Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 20 Mensuration Ii

Exercise 20(A)

Answer 1.

i) Height = 12 cm, radius = 5 cm
Answer:
Step 1: Find curved surface area.
Curved surface area = \( \pi r \sqrt{h^2 + r^2} \)
\( = \frac{22}{7} \times 5 \times \sqrt{12^2 + 5^2} \)
\( = \frac{22}{7} \times 5 \times \sqrt{169} \)
\( = \frac{22}{7} \times 5 \times 13 \)
\( = 204.29 \)

Curved surface area = 204.29 cm²

Step 2: Find total surface area.
Total surface area = area of circular base + curved surface area
\( = \pi r^2 + \pi r\sqrt{h^2 + r^2} \)
\( = \frac{22}{7} \times 5 \times 5 + 204.29 \)
\( = 78.57 + 204.29 \)
\( = 282.86 \)

Total surface area = 282.86 cm²

Step 3: Find volume.
Volume = \( \frac{1}{3} \times (\pi r^2) \times h \)
\( = \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12 \)
\( = 314.29 \)

Volume of the cone = 314.29 cm³
In simple words: We found three things: the curved surface (like wrapping paper around the cone), the total surface (including the bottom circle), and the volume (how much space inside the cone).

📝 Teacher's Note: Use the slant height formula first. Many students forget to find slant height before curved surface area. Show them a real cone or ice cream cone to make it clear.

🎯 Exam Tip: Write all three formulas clearly. Always find slant height first. Show each step and write units in your final answer.

ii) Height = 15 cm, radius = 8 cm
Answer:
Step 1: Find curved surface area.
Curved surface area = \( \pi r \sqrt{h^2 + r^2} \)
\( = \frac{22}{7} \times 8 \times \sqrt{15^2 + 8^2} \)
\( = \frac{22}{7} \times 8 \times \sqrt{289} \)
\( = \frac{22}{7} \times 8 \times 17 \)
\( = 427.43 \)

Curved surface area = 427.43 cm²

Step 2: Find total surface area.
Total surface area = area of circular base + curved surface area
\( = \pi r^2 + \pi r\sqrt{h^2 + r^2} \)
\( = \frac{22}{7} \times 8 \times 8 + 427.43 \)
\( = 201.14 + 427.43 \)
\( = 628.57 \)

Total surface area = 628.57 cm²

Step 3: Find volume.
Volume = \( \frac{1}{3} \times (\pi r^2) \times h \)
\( = \frac{1}{3} \times \frac{22}{7} \times 8 \times 8 \times 15 \)
\( = 1005.71 \)

Volume of the cone = 1005.71 cm³
In simple words: This cone is bigger than the first one, so all the areas and volume are larger numbers.

📝 Teacher's Note: Compare the answers with part (i). Students can see that bigger radius and height give bigger surface area and volume. This helps them check if their answer makes sense.

🎯 Exam Tip: Calculate slant height carefully. 15² + 8² = 225 + 64 = 289. The square root of 289 is 17. This is a common calculation mistake point.

iv) Height = 8 cm, diameter = 12 cm
Answer:
Diameter = 12 cm → r = 6 cm

Step 1: Find curved surface area.
Curved surface area = \( \pi r \sqrt{h^2 + r^2} \)
\( = \frac{22}{7} \times 6 \times \sqrt{8^2 + 6^2} \)
\( = \frac{22}{7} \times 6 \times \sqrt{100} \)
\( = \frac{22}{7} \times 6 \times 10 \)
\( = 188.57 \)

Curved surface area = 188.57 cm²

Step 2: Find total surface area.
Total surface area = area of circular base + curved surface area
\( = \pi r^2 + \pi r\sqrt{h^2 + r^2} \)
\( = \frac{22}{7} \times 6 \times 6 + 188.57 \)
\( = 113.14 + 188.57 \)
\( = 301.71 \)

Total surface area = 301.71 cm²

Step 3: Find volume.
Volume = \( \frac{1}{3} \times (\pi r^2) \times h \)
\( = \frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 8 \)
\( = 301.71 \)

Volume of the cone = 301.71 cm³
In simple words: When given diameter, we first divide by 2 to get radius. Then we follow the same steps as before.

📝 Teacher's Note: Emphasize that diameter ÷ 2 = radius. Many students forget this step and use diameter directly in the formula. Make them write "r = 6 cm" clearly.

🎯 Exam Tip: Always write "r = diameter ÷ 2" when given diameter. Check that 8² + 6² = 64 + 36 = 100, so slant height = 10.

Answer 2.
Answer:
[Diagram: This shows a cone with height marked as 12 cm and volume given as 154 cm³]

Given:
Volume of the cone = 154 cm³
Height = 12 cm

Step 1: Use volume formula to find radius.
\( \frac{1}{3} \times (\pi r^2) \times h = 154 \)
\( \frac{1}{3} \times (\pi r^2) \times 12 = 154 \)

Step 2: Solve for r².
\( r^2 = \frac{154 \times 3 \times 7}{12 \times 22} \)
\( r^2 = 12.25 \)
\( r = 3.5 \text{ cm} \)

Radius of the circular base of the cone is 3.5 cm
In simple words: We worked backwards from the volume to find the radius. We know volume and height, so we can find radius using the volume formula.

📝 Teacher's Note: Show students how to rearrange the formula. Volume = (1/3)πr²h, so r² = (Volume × 3)/(πh). This is working backwards from a known answer.

