}^3 \)

(ii) When the paper is rolled along its longer side:
Height of cylinder = h = 33 cm
Circumference of cross-section = 2πr = 44 cm
\( \Rightarrow 2 \times \frac{22}{7} \times r = 44 \)
\( \Rightarrow r = \frac{44 \times 7}{2 \times 22} = 7 \text{ cm} \)

Volume of cylinder = \( \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 33 = 5082 \text{ cm}^3 \)

(i) Volume when rolled along shorter side = 3811.5 cm³
(ii) Volume when rolled along longer side = 5082 cm³
In simple words: When we roll paper to make a cylinder, one side becomes the height and the other becomes the circumference (distance around the circle). Larger radius gives bigger volume.

📝 Teacher's Note: Use a sheet of paper to show students how rolling along different sides gives different cylinders. The side we roll around becomes the circumference.

🎯 Exam Tip: Remember: the side you roll along becomes the circumference (2πr). The other side becomes the height. Always calculate both cases separately.

Question 17. A metal cube of side 11 cm is completely submerged in water contained in a cylindrical vessel with diameter 28 cm. Find the rise in the level of water.
Answer:
Given:
Side of cube = 11 cm
Diameter of cylindrical vessel = 28 cm
Radius = 14 cm

Step 1: Apply displacement principle.
Volume of cube = Volume of water displaced in the cylinder
\( \Rightarrow (11)^3 = \pi r^2 h \)

Step 2: Calculate rise in water level.
\( \Rightarrow 11 \times 11 \times 11 = \frac{22}{7} \times \frac{28}{2} \times \frac{28}{2} \times h \)
\( \Rightarrow h = \frac{11 \times 11 \times 11 \times 7 \times 2 \times 2}{22 \times 28 \times 28} = \frac{121}{56} = 2.16 \text{ cm} \)

Rise in water level = 2.16 cm
In simple words: When we put the cube in water, it pushes away some water. The volume of pushed water equals the volume of the cube. This makes the water level go up.

📝 Teacher's Note: Show students with a glass of water and a stone. When the stone goes in, water level rises. The rise depends on the volume of the stone.

🎯 Exam Tip: Write the displacement formula: Volume of object = πr²h where h is the rise. Always convert diameter to radius first.

Question 18. A circular tank of diameter 2 m is dug and the earth removed is spread uniformly all around the tank to form an embankment 2 m in width and 1.6 m in height. Find the depth of the circular tank.
Answer:
Given:
Diameter of tank = 2 m, so radius = 1 m
Width of embankment = 2 m
Height of embankment = 1.6 m

Let the depth of the circular tank be h m.

Step 1: Find volume of tank.
Volume of the tank = πr²h = π × 1 × h = πh m³

Step 2: Find volume of embankment.
Volume of the embankment = Volume of hollow cylinder having height 1.6 m
= π(R² - r²)H
= π[(1 + 2)² - (1)²] × 1.6
= π(9 - 1) × 1.6
= 12.8π m³

Step 3: Apply conservation of earth.
Volume of tank = Volume of embankment
\( \Rightarrow \pi h = 12.8\pi \)
\( \Rightarrow h = 12.8 \text{ m} \)

Depth of circular tank = 12.8 m
In simple words: All the earth dug from the tank is used to make the embankment around it. So volume of hole = volume of embankment.

📝 Teacher's Note: Draw a circle for the tank and a ring around it for the embankment. Show that outer radius = tank radius + embankment width = 1 + 2 = 3 m.

🎯 Exam Tip: Remember: Volume of embankment = π(R² - r²)h where R is outer radius, r is inner radius. Always state that earth removed = earth in embankment.

Question 19. The sum of the inner and the outer curved surfaces of a hollow metallic cylinder is 1056 cm² and the volume of material in it is 1056 cm³. Find its internal and external radii. Given that the height of the cylinder is 21 cm.
Answer:
Given:
Sum of inner and outer curved surfaces = 1056 cm²
Volume of material = 1056 cm³
Height = 21 cm

Let R and r be the outer and inner radii of hollow metallic cylinder.
Let h be the height of the metallic cylinder.

Step 1: Set up first equation from surface areas.
Outer curved surface area + Inner curved surface area = 1056
\( \Rightarrow 2\pi Rh + 2\pi rh = 1056 \)
\( \Rightarrow 2\pi h(R + r) = 1056 \)
\( \Rightarrow 2 \times \frac{22}{7} \times 21(R + r) = 1056 \)
\( \Rightarrow R + r = \frac{1056 \times 7}{2 \times 22 \times 21} \)
\( \Rightarrow R + r = 8 \) ....(i)

Step 2: Set up second equation from volume of material.
Volume of material in it = 1056 cm³
\( \Rightarrow \pi R^2 h - \pi r^2 h = 1056 \)
\( \Rightarrow \pi h(R^2 - r^2) = 1056 \)
\( \Rightarrow \frac{22}{7} \times 21(R^2 - r^2) = 1056 \)
\( \Rightarrow R^2 - r^2 = \frac{1056 \times 7}{22 \times 21} \)
\( \Rightarrow (R + r)(R - r) = 16 \)
\( \Rightarrow 8 \times (R - r) = 16 \)
\( \Rightarrow R - r = 2 \) ....(ii)

Step 3: Solve the equations.
Adding (i) and (ii), we get
2R = 10 ⇒ R = 5 cm
\( \Rightarrow 5 - r = 2 \)
\( \Rightarrow r = 3 \text{ cm} \)

Internal radius = 3 cm and External radius = 5 cm
In simple words: We have a hollow cylinder (like a pipe). We used the total curved surface area and the volume of metal to find both radii.

📝 Teacher's Note: Draw a hollow cylinder cross-section showing inner radius r and outer radius R. Explain that volume of material = volume of big cylinder - volume of small cylinder.

🎯 Exam Tip: Set up two equations: one from surface areas (R + r) and one from volume of material (R² - r²). Use R² - r² = (R + r)(R - r) to solve easily.

Question 20. The difference between the outer curved surface area and the inner curved surface area of a hollow cylinder is 352 cm². If its height is 28 cm and the volume of material in it is 704 cm³; find its external curved surface area.
Answer:
Given:
Difference between outer and inner curved surface areas = 352 cm²
Height = 28 cm
Volume of material = 704 cm³

Let R and r be the outer and inner radii of hollow metallic cylinder.
Let h be the height of the metallic cylinder.

Step 1: Set up first equation from surface area difference.
Outer curved surface area - Inner curved surface area = 352
\( \Rightarrow 2\pi Rh - 2\pi rh = 352 \)
\( \Rightarrow 2\pi h(R - r) = 352 \)
\( \Rightarrow 2 \times \frac{22}{7} \times 28(R - r) = 352 \)
\( \Rightarrow R - r = \frac{352 \times 7}{2 \times 22 \times 28} \)
\( \Rightarrow R - r = 2 \) ....(i)

Step 2: Set up second equation from volume of material.
Volume of material = 704 cm³
\( \Rightarrow \pi R^2 h - \pi r^2 h = 704 \)
\( \Rightarrow \pi h(R^2 - r^2) = 704 \)
\( \Rightarrow \frac{22}{7} \times 28(R^2 - r^2) = 704 \)
\( \Rightarrow R^2 - r^2 = \frac{704 \times 7}{22 \times 28} = 8 \)
\( \Rightarrow (R + r)(R - r) = 8 \)
\( \Rightarrow (R + r) \times 2 = 8 \)
\( \Rightarrow R + r = 4 \) ....(ii)

Step 3: Solve for R and r.
From (i): R - r = 2
From (ii): R + r = 4
Adding: 2R = 6 ⇒ R = 3
Subtracting: 2r = 2 ⇒ r = 1

Step 4: Find external curved surface area.
External curved surface area = 2πRh
= \( 2 \times \frac{22}{7} \times 3 \times 28 \)
= \( 2 \times 22 \times 3 \times 4 \)
= 528 cm²

External curved surface area = 528 cm²
In simple words: We used the difference in curved surface areas and the volume of material to find the outer and inner radii. Then we calculated the outer curved surface area.

📝 Teacher's Note: Remind students that curved surface area of cylinder = 2πrh. For hollow cylinder, we consider both inner and outer surfaces. Volume of material = π(R² - r²)h.

🎯 Exam Tip: Always solve for both R and r first using two equations. Then substitute R in the formula for external curved surface area. Show all working clearly.

Question 21. The sum of the heights and the radius of a solid cylinder is 35 cm and its total surface area is 3080 cm², find the volume of the cylinder.
Answer:
Given:
r + h = 35 cm
Total surface area = 3080 cm²

Step 1: Write the formula for total surface area of a cylinder.
Total surface area = 2πr(h + r) = 3080

Step 2: Substitute r + h = 35.
2πr × 35 = 3080
2 × \( \frac{22}{7} \) × r × 35 = 3080
2 × 22 × r × 5 = 3080
220r = 3080
\( r = \frac{3080}{220} = 14 \) cm

Step 3: Find height.
h = 35 - r = 35 - 14 = 21 cm

Step 4: Calculate volume.
Volume = πr²h = \( \frac{22}{7} \) × 14 × 14 × 21 = 12936 cm³

Volume of cylinder = 12936 cm³
In simple words: We used the given information about sum of radius and height, plus the surface area to find the radius first. Then we found height and calculated volume using the volume formula.

📝 Teacher's Note: Show students that when we know r + h and the surface area formula, we can find one unknown first. Always substitute step by step.

🎯 Exam Tip: Write "Given" first, then show each step clearly. The formula for total surface area is 2πr(h + r). Don't forget units in the final answer.

Question 22. The total surface area of a solid cylinder is 616 cm². If the ratio between its curved surface area and total surface area is 1 : 2; find the volume of the cylinder.
Answer:
Given:
Total surface area = 616 cm²
Curved surface area : Total surface area = 1 : 2

Step 1: Find the relationship between h and r.
Let r and h be the radius and height of the cylinder.
Total surface area = 2πr(h + r) = 616 ... (i)
Curved surface area = 2πrh

Step 2: Use the given ratio.
\( \frac{\text{Curved surface area}}{\text{Total surface area}} = \frac{1}{2} \)
\( \frac{2πrh}{2πr(h + r)} = \frac{1}{2} \)
\( \frac{h}{h + r} = \frac{1}{2} \)
2h = h + r
h = r

Step 3: Substitute h = r in equation (i).
2πr(r + r) = 616
2 × \( \frac{22}{7} \) × 2r² = 616
\( r² = \frac{616 × 7}{2 × 22 × 2} = 49 \)
r = 7 = h

Step 4: Calculate volume.
Volume = πr²h = \( \frac{22}{7} \) × 7 × 7 × 7 = 1078 cm³

Volume of cylinder = 1078 cm³
In simple words: The ratio tells us that curved area is half of total area. This happens when height equals radius. We used this to solve the problem.

📝 Teacher's Note: Explain that when curved area is half of total area, it means the two circular bases together equal the curved surface. This only happens when h = r.

🎯 Exam Tip: Always use the ratio to find the relationship between h and r first. Then substitute in the surface area formula. Show the ratio calculation step clearly.

Question 23. A cylindrical vessel of height 24 cm and diameter 40 cm is full of water. Find the exact number of small cylindrical bottles, each of height 10 cm and diameter 8 cm, which can be filled with this water.
Answer:
Given:
Large vessel: Height = 24 cm, Diameter = 40 cm
Small bottles: Height = 10 cm, Diameter = 8 cm

Step 1: Find volume of large vessel.
Radius of large vessel = \( \frac{40}{2} = 20 \) cm
Volume of large vessel = πR²H = π × 20 × 20 × 24 = 9600π cm³

Step 2: Find volume of each small bottle.
Radius of small bottle = \( \frac{8}{2} = 4 \) cm
Volume of each small bottle = πr²h = π × 4 × 4 × 10 = 160π cm³

Step 3: Calculate number of bottles.
Number of bottles = \( \frac{\text{Volume of large vessel}}{\text{Volume of each small bottle}} \)
= \( \frac{9600π}{160π} = \frac{9600}{160} = 60 \)

Number of small bottles = 60
In simple words: We found how much water the big vessel holds, then how much each small bottle holds. Then we divided to see how many small bottles we can fill.

📝 Teacher's Note: Show students that π cancels out when we divide volumes. This makes calculation easier. Use real bottles to demonstrate the concept.

🎯 Exam Tip: Always convert diameter to radius first. The π will cancel when you divide volumes of similar shapes. Show the division step clearly.

Question 24. Two solid cylinders, one with diameter 60 cm and height 30 cm and the other with radius 30 cm and height 60 cm, are melted and recanted into a third solid cylinder of height 10 cm. Find the diameter of the cylinder formed.
Answer:
Given:
Cylinder 1: Diameter = 60 cm, Height = 30 cm
Cylinder 2: Radius = 30 cm, Height = 60 cm
New cylinder: Height = 10 cm

Step 1: Find volume of cylinder 1.
Radius of cylinder 1 = \( \frac{60}{2} = 30 \) cm
Volume V₁ = πr₁²h₁ = π × 30 × 30 × 30 = 27000π cm³

Step 2: Find volume of cylinder 2.
Volume V₂ = πr₂²h₂ = π × 30 × 30 × 60 = 54000π cm³

Step 3: Find total volume.
Let r be the radius of the third cylinder.
Height = h = 10 cm
Volume V = πr²h = πr² × 10

Step 4: Use conservation of volume.
V = V₁ + V₂
πr² × 10 = 27000π + 54000π
πr² × 10 = 81000π
r² = 8100
r = 90
Diameter = 2r = 180 cm

Diameter of new cylinder = 180 cm
In simple words: When we melt and reshape metal, the total amount stays the same. So we added the volumes of both old cylinders to find the volume of the new one.

📝 Teacher's Note: Demonstrate with clay or playdough. Show that when you reshape, volume stays same but shape changes. This is conservation of volume.

🎯 Exam Tip: Always write "conservation of volume" or "total volume remains same". Calculate each volume separately first, then add them. Don't forget to convert radius to diameter at the end.

Question 25. The total surface area of a hollow cylinder, which is open from both sides, is 3575 cm², area of the base ring is 357.5 cm² and height is 14 cm. Find the thickness of the cylinder.
Answer:
Given:
Total surface area = 3575 cm²
Area of base ring = 357.5 cm²
Height = 14 cm

Step 1: Set up variables.
Let external radius = R and internal radius = r
Let thickness = d = (R - r)

Step 2: Write formula for total surface area of hollow cylinder.
Total surface area = 2πRh + 2πrh + 2π(R² - r²)
= 2πh(R + r) + 2π(R + r)(R - r)
= 2π(R + r)[h + R - r]
= 2π(R + r)[h + d]
= 2π(R + r)(14 + d) = 3575 ... (i)

Step 3: Use area of base ring.
Area of base ring = π(R² - r²) = π(R + r)(R - r) = π(R + r)d = 357.5 ... (ii)

Step 4: Divide equation (i) by equation (ii).
\( \frac{2π(R + r)(14 + d)}{π(R + r)d} = \frac{3575}{357.5} \)
\( \frac{2(14 + d)}{d} = 10 \)
2(14 + d) = 10d
28 + 2d = 10d
8d = 28
d = 3.5 cm

Thickness of cylinder = 3.5 cm
In simple words: A hollow cylinder is like a pipe. The base ring is the area between outer and inner circles. We used both the total area and ring area to find the thickness.

📝 Teacher's Note: Use a real pipe or hollow tube to show students. Point out the difference between outer radius, inner radius, and thickness. Draw cross-section diagrams.

🎯 Exam Tip: For hollow cylinders, always define R (outer) and r (inner) clearly. The base ring area formula is π(R² - r²). Show the division step to eliminate π(R + r).

Question 26. The given figure shows a solid formed of a solid cube of side 40cm and a solid cylinder of radius 20 cm and height 50 cm attached to the cube as shown. Find the volume and the total surface area of the whole solid (Take π = 3.14)
Answer:
Given:
Cube: Side = 40 cm
Cylinder: Radius = 20 cm, Height = 50 cm
π = 3.14

Step 1: Find volume of cube.
Volume of cube = l³ = (40)³ = 64000 cm³

Step 2: Find volume of cylinder.
Volume of cylinder = πr²h = 3.14 × 20 × 20 × 50 = 62800 cm³

Step 3: Find total volume.
Total volume = Volume of cube + Volume of cylinder
= 64000 + 62800 = 126800 cm³

Step 4: Find total surface area.
Surface area of cube (5 faces, since one face is covered) = 5 × 40² = 8000 cm²
Curved surface area of cylinder = 2πrh = 2 × 3.14 × 20 × 50 = 6280 cm²
Area of top of cylinder = πr² = 3.14 × 20² = 1256 cm²
Area covered by cylinder on cube = πr² = 1256 cm² (subtract this)

Total surface area = 8000 + 6280 + 1256 - 1256 = 14280 cm²

Total volume = 126800 cm³
Total surface area = 14280 cm²

In simple words: We added volumes of both shapes. For surface area, we counted all outside surfaces but not the hidden parts where cylinder sits on cube.

📝 Teacher's Note: Use building blocks to show how shapes join. Explain that hidden surfaces (where cylinder touches cube) are not counted in total surface area.

🎯 Exam Tip: For combined solids, add volumes directly. For surface area, be careful about hidden surfaces - subtract the area where shapes touch each other. Always state what you're subtracting and why.

Question 27. Two right circular solid cylinders have radii in the ratio 3 : 5 and heights in the ratio 2 : 3, Find the ratio between their :
(i) curved surface areas.
(ii) volumes.

Solution:
Let the radii and height of two right circular cylinders be \( r_1, r_2 \) and \( h_1, h_2 \) respectively.
It is given that,
\( \frac{r_1}{r_2} = \frac{3}{5} \) and \( \frac{h_1}{h_2} = \frac{2}{3} \)

(i) \( \frac{\text{Curved surface area of cylinder 1}}{\text{Curved surface area of cylinder 2}} = \frac{2\pi r_1 h_1}{2\pi r_2 h_2} = \left(\frac{r_1}{r_2}\right) \times \left(\frac{h_1}{h_2}\right) = \frac{3}{5} \times \frac{2}{3} = \frac{2}{5} \)
∴ Ratio between their curved surface areas is 2 : 5.

(ii) \( \frac{\text{Volume of cylinder 1}}{\text{Volume of cylinder 2}} = \frac{\pi (r_1)^2 h_1}{\pi (r_2)^2 h_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right) = \left(\frac{3}{5}\right)^2 \times \frac{2}{3} = \frac{9}{25} \times \frac{2}{3} = \frac{18}{75} = \frac{6}{25} \)
∴ Ratio between their volumes is 6 : 25.

📝 Teacher's Note: Teach students that ratios multiply when we need surface area and ratios get squared for volume. Draw two cylinders on board with different sizes to show this clearly.

🎯 Exam Tip: Write the formulas first: curved surface area = 2πrh and volume = πr²h. Then substitute the given ratios step by step. Show each calculation clearly.