🎯 Exam Tip: Write "Given" and "To find" clearly. Show the formula first, then substitute values. Check your answer by putting r = 3.5 back into the volume formula.

Answer 3.
Answer:
[Diagram: This shows a cone with slant length marked as 15 cm and height marked as 12 cm]

Given:
Slant length = l = 15 cm
Height = h = 12 cm
Radius of the base = r

Step 1: Find radius using Pythagoras theorem.
We know, \( l^2 = h^2 + r^2 \)
\( r^2 = l^2 - h^2 \)
\( r = \sqrt{l^2 - h^2} \)
\( r = \sqrt{15^2 - 12^2} \)
\( r = \sqrt{225 - 144} \)
\( r = \sqrt{81} \)
\( r = 9 \text{ cm} \)

Radius = 9 cm

Step 2: Find volume.
Volume = \( \frac{1}{3} \times (\pi r^2) \times h \)
\( = \frac{1}{3} \times 3.14 \times 9 \times 9 \times 12 \)
\( = 1017.36 \text{ cm}^3 \)

Volume of the cone = 1017.36 cm³
In simple words: We used the slant height and vertical height to find the base radius. Then we found how much space the cone takes up.

📝 Teacher's Note: Draw a right triangle showing height, radius, and slant height. This makes Pythagoras theorem very clear. The slant height is like the hypotenuse of the triangle.

🎯 Exam Tip: Write the Pythagoras formula clearly: l² = h² + r². Check that 15² - 12² = 225 - 144 = 81, and √81 = 9. Show this calculation step by step.

Answer 4.
Answer:

[Diagram: A cone with diameter 12 cm at the base and height 8 cm]

Given:
Diameter = 12 cm
\( \Rightarrow \) radius = 6 cm
Height = 8 cm

Step 1: Find the slant height using Pythagoras theorem.
\( l^2 = r^2 + h^2 \)
\( l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \) cm

Step 2: Calculate curved surface area.
Curved surface area = \( \pi r l \)
\( = \frac{22}{7} \times 6 \times 10 \)
\( = \frac{22 \times 6 \times 10}{7} \)
\( = \frac{1320}{7} \)
\( = 188.57 \) cm²

Curved surface area = 188.57 cm²
In simple words: We found the slant height (like the side of the cone) using Pythagoras theorem. Then we used the formula for curved surface area which is like finding the area of the curved part of the cone.

📝 Teacher's Note: Show students how to find slant height using a right triangle. The radius, height and slant height make a right triangle. Draw this clearly on the board.

🎯 Exam Tip: Always find slant height first for cone problems. Write the formula \( l = \sqrt{r^2 + h^2} \) clearly. Then use curved surface area = \( \pi r l \).

Answer 5.
Answer:

[Diagram: A cylinder with diameter 12 m and height 10 m]

Given:
Diameter = 12 m
\( \Rightarrow \) radius = 6 m
Height = 10 m

Step 1: Find curved surface area.
Curved surface area = circumference of base × height
\( = 2\pi r \times h \)
\( = 2 \times \frac{22}{7} \times 6 \times 10 \)
\( = 377.14 \) m²

Step 2: Find total surface area.
Total surface area = Curved surface area + (2 × base area)
\( = 2\pi rh + 2\pi r^2 \)
\( = 2\pi r(h + r) \)
\( = 2 \times \frac{22}{7} \times 6 \times (10 + 6) \)
\( = 2 \times \frac{22}{7} \times 6 \times 16 \)
\( = 603.42 \) m²

Curved Surface Area = 377.14 m²
Total Surface Area = 603.42 m²
In simple words: Curved surface area is like wrapping paper around the side of a can. Total surface area includes the top and bottom circles too.

📝 Teacher's Note: Use a real cylinder like a can or bottle. Show students the curved part (side) and the flat parts (top and bottom). This makes the concept very clear.

🎯 Exam Tip: For cylinders, curved surface area = \( 2\pi rh \) and total surface area = \( 2\pi r(h + r) \). Remember to add both circular bases for total area.

Answer 6.
Answer:
Let radius of first cone be 3r and height be h. Then radius of second cone will be r and height will be 3h.

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

Ratio of volumes = \( \frac{\text{Volume of first cone}}{\text{Volume of second cone}} \)

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

\( = \frac{\frac{1}{3} \pi 9r^2 h}{\frac{1}{3} \pi r^2 3h} \)

\( = \frac{9r^2h}{3r^2h} \)

\( = \frac{9}{3} = \frac{3}{1} \)

Ratio of volumes of cone = 3:1
In simple words: We compared the volumes by putting one cone's measurements over the other's. The first cone has a bigger base but same height, so it has more volume.

📝 Teacher's Note: Explain that when radius gets 3 times bigger, the area becomes 9 times bigger (because area uses r²). This is why the ratio is 3:1, not 1:1.

🎯 Exam Tip: When comparing volumes, set up the ratio as a fraction. Cancel out common terms like π and 1/3. Remember that radius is squared in the volume formula.

Answer 7.
Answer:
The base circumferences of the cones are equal, therefore the radius of base are equal.

Let radius be r.

Ratio between slant heights = 5:4

Let slant height of first cone = 5x and of second cone = 4x

Curved surface area of cone = πrl (where l = slant height)

Ratio of curved surface areas =
\( = \frac{\pi r \times 5x}{\pi r \times 4x} \)
\( = \frac{5x}{4x} \)
\( = \frac{5}{4} \)

Ratio of curved surface areas = 5:4
In simple words: Since both cones have the same base size, the ratio of their curved areas is the same as the ratio of their slant heights.

📝 Teacher's Note: Emphasize that when the base radius is the same, curved surface area depends only on slant height. Draw two cones with same base but different slant heights.

🎯 Exam Tip: When radius is same, curved surface area ratio equals slant height ratio. Write this clearly: "Since r is same, ratio = l₁:l₂".

Answer 8.
Answer:
Volume of cone = \( \frac{1}{3} \times \pi r^2 \times h \)

\( 75\pi = \frac{1}{3} \times \pi \times 5 \times 5 \times h \)

\( h = \frac{225}{25} \)

\( h = 9 \) cm

Height of the cone = 9 cm
In simple words: We used the volume formula and put in the known values. Then we solved for the unknown height.

📝 Teacher's Note: Show students how to substitute known values into the formula step by step. Remind them that radius = 5 cm, so r² = 25 cm².

🎯 Exam Tip: Always write the volume formula first. Substitute all known values, then solve for the unknown. Don't forget to cancel π from both sides.

Answer 9.
Answer:
Curved surface area = 710 cm²
Radius (r) of base = 11.3 cm
Let slant height be l.

\( \pi rl = 710 \)
\( \frac{22}{7} \times 11.3 \times l = 710 \)
\( l = \frac{710 \times 7}{11.3 \times 22} \)
\( l = \frac{4970}{248.6} \)
\( l = 19.99 \) cm ≈ 20 cm

The slant height is 20 cm.
In simple words: We used the curved surface area formula and solved for slant height. Slant height is the distance from the tip of the cone to the edge of the base.

📝 Teacher's Note: Explain what slant height means by drawing a cone and showing the line from apex to the circumference. It's like the side edge of the cone.

🎯 Exam Tip: For finding slant height, use curved surface area = πrl. Substitute known values and solve for l. Round off the final answer appropriately.

Answer 10.
Answer:
Curved surface area of the tent = 264 m²
Slant height (l) = 12 m

\( \pi rl = 264 \)
\( \frac{22}{7} \times r \times 12 = 264 \)
\( r = \frac{264 \times 7}{22 \times 12} \)
\( r = \frac{1848}{264} \)
\( r = 7 \) cm

Radius of cone = 7 m

Let h be the vertical height.

We know, \( l^2 = r^2 + h^2 \)
\( h = \sqrt{l^2 - r^2} \)
\( h = \sqrt{12^2 - 7^2} \)
\( h = \sqrt{144 - 49} = \sqrt{95} \)
\( h = 9.75 \) m

Vertical height of cone = 9.75 m
In simple words: First we found the radius using the curved surface area. Then we used Pythagoras theorem to find the vertical height from the slant height and radius.

📝 Teacher's Note: Draw a right triangle inside the cone showing radius, height and slant height. This helps students see why we use Pythagoras theorem.

🎯 Exam Tip: For cone problems with slant height, always use \( l^2 = r^2 + h^2 \) to find the missing dimension. Show the right triangle clearly in your working.

Exercise 20.2

Answer 4.
Answer:
Given: Surface area = volume

Step 1: Write the formulas for surface area and volume of sphere.
Surface area = \( 4\pi r^2 \)
Volume = \( \frac{4}{3}\pi r^3 \)

Step 2: Apply the given condition.
\( 4\pi r^2 = \frac{4}{3}\pi r^3 \)

Step 3: Simplify by canceling common terms.
\( 3r^2 = r^3 \)

Step 4: Solve for r.
\( r = 3 \)

Therefore, radius of the sphere = 3 units
In simple words: We set the surface area formula equal to the volume formula. Then we solve the equation to find the radius. When radius is 3 units, the surface area equals the volume.

📝 Teacher's Note: Make students write both formulas first. Show them how to cancel \( 4\pi \) from both sides. Students often forget to divide by r² on both sides.

🎯 Exam Tip: Always write "Given:" first. Show each step clearly. The final answer is radius = 3 units. Don't forget to write units.

Answer 5.
Answer:
Given:
Diameter of circle = 2.8 cm
Radius = r = 1.4 cm

Step 1: Find area of the circle.
Area of circle = \( \pi r^2 \)
= \( \pi (1.4)^2 \)
= \( 1.96\pi \)

Step 2: Apply the given condition.
Surface area of sphere = Area of the circle
\( 4\pi r^2 = 1.96\pi \)

Step 3: Solve for r².
\( r^2 = \frac{1.96}{4} \)
\( r^2 = 0.49 \)

Step 4: Find radius.
\( r = 0.7 \text{ cm} \)

Therefore, radius of the sphere = 0.7 cm
In simple words: The sphere's surface area equals the circle's area. We use this fact to find how big the sphere is. The sphere's radius is smaller than the circle's radius.

📝 Teacher's Note: Students often confuse radius and diameter. Make them write r = d/2 first. Also show that 0.7² = 0.49 to check the answer.

🎯 Exam Tip: Write the condition clearly: "Surface area of sphere = Area of circle." This shows you understand the problem. Always check your answer by substituting back.