Question 28. A closed cylindrical tank, made of thin iron sheet, has diameter = 8.4 m and height 5.4 m. How much metal sheet, to the nearest m², is used in making this tank, if \( \frac{1}{15} \) of the sheet actually used was wasted in making the tank?

Solution:
Radius of the cylindrical tank = r = \( \frac{8.4}{2} = 4.2 \) m
Height of the cylindrical tank = h = 5.4 m
∴ Total surface area of the cylindrical tank = 2πr(h + r)
= \( 2 \times \frac{22}{7} \times 4.2(5.4 + 4.2) \)
= \( 2 \times 22 \times 0.6 \times 9.6 \)
= 253.44 m²

Area of sheet wasted in making the tank = \( \frac{1}{15} \times 253.44 = 16.896 \) m²
Hence, total sheet required = 253.44 + 16.896 = 270.34 m²

📝 Teacher's Note: Explain that "closed" means we need both curved surface and two circular ends. Show students that waste means extra sheet needed beyond the calculated area.

🎯 Exam Tip: For closed cylinder, use formula 2πr(h + r). Don't forget to add the wasted material. Round to nearest m² as asked in question.

Exercise 20 B

Question 1. Find the volume of a cone whose slant height is 17 cm and radius of base is 8 cm.

Solution:
Slant height (ℓ) = 17 cm
Radius (r) = 8 cm
But,
\( ℓ^2 = r^2 + h^2 \)
\( ⇒ h^2 = ℓ^2 - r^2 \)
\( ⇒ h^2 = 17^2 - 8^2 \)
\( ⇒ h^2 = 289 - 64 = 225 = (15)^2 \)
∴ h = 15

Now, volume of cone = \( \frac{1}{3} \pi r^2 h \)
= \( \frac{1}{3} \times \frac{22}{7} \times 8 \times 8 \times 15 \) cm³
= \( \frac{7040}{7} \) cm³
= 1005.71 cm³

📝 Teacher's Note: Draw a right triangle to show how slant height, radius and height are related. This helps students remember the Pythagoras theorem formula.

🎯 Exam Tip: Always find height first using \( h^2 = ℓ^2 - r^2 \). Then use volume formula \( \frac{1}{3}\pi r^2 h \). Show all working steps.

Question 2. The curved surface area of a cone is 12320 cm². If the radius of its base is 56 cm, find its height.

Solution:
Curved surface area = 12320 cm².
Radius of base (r) = 56 cm
Let slant height = ℓ
∴ πrℓ = 12320
\( ⇒ \frac{22}{7} \times 56 \times ℓ = 12320 \)
\( ⇒ ℓ = \frac{12320 \times 7}{56 \times 22} \)
\( ⇒ ℓ = 70 \text{ cm} \)

Height of the cone = \( \sqrt{ℓ^2 - r^2} \)
= \( \sqrt{(70)^2 - (56)^2} \)
= \( \sqrt{4900 - 3136} \)
= \( \sqrt{1764} \)
= 42 cm

📝 Teacher's Note: Remind students that curved surface area uses slant height, not actual height. First find slant height, then use Pythagoras to find actual height.

🎯 Exam Tip: Write curved surface area formula: πrℓ. Find slant height first, then use \( h = \sqrt{ℓ^2 - r^2} \) to get height.

Question 3. The circumference of the base of a 12 m high conical tent is 66 m. Find the volume of the air contained in it.

Solution:
Circumference of the conical tent = 66 m
and height (h) = 12 m
∴ Radius = \( \frac{C}{2\pi} = \frac{66 \times 7}{2 \times 22} = 10.5 \text{ m} \)

Therefore, volume of air contained in it = \( \frac{1}{3} \pi r^2 h \)
= \( \frac{1}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times 12 \text{ m}^3 \)
= 1386 m³

📝 Teacher's Note: Show students that circumference = 2πr, so radius = circumference ÷ 2π. This is a common step students forget in tent problems.

🎯 Exam Tip: First find radius from circumference using r = C/2π. Then use volume formula. Volume of air = volume of cone.

Question 4. The radius and height of a right circular cone are in the ratio 5:12 and its volume is 2512 cubic cm. Find the radius and slant height of the cone. (Take π = 3.14)

Solution:
The ratio between radius and height = 5:12
Volume = 2512 cubic cm
Let radius (r) = 5x, height (h) = 12x and slant height = ℓ

[Diagram: This diagram shows a cone with height h, radius r, and slant height ℓ marked.]

∴ \( ℓ^2 = r^2 + h^2 \)
\( ⇒ ℓ^2 = (5x)^2 + (12x)^2 \)
\( ⇒ ℓ^2 = 25x^2 + 144x^2 \)
\( ⇒ ℓ^2 = 169x^2 \)
\( ⇒ ℓ = 13x \)

Now Volume = \( \frac{1}{3} \pi r^2 h \)
\( ⇒ \frac{1}{3} \pi r^2 h = 2512 \)
\( ⇒ \frac{1}{3}(3.14)(5x)^2(12x) = 2512 \)
\( ⇒ \frac{1}{3}(3.14)(300x^3) = 2512 \)
∴ \( x^3 = \frac{2512 \times 3}{3.14 \times 300} = \frac{2512 \times 3 \times 100}{314 \times 300} = 8 \)
\( ⇒ x = 2 \)
∴ Radius = 5x = 5 × 2 = 10 cm
Height = 12x = 12 × 2 = 24 cm
Slant height = 13x = 13 × 2 = 26 cm

📝 Teacher's Note: Use x as the common factor when ratios are given. This makes calculations easier. Show the right triangle inside the cone to explain slant height.

🎯 Exam Tip: Let radius = 5x and height = 12x from the ratio. Find slant height using Pythagoras. Substitute in volume formula to find x. Then calculate actual values.

Question 5. Two right circular cones x and y are made, x having three times the radius of y and y having half the volume of x. Calculate the ratio between the heights of x and y.

Solution:
Let radius of cone y = r
Therefore, radius of cone x = 3r
Let volume of cone y = V
then volume of cone x = 2V
Let h₁ be the height of x and h2 be the height of y.

[Diagram: This diagram shows a cone with height h and radius r marked.]

Therefore, Volume of cone = \( \frac{1}{3} \pi r^2 h \)

Volume of cone x = \( \frac{1}{3} \pi (3r)^2 h_1 = \frac{1}{3} \pi 9r^2 h_1 = 3\pi r^2 h_1 \)

Volume of cone y = \( \frac{1}{3} \pi r^2 h_2 \)

∴ \( \frac{2V}{V} = \frac{3\pi r^2 h_1}{\frac{1}{3} \pi r^2 h_2} \)

\( ⇒ \frac{2}{1} = \frac{3h_1 \times 3}{h_2} = \frac{9h_1}{h_2} \)

\( ⇒ \frac{h_1}{h_2} = \frac{2}{1} \times \frac{1}{9} = \frac{2}{9} \)

∴ h₁ : h₂ = 2 : 9

📝 Teacher's Note: Draw two cones of different sizes. Explain that when radius increases 3 times, area increases 9 times (3²). This helps students see why the calculation works this way.

🎯 Exam Tip: Set up the volume ratio equation carefully. Remember that volume depends on r² so when radius triples, the r² part becomes 9 times bigger.

Question 6. The diameters of two cones are equal. If their slant heights are in the ratio 5:4, find the ratio of their curved surface areas.

Solution:
Since diameters are equal, radii are also equal.
Let radius of both cones = r
Let slant heights be ℓ₁ and ℓ₂
Given: ℓ₁ : ℓ₂ = 5 : 4

Curved surface area of cone 1 = πrℓ₁
Curved surface area of cone 2 = πrℓ₂

Ratio = \( \frac{\pi r ℓ_1}{\pi r ℓ_2} = \frac{ℓ_1}{ℓ_2} = \frac{5}{4} \)

Therefore, ratio of curved surface areas = 5 : 4

📝 Teacher's Note: When radii are equal, the curved surface area ratio equals the slant height ratio. This is because curved surface area = πrℓ, and πr cancels out.

🎯 Exam Tip: If radii are equal, curved surface area ratio = slant height ratio. Write this clearly and show the cancellation of πr terms.

Question 7. There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii.
Answer:
Given:
Let slant height of first cone = \( l \)
Then slant height of second cone = \( 2l \)
Let radius of first cone = \( r_1 \)
Let radius of second cone = \( r_2 \)

Step 1: Write curved surface areas.
Curved surface area of first cone = \( \pi r_1 l \)
Curved surface area of second cone = \( \pi r_2 (2l) = 2\pi r_2 l \)

Step 2: Use the given condition.
According to the problem, curved surface area of first cone is twice that of second cone.
\( \pi r_1 l = 2(2\pi r_2 l) \)
\( \pi r_1 l = 4\pi r_2 l \)
\( r_1 = 4r_2 \)

Step 3: Find the ratio.
\( \frac{r_1}{r_2} = \frac{4}{1} \)

Therefore, \( r_1 : r_2 = 4 : 1 \)
In simple words: We used the curved surface area formula for cones. The first cone has 4 times bigger radius than the second cone.

📝 Teacher's Note: Draw two cones on the board. Show students that curved surface area = πrl. When one area is twice the other, we can set up an equation to find the ratio.

🎯 Exam Tip: Always write the curved surface area formula first. Set up the equation using the given condition. Show all steps clearly to get full marks.

Question 8. A heap of wheat is in the form of a cone of diameter 16.8 m and height 3.5 m. Find its volume. How much cloth is required to just cover the heap?
Answer:
Given:
Diameter = 16.8 m
Therefore, radius (r) = 8.4 m
Height (h) = 3.5 m

Step 1: Find volume of the heap.
Volume = \( \frac{1}{3}\pi r^2 h \)
= \( \frac{1}{3} \times \frac{22}{7} \times 8.4 \times 8.4 \times 3.5 \)
= 258.72 m³

Step 2: Find slant height.
Slant height \( l = \sqrt{r^2 + h^2} \)
= \( \sqrt{(8.4)^2 + (3.5)^2} \)
= \( \sqrt{70.56 + 12.25} \)
= \( \sqrt{82.81} \)
= 9.1 m

Step 3: Find cloth required (curved surface area).
Curved surface area = \( \pi r l \)
= \( \frac{22}{7} \times 8.4 \times 9.1 \)
= 240.24 m²

Volume = 258.72 m³, Cloth required = 240.24 m²
In simple words: We found how much space the wheat heap takes up (volume) and how much cloth we need to cover its curved surface.

📝 Teacher's Note: Explain that cloth covers only the curved surface, not the base (since it's on the ground). Use Pythagoras theorem to find slant height.

🎯 Exam Tip: Remember the volume formula has 1/3. For cloth, use curved surface area formula πrl, not total surface area. Show slant height calculation clearly.

Question 9. Find what length of canvas, 1.5 m in width, is required to make a conical tent 48 m in diameter and 7 m in height. Given that 10% of the canvas is used in folds and stitching. Also, find the cost of the canvas at the rate of Rs. 24 per meter.
Answer:
Given:
Diameter of tent = 48 m
Therefore, radius (r) = 24 m
Height (h) = 7 m
Canvas width = 1.5 m

Step 1: Find slant height.
Slant height \( l = \sqrt{r^2 + h^2} \)
= \( \sqrt{(24)^2 + (7)^2} \)
= \( \sqrt{576 + 49} \)
= \( \sqrt{625} \)
= 25 m

Step 2: Find curved surface area of tent.
Curved surface area = \( \pi r l \)
= \( \frac{22}{7} \times 24 \times 25 \)
= \( \frac{13200}{7} \) m²

Step 3: Find extra canvas for folds and stitching.
Extra canvas = \( \frac{13200}{7} \times \frac{10}{100} = \frac{1320}{7} \) m²

Step 4: Find total canvas area required.
Total area = \( \frac{13200}{7} + \frac{1320}{7} = \frac{14520}{7} \) m²

Step 5: Find length of canvas.
Length = \( \frac{\text{Total area}}{\text{Width}} = \frac{14520/7}{1.5} = \frac{14520}{7} \times \frac{2}{3} = \frac{9680}{7} \) = 1382.86 m

Step 6: Find cost.
Total cost = \( \frac{9680}{7} \times 24 \) = Rs. 33,188.64

Length required = 1382.86 m, Cost = Rs. 33,188.64
In simple words: We found the tent's surface area, added 10% extra for sewing, then divided by canvas width to get length needed.

📝 Teacher's Note: Explain that canvas comes in rolls of fixed width. We need to find how much length to buy. Don't forget to add the 10% extra.

🎯 Exam Tip: Always add the extra percentage for folds. Divide total area by width to get length. Show the cost calculation step by step.

Question 10. A solid cone of height 8 cm and base radius 6 cm is melted and re-casted into identical cones, each of height 2 cm and diameter 1 cm. Find the number of cones formed.
Answer:
Given:
Large cone: Height = 8 cm, Radius = 6 cm
Small cones: Height = 2 cm, Diameter = 1 cm, so Radius = 0.5 cm

Step 1: Find volume of large cone.
Volume = \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times 6 \times 6 \times 8 = 96\pi \) cm³

Step 2: Find volume of one small cone.
Volume = \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times \frac{1}{2} \times \frac{1}{2} \times 2 = \frac{1}{6}\pi \) cm³

Step 3: Find number of small cones.
Number of cones = \( \frac{\text{Volume of large cone}}{\text{Volume of one small cone}} = \frac{96\pi}{\frac{1}{6}\pi} = 96\pi \times \frac{6}{\pi} = 576 \)

Number of cones formed = 576
In simple words: We found how much material is in the big cone, then divided by the material in each small cone to get the count.

📝 Teacher's Note: Emphasize that when material is melted and re-cast, volume stays the same. Show students that diameter 1 cm means radius 0.5 cm.

🎯 Exam Tip: Volume of material remains constant when melted and re-cast. Calculate both volumes using the same formula, then divide. Don't forget to convert diameter to radius.

Question 11. The total surface area of a right circular cone of slant height 13 cm is 90π cm². Calculate: (i) its radius in cm (ii) its volume in cm³. Take π = 3.14
Answer:
Given:
Total surface area = 90π cm²
Slant height (l) = 13 cm

(i) Finding radius:
Total surface area = \( \pi r l + \pi r^2 = \pi r(l + r) \)
∴ \( \pi r(l + r) = 90\pi \)
⇒ \( r(13 + r) = 90 \)
⇒ \( r^2 + 13r - 90 = 0 \)
⇒ \( r^2 + 13r - 5r - 90 = 0 \)
⇒ \( r(r + 18) - 5(r + 18) = 0 \)
⇒ \( (r + 18)(r - 5) = 0 \)

Either r + 18 = 0, then r = -18 which is not possible
or r - 5 = 0, then r = 5

Therefore, radius = 5 cm

(ii) Finding volume:
First find height using \( l^2 = r^2 + h^2 \)
\( 13^2 = 5^2 + h^2 \)
\( 169 = 25 + h^2 \)
\( h^2 = 144 \)
\( h = 12 \) cm

Volume = \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 5^2 \times 12 = \frac{1}{3} \times 3.14 \times 25 \times 12 = 314 \) cm³

Radius = 5 cm, Volume = 314 cm³
In simple words: We used the total surface area formula to make an equation and solve for radius. Then we found height using Pythagoras theorem and calculated volume.

📝 Teacher's Note: Show students how to solve the quadratic equation by factoring. Explain that negative radius is impossible, so we take the positive value.

🎯 Exam Tip: Total surface area = πrl + πr². Solve the quadratic equation carefully. Use Pythagoras theorem to find height from slant height and radius. Show all steps clearly.

Question 12. The area of the base of a conical solid is 38.5 cm² and its volume is 154 cm³. Find the curved surface area of the solid.
Answer:
Given:
Area of the base, \( \pi r^2 = 38.5 \text{ cm}^2 \)
Volume of the solid, \( V = 154 \text{ cm}^3 \)
Curved surface area of the solid = \( \pi r^2 h \)

Step 1: Find the height using volume formula.
Volume, \( V = \frac{1}{3} \pi r^2 h \)
\( 154 = \frac{1}{3} \pi r^2 h \)
\( h = \frac{154 \times 3}{\pi r^2} \)
\( h = \frac{154 \times 3}{38.5} = 12 \text{ cm} \)

Step 2: Find the radius from base area.
Area = 38.5
\( \pi r^2 = 38.5 \)
\( r^2 = \frac{38.5}{3.14} \)
\( r = \sqrt{\frac{38.5}{3.14}} = 3.5 \text{ cm} \)

Step 3: Find curved surface area using formula.
Curved surface area of solid = \( \pi r l \)
= \( \pi r \sqrt{r^2 + h^2} \)
= \( \pi \times 3.5 \times \sqrt{3.5^2 + 12^2} \)
= \( \pi \times 3.5 \times 12.5 \)
= \( 137.44 \text{ cm}^2 \)

In simple words: We found the height first using volume formula. Then we found radius from base area. Finally we used the slant height formula to get curved surface area.

📝 Teacher's Note: Show students that slant height is like the side of a triangle. Use Pythagoras theorem to find it. Draw a cone and show the right triangle inside it.

🎯 Exam Tip: Always find slant height using \( l = \sqrt{r^2 + h^2} \). Write this formula clearly. The curved surface area formula is \( \pi r l \), not \( \pi r^2 l \).

Question 13. A vessel, in the form of an inverted cone, is filled with water to the brim. Its height is 32 cm and diameter of the base is 25.2 cm. Six equal solid cones are dropped in it, so that they are fully submerged. As a result, one-fourth of water in the original cone overflows. What is the volume of each of the solid cones submerged?
Answer:
[Diagram: An inverted cone with diameter 25.2 cm at the top and height 32 cm, showing the vessel shape.]

Volume of vessel = volume of water = \( \frac{1}{3} \pi r^2 h \)

diameter = 25.2 cm, therefore radius = 12.6 cm
height = 32 cm

Volume of water in the vessel = \( \frac{1}{3} \pi r^2 h \)

= \( \frac{1}{3} \times \frac{22}{7} \times 12.6 \times 12.6 \times 32 \)
= \( 5322.24 \text{ cm}^3 \)

On submerging six equal solid cones into it, one-fourth of the water overflows.

Therefore, volume of the equal solid cones submerged
= Volume of water that overflows

= \( \frac{1}{4} \times 5322.24 \)
= \( 1330.56 \text{ cm}^3 \)

Now, volume of each cone submerged
= \( \frac{1330.56}{6} = 221.76 \text{ cm}^3 \)

In simple words: When we put cones in water, they push out water equal to their own volume. One-fourth of original water overflows, so total volume of 6 cones = one-fourth of water volume.

📝 Teacher's Note: Use a bucket of water and some stones to show displacement. When stones go in, water level rises and may overflow. The overflow volume equals stone volume.

🎯 Exam Tip: Remember: volume of objects dropped = volume of water that overflows. Then divide by number of objects to get volume of each one.