Answer 11.
Answer:
Given:
Circular base = 160 m²
Capacity = 600 m³

Step 1: Find radius from circular base area.
\( \pi r^2 = 160 \)
\( r = \sqrt{\frac{160 \times 7}{22}} \)
\( r = \sqrt{50.909} = 7.134 \text{ m} \)

Step 2: Find height using volume formula.
\( \frac{1}{3}\pi r^2 h = 600 \)
\( \frac{1}{3} \times \frac{22}{7} \times 7.13 \times 7.13 \times h = 600 \)
\( h = \frac{600 \times 3 \times 7}{22 \times 7.13 \times 7.13} \)
\( h = 11.265 \text{ m} \)

Step 3: Find slant height.
\( l = \sqrt{r^2 + h^2} \)
\( l = \sqrt{7.134^2 + 11.265^2} \)
\( l = \sqrt{177.624} = 13.327 \text{ m} \)

Step 4: Find curved surface area.
Curved surface area = \( \pi rl \)
= \( \frac{22}{7} \times 7.134 \times 13.327 = 298.9 \text{ m}^2 \)

Therefore, area of canvas = 298.9 m²
In simple words: We found the tent's radius from its base area. Then we found height from volume. Finally we calculated the slant side and curved area. This gives us the canvas needed.

📝 Teacher's Note: Draw a cone and label r, h, and l. Show students that slant height is like the hypotenuse of a triangle. Use Pythagoras theorem: l² = r² + h².

🎯 Exam Tip: Always find radius first, then height, then slant height. Write each step clearly. The curved surface area formula is πrl, not πr².

Answer 12.
Answer:
Given:
Internal radius = r = 3.5 cm
Height = h = 21 cm
Thickness = 0.5 cm

Step 1: Find outer radius.
Outer radius = R = (3.5 + 0.5) cm = 4 cm

Step 2: Find volume of metal used.
Volume = \( \pi h(R^2 - r^2) \)
= \( \frac{22}{7} \times 21 \times (4^2 - 3.5^2) \)
= \( \frac{22}{7} \times 21 \times (16 - 12.25) \)
= \( \frac{22}{7} \times 21 \times 3.75 \)
= 247.5 cm³

Step 3: This metal is recast into a cone.
Volume of cone = 247.5 cm³
Height of cone = 7 cm

Step 4: Find radius of cone.
\( \frac{1}{3}\pi r_1^2 h = 247.5 \)
\( \frac{1}{3} \times \frac{22}{7} \times r_1^2 \times 7 = 247.5 \)
\( r_1^2 = \frac{247.5 \times 3 \times 7}{22 \times 7} = 33.75 \)
\( r_1 = 5.8 \text{ cm} \)

Therefore, radius of cone = 5.8 cm
In simple words: We found how much metal is in the hollow cylinder walls. This same amount of metal makes a solid cone. We used the cone volume formula to find its radius.

📝 Teacher's Note: Show students that volume of hollow cylinder = volume of big cylinder - volume of small cylinder. Draw both shapes to make it clear.

🎯 Exam Tip: Write the volume formula for hollow cylinder as π(R² - r²)h. Don't forget that when metal is recast, the volume stays the same.

Answer 13.
Answer:
Given:
Height of cylinder = H = 8 m
Height of cone = h = 4 m
Diameter = 14 m, so radius = r = 7 m

Step 1: Find slant height of cone.
\( l = \sqrt{r^2 + h^2} \)
\( l = \sqrt{7^2 + 4^2} \)
\( l = \sqrt{65} = 8.06 \text{ m} \)

Step 2: Find area of canvas used.
Area = Curved surface area of cylinder + Curved surface area of cone
= \( 2\pi rH + \pi rl \)
= \( \left(2 \times \frac{22}{7} \times 7 \times 8\right) + \left(\frac{22}{7} \times 7 \times 8.06\right) \)
= 352 + 177.32
= 529.32 m²

Therefore, area of canvas used = 529.32 m²
In simple words: The tent has a cylinder part and a cone part on top. We find the curved area of both parts and add them. This gives the total canvas needed.

📝 Teacher's Note: Draw the tent shape - cylinder with cone on top. Show students that we only need curved areas, not the flat circular parts where they join.

🎯 Exam Tip: Always find slant height first for the cone part. Add both curved surface areas. Don't include the circular base where cylinder and cone meet.

Answer 14.
Answer:
Given:
Height of cylinder = h = 5 m
Slant height of cone = l = 53 m
Diameter = 42 m, so radius = r = 21 m

Step 1: Find area of canvas used.
Area = Curved surface area of cylinder + Curved surface area of cone
= \( 2\pi rh + \pi rl \)
= \( \left(2 \times \frac{22}{7} \times 21 \times 5\right) + \left(\frac{22}{7} \times 21 \times 53\right) \)
= 660 + 3498
= 4158 m²

Therefore, area of canvas required = 4158 m²
In simple words: This is another tent with cylinder and cone parts. We add the curved areas of both parts to get total canvas needed.

📝 Teacher's Note: This problem gives slant height directly. Students don't need to calculate it. Remind them to read the given information carefully.

🎯 Exam Tip: When slant height is given directly, use it in the formula πrl. Don't try to calculate it again using Pythagoras theorem.

Answer 15.
Answer:
Given:
Height of cylinder = h1 = 32 cm
Radius of bucket = r1 = 18 cm
Height of cone = h2 = 24 cm

Step 1: Apply volume condition.
Volume of sand in bucket = Volume of sand in cone
\( \pi r_1^2 h_1 = \frac{1}{3}\pi r_2^2 h_2 \)

Step 2: Substitute values and solve.
\( 18 \times 18 \times 32 = \frac{1}{3} \times r_2^2 \times 24 \)
\( r_2^2 = \frac{10368 \times 3}{24} = 1296 \)
\( r_2 = 36 \text{ cm} \)

Therefore, radius of conical heap = 36 cm
In simple words: When sand is poured from the bucket, it forms a cone shape. The amount of sand stays the same, so volumes are equal.

📝 Teacher's Note: Explain that sand takes the shape of its container. Show students how to set up the equation: cylinder volume = cone volume.

🎯 Exam Tip: Write the volume equality first. Don't forget the 1/3 factor in cone volume formula. Check your answer by calculating both volumes.

Answer 6.
Answer:
Given:
Radius of sphere = 9 m

Step 1: Find volume of sphere
\( \text{Volume of sphere} = \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times 9 \times 9 \times 9 \)
\( = 3054.857 \text{ m}^3 \) .......... (i)

Step 2: Find dimensions of cylindrical wire
Diameter of cylindrical wire = 4 m
Therefore, radius = 2 m
Let length of wire be h

Step 3: Find volume of wire in terms of h
\( \text{Volume} = \pi r^2 h \)
\( = \frac{22}{7} \times 2 \times 2 \times h \)
\( = \frac{88h}{7} \text{ m}^3 \) .......... (ii)

Step 4: Equate volumes from (i) and (ii)
\( \frac{88h}{7} = 3054.857 \)
\( h = \frac{3054.857 \times 7}{88} \)
\( h = 243 \text{ m} \)

Length of the wire = 243 m
In simple words: When we melt the sphere and make it into a wire, the volume stays the same. We used this fact to find the length of the wire.

📝 Teacher's Note: Explain that when we change shape, volume stays the same. Use clay to show students — make a ball, then roll it into a thin rope. Same amount of clay.

🎯 Exam Tip: Always write "Volume of sphere = Volume of wire" first. Then substitute the formulas. Show each step clearly with units.

Answer 7.
Answer:
Given:
Radius of sphere = 9 cm

Step 1: Find volume of sphere
\( \text{Volume of sphere} = \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times 9 \times 9 \times 9 \)
\( = 3054.657 \text{ cm}^3 = 30.55 \times 10^{-4} \text{ m}^3 \) .......... (i)

Step 2: Find dimensions of cylindrical wire
Diameter of cylindrical wire = 2 mm
Therefore, radius = 1 mm = 0.001 m
Let length of wire be h

Step 3: Find volume of wire in terms of h
\( \text{Volume} = \pi r^2 h \)
\( = \frac{22}{7} \times 0.001 \times 0.001 \times h \)
\( = 3.142 \times 10^{-6} h \text{ m}^3 \) .......... (ii)

Step 4: Equate volumes from (i) and (ii)
\( 3.142 \times 10^{-6} h = 30.55 \times 10^{-4} \)
\( h = \frac{30.55 \times 10^{-4}}{3.142 \times 10^{-6}} \)
\( h = 972 \text{ m} \)

Length of the wire = 972 m
In simple words: The sphere becomes a very thin wire that is almost 1 km long! When we make things thinner, they become much longer.

📝 Teacher's Note: Show students that when diameter becomes very small (2 mm), the length becomes very big (972 m). This helps them understand the relationship.

🎯 Exam Tip: Be careful with unit conversion. Convert everything to the same units before calculating. Check your answer — it should make sense.

Answer 8.
Answer:
Given:
Let r be the radii of sphere and cone

Step 1: Write volume formulas
\( \text{Volume of sphere} = \frac{4}{3}\pi r^3 = \frac{1}{3}\pi r^2 h \) (h = 2r for sphere)
\( \text{Volume of cone} = \frac{1}{3}\pi r^2 h \)

Step 2: Use the condition
But h = 2r for sphere
Therefore, h = 2r for cone also

Step 3: Compare volumes
Both have the same formula when h = 2r
Hence, proved
In simple words: When a cone has the same radius and height as the sphere's diameter, they hold the same amount of space inside.

📝 Teacher's Note: Draw both shapes on the board. Show that the sphere's diameter becomes the cone's height. This makes the volumes equal.

🎯 Exam Tip: Write the condition clearly: "height of cone = diameter of sphere = 2r". Then substitute in the volume formulas to prove equality.

Answer 9.
Answer:
Given:
Let r, h be the radius and height of cylinder, cone and sphere

Step 1: Write volume formulas
\( \text{Volume of cylinder} = \pi r^2 h \)
\( \text{Volume of sphere} = \frac{4}{3}\pi r^3 \) (h = 2r for sphere)
\( \text{Volume of cone} = \frac{1}{3}\pi r^2 h \)

Step 2: Find the ratio
\( \pi r^2 h : \frac{1}{3}\pi r^2 h : \frac{4}{3}\pi r^3 \)

Step 3: Simplify the ratio
The volume of a cylinder is three times the volume of a cone with equal height and radius. The volume of a sphere is two times the volume of a cone with equal height and radius.

So the ratio of volumes is 3:1:2
In simple words: If cylinder, cone, and sphere have the same size, the cylinder holds the most, then sphere, then cone. The ratio is 3:1:2.

📝 Teacher's Note: Use three containers of the same size — cylinder, cone, and sphere. Fill them with rice to show the volume difference. Students can see which holds more.

🎯 Exam Tip: Remember the ratio 3:1:2 for cylinder:cone:sphere. This is a standard result that appears in many exam questions.