Question 14. The volume of a conical tent is 1232 m³ and the area of the base floor is 154 m². Calculate the:
(i) radius of the floor
(ii) height of the tent
(iii) length of the canvas required to cover this conical tent if its width is 2 m.
Answer:

(i) Let r be the radius of the base of the conical tent, then area of the base floor = \( \pi r^2 \text{ m}^2 \)

\( \pi r^2 = 154 \)
\( \frac{22}{7} \times r^2 = 154 \)
\( r^2 = \frac{154 \times 7}{22} = 49 \)
\( r = 7 \)

Hence, radius of the base of the conical tent i.e. the floor = 7 m

(ii) Let h be the height of the conical tent, then the volume = \( \frac{1}{3} \pi r^2 h \text{ m}^3 \)

\( \frac{1}{3} \pi r^2 h = 1232 \)

\( \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times h = 1232 \)

\( h = \frac{1232 \times 3}{22 \times 7} = 24 \)

Hence, the height of the tent = 24 m

(iii) Let l be the slant height of the conical tent, then \( l = \sqrt{h^2 + r^2} \text{ m} \)

\( l = \sqrt{r^2 + r^2} = \sqrt{(24)^2 + (7)^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ m} \)

The area of the canvas required to make the tent = \( \pi r l \text{ m}^2 \)

\( \pi r l = \frac{22}{7} \times 7 \times 25 \text{ m}^2 = 550 \text{ m}^2 \)

Length of the canvas required to cover the conical tent of its width 2 m = \( \frac{550}{2} = 275 \text{ m} \)

In simple words: First we found radius from base area. Then height from volume. Then slant height using Pythagoras. Finally, we found how long the cloth should be if it is 2 m wide.

📝 Teacher's Note: Show students that a cone opens into a sector of a circle. The curved surface area is like the area of that sector. Use a paper cone to demonstrate this.

🎯 Exam Tip: Do parts in order: (i) radius from base area, (ii) height from volume, (iii) slant height, then canvas area. Show all formulas clearly.

Exercise 20 C

Question 1. The surface area of a sphere is 2464 cm², find its volume.
Answer:

Surface area of the sphere = 2464 cm²
Let radius = r, then

\( 4\pi r^2 = 2464 \)
\( 4 \times \frac{22}{7} \times r^2 = 2464 \)
\( r^2 = \frac{2464 \times 7}{4 \times 22} = 196 \)
\( r = 14 \text{ cm} \)

Volume = \( \frac{4}{3} \pi r^3 \)

\( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 = 11498.67 \text{ cm}^3 \)

In simple words: We found radius first using surface area formula. Then we used volume formula with that radius to get the answer.

📝 Teacher's Note: Remind students that sphere surface area is \( 4\pi r^2 \) and volume is \( \frac{4}{3}\pi r^3 \). Make them remember these two formulas well.

🎯 Exam Tip: Always find radius first. Then use it in volume formula. Write both formulas clearly at the start.

Question 2. The volume of a sphere is 38808 cm³; find its diameter and the surface area.
Answer:

Volume of the sphere = 38808 cm³
Let radius of sphere = r

\( \frac{4}{3} \pi r^3 = 38808 \)

\( \frac{4}{3} \times \frac{22}{7} \times r^3 = 38808 \)

\( r^3 = \frac{38808 \times 7 \times 3}{4 \times 22} = 9261 \)

\( r = 21 \text{ cm} \)

Diameter = 2r = 21 × 2 cm = 42 cm

Surface area = \( 4\pi r^2 = 4 \times \frac{22}{7} \times 21 \times 21 \text{ cm}^2 = 5544 \text{ cm}^2 \)

In simple words: We found radius from volume using cube root. Then diameter is double the radius. Surface area uses the radius we found.

📝 Teacher's Note: Students often forget that diameter = 2 × radius. Make them write this step clearly. Also help them with cube root calculation.

🎯 Exam Tip: When finding cube root, try common values like 20³, 21³ etc. Write diameter = 2r separately, don't skip this step.

Question 3. A spherical ball of lead has been melted and made into identical smaller balls with radius equal to half the radius of the original one. How many such balls can be made?
Answer:

Let the radius of spherical ball = r

Volume = \( \frac{4}{3} \pi r^3 \)

Radius of smaller ball = \( \frac{r}{2} \)

Volume of smaller ball = \( \frac{4}{3} \pi \left(\frac{r}{2}\right)^3 = \frac{4}{3} \pi \frac{r^3}{8} = \frac{\pi r^3}{6} \)

Therefore, number of smaller balls made out of the given ball =

\( \frac{\frac{4}{3} \pi r^3}{\frac{\pi r^3}{6}} = \frac{4}{3} \times 6 = 8 \)

In simple words: When radius becomes half, volume becomes 1/8 of original. So we can make 8 smaller balls from one big ball.

📝 Teacher's Note: Show students that when radius is halved, volume becomes (1/2)³ = 1/8 times. This is why we get 8 balls.

🎯 Exam Tip: Remember: radius half means volume becomes 1/8. So number of balls = 8. Always check your answer makes sense.

Question 4. How many balls each of radius 1 cm can be made by melting a bigger ball whose diameter is 8 cm.
Answer:

Diameter of bigger ball = 8 cm
Therefore, Radius of bigger ball = 4 cm

Volume = \( \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times 4 \times 4 \times 4 = \frac{256\pi}{3} \text{ cm}^3 \)

Radius of small ball = 1 cm

Volume = \( \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times 1 \times 1 \times 1 = \frac{4\pi}{3} \text{ cm}^3 \)

Number of balls = \( \frac{\frac{256\pi}{3}}{\frac{4\pi}{3}} = \frac{256\pi}{3} \times \frac{3}{4\pi} = 64 \)

In simple words: We found volume of big ball and small ball. Then divided big volume by small volume to get how many small balls we can make.

📝 Teacher's Note: Make students remember that diameter = 2 × radius. Also show them that when we divide volumes, π cancels out, making calculation easier.

🎯 Exam Tip: Always convert diameter to radius first. Number of balls = volume of big ball ÷ volume of small ball. The π will cancel out.

Question 5. 8 metallic spheres each of radius 2 mm, are melted and cast into a single sphere. Calculate the radius of the new sphere.
Solution:
Given:
Number of small spheres = 8
Radius of each small sphere = 2 mm = \( \frac{1}{5} \) cm

Step 1: Find volume of one small sphere.
\[ \text{Volume} = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = \frac{88}{21 \times 125} \text{ cm}^3 \]

Step 2: Find total volume of 8 spheres.
\[ \text{Volume of 8 spheres} = \frac{88 \times 8}{21 \times 125} = \frac{704}{21 \times 125} \text{ cm}^3 \]

Step 3: Let radius of new sphere be R.
\[ \text{Volume} = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} R^3 = \frac{88}{21} R^3 \]

Step 4: Equate volumes.
\[ \frac{88}{21} R^3 = \frac{704}{21 \times 125} \]
\[ R^3 = \frac{704}{21 \times 125} \times \frac{21}{88} = \frac{8}{125} \]
\[ R = \frac{2}{5} = 0.4 \text{ cm} = 4 \text{ mm} \]
In simple words: When we melt many small spheres and make one big sphere, the total volume stays the same. We use the volume formula to find how big the new sphere is.

📝 Teacher's Note: Show students that volume is always conserved when melting and recasting. The key is to find total volume first, then use it to find the new radius.

🎯 Exam Tip: Always write "Given" first, then show volume formula clearly. Remember that 1 mm = 1/10 cm. Show all calculation steps to get full marks.

Question 6. The volume of one sphere is 27 times that of another sphere. Calculate the ratio of their:
(i) radii
(ii) surface areas
Solution:
Given:
Volume of first sphere = 27 × volume of second sphere

Let radius of first sphere = \( r_1 \)
Let radius of second sphere = \( r_2 \)

Step 1: Write volume formulas.
Volume of first sphere = \( \frac{4}{3}\pi r_1^3 \)
Volume of second sphere = \( \frac{4}{3}\pi r_2^3 \)

Step 2: Use the given condition.
\[ \frac{4}{3}\pi r_1^3 = 27 \times \frac{4}{3}\pi r_2^3 \]
\[ r_1^3 = 27r_2^3 = (3r_2)^3 \]
\[ r_1 = 3r_2 \]
\[ \frac{r_1}{r_2} = \frac{3}{1} \]

(i) Ratio of radii = 3 : 1

Step 3: Find ratio of surface areas.
Surface area of first sphere = \( 4\pi r_1^2 \)
Surface area of second sphere = \( 4\pi r_2^2 \)

\[ \text{Ratio} = \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{r_1^2}{r_2^2} = \frac{3^2}{1^2} = \frac{9}{1} \]

(ii) Ratio of surface areas = 9 : 1
In simple words: If one sphere has 27 times more volume, its radius is 3 times bigger. Its surface area becomes 9 times bigger because area depends on radius squared.

📝 Teacher's Note: Explain that volume depends on r³, so if volume ratio is 27:1, radius ratio is 3:1. Surface area depends on r², so it becomes 9:1.

🎯 Exam Tip: Remember the pattern: if volume ratio is a³:b³, then radius ratio is a:b and surface area ratio is a²:b². This saves time in exams.

Question 7. If the number of square centimeters on the surface of a sphere is equal to the number of cubic centimeters in the volume, what is the diameter of the sphere?
Solution:
Let r be the radius of the sphere.

Surface area = \( 4\pi r^2 \) and volume = \( \frac{4}{3}\pi r^3 \)

According to the condition:
\[ 4\pi r^2 = \frac{4}{3}\pi r^3 \]

Step 1: Solve for r.
\[ \frac{r^3}{r^2} = 4\pi \times \frac{3}{4\pi} \]
\[ r = 3 \text{ cm} \]

Step 2: Find diameter.
Diameter of sphere = 2 × 3 cm = 6 cm
In simple words: We set the surface area number equal to the volume number. When we solve this equation, we get radius = 3 cm, so diameter = 6 cm.

📝 Teacher's Note: This is a special case where numerical values of surface area and volume are equal. Emphasize that we are comparing numbers, not actual quantities with units.

🎯 Exam Tip: Set up the equation 4πr² = (4/3)πr³ and cancel common terms carefully. Always double-check by substituting the answer back.

Question 8. A solid metal sphere is cut through its centre into 2 equal parts. If the diameter of the sphere is \( 3\frac{1}{2} \) cm, find the total surface area of each part correct to 2 decimal places.
Solution:
Given:
Diameter of sphere = \( 3\frac{1}{2} \) cm = \( \frac{7}{2} \) cm

Step 1: Find radius.
Radius of sphere = \( \frac{7}{4} \) cm

Step 2: Find total curved surface area of each hemisphere.
Total curved surface area of each hemisphere = \( 2\pi r^2 + \pi r^2 = 3\pi r^2 \)

\[ = 3 \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \]
\[ = 28.88 \text{ cm}^2 \]
In simple words: When we cut a sphere in half, each piece has curved surface plus a flat circular base. The total area includes both.

📝 Teacher's Note: Explain that cutting a sphere creates two hemispheres. Each hemisphere has curved surface area (2πr²) plus the flat circular area (πr²), giving total 3πr².

🎯 Exam Tip: Remember hemisphere total surface area = 3πr². Don't forget the flat circular surface that gets exposed when cutting. Round to required decimal places.

Question 9. The internal and external diameters of a hollow hemi-spherical vessel are 21 cm and 28 cm respectively. Find:
(i) internal curved surface area
(ii) external curved surface area
(iii) total surface area
(iv) volume of material of the vessel.
Solution:
Given:
External diameter = 28 cm, so External radius (R) = 14 cm
Internal diameter = 21 cm, so Internal radius (r) = \( \frac{21}{2} \) cm

(i) Internal curved surface area:
\[ = 2\pi r^2 = 2 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = 693 \text{ cm}^2 \]

(ii) External curved surface area:
\[ = 2\pi R^2 = 2 \times \frac{22}{7} \times 14 \times 14 = 1232 \text{ cm}^2 \]

(iii) Total surface area:
\[ = 2\pi R^2 + 2\pi r^2 + \pi(R^2 - r^2) \]
\[ = 693 + 1232 + \frac{22}{7} \left(14^2 - \left(\frac{21}{2}\right)^2\right) \]
\[ = 1925 + \frac{22}{7}(196 - \frac{441}{4}) \]
\[ = 1925 + \frac{22}{7} \times \frac{343}{4} \]
\[ = 1925 + 269.5 = 2194.5 \text{ cm}^2 \]

(iv) Volume of material used:
\[ = \frac{2}{3}\pi(R^3 - r^3) \]
\[ = \frac{2}{3} \times \frac{22}{7} \left(14^3 - \left(\frac{21}{2}\right)^3\right) \]
\[ = \frac{44}{21}(2744 - 1157.625) \]
\[ = \frac{44}{21} \times 1586.375 = 3323.83 \text{ cm}^3 \]
In simple words: A hollow hemisphere has inner and outer curved surfaces plus a ring-shaped flat area at the base. We find each area separately and add them up.

📝 Teacher's Note: Draw a cross-section of the hollow hemisphere to show students the three surfaces: inner curved, outer curved, and the ring-shaped base area.

🎯 Exam Tip: For hollow hemispheres, total surface area includes inner curved + outer curved + ring area. Ring area = π(R² - r²). Show all three parts clearly.

Question 10. A solid sphere and a solid hemi-sphere have the same total surface area. Find the ratio between their volumes.
Solution:
Let the radius of the sphere be 'r₁'
Let the radius of the hemisphere be 'r₂'

Given condition:
TSA of sphere = TSA of hemi-sphere
\[ 4\pi r_1^2 = 3\pi r_2^2 \]
\[ r_2^2 = \frac{4}{3}r_1^2 \]
\[ r_2 = \frac{2}{\sqrt{3}}r_1 \]

Step 1: Find volumes.
Volume of sphere, V₁ = \( \frac{4}{3}\pi r_1^3 \)
Volume of hemisphere, V₂ = \( \frac{2}{3}\pi r_2^3 \)

\[ V_2 = \frac{2}{3}\pi r_2^3 = \frac{2}{3}\pi \left(\frac{r_1 2}{\sqrt{3}}\right)^3 = \frac{2}{3}\pi \frac{r_1^3 8}{3\sqrt{3}} \]

Step 2: Find ratio V₁/V₂.
\[ \frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{2}{3}\pi \frac{8}{3\sqrt{3}}r_1^3} = \frac{4}{3} \times \frac{3\sqrt{3}}{2 \times 8} = \frac{4}{3} \times \frac{9\sqrt{3}}{16} = \frac{3\sqrt{3}}{4} \]
In simple words: When a sphere and hemisphere have the same total surface area, the hemisphere has a bigger radius. We use this to find the volume ratio.

📝 Teacher's Note: Remind students that hemisphere total surface area = 3πr² (curved + flat base), while sphere surface area = 4πr². This gives the radius relationship.

🎯 Exam Tip: First find the radius relationship from equal surface areas, then substitute this into the volume ratio. Keep the final answer in surd form unless asked to simplify.

Question 11. Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted and recasted
[Note: This question appears incomplete in the provided content.]

Exercise 20 D

Question 1. A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.
Answer:
Given:
Radius of solid sphere = R = 15 cm
∴ Volume of sphere melted = \( \frac{4}{3} \pi R^3 = \frac{4}{3} \times \pi \times 15 \times 15 \times 15 \)

Radius of each cone recasted = r = 2.5 cm
Height of each cone recasted = h = 8 cm
∴ Volume of each cone recasted = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times 2.5 \times 2.5 \times 8 \)

∴ Number of cones recasted = \( \frac{\text{Volume of sphere melted}}{\text{Volume of each cone formed}} \)
= \( \frac{\frac{4}{3} \times \pi \times 15 \times 15 \times 15}{\frac{1}{3} \times \pi \times 2.5 \times 2.5 \times 8} \)
= 270

In simple words: We find how much material the sphere has. Then we find how much material one cone needs. We divide to see how many cones we can make from the sphere's material.

📝 Teacher's Note: Show students that when we melt and recast, the volume stays the same. Only the shape changes. This is a key idea they must remember.

🎯 Exam Tip: Always write the volume formulas first. Then set up the equation "Volume of sphere = Number of cones × Volume of one cone". This gets you marks.

Question 2. A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.
Answer:
Given:
External diameter = 8 cm
Therefore, radius (R) = 4 cm
Internal diameter = 4 cm
Therefore, radius (r) = 2 cm

Step 1: Find volume of metal used in hollow sphere.
Volume of metal used in hollow sphere = \( \frac{4}{3} \pi (R^3 - r^3) = \frac{4}{3} \times \frac{22}{7} \times (4^3 - 2^3) = \frac{88}{21} (64 - 8) = \frac{88}{21} \times 56 \) cm³ ...(i)

Step 2: Find volume of cone.
Diameter of cone = 8 cm
Therefore, radius = 4 cm
Let height of cone = h
∴ Volume = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 4 \times 4 \times h = \frac{352}{21} h \) ...(ii)

Step 3: Equate volumes.
From (i) and (ii)
\( \frac{352}{21} h = \frac{88}{21} \times 56 \)
⇒ h = \( \frac{88 \times 56 \times 21}{21 \times 352} = 14 \) cm

Height of the cone = 14 cm

In simple words: A hollow sphere is like a ball with air inside. We only use the solid part (the shell) to make the cone. So we subtract the inner volume from outer volume.

📝 Teacher's Note: Use a hollow ball and a solid ball to show the difference. Explain that we only melt the material, not the air inside the hollow sphere.

🎯 Exam Tip: For hollow sphere, always write "Volume = Volume of outer sphere - Volume of inner sphere". This formula is very important for marks.

Question 3. The radii of the internal and external surfaces of a metallic spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid right circular cone of height 32 cm. Find the diameter of the base of the cone.
Answer:
Given:
Internal radius = 3cm
External radius = 5 cm
Height of cone = 32 cm

Step 1: Find volume of spherical shell.
Volume of spherical shell = \( \frac{4}{3} \pi (5^3 - 3^3) \)
= \( \frac{4}{3} \times \frac{22}{7} (125 - 27) \)
= \( \frac{4}{3} \times \frac{22}{7} \times 98 \)

Step 2: Find volume of solid circular cone.
Volume of solid circular cone = \( \frac{1}{3} \pi r^2 h \)
= \( \frac{1}{3} \times \frac{22}{7} \times r^2 \times 32 \)

Step 3: Equate volumes.
Vol. of Cone = Vol. of sphere
⇒ \( \frac{1}{3} \times \frac{22}{7} \times r^2 \times 32 = \frac{4}{3} \times \frac{22}{7} \times 98 \)
⇒ \( r^2 = \frac{4 \times 98}{32} \)
∴ r = \( \frac{7}{2} = 3.5 \)cm

Hence, diameter = 2r = 7 cm

In simple words: We find how much metal is in the hollow sphere shell. Then we use all that metal to make a cone. The volumes must be equal because no material is lost.

📝 Teacher's Note: Show students a metal bowl (hollow sphere) and explain that we melt only the metal part, not the empty space inside. The cone uses all this metal.

🎯 Exam Tip: Always find radius first, then double it to get diameter. Don't forget this last step - many students lose marks here.

Question 12. The surface area of a solid sphere is increased by 21% without changing its shape. Find the percentage increase in its: (i) radius (ii) volume
Answer:
Given:
Surface area is increased by 21%

Step 1: Find new surface area.
Let the radius of the sphere be 'r'.
Total surface area the sphere, S = \( 4\pi r^2 \)
New surface area of the sphere, S' = \( 4\pi r^2 + \frac{21}{100} \times 4\pi r^2 \)
= \( \frac{121}{100} 4\pi r^2 \)

(i) Find percentage change in radius.
Let the new radius be \( r_1 \)
S' = \( 4\pi r_1^2 \)
S' = \( \frac{121}{100} 4\pi r^2 \)
⇒ \( 4\pi r_1^2 = \frac{121}{100} 4\pi r^2 \)
⇒ \( r_1^2 = \frac{121}{100} r^2 \)
⇒ \( r_1 = \frac{11}{10} r \)
⇒ \( r_1 = r + \frac{r}{10} \)
⇒ \( r_1 - r = \frac{r}{10} \)
⇒ Change in radius = \( \frac{r}{10} \)

Percentage change in radius = \( \frac{\text{Change in radius}}{\text{Original radius}} \times 100 \)
= \( \frac{r/10}{r} \times 100 \)
= 10

Percentage change in radius = 10%

(ii) Find percentage change in volume.
Let the volume of the sphere be V.
Let the new volume of the sphere be V'.
V = \( \frac{4}{3} \pi r^3 \)
V' = \( \frac{4}{3} \pi r_1^3 \)
⇒ V' = \( \frac{4}{3} \pi \left(\frac{11r}{10}\right)^3 \)
⇒ V' = \( \frac{4}{3} \pi \frac{1331}{1000} r^3 \)
⇒ V' = \( \frac{4}{3} \pi r^3 \frac{1331}{1000} \)
⇒ V' = \( \frac{1331}{1000} V \)
⇒ V' = V + \( \frac{331}{1000} V \)
⇒ V' - V = \( \frac{331}{1000} V \)
∴ Change in volume = \( \frac{331}{1000} V \)

Percentage change in volume = \( \frac{\text{Change in volume}}{\text{Original Volume}} \times 100 \)
= \( \frac{331V/1000}{V} \times 100 \)
= \( \frac{331}{10} \)
= 33.1

Percentage change in volume = 33.1%

In simple words: When surface area increases by 21%, the radius increases by 10%. But volume depends on radius cubed, so it increases much more - by 33.1%.

📝 Teacher's Note: Explain that surface area depends on r², but volume depends on r³. So a small change in radius gives a big change in volume. Use balloons to show this.

🎯 Exam Tip: Remember the formulas: Surface area = 4πr², Volume = (4/3)πr³. When radius changes, volume changes by the cube of that change. Show all working steps clearly.

Question 4. Total volume of three identical cones is the same as that of a bigger cone whose height is 9 cm and diameter 40 cm. Find the radius of the base of each smaller cone, if height of each is 108 cm.
Answer:
Given:
Bigger cone: height = 9 cm, diameter = 40 cm, so radius = 20 cm
Smaller cones: height = 108 cm each, radius = r cm each
Total volume of 3 smaller cones = Volume of 1 bigger cone

Step 1: Find volume of bigger cone.
Volume of larger cone = \( \frac{1}{3} \pi \times 20^2 \times 9 \)

Step 2: Find volume of one smaller cone.
Volume of smaller cone = \( \frac{1}{3} \pi \times r^2 \times 108 \)

Step 3: Set up the equation.
Volume of larger cone = 3 × Volume of smaller cone
\( \frac{1}{3} \pi \times 20^2 \times 9 = \frac{1}{3} \pi \times r^2 \times 108 \times 3 \)

Step 4: Solve for r.
\( r^2 = \frac{20^2 \times 9}{108 \times 3} \)
\( r^2 = \frac{400 \times 9}{324} = \frac{3600}{324} = \frac{100}{9} \)
\( r = \frac{10}{3} = 3\frac{1}{3} \text{ cm} \)

Radius of each smaller cone = \( 3\frac{1}{3} \) cm
In simple words: We found that the big cone has the same volume as all three small cones together. Using the cone volume formula, we calculated how big each small cone's base must be.

📝 Teacher's Note: Show students that when we have 3 cones with the same volume as 1 big cone, each small cone has 1/3 the volume. This makes the math easier to understand.

🎯 Exam Tip: Always write the formula first, then substitute values clearly. Show each step of solving the equation. Don't forget to convert improper fractions to mixed numbers in the final answer.

Question 5. A solid rectangular block of metal 49 cm by 44 cm by 18 cm is melted and formed into a solid sphere. Calculate the radius of the sphere.
Answer:
Given:
Rectangular block dimensions: 49 cm × 44 cm × 18 cm
When melted, it forms a sphere with radius r cm

Step 1: Find volume of rectangular block.
Volume of rectangular block = 49 × 44 × 18 = 38808 cm³

Step 2: Set up equation using sphere volume formula.
Let r be the radius of sphere
Volume of sphere = \( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times r^3 = \frac{88}{21} r^3 \)

Step 3: Equate the volumes.
Volume of block = Volume of sphere
\( \frac{88}{21} r^3 = 38808 \)

Step 4: Solve for r.
\( r^3 = 38808 \times \frac{21}{88} = 441 \times 21 = 9261 \)
\( r^3 = 9261 \)
\( r = 21 \text{ cm} \)

Radius of sphere = 21 cm
In simple words: When we melt the block, all the metal stays the same. So the rectangular block and the sphere have the same volume. We used this fact to find the sphere's radius.

📝 Teacher's Note: Remind students that melting doesn't change the amount of material, only the shape. The volume stays exactly the same. This is a key concept in these problems.

🎯 Exam Tip: Write "Volume remains same when melted" at the start. Calculate the rectangular volume first, then equate to sphere volume formula. Show all substitution steps clearly.

Question 6. A hemi-spherical bowl of internal radius 9 cm is full of liquid. This liquid is to be filled into conical shaped small containers each of diameter 3 cm and height 4 cm. How many containers are necessary to empty the bowl?
Answer:
Given:
Hemispherical bowl: radius = 9 cm
Small conical containers: diameter = 3 cm, so radius = 1.5 cm, height = 4 cm

Step 1: Find volume of hemispherical bowl.
Volume = \( \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi \times 9^3 = \frac{2}{3} \pi \times 729 = 486\pi \text{ cm}^3 \)

Step 2: Find volume of one conical container.
Volume of bottle = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times \left(\frac{3}{2}\right)^2 \times 4 = 3\pi \text{ cm}^3 \)

Step 3: Calculate number of containers needed.
Number of bottles = \( \frac{486\pi}{3\pi} = 162 \)

Number of containers = 162
In simple words: We found how much liquid the bowl holds, then how much each small container holds. Dividing gives us the number of containers needed.

📝 Teacher's Note: Explain that hemisphere means half a sphere. So its volume is half of a full sphere's volume. Students often forget this and use the full sphere formula.

🎯 Exam Tip: Write the hemisphere volume formula as (1/2) × (4/3)πr³. For containers, use the cone volume formula (1/3)πr²h. The final answer should be a whole number.

Question 7. A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone if it is completely filled.
Answer:
Given:
Hemispherical bowl: diameter = 7.2 cm, so radius = 3.6 cm
Inverted cone: radius = 4.8 cm, height = h cm

Step 1: Find volume of hemispherical bowl.
Volume of sauce in hemispherical bowl = \( \frac{2}{3} \pi r^3 = \frac{2}{3} \pi \times (3.6)^3 \)

Step 2: Find volume of cone.
Volume of cone = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times (4.8)^2 \times h \)

Step 3: Equate the volumes.
Volume of sauce in hemispherical bowl = Volume of cone
\( \frac{2}{3} \pi \times (3.6)^3 = \frac{1}{3} \pi \times (4.8)^2 \times h \)

Step 4: Solve for h.
\( h = \frac{2 \times 3.6 \times 3.6 \times 3.6}{4.8 \times 4.8} = \frac{2 \times 46.656}{23.04} = 4.05 \text{ cm} \)

Height of the cone = 4.05 cm
In simple words: The sauce amount stays the same when poured from bowl to cone. We used this to find how tall the cone must be to hold all the sauce.

📝 Teacher's Note: Emphasize that "completely filled" means all the sauce fits exactly. The volumes must be equal. Use decimal calculations carefully - show each step.

🎯 Exam Tip: Set volumes equal to each other. Cancel π from both sides first. Be careful with decimal calculations and show your working clearly.

Question 8. A solid cone of radius 5 cm and height 8 cm is melted and made into small spheres of radius 0.5 cm. Find the number of spheres formed.
Answer:
Given:
Original cone: radius = 5 cm, height = 8 cm
Small spheres: radius = 0.5 cm each

Step 1: Find volume of original cone.
Volume of cone = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 5 \times 5 \times 8 = \frac{200\pi}{3} \text{ cm}^3 \)

Step 2: Find volume of one small sphere.
Volume of one sphere = \( \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times (0.5)^3 = \frac{4}{3} \pi \times \frac{1}{8} = \frac{\pi}{6} \text{ cm}^3 \)

Step 3: Calculate number of spheres.
Number of spheres = \( \frac{\text{Total volume}}{\text{Volume of one sphere}} = \frac{\frac{200\pi}{3}}{\frac{\pi}{6}} = \frac{200\pi}{3} \times \frac{6}{\pi} = \frac{200 \times 6}{3} = 400 \)

Number of spheres formed = 400
In simple words: We found how much material the big cone has, then how much material each small sphere needs. Dividing tells us how many small spheres we can make.

📝 Teacher's Note: Show students that (0.5)³ = 1/8. This fraction calculation is where many students make mistakes. Practice these cube calculations separately.

🎯 Exam Tip: Write both volume formulas clearly. When dividing fractions, multiply by the reciprocal. Show the π cancellation step clearly.

Question 9. The total area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate: (i) the radius of the solid sphere (ii) the number of cones recasted [π = 3.14]
Answer:
Given:
Total surface area of sphere = 1256 cm²
Small cones: radius = 2.5 cm, height = 8 cm
π = 3.14

Part (i): Find radius of sphere
Let radius of sphere = r cm

Step 1: Use surface area formula.
\( 4\pi r^2 = 1256 \)
\( 4 \times \frac{22}{7} r^2 = 1256 \)

Step 2: Solve for r.
\( r^2 = \frac{1256 \times 7}{4 \times 22} = \frac{8792}{88} = 99.909 \approx 100 \)
\( r = 10 \text{ cm} \)

Part (ii): Find number of cones
Step 3: Find volume of sphere.
Volume of sphere = \( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 10^3 = \frac{88000}{21} \text{ cm}^3 \)

Step 4: Find volume of one cone.
Volume of one cone = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (2.5)^2 \times 8 = \frac{1100}{21} \text{ cm}^3 \)

Step 5: Calculate number of cones.
Number of cones = \( \frac{88000/21}{1100/21} = \frac{88000}{1100} = 80 \)

(i) Radius of sphere = 10 cm
(ii) Number of cones = 80
In simple words: First we found the sphere's radius using its surface area. Then we found how many small cones can be made from the same amount of material.

📝 Teacher's Note: Students often confuse surface area formula (4πr²) with volume formula (4/3πr³). Write both formulas clearly on the board and explain the difference.

🎯 Exam Tip: For part (i), use surface area formula 4πr². For part (ii), use volume formulas and remember that material amount stays same when melted. Show all fraction calculations step by step.

Question 10. A solid metallic cone, with radius 6 cm and height 10 cm, is made of some heavy metal A. In order to reduce weight, a conical hole is made in the cone as shown and it is completely filled with a lighter metal B. The conical hole has a diameter of 6 cm and depth 4 cm. Calculate the ratio of the volume of the metal A to the volume of metal B in the solid.
Answer:
Given:
Original cone (metal A): radius = 6 cm, height = 10 cm
Conical hole (metal B): diameter = 6 cm, so radius = 3 cm, depth = 4 cm

Step 1: Find volume of original cone.
Volume of original cone = \( \frac{1}{3} \pi \times 6^2 \times 10 = \frac{1}{3} \pi \times 36 \times 10 = 120\pi \text{ cm}^3 \)

Step 2: Find volume of conical hole.
Volume of conical hole = \( \frac{1}{3} \pi \times 3^2 \times 4 = \frac{1}{3} \pi \times 9 \times 4 = 12\pi \text{ cm}^3 \)

Step 3: Find volume of metal A remaining.
Volume of metal A = Volume of original cone - Volume of hole
Volume of metal A = \( 120\pi - 12\pi = 108\pi \text{ cm}^3 \)

Step 4: Find volume of metal B.
Volume of metal B = Volume of conical hole = \( 12\pi \text{ cm}^3 \)

Step 5: Find the ratio.
Ratio = \( \frac{\text{Volume of metal A}}{\text{Volume of metal B}} = \frac{108\pi}{12\pi} = \frac{108}{12} = 9:1 \)

Ratio of volume of metal A to metal B = 9:1
In simple words: We found how much space the hole takes up, then subtracted it from the total cone volume to get metal A's volume. Metal B fills the hole completely.

📝 Teacher's Note: Draw a diagram showing the original cone and the hole inside it. Explain that metal A is the remaining part after removing the hole. Metal B fills the hole exactly.

🎯 Exam Tip: Find both volumes separately first. Remember that metal A = original volume minus hole volume. Metal B = hole volume. Express the final ratio in simplest form like 9:1.

[Diagram: This diagram shows a large cone with a smaller cone-shaped hole cut out from the top, filled with different metal.]

Question 11. A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 3 cm and height 8 cm. Find the number of cones.
Answer:
Given:
Internal radius of hollow sphere = 6 cm
External radius of hollow sphere = 8 cm
Base radius of each cone = 3 cm
Height of each cone = 8 cm

Step 1: Find volume of hollow sphere.
Volume of hollow sphere = Volume of outer sphere - Volume of inner sphere
\[ = \frac{4}{3}\pi(8^3 - 6^3) \]
\[ = \frac{4}{3}\pi(512 - 216) \]
\[ = \frac{4}{3}\pi \times 296 \]
\[ = \frac{1184\pi}{3} \]

Step 2: Find volume of one small cone.
Volume of one cone = \( \frac{1}{3}\pi r^2 h \)
\[ = \frac{1}{3}\pi \times 3^2 \times 8 \]
\[ = \frac{1}{3}\pi \times 9 \times 8 \]
\[ = 24\pi \]

Step 3: Find number of cones.
Number of cones = \( \frac{\text{Volume of hollow sphere}}{\text{Volume of one cone}} \)
\[ = \frac{\frac{1184\pi}{3}}{24\pi} \]
\[ = \frac{1184}{3 \times 24} \]
\[ = \frac{1184}{72} \]
\[ = 16.44 \approx 16 \]

Number of cones = 16
In simple words: We found how much metal is in the hollow sphere. Then we found how much metal is needed for one cone. We divided to get the number of cones.

📝 Teacher's Note: Show students that hollow sphere means there is empty space inside. We only count the metal part. Use a tennis ball cut in half to show this idea.

🎯 Exam Tip: Always write the formula for hollow sphere volume first. Remember to subtract inner volume from outer volume. Show all steps clearly.

Question 12. The surface area of a solid metallic sphere is 2464 cm². It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate: (i) the radius of the sphere. (ii) the number of cones recast. (Take π = 22/7)
Answer:
Given:
Surface area of sphere = 2464 cm²
Radius of each cone = 3.5 cm
Height of each cone = 7 cm
π = 22/7

(i) Finding the radius of sphere:
Step 1: Use surface area formula for sphere.
Surface area of sphere = \( 4\pi R^2 \)
\( 4\pi R^2 = 2464 \)

Step 2: Substitute value of π.
\( 4 \times \frac{22}{7} \times R^2 = 2464 \)
\( \frac{88}{7} \times R^2 = 2464 \)

Step 3: Solve for R².
\( R^2 = \frac{2464 \times 7}{88} = \frac{17248}{88} = 196 \)

Step 4: Find R.
\( R = \sqrt{196} = 14 \text{ cm} \)

(ii) Finding number of cones:
Step 5: Find volume of sphere.
Volume of sphere = \( \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 \)

Step 6: Find volume of each cone.
Volume of each cone = \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 7 \)

Step 7: Calculate number of cones.
Number of cones = \( \frac{\text{Volume of sphere}}{\text{Volume of each cone}} \)
\[ = \frac{\frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14}{\frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 7} \]
\[ = \frac{4 \times 14 \times 14 \times 14}{3.5 \times 3.5 \times 7} \]
\[ = \frac{4 \times 2744}{86.625} = 128 \]

Answer: (i) Radius = 14 cm (ii) Number of cones = 128
In simple words: First we used the surface area to find how big the sphere is. Then we calculated how many small cones can be made from all that metal.

📝 Teacher's Note: Remind students that surface area formula is 4πR² and volume formula is (4/3)πR³. These are different formulas for different purposes.

🎯 Exam Tip: Always write "Given" first and list all values. Show the formula before substituting numbers. Check your answer - 128 cones from one sphere seems reasonable.

Exercise 20 E

Question 1. A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.
Answer:
Given:
Height of cone = 15 cm
Diameter = 7 cm
Radius = 7/2 = 3.5 cm

Step 1: Find volume of cone.
Volume of cone = \( \frac{1}{3}\pi r^2 h \)
\[ = \frac{1}{3} \times \pi \times (3.5)^2 \times 15 \]
\[ = \frac{1}{3} \times \pi \times 12.25 \times 15 \]
\[ = \frac{183.75\pi}{3} = 61.25\pi \]

Step 2: Find volume of hemisphere.
Volume of hemisphere = \( \frac{2}{3}\pi r^3 \)
\[ = \frac{2}{3} \times \pi \times (3.5)^3 \]
\[ = \frac{2}{3} \times \pi \times 42.875 \]
\[ = \frac{85.75\pi}{3} \]

Step 3: Find total volume.
Total volume = Volume of cone + Volume of hemisphere
\[ = 61.25\pi + \frac{85.75\pi}{3} \]
\[ = \frac{183.75\pi + 85.75\pi}{3} \]
\[ = \frac{269.5\pi}{3} \]
\[ = 89.83\pi \text{ cm}^3 \]

Using π = 22/7:
\[ = 89.83 \times \frac{22}{7} = 282.33 \text{ cm}^3 \]

Volume of solid = 282.33 cm³
In simple words: We added the volume of the cone on top and the hemisphere at the bottom. Think of it like an ice cream cone with a scoop on top.