Answer 10.
Answer:
Given:
Diameter of spherical marble = 1.4 cm
Therefore, radius = 0.7 cm

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

Step 2: Find volume of water in beaker
Diameter of beaker = 7 cm
Therefore, radius = 3.5 cm
Height of water = 5.6 cm
\( \text{Volume of water} = \pi r^2 h \)
\( = \pi \times (3.5)^2 \times 5.6 \text{ cm}^3 \) .......... (ii)

Step 3: Calculate number of balls
\( \text{No. of balls dropped} = \frac{\text{Volume of water}}{\text{Volume of ball}} \)
\( = \frac{\pi \times (3.5)^2 \times 5.6}{\frac{4}{3} \times \pi \times (0.7)^3} \)
\( = \frac{3 \times (3.5)^2 \times 5.6}{4 \times (0.7)^3} \)
\( = 150 \)

No. of balls dropped = 150
In simple words: We found how much water was displaced. Then we divided by the volume of one ball to get the number of balls.

📝 Teacher's Note: Explain Archimedes' principle — when you drop something in water, the water level rises by exactly the volume of the object dropped.

🎯 Exam Tip: Set up the equation: Volume displaced = Number of balls × Volume of one ball. Then solve for number of balls. Show all calculation steps clearly.

Answer 11.
Answer:
Given:
Radius of sphere = 10 cm

Step 1: Find volume of sphere
\( \text{Volume of sphere} = \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times 10 \times 10 \times 10 \text{ cm}^3 \)
\( = 4190.476 \text{ cm}^3 \)

Therefore, volume of water = 4190.476 cm³

Step 2: Find height of water in cylinder
Radius of base of cylinder = 20 cm
Let h be the height of the water
\( \pi r^2 h = 4190.476 \)
\( \frac{22}{7} \times 20 \times 20 \times h = 4190.476 \)
\( 1257.143h = 4190.476 \)
\( h = 3.33 \text{ cm} \)

Increase in water level = 3.33 cm
In simple words: The sphere pushes away water equal to its own volume. This water spreads in the cylinder and raises the level by 3.33 cm.

📝 Teacher's Note: Demonstrate with a ball and a wide container of water. When you put the ball in, the water level goes up. The rise depends on the ball's size and container's width.

🎯 Exam Tip: Volume of sphere = Volume of water displaced = Area of cylinder base × height rise. Use this equation to find the height rise. Units must match throughout.

Answer 12.
Answer:
Given:
One = 8 cm
Radius = 5 cm

Step 1: Find volume of cone.
Volume = \( \frac{1}{3}\pi r^2h \)
= \( \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 8 \text{ cm}^3 \)
= \( \frac{4400}{21} \text{ cm}^3 \)

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

Step 3: Find radius and volume of each ball.
Radius of each ball = 0.5 cm = \( \frac{1}{2} \) cm

Volume of a ball = \( \frac{4}{3}\pi r^3 \)
= \( \frac{4}{3} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \text{ cm}^3 \)
= \( \frac{11}{21} \text{ cm}^3 \)

Step 4: Find number of balls.
Therefore, No. of balls = \( \frac{1100}{21} \div \frac{11}{21} = 100 \)

Hence, number of lead balls = 100
In simple words: We found how much water flowed out. Then we found how big each ball is. We divided the water volume by ball volume to get the number of balls.

📝 Teacher's Note: Show students that when balls are dropped in water, they push out water equal to their own volume. This is called displacement. Use real balls and water to demonstrate.

🎯 Exam Tip: Always write "volume displaced = volume of balls" clearly. Show each step - cone volume, water flowed out, ball volume, then division to get the answer.

Answer 13.
Answer:
Given:
Radius = 10 cm

Step 1: Find total surface area of hemisphere.
Total surface area = \( 3\pi r^2 \)
= \( 3 \times \frac{22}{7} \times 10 \times 10 \text{ cm}^2 \)
= 942.86 cm²

Step 2: Find volume of hemisphere.
Volume of hemisphere = \( \frac{2}{3}\pi r^3 \)
= \( \frac{2}{3} \times 3.14 \times 10 \times 10 \times 10 \text{ cm}^3 \)
= 2093.3 cm³

Total surface area = 942.86 cm² and volume = 2093.3 cm³
In simple words: A hemisphere is half a ball. Its surface area includes the curved part plus the flat circular base. Volume is how much space it takes up inside.

📝 Teacher's Note: Cut an orange in half. The curved part plus the flat circle is the total surface area. Students can see and touch this easily.

🎯 Exam Tip: Remember hemisphere formulas: Surface area = 3πr² (not 2πr²). Volume = (2/3)πr³. Write both formulas first, then substitute values.

Answer 14.
Answer:
Given:
Diameter of the hemispherical dome = 10 m
Therefore, radius of dome = 5 m

Step 1: Find curved surface area.
Curved surface area = \( 2\pi r^2 \)
= \( 2 \times \frac{22}{7} \times 5 \times 5 \)
= 157.14 m²

Step 2: Calculate painting cost.
Cost of painting one sq. metre = Rs. 1.40
Cost of painting 157.14 m² = Rs.(1.40 × 157.14)
= Rs. 219.99 = Rs 220

Therefore, cost of painting the dome = Rs 220
In simple words: We found how much area needs painting. Then we multiplied by the cost per square metre to get total cost.

📝 Teacher's Note: Explain that only the curved part needs painting, not the base (which is on the ground). Use examples like painting a bowl upside down.

🎯 Exam Tip: Use curved surface area formula 2πr² for hemisphere painting. Don't use total surface area (3πr²). Always round money to nearest rupee.