📝 Teacher's Note: Draw a simple diagram showing cone sitting on hemisphere. Students can see this shape in real life - like a rocket or an ice cream cone.

🎯 Exam Tip: Remember hemisphere volume is (2/3)πr³, not (4/3)πr³. That's for full sphere. Add both volumes for composite solid.

Question 2. A buoy is made in the form of a hemisphere surmounted by a right circular cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 m and its volume is two-third the volume of hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two decimal places.
Answer:
Given:
Radius of hemisphere and cone base = 3.5 m
Volume of cone = (2/3) × Volume of hemisphere

Step 1: Find volume of hemisphere.
Volume of hemisphere = \( \frac{2}{3}\pi r^3 \)
\[ = \frac{2}{3} \times \frac{22}{7} \times (3.5)^3 \]
\[ = \frac{2}{3} \times \frac{22}{7} \times 42.875 \]
\[ = \frac{539}{6} \text{ m}^3 \]

Step 2: Find volume of cone.
Volume of cone = \( \frac{2}{3} \times \frac{539}{6} = \frac{2 \times 539}{3 \times 6} = \frac{1078}{18} \text{ m}^3 \)

Step 3: Find height of cone.
Let height = h
\( \frac{1}{3}\pi r^2 h = \frac{1078}{18} \)
\[ \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times h = \frac{1078}{18} \]
\[ h = \frac{1078 \times 3 \times 7}{18 \times 22 \times 12.25} \]
\[ h = \frac{14}{3} = 4\frac{2}{3} = 4.67 \text{ m} \]

Step 4: Find surface area of buoy.
Surface area = Curved surface of hemisphere + Curved surface of cone
Surface area = \( 2\pi r^2 + \pi rl \)

Step 5: Find slant height l.
\( l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + (4.67)^2} \)
\[ = \sqrt{12.25 + 21.81} = \sqrt{34.06} = 5.83 \text{ m} \]

Step 6: Calculate total surface area.
Surface area = \( 2 \times \frac{22}{7} \times (3.5)^2 + \frac{22}{7} \times 3.5 \times 5.83 \)
\[ = 77 + 113.17 = 190.17 \text{ m}^2 \]

Height of cone = 4.67 m, Surface area = 190.17 m²
In simple words: A buoy is like a ball with a cone on top. We used the given volume relationship to find the cone's height. Then we added the curved surfaces of both parts.

📝 Teacher's Note: Explain that buoy floats on water to mark safe areas for boats. The cone shape helps it stay stable in water. Draw both shapes joined together.

🎯 Exam Tip: For surface area of composite solids, only count the outer surfaces that touch air or water. Don't count the flat circular base where cone and hemisphere join.

Question 3. From a rectangular solid of metal 42 cm by 30 cm by 20 cm, a conical cavity of diameter 14 cm and depth 24 cm is scooped out. Find
(i) the surface area of the remaining solid
(ii) the volume of remaining solid
(iii) the weight of the material drilled out if it weighs 7 gm per cm³.
Answer:
Step 1: Find total surface area of cuboid.
Total surface area of cuboid = \( 2(lb + bh + lh) \)
= \( 2(42 \times 30 + 30 \times 20 + 20 \times 42) \)
= \( 2(1260 + 600 + 840) \)
= \( 2 \times 2700 \)
= \( 5400 \text{ cm}^2 \)

Step 2: Find area of conical cavity.
Diameter of cone = 14 cm
\( \implies \) Radius of cone = \( \frac{14}{2} = 7 \) cm
Area of circular base = \( \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154 \text{ cm}^2 \)
Area of curved surface = \( \pi r l = \frac{22}{7} \times 7 \times \sqrt{7^2 + 24^2} = 22\sqrt{49 + 576} = 22 \times 25 = 550 \text{ cm}^2 \)
Surface area of remaining part = \( 5400 + 550 - 154 = 5796 \text{ cm}^2 \)

Step 3: Find volume of remaining solid.
Dimensions of rectangular solid = (42 × 30 × 20) cm
Volume = \( 42 \times 30 \times 20 = 25200 \text{ cm}^3 \)
Radius of conical cavity = 7 cm, height = 24 cm
Volume of cone = \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 = 1232 \text{ cm}^3 \)
Volume of remaining solid = \( 25200 - 1232 = 23968 \text{ cm}^3 \)

Step 4: Find weight of material drilled out.
Weight = \( 1232 \times 7 = 8624 \text{ g} = 8.624 \text{ kg} \)

In simple words: We found the surface area by adding the cuboid surface and cone curved surface, then removing the base circle area. Volume is cuboid volume minus cone volume. Weight is volume times density.

[Diagram: This diagram shows a rectangular block with a conical hole cut into it from the top. The cone goes down into the block.]

📝 Teacher's Note: Draw a soap bar and show how scooping out a cone shape increases surface area because we add curved surface inside. Students often forget to subtract the base circle area.

🎯 Exam Tip: Always write all three parts separately with clear labels (i), (ii), (iii). Show all formula substitutions step by step. Remember slant height formula for cone surface area.

Question 4. The cubical block of side 7 cm is surmounted by a hemisphere of the largest size. Find the surface area of the resulting solid.
Answer:
Step 1: Find hemisphere dimensions.
The diameter of the largest hemisphere that can be placed on a face of a cube of side 7 cm will be 7 cm.
Therefore, radius = \( r = \frac{7}{2} \) cm

Step 2: Find curved surface area of hemisphere.
Its curved surface area = \( 2\pi r^2 = 2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 77 \text{ cm}^2 \)

Step 3: Find surface area of top face of cube after placing hemisphere.
Surface area of the top of the resulting solid = Surface area of the top face of the cube - Area of the base of the hemisphere
= \( (7 \times 7) - \left(\frac{22}{7} \times \frac{49}{4}\right) = 49 - \frac{77}{2} = \frac{98 - 77}{2} = \frac{21}{2} = 10.5 \text{ cm}^2 \)

Step 4: Find total surface area.
Surface area of the cube = \( 5 \times (\text{side})^2 = 5 \times 49 = 245 \text{ cm}^2 \)
Total area of resulting solid = \( 245 + 10.5 + 77 = 332.5 \text{ cm}^2 \)

In simple words: We put a half-ball on top of a cube. The total surface is: 5 faces of cube + curved part of half-ball - the circle where they touch.

📝 Teacher's Note: Use a dice and a half tennis ball to show this. Students often forget that one face of the cube gets partially covered by the hemisphere base.

🎯 Exam Tip: Always mention "largest hemisphere" means diameter equals cube side. Subtract the circular base area from cube's top face. Show all three parts clearly.

Question 5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of the top which is open is 5 cm. It is filled with water. 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:
Step 1: Find volume of conical vessel.
Height of cone = 8 cm
Radius = 5 cm
Volume = \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 8 = \frac{4400}{21} \text{ cm}^3 \)

Step 2: Find volume of water that flowed out.
Volume of water that flowed out = \( \frac{1}{4} \times \frac{4400}{21} = \frac{1100}{21} \text{ cm}^3 \)

Step 3: Find volume of one lead ball.
Radius of each ball = 0.5 cm = \( \frac{1}{2} \) cm
Volume of a ball = \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{11}{21} \text{ cm}^3 \)

Step 4: Find number of lead balls.
Number of balls = \( \frac{\text{Volume of water flowed out}}{\text{Volume of one ball}} = \frac{1100/21}{11/21} = 100 \)

Hence, number of lead balls = 100

In simple words: The lead balls take up space in the water. This makes some water overflow. The volume of balls equals the volume of water that overflowed.

[Diagram: This diagram shows an upside-down cone (like an ice cream cone turned over) filled with water, with small round balls being dropped into it.]

📝 Teacher's Note: Fill a transparent cone-shaped container with water and drop marbles. Students can see water overflowing. The displaced water volume equals the marble volume.

🎯 Exam Tip: Key idea: Volume of balls = Volume of displaced water. Write this clearly at the start. One-fourth means divide total volume by 4.

Question 6. A hemispherical bowl has negligible thickness and the length of its circumference is 198 cm. Find the capacity of the bowl.
Answer:
Step 1: Find radius from circumference.
Let r be the radius of the bowl.
\( 2\pi r = 198 \)
\( \implies r = \frac{198 \times 7}{2 \times 22} = \frac{1386}{44} = 31.5 \) cm

Step 2: Find capacity of hemispherical bowl.
Capacity of the bowl = \( \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (31.5)^3 = 65488.5 \text{ cm}^3 \)

In simple words: The circumference is the distance around the rim of the bowl. We use this to find how wide the bowl is, then calculate how much it can hold.

📝 Teacher's Note: Measure the rim of a round bowl with string to find circumference. Show that hemisphere volume is half of sphere volume (but we use 2/3, not 1/2, because of the formula).

🎯 Exam Tip: Capacity means volume. Use hemisphere volume formula (2/3)πr³, not sphere formula (4/3)πr³. Always find radius first from circumference 2πr.

Question 7. Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r cm.
Answer:
For the volume of cone to be largest, h = r cm
Volume of the cone = \( \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times r^2 \times r = \frac{1}{3}\pi r^3 \)

In simple words: The biggest cone we can cut from a hemisphere has its tip at the flat center and its base is the curved edge. Its height equals the hemisphere radius.

[Diagram: This diagram shows a hemisphere (half ball) with a cone carved inside it. The cone's base sits on the flat circular face of the hemisphere.]

📝 Teacher's Note: Cut an orange in half. The biggest cone you can carve will have its point at the center of the flat face and its circular edge touching the curved boundary.

🎯 Exam Tip: The maximum cone has height = radius of hemisphere. The cone's base radius also equals hemisphere radius. Write both these facts clearly.

Question 8. The radii of the bases of two solid right circular cones of same height are r₁ and r₂ respectively. The cones are melted and recast into a solid sphere of radius R. Find the height of each cone in terms r₁, r₂ and R.
Answer:
Let the height of the solid cones be 'h'

Step 1: Write volumes of both cones.
Volume of solid circular cones:
\( V_1 = \frac{1}{3}\pi r_1^2 h \)
\( V_2 = \frac{1}{3}\pi r_2^2 h \)

Step 2: Write volume of sphere.
Volume of sphere = \( \frac{4}{3}\pi R^3 \)

Step 3: Use conservation of volume.
Volume of sphere = Volume of cone 1 + Volume of cone 2
\( \frac{4}{3}\pi R^3 = \frac{1}{3}\pi r_1^2 h + \frac{1}{3}\pi r_2^2 h \)
\( \implies 4R^3 = r_1^2 h + r_2^2 h \)
\( \implies h(r_1^2 + r_2^2) = 4R^3 \)
\( \implies h = \frac{4R^3}{(r_1^2 + r_2^2)} \)

In simple words: When we melt two cones and make a sphere, the total volume stays the same. We use this fact to find the height.

📝 Teacher's Note: Use two paper cones and a ball of clay. Show that when you melt and reshape, the amount of material (volume) never changes.

🎯 Exam Tip: Key principle: Total volume before melting = Total volume after melting. Write this equation first, then solve for h. Keep R³ in the numerator.

Question 9. A solid hemisphere of diameter 28 cm is melted and recast into a number of identical solid cones, each of diameter 14 cm and height 8 cm. Find the number of cones so formed.
Answer:
Step 1: Find volume of hemisphere.
Diameter of hemisphere = 28 cm \( \implies \) Radius = 14 cm
Volume of hemisphere = \( \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 14^3 = \frac{2}{3} \times \frac{22}{7} \times 2744 = \frac{120736}{21} \text{ cm}^3 \)

Step 2: Find volume of one cone.
Diameter of cone = 14 cm \( \implies \) Radius = 7 cm, Height = 8 cm
Volume of one cone = \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 8 = \frac{1}{3} \times \frac{22}{7} \times 49 \times 8 = \frac{8624}{21} \text{ cm}^3 \)

Step 3: Find number of cones.
Number of cones = \( \frac{\text{Volume of hemisphere}}{\text{Volume of one cone}} = \frac{120736/21}{8624/21} = \frac{120736}{8624} = 14 \)

In simple words: We melt a big half-ball and make many small cones from it. To find how many cones, we divide the big volume by the small volume.

📝 Teacher's Note: Use play dough. Make a hemisphere, then break it into small pieces to form cones. Count them. This shows division of volumes gives the number.

🎯 Exam Tip: Always write: Number of cones = Total volume ÷ Volume of one cone. Calculate both volumes first, then divide. Check that your answer is a whole number.

Question 10. A cone and a hemisphere have the same base and same height. Find the ratio between their volumes.
Answer:
Given:
Let radius of base = r
Let height = h
For hemisphere: height = radius = r

Step 1: Find volume of hemisphere
Volume of hemisphere = \( \frac{2}{3}\pi r^3 \)

Step 2: Find volume of cone
Volume of cone = \( \frac{1}{3}\pi r^2 h \)
Since height of cone = height of hemisphere = r
Volume of cone = \( \frac{1}{3}\pi r^2 \times r = \frac{1}{3}\pi r^3 \)

Step 3: Find the ratio
Ratio = \( \frac{\text{Volume of cone}}{\text{Volume of hemisphere}} \)

\( = \frac{\frac{1}{3}\pi r^3}{\frac{2}{3}\pi r^3} \)

\( = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2} \)

Therefore, the ratio of volumes = 1:2
In simple words: The cone's volume is half of the hemisphere's volume. Think of it like filling two containers - the round bowl (hemisphere) holds twice as much as the pointed cone.

📝 Teacher's Note: Show students that both shapes have same base circle and same height. The hemisphere is like half a ball, the cone is pointed. This makes the hemisphere bigger inside.

🎯 Exam Tip: Write the ratio as 1:2, not just 1/2. Also write "Volume of cone : Volume of hemisphere = 1:2" for full marks.

Exercise 20 F

Question 1. From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base are removed. Find the volume of the remaining solid.
Answer:
Given:
Height of cylinder (h) = 10 cm
Radius of base (r) = 6 cm
Height of cone = 10 cm
Radius of base of cone = 6 cm

Step 1: Find volume of cylinder
Volume of cylinder = \( \pi r^2 h \)
= \( \pi \times 6^2 \times 10 \)
= \( 360\pi \) cm³

Step 2: Find volume of cone
Volume of cone = \( \frac{1}{3}\pi r^2 h \)
= \( \frac{1}{3} \times \pi \times 6^2 \times 10 \)
= \( 120\pi \) cm³

Step 3: Find volume of remaining solid
Volume of remaining solid = Volume of cylinder - Volume of cone
= \( 360\pi - 120\pi \)
= \( 240\pi \)
= \( \frac{2}{3} \times 360\pi \)
= \( \frac{2}{3} \times \frac{22}{7} \times 360 \)
= \( \frac{5280}{7} \)
= \( 754\frac{2}{7} \) cm³
In simple words: We take away the cone from inside the cylinder. What's left is like a cylinder with a cone-shaped hole inside it.

📝 Teacher's Note: Use a real cylinder (like a can) and show how removing a cone shape from it leaves a hollow space. Students can visualize this easily.

🎯 Exam Tip: Always subtract cone volume from cylinder volume. Write the final answer as both fraction and decimal form. Show all steps clearly.

Question 2. From a solid cylinder whose height is 16 cm and radius is 12 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume and total surface area of the remaining solid.
Answer:
Given:
Cylinder: Height = 16 cm, Radius = 12 cm
Conical cavity: Height = 8 cm, Radius = 6 cm
 

[Diagram: This diagram shows a cylinder with a conical cavity (hole) cut out from the top. The cylinder is 16 cm tall with 12 cm radius. The cone-shaped hole is 8 cm deep with 6 cm radius at the top.]

Part (i): Volume of remaining solid

Step 1: Find volume of cylinder
Volume = \( \pi r^2 h = \pi \times 12^2 \times 16 \)
= \( \frac{22}{7} \times 144 \times 16 = \frac{50688}{7} \) cm³

Step 2: Find volume of conical cavity
Volume = \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 6^2 \times 8 \)
= \( \frac{2112}{7} \) cm³

Step 3: Find volume of remaining solid
Volume = \( \frac{50688}{7} - \frac{2112}{7} = \frac{48576}{7} \)
= 6939.43 cm³

Part (ii): Total surface area

Step 4: Find slant height of cone
l = \( \sqrt{h^2 + r^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \) cm

Step 5: Calculate total surface area
Surface area = Curved surface area of cylinder + Base area of cylinder + Curved surface area of conical cavity + Area of circular ring
= \( 2\pi Rh + \pi R^2 + \pi rl + \pi(R^2 - r^2) \)
= \( 2 \times \frac{22}{7} \times 12 \times 16 + \frac{22}{7} \times 6 \times 10 + \frac{22}{7} \times 12^2 + \frac{22}{7}(144 - 36) \)
= \( \frac{8448}{7} + \frac{1320}{7} + \frac{3168}{7} + \frac{2376}{7} \)
= \( \frac{15312}{7} \) = 2187.43 cm²
In simple words: We dig out a cone-shaped hole from the cylinder. The remaining solid is like a cylinder with a cone-shaped cavity inside.

📝 Teacher's Note: Use a clay cylinder and scoop out a cone shape to show students. The surface area includes the inside walls of the cone-shaped hole.

🎯 Exam Tip: For surface area, don't forget to add the curved surface of the conical cavity. Many students miss this part and lose marks.

Question 3. A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 80 m, calculate the total area of canvas required. Also, find the total cost of canvas used at Rs 15 per meter if the width is 1.5 m
Answer:
Given:
Height of cylindrical part = 4 m
Diameter = 105 m, so radius = 105/2 = 52.5 m
Slant height of cone = 80 m
 

[Diagram: This diagram shows a circus tent with a cylindrical bottom part (4 m tall) topped by a conical part. The base diameter is 105 m and the slant height of the cone is 80 m.]

Step 1: Find total curved surface area
Total surface area = Curved surface area of cylinder + Curved surface area of cone
= \( 2\pi rh + \pi rl \)
= \( 2 \times \frac{22}{7} \times \frac{105}{2} \times 4 + \frac{22}{7} \times \frac{105}{2} \times 80 \)
= 1320 + 13200
= 14520 m²

Step 2: Find length of canvas needed
Width of canvas = 1.5 m
Length of canvas = \( \frac{14520}{1.5} = 9680 \) m

Step 3: Find total cost
Cost = 9680 × 15 = Rs 145200
In simple words: The tent is like a cylinder with a cone on top. We need canvas to cover the curved surfaces of both parts. Then we find how much this canvas costs.

📝 Teacher's Note: Show students a real tent or draw one on the board. Explain that we don't need canvas for the bottom (ground) - only the walls and roof.

🎯 Exam Tip: Don't include the base area in surface area calculations for tents. Only curved surfaces need canvas. Always show cost calculation clearly.

Question 4. A circus tent is cylindrical to a height of 8 m surmounted by a conical part. If total height of the tent is 13 m and the diameter of its base is 24 m; calculate:
(i) total surface area of the tent
(ii) area of canvas, required to make this tent allowing 10% of the canvas used for folds and stitching.
Answer:
Given:
Height of cylindrical part = 8 m
Total height = 13 m
Height of conical part = 13 - 8 = 5 m
Diameter = 24 m, so radius = 12 m

Step 1: Find slant height of cone
l = \( \sqrt{h^2 + r^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \) m

Step 2: Find total surface area
Surface area = Curved surface area of cylinder + Curved surface area of cone
= \( 2\pi rh + \pi rl \)
= \( 2 \times \frac{22}{7} \times 12 \times 8 + \frac{22}{7} \times 12 \times 13 \)
= \( \frac{4224}{7} + \frac{3432}{7} \)
= \( \frac{7656}{7} \) = 1093.71 m²

Step 3: Find canvas area with 10% extra
Extra canvas for folds and stitching = 10% of 1093.71 = 109.37 m²
Total canvas required = 1093.71 + 109.37 = 1203.08 m²
In simple words: This tent has a cylinder bottom and cone top. We calculate canvas for both parts, then add 10% extra for stitching the pieces together.

📝 Teacher's Note: Explain to students why we need extra canvas - when we stitch pieces together, some material is used up in the folds and seams.

🎯 Exam Tip: Always add the percentage for folds and stitching. Write it as: "Canvas needed = Surface area + 10% extra". Don't forget this step.

Question 4. A tent is in the shape of a cylinder with diameter 24 m surmounted by a conical top of height 5 m and slant height 13 m. Find the area of canvas used in stitching the tent if the cylindrical part is 8 m high.
Answer:
Given:
Height of the cylindrical part = H = 8 m
Height of the conical part = h = (13 – 8)m = 5 m
Diameter = 24 m → radius = r = 12 m
Slant height of the cone = l

Step 1: Find slant height of the cone.
\( l = \sqrt{r^2 + h^2} \)
\( l = \sqrt{12^2 + 5^2} \)
\( l = \sqrt{169} = 13 \) m

Step 2: Find total surface area of the tent.
Total surface area = \( 2\pi rh + \pi rl = \pi r(2h + l) \)
\( = \frac{22}{7} \times 12 \times (2 \times 8 + 13) \)
\( = \frac{264}{7}(16 + 13) \)
\( = \frac{264}{7} \times 29 \)
\( = \frac{7656}{7} \) m²
\( = 1093.71 \) m²

Step 3: Find area of canvas used in stitching.
Area of canvas used in stitching = total area
Total area of canvas = \( \frac{7656}{7} + \frac{\text{Total area of canvas}}{10} \)
\( \Rightarrow \text{Total area of canvas} - \frac{\text{Total area of canvas}}{10} = \frac{7656}{7} \)
\( \Rightarrow \text{Total area of canvas}(1 - \frac{1}{10}) = \frac{7656}{7} \)
\( \Rightarrow \text{Total area of canvas} \times \frac{9}{10} = \frac{7656}{7} \)
\( \Rightarrow \text{Total area of canvas} = \frac{7656}{7} \times \frac{10}{9} \)
\( \Rightarrow \text{Total area of canvas} = \frac{76560}{63} = 1215.23 \) m²

In simple words: We found the tent surface area by adding cylinder curved surface and cone curved surface. Then we added 10% extra for stitching waste.

📝 Teacher's Note: Students often forget that slant height is different from vertical height. Draw a right triangle to show how we use Pythagoras theorem to find slant height.

🎯 Exam Tip: Always show the formula first, then substitute values. Don't forget to add the extra material for stitching as mentioned in the question.

Question 5. A cylindrical boiler, 2 m high, is 3.5 m in diameter. It has a hemispherical lid. Find the volume of its interior, including the part covered by the lid.
Answer:
Given:
Diameter of cylindrical boiler = 3.5 m
Therefore, radius = \( \frac{3.5}{2} = \frac{35}{20} = \frac{7}{4} \) m
Height (h) = 2 m
Radius of hemisphere (R) = \( \frac{7}{4} \) m

Step 1: Find total volume of the boiler.
Total volume = \( \pi r^2 h + \frac{2}{3} \pi r^3 \)
\( = \pi r^2(h + \frac{2}{3}r) \)
\( = \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4}(2 + \frac{2}{3} \times \frac{7}{4}) \)
\( = \frac{77}{8}(2 + \frac{7}{6}) \)
\( = \frac{77}{8} \times \frac{19}{6} \)
\( = \frac{1463}{48} \)
\( = 30.48 \) m³

In simple words: We added the volume of a cylinder and half of a sphere. The lid is like half a ball on top of a can.

📝 Teacher's Note: Show students a real can with a dome-shaped lid. The volume is cylinder plus hemisphere. Make sure they use the hemisphere formula, not full sphere.

🎯 Exam Tip: Write the hemisphere volume formula as (2/3)πr³, not (4/3)πr³. This is a common mistake that costs marks.

Question 6. A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylindrical part is \( 4\frac{2}{3} \) m and the diameter of hemisphere is 3.5 m. Calculate the capacity and the internal surface area of the vessel.
Answer:
Given:
Diameter of the base = 3.5 m
Therefore, radius = \( \frac{3.5}{2} \) m = 1.75 m = \( \frac{7}{4} \) m
Height of cylindrical part = \( 4\frac{2}{3} = \frac{14}{3} \) m

Step 1: Find capacity (volume) of the vessel.
Capacity = \( \pi r^2 h + \frac{2}{3} \pi r^3 = \pi r^2(h + \frac{2}{3}r) \)
\( = \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4}(\frac{14}{3} + \frac{2}{3} \times \frac{7}{4}) \)
\( = \frac{77}{8}(\frac{14}{3} + \frac{7}{6}) \)
\( = \frac{77}{8}(\frac{28 + 7}{6}) \)
\( = \frac{77}{8} \times \frac{35}{6} \)
\( = \frac{2695}{48} \)
\( = 56.15 \) m³

Step 2: Find internal curved surface area.
Internal curved surface area = \( 2\pi rh + 2\pi r^2 = 2\pi r(h + r) \)
\( = 2 \times \frac{22}{7} \times \frac{7}{4}(\frac{14}{3} + \frac{7}{4}) \)
\( = 11(\frac{56 + 21}{12}) \)
\( = 11 \times \frac{77}{12} \)
\( = \frac{847}{12} \)
\( = 70.58 \) m²

In simple words: This vessel is like a bucket with a rounded bottom. We found how much water it can hold and how much surface is inside.

📝 Teacher's Note: Draw a glass with a rounded bottom to help students visualize. The curved surface area includes cylinder wall plus hemisphere surface, but not the top opening.

🎯 Exam Tip: For hemisphere, use 2πr² for curved surface area, not 4πr². The flat circular base is not counted in internal surface area.

Question 7. A wooden toy is in the shape of a cone mounted on a cylinder. If the height of the cone is 24 cm, the total height of the toy is 60 cm and the radius of the base of the cone is twice the radius of the base of the cylinder = 10 cm, find the total surface area of the toy. [Take π = 3.14]
Answer:
Given:
Height of the cone = 24 cm
Height of the cylinder = 36 cm
Radius of the cone = twice the radius of the cylinder = 10 cm
Radius of the cylinder = 5 cm

Step 1: Find slant height of the cone.
Slant height = \( \sqrt{r^2 + h^2} \)
\( = \sqrt{10^2 + 24^2} \)
\( = \sqrt{100 + 576} \)
\( = \sqrt{676} \)
\( = 26 \) cm

Step 2: Find total surface area of the toy.
Surface area = curved area of the conical point + curved area of the cylinder
\( = \pi l r + \pi r^2 + 2\pi RH \)
\( = \pi[l r + r^2 + 2RH] \)
\( = 3.14[10 \times 26 + (10)^2 + 2 \times 5 \times 36] \)
\( = 3.14[260 + 100 + 360] \)
\( = 3.14[720] \)
\( = 2260.8 \) cm²

In simple words: This toy looks like an ice cream cone on top of a can. We added the curved surfaces of both parts, plus the bottom of the cylinder.

📝 Teacher's Note: Use an actual ice cream cone to show this shape. The flat top of cylinder is covered by the cone base, so we don't count it twice.

🎯 Exam Tip: Don't add the top circular area of cylinder because it's covered by the cone base. Only count exposed surfaces.

Question 8. A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm x 10.5 cm. Find the rise in level of the water when the solid is submerged.
Answer:
Given:
Diameter of cylindrical container = 42 cm
Radius = 21 cm
Dimensions of rectangular solid = 22 cm × 14 cm × 10.5 cm

Step 1: Find volume of the rectangular solid.
Volume of solid = 22 × 14 × 10.5 = 3234 cm³

Step 2: Find rise in water level.
When solid is submerged, it displaces water equal to its volume.
Volume displaced = Area of base × Rise in level
3234 = π × 21² × h
3234 = \( \frac{22}{7} \) × 441 × h
3234 = 22 × 63 × h
3234 = 1386 × h
h = \( \frac{3234}{1386} \) = 2.33 cm

Rise in water level = 2.33 cm

In simple words: When you put the iron block in water, it pushes water up. The volume of water pushed up equals the volume of the iron block.

📝 Teacher's Note: Demonstrate with a measuring cylinder and a stone. When stone goes in, water level rises. The rise shows the stone's volume.

🎯 Exam Tip: Remember Archimedes principle: Volume displaced = Volume of object submerged. Use this to set up your equation for rise in level.

Question 9. Spherical marbles of diameter 1.4 cm are dropped into beaker containing some water and are fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm.
Answer:
Given:
Diameter of spherical marble = 1.4 cm
Therefore, radius = 0.7 cm
Diameter of beaker = 7 cm
Therefore, radius = \( \frac{7}{2} \) cm
Height of water rise = 5.6 cm

Step 1: Find volume of one marble.
\( \text{Volume of one ball} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.7)^3 \text{ cm}^3 \)

Step 2: Find volume of water displaced.
\( \text{Volume of water} = \pi r^2 h = \pi \times \left(\frac{7}{2} \times \frac{7}{2} \times 5.6\right) \text{cm}^3 = \pi \times \frac{49 \times 56}{4 \times 10} \text{cm}^3 \)

Step 3: Find number of balls dropped.
\( \text{No. of balls} = \frac{\pi \times 49 \times 56 \times 3}{4 \times 10 \times 4\pi \times (0.7)^3} = 150 \)

Number of marbles dropped = 150
In simple words: When marbles go into water, they push the water up. The volume of water pushed up equals the total volume of all marbles. We divide total water volume by one marble volume to get the count.

📝 Teacher's Note: Show students with a glass of water and some stones. When stones go in, water level rises. This rise tells us how much space the stones take.

🎯 Exam Tip: Always find volume of one marble first, then total water displaced. Write the formula for sphere volume clearly. Check your arithmetic twice.

Question 10. The cross-section of a railway tunnel is a rectangle 6 m broad and 8 m high surmounted by a semi-circle as shown in the figure. The tunnel is 35 m long. Find the cost of plastering the internal surface of the tunnel (excluding the floor) at the rate of Rs 2.25 per m².
Answer:
Given:
Breadth of the tunnel = 6 m
Height of the tunnel = 8 m
Length of the tunnel = 35 m
Radius of the semi-circle = 3 m

Step 1: Find circumference of semi-circle.
\( \text{Circumference of semi-circle} = \pi r = \frac{22}{7} \times 3 = \frac{66}{7} \text{ m} \)

Step 2: Find internal surface area (excluding floor).
\( \text{Internal surface area} = 35\left(8 + 8 + \frac{66}{7}\right) \)
\( = 35\left(16 + \frac{66}{7}\right) \)
\( = 35\left(\frac{112 + 66}{7}\right) \)
\( = 35 \times \frac{178}{7} = 890 \text{ m}^2 \)

Step 3: Calculate total cost.
Rate of plastering = Rs 2.25 per m²
\( \text{Total cost} = \text{Rs } 890 \times \frac{225}{100} = \text{Rs } 890 \times \frac{9}{4} = \text{Rs } \frac{8010}{4} = \text{Rs } 2002.50 \)

Cost of plastering = Rs 2002.50
In simple words: The tunnel has walls and a curved roof but no floor to plaster. We find the area of two side walls plus curved roof, then multiply by length and cost per square meter.

📝 Teacher's Note: Draw the tunnel cross-section on board. Show students that we need two rectangles (side walls) plus one semi-circle (curved roof). Floor is not plastered.

🎯 Exam Tip: Remember "excluding floor" means only walls and roof. Calculate perimeter first, then multiply by length. Always write units - Rs for money, m² for area.

Question 11. The horizontal cross-section of a water tank is in the shape of a rectangle with semicircle at one end, as shown in the following figure. The water is 2.4 metres deep in the tank. Calculate the volume of water in the tank in gallons. (Given: 1 gallon = 4.5 litres)
Answer:
[Diagram: This diagram shows the top view of a water tank with rectangular part (21m × 7m) and semicircular part at one end with radius 3.5m]
Given:
Length = 21 m
Depth of water = 2.4 m
Breadth = 7 m
Therefore, radius of semicircle = \( \frac{7}{2} \) m

Step 1: Find area of cross-section.
\( \text{Area} = l \times b + \frac{1}{2} \pi r^2 \)
\( = 21 \times 7 + \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
\( = 147 + \frac{77}{4} \)
\( = \frac{588 + 77}{4} = \frac{665}{4} \text{ m}^2 \)

Step 2: Find volume of water.
\( \text{Volume} = \frac{665}{4} \times 2.4 \text{ m}^3 = 665 \times 0.6 = 399 \text{ m}^3 \)

Step 3: Convert to gallons.
\( = 399 \times 100^3 \text{ cm}^3 \)
\( = \frac{399 \times 100 \times 100 \times 100}{1000} \text{ litres} \)
\( = \frac{399 \times 100 \times 100 \times 100}{1000 \times 4.5} \text{ gallons} \)
\( = \frac{399 \times 100 \times 100 \times 100 \times 10}{1000 \times 45} \text{ gallons} \)
\( = \frac{1330000}{15} \text{ gallons} = \frac{266000}{3} \text{ gallons} = 88666.67 \text{ gallons} \)

Volume of water = 88666.67 gallons
In simple words: The tank has a rectangle plus half-circle shape when seen from top. We find this area, multiply by water depth to get volume, then convert cubic meters to gallons using the given conversion.

📝 Teacher's Note: Help students see that this is rectangle area plus semicircle area. Draw it clearly. The depth is the same everywhere - 2.4m deep water.

🎯 Exam Tip: First find area of cross-section (rectangle + semicircle). Then multiply by depth. Convert units step by step: m³ → cm³ → litres → gallons. Show all conversion steps.

Question 12. The given figure shows the cross-section of a water channel consisting of a rectangle and a semicircle. Assuming that the channel is always full, find the volume of water discharged through it in one minute if water is flowing at the rate of 20 cm per second. Give your answer in cubic meters correct to one place of decimal.
Answer:
[Diagram: This diagram shows a water channel cross-section with rectangular part (21cm wide, 7cm high) and semicircular part at bottom with radius 10.5cm]
Given:
Length = 21 cm, Breadth = 7 cm
Radius of semicircle = \( \frac{21}{2} \) cm
Flow rate = 20 cm per second

Step 1: Find area of cross-section.
\( \text{Area} = l \times b + \frac{1}{2} \pi r^2 \)
\( = 21 \times 7 + \frac{1}{2} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \)
\( = 147 + \frac{693}{4} \)
\( = \frac{588 + 693}{4} = \frac{1281}{4} \text{ cm}^2 \)

Step 2: Find length of water column in one minute.
Flow rate = 20 cm per second
Length of water column = 20 × 60 = 1200 cm

Step 3: Find volume of water discharged.
\( \text{Volume} = \frac{1281}{4} \times 1200 \text{ cm}^3 = 384300 \text{ cm}^3 \)
\( = \frac{384300}{100 \times 100 \times 100} \text{ m}^3 = 0.3843 \text{ m}^3 = 0.4 \text{ m}^3 \)

Volume of water discharged = 0.4 m³
In simple words: Water flows through a tube that has rectangle plus semicircle shape. In one minute, a column of water 1200 cm long flows through. We multiply this length by the cross-section area to get volume.

📝 Teacher's Note: Use a garden hose example. If water flows at 20 cm/sec, in 60 seconds it travels 1200 cm. The volume is like a long cylinder with special cross-section shape.

🎯 Exam Tip: Find cross-section area first (rectangle + semicircle). Then find how far water travels in one minute. Multiply area × length. Convert cm³ to m³ by dividing by 1000000.

Question 13. An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is \( 3\frac{1}{2} \) cm and height 8 cm. Find the volume of water required to fill the vessel. If this cone is replaced by another cone, whose height is \( 1\frac{3}{4} \) cm and radius of whose base is 2 cm, find the drop in the water level.
Answer:
Given:
Cylindrical vessel: diameter = 7 cm, height = 8 cm
First cone: base diameter = \( 3\frac{1}{2} \) cm = \( \frac{7}{2} \) cm, height = 8 cm
Second cone: height = \( 1\frac{3}{4} \) cm = \( \frac{7}{4} \) cm, radius = 2 cm

Step 1: Find radius of cylinder
Radius of cylinder = \( \frac{7}{2} \) cm

Step 2: Find volume of cylinder
Volume of cylinder = \( \pi r^2 h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 8 = 308 \text{ cm}^3 \)

Step 3: Find volume of first cone
Base diameter = \( \frac{7}{2} \) cm
Radius of first cone = \( \frac{7}{4} \) cm
Volume of first cone = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times 8 = \frac{77}{3} \text{ cm}^3 \)

Step 4: Find remaining volume for water
Volume for water = \( 308 - \frac{77}{3} = \frac{924 - 77}{3} = \frac{847}{3} = 282.33 \text{ cm}^3 \)

Step 5: Find volume of second cone
Height = \( \frac{7}{4} \) cm, radius = 2 cm
Volume of second cone = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 2 \times 2 \times \frac{7}{4} = \frac{22}{3} \text{ cm}^3 \)

Step 6: Find volume of water displaced
Volume displaced = \( \frac{77}{3} - \frac{22}{3} = \frac{55}{3} \text{ cm}^3 \)

Step 7: Find drop in water level
Let h be the height of water dropped
Volume displaced = Area of base × h
\( \frac{55}{3} = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h = \frac{77}{2} h \)
Therefore: \( h = \frac{55}{3} \times \frac{2}{77} = \frac{10}{21} \text{ cm} \)

Water required to fill vessel = 282.33 cm³
Drop in water level = \( \frac{10}{21} \) cm
In simple words: First we find how much space is left in the cylinder after putting the first cone. This space is filled with water. When we replace the first cone with second cone, some water comes down because the second cone is smaller.

📝 Teacher's Note: Show students with actual containers and objects. When we put a small object in place of a big object, the water level goes down. The difference in volumes tells us how much.

🎯 Exam Tip: Always find volume of cylinder first, then volume of cone, then subtract. For water level change, use the formula: Volume change = Base area × height change.