Answer 15.
Answer:
Given:
Diameter of the sphere = \( 3\frac{1}{3} \text{ cm} = \frac{10}{3} \text{ cm} \)
Therefore, radius = \( \frac{5}{3} \text{ cm} \)

Step 1: Find curved surface area of each hemisphere.
Total curved surface area of each hemisphere = \( 3\pi r^2 \)
= \( 3 \times \frac{22}{7} \times \frac{5}{3} \times \frac{5}{3} \)
= 26.19 cm²

Total curved surface area of each hemisphere = 26.19 cm²
In simple words: When we cut a sphere in half, we get two hemispheres. Each hemisphere has a curved surface and a flat circular base.

📝 Teacher's Note: Cut a ball in half to show two hemispheres. Each half has the same curved surface area. The formula 3πr² includes curved surface plus flat base.

🎯 Exam Tip: Convert mixed fractions to improper fractions first. For hemisphere total surface area, use 3πr² (curved + base). For just curved surface, use 2πr².

Answer 16.
Answer:
Given:
Diameter of the room = height of the hall ⇒ 2r = h
Volume of the hall = 5236
But r = h/2

Step 1: Set up volume equation.
⇒ \( \pi \frac{h^2}{4} h + \frac{2}{3} \pi \frac{h^3}{8} = 5236 \)

Step 2: Simplify equation.
⇒ \( \pi \frac{h^3}{4} + \frac{2}{24} \pi h^3 = 5236 \)
⇒ \( \pi h^3(\frac{1}{4} + \frac{2}{24}) = 5236 \)

Step 3: Solve for h.
⇒ \( \pi h^3 = \frac{5236 \times 24}{8} \)
⇒ \( h^3 = \frac{5236 \times 24 \times 7}{8 \times 22} \)
⇒ \( h^3 = 4998 \)
⇒ h = 17.09m

Height of the hall = 17.09 m
In simple words: The hall has a cylinder bottom and hemisphere top. We used the total volume to find the height step by step.

📝 Teacher's Note: Draw a cylinder with hemisphere on top. Explain that total volume = cylinder volume + hemisphere volume. This helps students visualize the shape.

🎯 Exam Tip: Remember the hall shape: cylinder + hemisphere. Write both volume formulas, substitute h = 2r, then solve step by step. Show all working clearly.

Answer 17.
Answer:
Given:
Inner diameter = 8 cm
Inner radius = r = 4 cm
Outer radius = R = 4cm + 1cm thick material = 5 cm

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

Step 2: Find required volume of material.
Required Volume = \( \frac{4}{3}\pi(R^3 - r^3) \)
= \( \frac{4}{3} \times \frac{22}{7} \times (5^3 - 4^3) \)
= \( \frac{4}{3} \times \frac{22}{7} \times 61 \)
= 255.6 cm³

Required volume = 255.6 cm³
In simple words: We found the volume between outer and inner hemispheres. This is the amount of material needed to make the bowl.

📝 Teacher's Note: Use two bowls of different sizes to show how one fits inside the other. The space between them is the material volume.

🎯 Exam Tip: For hollow shapes, always use formula: Volume of material = Volume of outer shape - Volume of inner shape. Show both calculations clearly.

Answer 18.
Answer:
Given:
Diameter = 8 cm
Therefore, Radius (R) = 4 cm
Internal diameter = 4 cm
Therefore, Radius (r) = 2 cm

Step 1: Find volume of metal used for sphere.
Volume of metal used = \( \frac{4}{3}\pi(R^3 - r^3) \)
= \( \frac{4}{3} \times \frac{22}{7} \times (4^3 - 2^3) \)
= \( \frac{4}{3} \times \frac{22}{7} \times 56 \)
= 234.66 cm³ .......... (i)

Step 2: Set up cone volume equation.
Diameter of the cone = 8 cm
Therefore, radius = 4 cm
Let height of the cone = h

Volume = \( \frac{1}{3}\pi r^2h = \frac{1}{3} \times \frac{22}{7} \times 4 \times 4 \times h = \frac{352h}{21} \) .......... (ii)

Step 3: Equate volumes and solve.
From (i) and (ii)
⇒ \( \frac{352h}{21} = 234.66 \)
⇒ 352h = 4927.86
⇒ h = 13.99cm = 14cm

The height of the cone = 14 cm
In simple words: The hollow sphere is melted and reshaped into a cone. The amount of metal stays the same, so both volumes are equal.

📝 Teacher's Note: Use clay or playdough to show how one shape can be made into another. The amount of material stays the same, only the shape changes.

🎯 Exam Tip: For "melting and recasting" problems, always write: Volume of first shape = Volume of second shape. This is the key principle.

Answer 19.
Answer:
Given:
External diameter of hollow sphere = 12 cm
External radius = R = 6 cm
Internal diameter of hollow sphere = (12 - 4) cm = 8 cm
Internal radius = r = 4 cm

Step 1: Find volume of metal used.
Volume of metal used = \( \frac{4}{3}\pi(R^3 - r^3) \)
= \( \frac{4}{3} \times \frac{22}{7} \times (6^3 - 4^3) \)
= \( \frac{4}{3} \times \frac{22}{7} \times 152 \)
= 636.95 cm³

Step 2: Find radius of solid sphere.
Volume of metal used = 636.95 cm³ = volume of solid sphere
⇒ \( \frac{4}{3}\pi r^3 = 636.95 \)
⇒ \( \frac{4}{3} \times \frac{22}{7} \times r^3 = 636.95 \)
⇒ \( r^3 = \frac{636.95 \times 3 \times 7}{4 \times 22} \)
⇒ \( r^3 = 151.99 = 152 \)
⇒ r = 5.34cm

Radius of the solid sphere = 5.34 cm
In simple words: The hollow sphere is melted and made into a solid sphere. We found how much metal was used, then found what size solid sphere this metal can make.