Question 14. A cylindrical can, whose base is horizontal and of radius 3.5 cm, contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate:
(i) the total surface area of the can in contact with water when the sphere is in it;
(ii) the depth of water in the can before the sphere was put into the can.
Answer:

[Diagram: A cylindrical can with a sphere fitting exactly inside it. The sphere touches the bottom and the curved sides of the cylinder, and water just covers the sphere.]

Given:
Radius of cylindrical can = 3.5 cm
Sphere just fits in the can (diameter of sphere = diameter of can = 7 cm)
Water just covers the sphere

Step 1: Find dimensions
Since sphere just fits: radius of sphere = 3.5 cm
Height of water = diameter of sphere = 7 cm

Step 2: Find total surface area in contact with water
Surface area = Base area + Curved surface area
= \( \pi r^2 + 2\pi rh \)
= \( \frac{22}{7} \times 3.5 \times 3.5 + 2 \times \frac{22}{7} \times 3.5 \times 7 \)
= \( \frac{77}{2} + 154 = 38.5 + 154 = 192.5 \text{ cm}^2 \)

Step 3: Find depth of water before sphere was added
Let depth of water = x cm
Volume of water + Volume of sphere = Volume of cylinder up to water level
\( \pi r^2 x + \frac{4}{3}\pi r^3 = \pi r^2 h \)
\( \pi r^2 x + \frac{4}{3}\pi r^3 = \pi r^2(x + \frac{4}{3}r) \)
\( h = x + \frac{4}{3}r \)
\( x = h - \frac{4}{3}r = 7 - \frac{4}{3} \times 3.5 = 7 - \frac{14}{3} = \frac{21-14}{3} = \frac{7}{3} = 2\frac{1}{3} \text{ cm} \)

(i) Total surface area = 192.5 cm²
(ii) Depth of water before sphere = 2⅓ cm
In simple words: The sphere pushes up the water level. Before the sphere was added, the water was lower. We calculate how much lower by using the volume of the sphere.

📝 Teacher's Note: Use a ball and a glass of water to show this. When you put the ball in, the water level rises. The ball takes up space, so there was less water before.

🎯 Exam Tip: Remember that sphere volume = (4/3)πr³. For surface area, don't forget both base and curved surface. Write all formulas clearly before substituting.

Question 15. A hollow cylinder has solid hemisphere inward at one end and on the other end it is closed with a flat circular plate. The height of water is 10 cm when flat circular surface is downward. Find the level of water, when it is inverted upside down, common diameter is 7 cm and height of the cylinder is 20 cm.
Answer:

[Diagram: A hollow cylinder with a hemisphere bulging inward at one end and flat at the other end. Shows the cylinder in both positions - flat end down, then hemisphere end down.]

Given:
Diameter = 7 cm, so radius = 3.5 cm
Height of cylinder = 20 cm
Water height when flat end is down = 10 cm

Step 1: Find volume of water
Volume of water = \( \pi r^2 h = \pi \left(\frac{7}{2}\right)^2 \times 10 = \frac{22}{7} \times \frac{49}{4} \times 10 = \frac{1540}{4} = 385 \text{ cm}^3 \)

Step 2: Find volume of hemisphere
Volume of hemisphere = \( \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 = \frac{2}{3} \times \frac{22}{7} \times \frac{343}{8} = \frac{539}{6} \text{ cm}^3 \)

Step 3: Find water level when inverted
Let new water height = h cm
When hemisphere end is down:
Volume of water = Volume of cylinder up to height h - Volume of hemisphere
\( 385 = \pi r^2 h - \frac{539}{6} \)
\( 385 = \frac{22}{7} \times \frac{49}{4} \times h - \frac{539}{6} \)
\( 385 + \frac{539}{6} = \frac{22 \times 49}{28} \times h \)
\( \frac{2310 + 539}{6} = \frac{77h}{2} \)
\( \frac{2849}{6} = \frac{77h}{2} \)
\( h = \frac{2849 \times 2}{6 \times 77} = \frac{5698}{462} = 12\frac{1}{3} \text{ cm} \)

Water level when inverted = 12⅓ cm
In simple words: When we flip the cylinder, the hemisphere takes up space at the bottom. So the same amount of water spreads higher because there is less space at the bottom.

📝 Teacher's Note: Use two identical bottles - one with a bump at the bottom, one flat. Put the same amount of water in both. The water level will be higher in the bottle with the bump.

🎯 Exam Tip: Volume of water stays the same in both positions. Calculate hemisphere volume carefully using (2/3)πr³. Set up the equation properly for the inverted case.

Exercise 20 G

Question 1. What is the least number of solid metallic spheres, each of 6 cm diameter, that should be melted and recast to form a solid metal cone whose height is 45 cm and diameter is 12 cm?
Answer:
Given:
Sphere diameter = 6 cm, so radius = 3 cm
Cone height = 45 cm, diameter = 12 cm, so radius = 6 cm

Step 1: Find volume of one sphere
Volume of 1 sphere = \( \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi \times 27 = 36\pi \text{ cm}^3 \)

Step 2: Find volume of cone
Volume of cone = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 6^2 \times 45 = \frac{1}{3} \pi \times 36 \times 45 = 540\pi \text{ cm}^3 \)

Step 3: Find number of spheres needed
Number of spheres = \( \frac{\text{Volume of cone}}{\text{Volume of 1 sphere}} = \frac{540\pi}{36\pi} = \frac{540}{36} = 15 \)

The least number of spheres needed = 15
In simple words: We melt 15 small spheres and reshape the metal to make one big cone. The total amount of metal stays the same.

📝 Teacher's Note: Explain that when we melt and recast, volume stays the same. It's like taking 15 small balls of clay and making one big cone from them.

🎯 Exam Tip: Write the volume formulas first. Sphere volume = (4/3)πr³, Cone volume = (1/3)πr²h. Divide cone volume by sphere volume to get the answer.

Question 2. A largest sphere is to be carved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere. (Answer correct to the nearest integer)
Answer:
Given:
Cylinder: radius = 7 cm, height = 14 cm

Step 1: Find radius of largest sphere
For the largest sphere that fits inside the cylinder:
Radius of sphere = radius of cylinder = 7 cm
(The sphere diameter = cylinder diameter, so sphere just touches the curved surface)

Step 2: Find volume of sphere
Volume of sphere = \( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 = \frac{4 \times 22 \times 343}{3 \times 7} = \frac{4312}{3} = 1437.33 \text{ cm}^3 \)

Volume of sphere = 1437 cm³ (nearest integer)
In simple words: The biggest ball that fits inside the cylinder has the same width as the cylinder. Its height also fits because 14 > 2×7.

📝 Teacher's Note: Show that the sphere touches the cylinder walls on all sides. The sphere diameter equals cylinder diameter. Check that sphere height (2r = 14) fits in cylinder height.

🎯 Exam Tip: For largest sphere in cylinder, radius of sphere = radius of cylinder. Use π = 22/7 for calculations. Round to nearest integer as asked.

Question 3. A right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled in identical cones of height 12 cm and diameter 6 cm having a semi-spherical shape on the top. Find the number of cones required.
Answer:
Given:
Cylinder: diameter = 12 cm, radius = 6 cm, height = 15 cm
Cone: diameter = 6 cm, radius = 3 cm, height = 12 cm
Each cone has a semi-spherical top

Step 1: Find volume of cylinder
Volume of cylinder = \( \pi r^2 h = \pi \times 6^2 \times 15 = 540\pi \) cm³

Step 2: Find volume of one cone with semi-sphere
Volume of cone = \( \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 3^2 \times 12 = 36\pi \) cm³
Volume of semi-sphere = \( \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi \times 3^3 = 18\pi \) cm³
Total volume of one cone = \( 36\pi + 18\pi = 54\pi \) cm³

Step 3: Find number of cones required
Number of cones = \( \frac{\text{Volume of cylinder}}{\text{Volume of one cone}} = \frac{540\pi}{54\pi} = 10 \)

Number of cones required = 10
In simple words: We divide the total ice-cream volume by the volume of one cone (including its round top). This gives us how many cones we can fill.

📝 Teacher's Note: Remind students that each cone has a semi-sphere on top. So the volume is cone + half of a sphere. Draw a cone with an ice-cream scoop on top to help them see.

🎯 Exam Tip: Always add the cone volume and semi-sphere volume together first. Then divide total cylinder volume by this combined volume. Show all steps clearly.

Question 4. A solid is in the form of a cone standing on a hemisphere with both their radii being equal to 8 cm and the height of cone is equal to its radius. Find in terms of π, the volume of the solid.
Answer:
Given:
Radius of hemisphere = radius of cone = 8 cm
Height of cone = radius = 8 cm

Step 1: Find volume of hemisphere
Volume of hemisphere = \( \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi \times 8^3 = \frac{2}{3}\pi \times 512 = \frac{1024\pi}{3} \) cm³

Step 2: Find volume of cone
Volume of cone = \( \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 8^2 \times 8 = \frac{1}{3}\pi \times 64 \times 8 = \frac{512\pi}{3} \) cm³

Step 3: Find total volume
Total volume = Volume of hemisphere + Volume of cone
= \( \frac{1024\pi}{3} + \frac{512\pi}{3} = \frac{1536\pi}{3} = 512\pi \) cm³

Volume of the solid = 512π cm³
In simple words: The solid has two parts - a round ball cut in half (hemisphere) and a cone sitting on top. We add both volumes together.

📝 Teacher's Note: Use a real ice-cream cone to show this shape. The cone sits on top of a half ball. Students can see this combination easily.

🎯 Exam Tip: Write the formula for hemisphere as half of a sphere volume. Always add the two volumes at the end. Keep π in the final answer when asked for.

Question 5. The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of wire.
Answer:
Given:
Sphere diameter = 6 cm, radius = 3 cm
Wire diameter = 0.2 cm, radius = 0.1 cm

Step 1: Find volume of sphere
Volume = \( \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times 3^3 = \frac{4}{3}\pi \times 27 = 36\pi \) cm³

Step 2: Find volume of wire in terms of length
Let length of wire = h cm
Volume of wire = \( \pi r^2 h = \pi \times (0.1)^2 \times h = \pi \times 0.01 \times h = 0.01\pi h \) cm³

Step 3: Use the fact that volumes are equal
Volume of sphere = Volume of wire
\( 36\pi = 0.01\pi h \)
\( h = \frac{36\pi}{0.01\pi} = \frac{36}{0.01} = 3600 \) cm = 36 m

Length of wire = 36 m
In simple words: When we melt the sphere, no material is lost. So the volume stays the same. We use this to find how long the thin wire will be.

📝 Teacher's Note: Show students how a ball of clay can be rolled into a long thin rope. The amount of clay stays the same, but the shape changes. This is the same idea.

🎯 Exam Tip: Always write "Volume of sphere = Volume of wire" clearly. Convert the final answer to meters if it's a very big number in cm. Show this conversion step.

Question 6. Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.
Answer:
Given:
A sphere fits exactly inside a cube
Let edge of cube = a

Step 1: Find volume of cube
Volume of cube = \( a^3 \)

Step 2: Find radius of sphere
When sphere fits exactly inside cube, diameter of sphere = edge of cube
So radius of sphere = \( \frac{a}{2} \)

Step 3: Find volume of sphere
Volume of sphere = \( \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times \left(\frac{a}{2}\right)^3 = \frac{4}{3}\pi \times \frac{a^3}{8} = \frac{\pi a^3}{6} \)

Step 4: Find the ratio
Ratio = \( \frac{\text{Volume of cube}}{\text{Volume of sphere}} = \frac{a^3}{\frac{\pi a^3}{6}} = \frac{a^3 \times 6}{\pi a^3} = \frac{6}{\pi} \)

Ratio = 6 : π
In simple words: The sphere touches all six faces of the cube from inside. The cube's volume is bigger than the sphere's volume by this ratio.

📝 Teacher's Note: Put a ball inside a box to show how it touches all faces. The ball's diameter equals the box's side length. This visual makes the concept clear.

🎯 Exam Tip: Remember that when a sphere fits exactly in a cube, diameter of sphere = edge of cube. Write this relationship first, then find the volumes.

Question 7. An iron pole consisting of a cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1 cm³ of iron has 8 gm of mass (approx). (Take π = 355/113)
Answer:
Given:
Cylindrical part: height = 110 cm, diameter = 12 cm, radius = 6 cm
Conical part: height = 9 cm, radius = 6 cm
Density of iron = 8 gm/cm³
π = 355/113

Step 1: Find volume of cylindrical part
Volume = \( \pi r^2 h = \frac{355}{113} \times 6^2 \times 110 = \frac{355}{113} \times 36 \times 110 \) cm³

Step 2: Find volume of conical part
Volume = \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{355}{113} \times 6^2 \times 9 = \frac{355}{113} \times 36 \times 3 \) cm³

Step 3: Find total volume
Total volume = \( \frac{355}{113} \times 36 \times 110 + \frac{355}{113} \times 36 \times 3 \)
= \( \frac{355}{113} \times 36 \times (110 + 3) = \frac{355}{113} \times 36 \times 113 \)
= 355 × 36 = 12780 cm³

Step 4: Find mass
Mass = Volume × Density = 12780 × 8 = 102240 gm = 102.24 kg

Mass of the pole = 102.24 kg
In simple words: The pole has two parts - a cylinder (main body) and a cone (top part). We find both volumes, add them, then multiply by the weight per cm³.

📝 Teacher's Note: Show students a flagpole or a pencil with a cone-shaped tip. This helps them visualize the cylinder + cone combination. The math is just adding two volumes.

🎯 Exam Tip: Always convert the final answer from grams to kilograms by dividing by 1000. Write "Mass = Volume × Density" clearly. Show the addition of both volumes in one step.

Question 8. In the following diagram a rectangular platform with a semicircular end on one side is 22 meters long from one end to the other end. If the length of the half circumference is 11 meters, find the cost of constructing the platform, 1.5 meters high at the rate of Rs 4 per cubic meters.
Answer:

[Diagram: This diagram shows a rectangular platform with a semicircular end attached to one side, total length 22 meters.]

Given:
Total length = 22 m
Half circumference of semicircle = 11 m
Height = 1.5 m
Rate = Rs 4 per m³

Step 1: Find radius of semicircle
Half circumference = \( \pi r = 11 \)
Radius = \( \frac{11}{\pi} = \frac{11 \times 7}{22} = \frac{7}{2} = 3.5 \) m

Step 2: Find breadth and length of rectangular part
Breadth of rectangle = diameter of semicircle = 2 × 3.5 = 7 m
Length of rectangle = 22 - 3.5 = 18.5 m

Step 3: Find area of platform
Area = Area of rectangle + Area of semicircle
= \( 18.5 \times 7 + \frac{1}{2}\pi r^2 = 129.5 + \frac{1}{2} \times \frac{22}{7} \times (3.5)^2 \)
= \( 129.5 + \frac{77}{4} = 129.5 + 19.25 = 148.75 \) m²

Step 4: Find volume
Volume = Area × Height = 148.75 × 1.5 = 223.125 m³

Step 5: Find cost
Cost = Volume × Rate = 223.125 × 4 = Rs 892.50

Cost of construction = Rs 892.50
In simple words: The platform is like a rectangle with half a circle stuck to one end. We find the total area, then multiply by height and rate.

📝 Teacher's Note: Draw a rectangle on the board and attach a semicircle to one end. This shape is common in stadiums and playgrounds. Students can visualize it easily.

🎯 Exam Tip: Always find the radius first from the given circumference. Then find the rectangular length by subtracting radius from total length. Show area calculation step by step.

Question 9. The cross-section of a tunnel is a square of side 7 m surmounted by a semicircle as shown in the following figure. The tunnel is 80 m long. Calculate: (i) its volume (ii) the surface area of the tunnel (excluding the floor) and (iii) its floor area
Answer:

[Diagram: This diagram shows a square with a semicircle on top, forming the cross-section of a tunnel, with side length 7 meters.]

Given:
Side of square = 7 m
Radius of semicircle = 3.5 m
Length of tunnel = 80 m

Step 1: Find area of cross-section
Area = Area of square + Area of semicircle
= \( 7^2 + \frac{1}{2}\pi r^2 = 49 + \frac{1}{2} \times \frac{22}{7} \times (3.5)^2 \)
= \( 49 + \frac{77}{4} = 49 + 19.25 = 68.25 \) m²

(i) Volume of tunnel
Volume = Area of cross-section × Length = 68.25 × 80 = 5460 m³

(ii) Surface area (excluding floor)
Surface area = Perimeter of cross-section × Length - Floor area
Perimeter = 3 sides of square + semicircle circumference
= \( 3 \times 7 + \pi \times 3.5 = 21 + 11 = 32 \) m
Surface area = 32 × 80 = 2560 m²

(iii) Floor area
Floor area = Width × Length = 7 × 80 = 560 m²

Answers: (i) 5460 m³ (ii) 2560 m² (iii) 560 m²
In simple words: The tunnel is like a long tube with a special cross-section. We find the area of the front face, then multiply by length to get volume.

📝 Teacher's Note: Use a cardboard tube or show a real tunnel picture. The cross-section is like looking at the tunnel from the front. The volume is this area stretched over the length.

🎯 Exam Tip: For surface area, count only 3 sides of the square (not the floor) plus the semicircle. For floor area, use only the bottom side of the square times the length.

Question 10. A cylindrical water tank of diameter 2.8m and height 4.2m is being fed by a pipe of diameter 7 cm through which water flows at the rate of 4m/s. Calculate, in minutes, the time it takes to fill the tank.
Answer:
Given:
Diameter of cylindrical tank = 2.8 m
Therefore, radius = 1.4 m
Height = 4.2 m

Step 1: Find volume of tank.
Volume of water filled in it = \( \pi r^2 h \)
\[ = \frac{22}{7} \times 1.4 \times 1.4 \times 4.2 \text{ m}^3 \]
\[ = \frac{181.104}{7} \text{ m}^3 \]
\[ = 25.872 \text{ m}^3 \]

Step 2: Find pipe details.
Diameter of pipe = 7 cm
Therefore, radius (r) = \( \frac{7}{2} \) cm
Let length of water in the pipe = h₁

Step 3: Find volume of water flowing per second.
Volume = \( \pi r^2 h_1 \)
\[ = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h_1 \]
\[ = \frac{77}{2} h_1 \text{ cm}^3 \]

Step 4: Use flow rate to find h₁.
From the given information:
\[ \frac{77}{2} h_1 \text{ cm}^3 = 25.872 \times 10^6 \text{ cm}^3 \]
\[ h_1 = \frac{25.872 \times 10^6 \times 2}{77} \text{ cm} \]
\[ h_1 = \frac{25.872 \times 10^6 \times 2}{77 \times 100} \text{ m} \]
\[ h_1 = 0.672 \times 10^2 \text{ m} \]
\[ h_1 = 6720 \text{ m} \]

Step 5: Calculate time.
Therefore, time taken at the speed of 4 m per second
\[ = \frac{6720}{4 \times 60} \text{ minutes} = 28 \text{ minutes} \]

In simple words: We find how much water the tank can hold. Then we find how fast water flows through the pipe. Finally we divide total water by flow rate to get time.