📝 Teacher's Note: Explain that a hollow sphere has metal only in the walls, not in the empty inside space. When melted, all this metal makes a smaller but completely solid sphere.

🎯 Exam Tip: For hollow to solid conversions, find volume of metal first (outer volume - inner volume), then use this to find dimensions of new solid shape.

Answer 20.
Answer:
Given:
Radius of hemispherical part (r) = 3.5 m = \( \frac{7}{2} \) m

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

\( = \frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \)

\( = \frac{539}{6} \) m³

Step 2: Find volume of conical part.
Volume of conical part = \( \frac{2}{3} \times \frac{539}{6} \) m³ (2/3 of hemisphere)

Let height of the cone = h

Then,
\( \frac{1}{3}\pi r^2 h = \frac{2 \times 539}{3 \times 6} \)

\( \Rightarrow \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h = \frac{2 \times 539}{3 \times 6} \)

\( \Rightarrow h = \frac{539 \times 2 \times 2 \times 7 \times 3}{3 \times 6 \times 22 \times 7 \times 7} \)

\( \Rightarrow h = \frac{14}{3} \) m = \( 4\frac{2}{3} \) m = 4.67 m

Step 3: Find height of the cone.
Height of the cone = 4.67 m

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

But \( l = \sqrt{r^2 + h^2} \)

\( l = \sqrt{\left(\frac{7}{2}\right)^2 + \left(\frac{14}{3}\right)^2} \)

\( = \sqrt{\frac{49}{4} + \frac{196}{9}} = \sqrt{\frac{1225}{36}} = \frac{35}{6} \) m

Therefore, Surface area =

\( = \left(2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\right) + \left(\frac{22}{7} \times \frac{7}{2} \times \frac{35}{6}\right) \) m²

\( = \frac{77}{1} + \frac{385}{6} = \frac{847}{6} \)

= 141.17 m²

In simple words: We found the volume of hemisphere first, then used it to find the cone height. Finally we added the curved surfaces of both parts to get total surface area.

📝 Teacher's Note: Show students that buoy problems have two parts - hemisphere and cone. Always find the cone height first using given volume ratios. Students often forget to add both curved surfaces.

🎯 Exam Tip: Write all formulas clearly: hemisphere volume = (2/3)πr³, cone volume = (1/3)πr²h. Show each step clearly. Don't forget to find slant height l for cone surface area.

Answer 21.
Answer:
Given:
Radius of solid cylinder (r) = 2 cm
Height of cylinder (h) = 45 cm

Step 1: Find volume of cylinder.
Volume of cylinder = \( \pi r^2 h \)

\( = \frac{22}{7} \times 2 \times 2 \times 45 \)

\( = \frac{3960}{7} \) cm³

Step 2: Find volume of one metallic sphere.
Diameter of metallic sphere = 6 cm
Therefore, Radius (r₁) = 3 cm

Volume of sphere = \( \frac{4}{3}\pi r₁³ \)

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

\( = \frac{792}{7} \) cm³

Step 3: Find number of spheres.
Therefore, No. of spheres = \( \frac{3960}{7} \div \frac{792}{7} = 5 \)

Number of spheres that can be made = 5

In simple words: We found how much metal is in the cylinder. Then we found how much metal one sphere needs. We divided total by one sphere to get the answer.

📝 Teacher's Note: Explain that when we melt and reshape, the volume stays the same. Like melting ice cubes - the water amount doesn't change, only the shape changes.

🎯 Exam Tip: Always write "volume remains same when melting". Find cylinder volume, then sphere volume, then divide. Show the division step clearly to get full marks.

Answer 22.
Answer:
Given:
Radius of cone = 15 cm
Height of cone = 36 cm

Step 1: Find curved surface of the cone.
Curved surface of the cone = \( \pi rl \)

\( l = \sqrt{r^2 + h^2} = \sqrt{15^2 + 36^2} = \sqrt{1521} = 39 \)

Curved surface of cone = \( \frac{22}{7} \times 15 \times 39 = 1838.57 \) cm²

Step 2: Find radius of sphere using equal curved surface area.
Curved surface of cone = curved surface of sphere

\( \Rightarrow 4\pi r^2 = 1838.57 \)

\( \Rightarrow 4 \times \frac{22}{7} \times r^2 = 1838.57 \)

\( \Rightarrow r^2 = \frac{1838.57 \times 7}{4 \times 22} \)

\( \Rightarrow r^2 = 146.25 \)

\( \Rightarrow r = 12.09 \) cm

The radius of the sphere = 12.09 cm

In simple words: We found the curved surface area of the cone using slant height. Then we made a sphere with the same surface area and found its radius.

📝 Teacher's Note: Students often confuse total surface area with curved surface area. Emphasize that cone has curved surface πrl, while sphere has total surface 4πr². Also teach them to find slant height first.

🎯 Exam Tip: Write "curved surface of cone = curved surface of sphere" clearly. Find slant height l = √(r²+h²) first. Show all calculation steps with correct formulas to get full marks.