📝 Teacher's Note: Make sure students convert all units to the same system first. A common mistake is mixing meters and centimeters without converting properly.

🎯 Exam Tip: Always write "Given:" clearly and list all values with units. Show each step of volume calculation separately. Write final answer with correct units (minutes).

Question 11. Water flows, at 9 km per hour, through a cylindrical pipe of cross-sectional area 25 cm². If this water is collected into a rectangular cistern of dimensions 7.5m by 5m by 4m; calculate the rise in level in the cistern in 1 hour 15 minutes.
Answer:
Given:
Rate of flow of water = 9 km/hr
Water flow in 1 hour 15 minutes
i.e. in \( \frac{5}{4} \) hr = \( 9 \times \frac{5}{4} = \frac{45}{4} \) km = \( \frac{45}{4} \times 1000 = 11250 \) m

Step 1: Find cross-sectional area in m².
Area of cross-section = 25 cm² = \( \frac{25}{10000} \) m² = \( \frac{1}{400} \) m²

Step 2: Find volume of water.
Therefore, volume of water = \( \frac{1}{400} \times 11250 = 28.125 \) m³

Step 3: Find tank dimensions.
Dimensions of water tank = 7.5m × 5m × 4m
Area of tank = l × b = 7.5 × 5 = 37.5 m²

Step 4: Calculate rise in level.
Let h be the height of water then,
37.5 × h = 28.125
\[ h = \frac{28.125}{37.5} = 0.75 \text{ m} = 75 \text{ cm} \]

In simple words: Water flows through a pipe and fills a tank. We find how much water flows in the given time, then divide by tank area to get height rise.

📝 Teacher's Note: Students often forget to convert km to m and cm² to m². Make them practice unit conversions separately before solving volume problems.

🎯 Exam Tip: Always convert all measurements to the same unit system first. Show the conversion steps clearly. Write "rise in level = 75 cm" as the final answer.

Question 12. The given figure shows the cross-section of a cone, a cylinder and a hemisphere all with the same diameter 10 cm and the other dimensions are as shown. Calculate: (i) the total surface area (ii) the total volume of the solid (iii) the density of the material if its total weight is 1.7 kg
Answer:

[Diagram: This diagram shows a composite solid with a cone, cylinder, and hemisphere, all with diameter 10 cm. The cone and cylinder each have height 12 cm.]

Given:
Diameter = 10 cm
Therefore, radius (r) = 5 cm
Height of the cone (h) = 12 cm
Height of the cylinder = 12 cm

Step 1: Find slant height of cone.
\( l = \sqrt{h^2 + r^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \) cm

Step 2: Calculate total surface area.
(i) Total surface area of the solid
= \( \pi r^2 + 2\pi rh + 2\pi r^2 \)
= \( \pi r(l + 2h + 2r) \)
\[ = \frac{22}{7} \times 5[13 + (2 \times 12) + (2 \times 5)] \]
\[ = \frac{110}{7}[13 + 24 + 10] \]
\[ = \frac{110}{7} \times 47 \]
\[ = \frac{5170}{7} \]
= 738.57 cm²

Step 3: Calculate total volume.
(ii) Total volume of the solid
\[ = \frac{1}{3}\pi r^2h + \pi r^2h + \frac{2}{3}\pi r^3 \]
\[ = \pi r^2\left[\frac{1}{3}h + h + \frac{2}{3}r\right] \]
\[ = \frac{22}{7} \times 5 \times 5\left[\frac{1}{3} \times 12 + 12 + \frac{2}{3} \times 5\right] \]
\[ = \frac{550}{7}\left[4 + 12 + \frac{10}{3}\right] \]
\[ = \frac{550}{7}\left[16 + \frac{10}{3}\right] \]
\[ = \frac{550}{7} \times \frac{58}{3} \]
\[ = \frac{31900}{21} \]
= 1519.0476 cm³

Step 4: Calculate density.
(iii) Total weight of the solid = 1.7 kg
\[ \text{Density} = \frac{1.7 \times 1000}{1519.0476} \text{ gm/cm}^3 = 1.119 \text{ gm/cm}^3 \]
⇒ Density = 1.12 gm/cm³

In simple words: This solid has three parts joined together. We add up surface areas and volumes of all three parts. Then we use weight to find density.

📝 Teacher's Note: Draw each shape separately first. Students get confused when shapes are combined. Remind them that hemisphere surface area is only the curved part, not the flat circle.

🎯 Exam Tip: For composite solids, always identify each part clearly. Write formulas for cone, cylinder, and hemisphere separately. Round final answers to 2 decimal places.

Question 13. A solid, consisting of a right circular cone, standing on a hemisphere, is placed upright, in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is 3 cm and its height is 6 cm; the radius of the hemisphere is 2 cm and the height of the cone is 4 cm. Give your answer to the nearest cubic centimeter.
Answer:

[Diagram: This diagram shows a cylinder containing a composite solid made of a hemisphere with a cone on top.]

Given:
Radius of cylinder = 3 cm
Height of cylinder = 6 cm
Radius of hemisphere = 2 cm
Height of cone = 4 cm

Step 1: Find volume of cylinder when full.
Volume of water in the cylinder when it is full =
\( \pi r^2h = \pi \times 3 \times 3 \times 6 = 54\pi \) cm³

Step 2: Find volume of displaced water.
Volume of water displaced = volume of cone + volume of hemisphere
\[ = \frac{1}{3}\pi r^2h + \frac{2}{3}\pi r^3 \]
\[ = \frac{1}{3}\pi r^2(h + 2r) \]
\[ = \frac{1}{3}\pi \times 2 \times 2(4 + 2 \times 2) \]
\[ = \frac{1}{3}\pi \times 4 \times 8 \]
\[ = \frac{32}{3}\pi \text{ cm}^3 \]

Step 3: Find volume of water left.
Therefore, volume of water which is left
\[ = 54\pi - \frac{32}{3}\pi \]
\[ = \frac{130}{3}\pi \text{ cm}^3 \]
\[ = \frac{130}{3} \times \frac{22}{7} \text{ cm}^3 \]
\[ = \frac{2860}{21} \text{ cm}^3 \]
= 136.19 cm³
= 136 cm³

In simple words: A solid is put into a full water cylinder. The solid pushes out some water. We find how much water is left by subtracting the solid's volume from the cylinder's volume.

📝 Teacher's Note: Use a real example with a glass of water and a stone. When you put the stone in, water overflows. The amount that overflows equals the stone's volume.

🎯 Exam Tip: Always write "Volume left = Volume of cylinder - Volume of solid". Show the calculation clearly. Round to nearest whole number when asked.

Question 14. A metal container in the form of a cylinder is surmounted by a hemisphere of the same radius. The internal height of the cylinder is 7 m and the internal radius is 3.5 m. Calculate:
(i) the total area of the internal surface, excluding the base;
(ii) the internal volume of the container in \( m^3 \).
Answer:

[Diagram: This diagram shows a cylinder with a hemisphere on top, with radius 3.5 m and cylinder height 7 m.]

Given:
Radius of the cylinder = 3.5 m
Height = 7 m

Step 1: Find total surface area excluding the base.
Total surface area of container excluding the base = Curved surface area of the cylinder + area of hemisphere
\[ = 2\pi rh + 2\pi r^2 \]
\[ = \left(2 \times \frac{22}{7} \times 3.5 \times 7\right) + \left(2 \times \frac{22}{7} \times 3.5 \times 3.5\right) \]
= 154 + 77 \( m^2 \)
= 231 \( m^2 \)

Step 2: Find volume of the container.
Volume of the container = \( \pi r^2 h + \frac{2}{3}\pi r^3 \)
\[ = \left(\frac{22}{7} \times 3.5 \times 3.5 \times 7\right) + \left(\frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5\right) \]
\[ = \frac{539}{2} + \frac{539}{6} \]
\[ = \frac{1617 + 539}{6} \]
\[ = \frac{2156}{6} \]
= 359.33 \( m^3 \)

In simple words: We found the surface area by adding curved cylinder surface and hemisphere surface. Then we found volume by adding cylinder volume and hemisphere volume.

📝 Teacher's Note: Draw a cylinder with a ball on top. Show students that we don't count the base because it's covered. The hemisphere is half a ball.

🎯 Exam Tip: Always write formulas first. For hemisphere, use \( \frac{2}{3}\pi r^3 \) for volume and \( 2\pi r^2 \) for surface area. Don't forget units.

Question 15. An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for fold and for stitching. Give your answer to nearest \( m^2 \).
Answer:

[Diagram: This diagram shows a cylinder with a cone on top, with total height 85 m, cylindrical part 50 m, and base diameter 168 m.]

Given:
Total height of the tent = 85 m
Diameter of the base = 168 m
Therefore, radius (r) = 84 m
Height of the cylindrical part = 50 m
Then height of the conical part = (85 - 50) = 35 m

Step 1: Find slant height of cone.
Slant height (l) = \( \sqrt{r^2 + h^2} = \sqrt{84^2 + 35^2} = \sqrt{7056 + 1225} = \sqrt{8281} = 91 \) m

Step 2: Find total surface area.
Total surface area of the tent = \( 2\pi rh + \pi rl = \pi r(2h + l) \)
\[ = \frac{22}{7} \times 84(2 \times 50 + 91) \]
= 264(100 + 91)
= 264 × 191
= 50424 \( m^2 \)

Step 3: Add 20% extra for folds and stitching.
Total area of canvas needed
\[ = 50424 \times \frac{120}{100} \]
= 60508.8
= 60509 \( m^2 \)

In simple words: We found the surface area of cylinder and cone together. Then we added 20% extra material for folding and sewing the tent.

📝 Teacher's Note: Explain that tents need extra cloth for folding edges and stitching. Show how to find slant height using Pythagoras theorem with a right triangle.

🎯 Exam Tip: Don't forget the 20% extra. Calculate slant height first for cone surface area. Write final answer to nearest whole number as asked.

Question 16. A test tube consists of a hemisphere and a cylinder of the same radius. The volume of the water required to fill the whole tube is \( \frac{5159}{6} cm^3 \) and \( \frac{4235}{6} cm^3 \) of water are required to fill the tube to a level which is 4 cm below the top of the tube. Find the radius of the tube and the length of its cylindrical part.
Answer:
Step 1: Set up the equations.
Let r = radius and h = height of cylindrical part
Volume of whole tube = \( \pi r^2 h + \frac{2}{3}\pi r^3 = \frac{5159}{6} \) ... (i)
Volume up to 4 cm below top = \( \pi r^2(h-4) + \frac{2}{3}\pi r^3 = \frac{4235}{6} \) ... (ii)

Step 2: Divide equation (i) by equation (ii).
\[ \frac{2r + 3h}{2r + 3h - 12} = \frac{5159}{4235} \] ... (iii)

Step 3: Subtract equation (ii) from equation (i).
\[ \pi r^2(12) = \frac{5159}{2} - \frac{4235}{2} = \frac{924}{2} \]
\[ 12 \times \frac{22}{7} \times r^2 = \frac{924}{2} \]
\[ r^2 = \frac{924 \times 7}{2 \times 12 \times 22} = \frac{7 \times 7}{2 \times 2} \]
\[ r^2 = \frac{49}{4} \]
\[ r = \frac{7}{2} = 3.5 \text{ cm} \]

Step 4: Find h using equation (iii).
\[ \frac{2 \times \frac{7}{2} + 3h}{2 \times \frac{7}{2} + 3h - 12} = \frac{5159}{4235} \]
\[ \frac{7 + 3h}{7 + 3h - 12} = \frac{5159}{4235} \]
\[ \frac{7 + 3h}{7 + 3h - 12} = \frac{469}{385} \]
2695 + 1155h = 1407h - 2345
252h = 5040
h = 20

Final Answer: Height = 20 cm and radius = 3.5 cm

In simple words: We used the difference in volumes to find how much water fits in the top 4 cm part. This helped us find the radius first, then the height.

📝 Teacher's Note: Show students that when water level drops by 4 cm, only the cylinder part loses volume. The hemisphere stays full. This makes the problem easier to solve.

🎯 Exam Tip: Set up two volume equations clearly. The key insight is that only cylinder volume changes when water level drops. Solve for radius first, then height.

Question 17. A solid is in the form of a right circular cone mounted on a hemisphere. The diameter of the base of the cone, which exactly coincides with hemisphere, is 7 cm and its height is 8 cm. The solid is placed in a cylindrical vessel of internal radius 7 cm and height 10 cm. How much water, in \( cm^3 \), will be required to fill the vessel completely?
Answer:

[Diagram: This diagram shows a cone sitting on top of a hemisphere, all placed inside a cylinder.]

Given:
Diameter of hemisphere = 7 cm
Diameter of the base of the cone = 7 cm
Therefore, radius (r) = 3.5 cm
Height (h) = 8 cm

Step 1: Find volume of the solid.
Volume of the solid = \( \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2(h + 2r) \)
\[ = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5(8 + 2 \times 3.5) \]
\[ = \frac{77}{6}(8 + 7) \]
\[ = \frac{385}{2} \]
= 192.5 \( cm^3 \)

Step 2: Find volume of cylindrical vessel.
Radius of cylindrical vessel (R) = 7 cm
Height (H) = 10 cm
Volume = \( \pi R^2 H \)
\[ = \frac{22}{7} \times 7 \times 7 \times 10 \]
= 1540 \( cm^3 \)

Step 3: Find water required.
Volume of water required to fill = 1540 - 192.5 = 1347.5 \( cm^3 \)

In simple words: We found how much space the solid takes up. Then we found the total space in the vessel. The difference is how much water we need.

📝 Teacher's Note: Use a ball and cone toy to show this shape. Put it in a jar to show how water fills the remaining space around the solid.

🎯 Exam Tip: Calculate solid volume first, then vessel volume. Subtract to find water needed. Don't mix up the radii - solid has radius 3.5 cm, vessel has radius 7 cm.

Question 18. Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.
Answer:
Step 1: Find total volume of melted spheres.
Volume of the cone = Sum of volumes of the two melted spheres
\[ \frac{1}{3}\pi r^2 \times 8 = \frac{4}{3}\pi \times (2)^3 + \frac{4}{3}\pi \times (4)^3 \]
8r² = 4 × 8 + 4 × 64
8r² = 32 + 256
8r² = 288
r² = 36
r = 6

Final Answer: The radius of the cone so formed is 6 cm.

In simple words: When we melt the two balls, all that metal becomes the cone. So the volume stays the same - just the shape changes.

📝 Teacher's Note: Show students two different sized balls. Explain that when melted, all the material goes into making the new cone. Volume is conserved but shape changes.

🎯 Exam Tip: Write "Volume is conserved" at the start. Set up equation: cone volume = sum of sphere volumes. Calculate step by step and don't forget to take square root.

Question 19. A certain number of metallic cones, each of radius 2 cm and height 3 cm, are melted and recast into a solid sphere of radius 6 cm. Find the number of cones used.
Answer:
Step 1: Set up the equation.
Let the number of cones melted be n.
Let the radius of sphere be \( r_s = 6 \) cm
Radius of cone be \( r_c = 2 \) cm
And, height of the cone be h = 3 cm
Volume of sphere = n (Volume of a metallic cone)
\[ \frac{4}{3}\pi r_s^3 = n\left(\frac{1}{3}\pi r_c^2 h\right) \]

Step 2: Substitute values.
\[ \frac{4}{3}\pi r_s^3 = n\left(\frac{1}{3}\pi r_c^2 h\right) \]
\[ \frac{4r_s^3}{r_c^2 h} = n \]
\[ n = \frac{4(6)^3}{(2)^2(3)} \]
\[ n = \frac{4 \times 216}{4 \times 3} \]
n = 72

Final Answer: The number of cones is 72.

In simple words: We found how many small cones have the same total volume as one big sphere. All the metal from 72 cones makes one sphere.

📝 Teacher's Note: Use small identical objects like coins to show how many small things can make one big thing. The total amount of material stays the same.

🎯 Exam Tip: Set up the equation clearly: "Volume of sphere = n × Volume of one cone." Substitute all values carefully and solve for n step by step.

Question 20. A conical tent is to accommodate 77 persons. Each person must have 16\( m^3 \) of air to breathe. Given the radius of the tent as 7m, find the height of the tent and also its curved surface area.
Answer:
Step 1: Find required volume of tent.
Total volume required = 77 × 16 = 1232 \( m^3 \)

Step 2: Find height using volume formula.
Volume of cone = \( \frac{1}{3}\pi r^2 h \)
\[ 1232 = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times h \]
\[ 1232 = \frac{22 \times 49 \times h}{21} \]
\[ h = \frac{1232 \times 21}{22 \times 49} = \frac{1232 \times 3}{22 \times 7} = \frac{3696}{154} = 24 \text{ m} \]

Step 3: Find slant height.
Slant height = \( \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ m} \)

Step 4: Find curved surface area.
Curved surface area = \( \pi rl = \frac{22}{7} \times 7 \times 25 = 550 \text{ m}^2 \)

Final Answer: Height = 24 m, Curved surface area = 550 \( m^2 \)

In simple words: We found how much air space all people need. Then we used that to find how tall the tent must be. Finally we found the cloth needed for the slanted sides.

📝 Teacher's Note: Explain that each person needs breathing space. Show students how to use Pythagoras theorem to find slant height from height and radius.

🎯 Exam Tip: First find total volume needed. Use cone volume formula to find height. Then find slant height using Pythagoras. Finally calculate curved surface area with πrl formula.

Solution:

According to the condition in the question,

\( 77 \times 16 = \frac{1}{3} \pi r^2 h \)

\( \Rightarrow 77 \times 16 = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times h \)

\( \Rightarrow h = \frac{77 \times 16 \times 3 \times 7}{22 \times 7 \times 7} \)

\( \Rightarrow h = \frac{11 \times 16 \times 3}{22} \)

\( \Rightarrow h = 24 \text{ m} \)

We know that,

\( l^2 = r^2 + h^2 \)

\( \Rightarrow l^2 = (7)^2 + (24)^2 \)

\( \Rightarrow l^2 = 49 + 576 \)

\( \Rightarrow l^2 = 625 \)

\( \Rightarrow l = 25 \text{ m} \)

\( \therefore \text{Curved Surface Area} = \pi r l = \frac{22}{7} \times 7 \times 25 = 550\text{m}^2 \)

Therefore the height of the tent is 24m and its curved surface area is 550m².

📝 Teacher's Note: This is a cone problem. Show students how we use volume formula first to find height, then use Pythagoras theorem to find slant height. Draw a cone and mark radius, height, and slant height clearly.

🎯 Exam Tip: Always write the formula first, then substitute values step by step. Remember that curved surface area of cone uses slant height (l), not vertical height (h). Don't forget to write units in your final answer.