NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 12: Surface Areas & Volumes

Exercise 12.4

Step 1: Understand the problem.

A metallic hemisphere is melted and reshaped into a cone. Since the material is the same, the volume of hemisphere = volume of cone.

Step 2: Write the formula for the volume of a hemisphere.

The volume of a sphere is \(V = \dfrac{4}{3}\pi r^3\).

A hemisphere is half a sphere, so:

\(V_{hemisphere} = \dfrac{1}{2} \times \dfrac{4}{3}\pi r^3 = \dfrac{2}{3}\pi r^3\).

Step 3: Substitute the radius of the hemisphere.

Here, \(r = 8\,\text{cm}\).

So, \(V_{hemisphere} = \dfrac{2}{3}\pi (8)^3 = \dfrac{2}{3}\pi (512) = \dfrac{1024}{3}\pi\,\text{cm}^3\).

Step 4: Write the formula for the volume of a cone.

The volume of a cone is \(V_{cone} = \dfrac{1}{3}\pi R^2 h\),

where \(R\) is the base radius, \(h\) is the height.

Step 5: Substitute the base radius of the cone.

Here, \(R = 6\,\text{cm}\).

So, \(V_{cone} = \dfrac{1}{3}\pi (6)^2 h = \dfrac{1}{3}\pi (36) h = 12\pi h\,\text{cm}^3\).

Step 6: Equate the volumes.

\(V_{hemisphere} = V_{cone}\)

\(\dfrac{1024}{3}\pi = 12\pi h\)

Step 7: Simplify the equation.

Cancel \(\pi\) on both sides:

\(\dfrac{1024}{3} = 12h\)

Multiply both sides by 3:

\(1024 = 36h\)

Divide both sides by 36:

\(h = \dfrac{1024}{36} = \dfrac{256}{9}\,\text{cm}\).

Step 8: Write the final answer in decimal form.

\(h \approx 28.44\,\text{cm}\).

Therefore, the height of the cone is \(28.44\,\text{cm}\).

Step 1: Write down the dimensions of the rectangular tank.

Length \(= 11\,\text{m}\), Breadth \(= 6\,\text{m}\), Height of water \(= 5\,\text{m}\).

Step 2: Find the volume of water in the rectangular tank.

\(V = \text{length} \times \text{breadth} \times \text{height}\)

\(V = 11 \times 6 \times 5 = 330\,\text{m}^3\).

Step 3: This water is poured into the cylindrical tank, so the volume of water in both tanks is the same.

Step 4: Write the formula for the volume of a cylinder.

\(V = \pi r^2 h\)

Here, radius \(r = 3.5\,\text{m}\), height = \(h\,\text{m}\) (to be found).

Step 5: Substitute values.

\(330 = \pi (3.5)^2 h\)

\(330 = \pi (12.25) h\)

\(330 = 12.25\pi h\)

Step 6: Solve for \(h\).

\(h = \dfrac{330}{12.25\pi}\)

\(h = \dfrac{330}{38.465} \approx 8.57\,\text{m}\).

Step 7: Write the exact simplified fraction.

\(h = \dfrac{60}{7}\,\text{m}\).

Therefore, the height of water in the cylindrical tank is \(\dfrac{60}{7}\,\text{m} \approx 8.57\,\text{m}.\)

Step 1: Understand the problem.

The box is made of iron. It is hollow inside (open box). To find the volume of iron used, we subtract the inside (internal) volume from the outside (external) volume.

Step 2: External dimensions of the box.

Length = 36 cm, Breadth = 25 cm, Height = 16.5 cm

So, external volume = \(36 \times 25 \times 16.5 = 14850\,\text{cm}^3\).

Step 3: Internal dimensions.

The thickness of the iron sheet = 1.5 cm. This thickness reduces the size on both sides of each dimension.

• Internal length = \(36 - 2 \times 1.5 = 33\,\text{cm}\)
• Internal breadth = \(25 - 2 \times 1.5 = 22\,\text{cm}\)
• Internal height = \(16.5 - 1.5 = 15\,\text{cm}\)

(For height, we subtract only once, because the top is open.)

Step 4: Internal volume.

Internal volume = \(33 \times 22 \times 15 = 10890\,\text{cm}^3\).

Step 5: Volume of iron used.

Iron volume = External volume − Internal volume

= \(14850 − 10890 = 3960\,\text{cm}^3\).

Step 6: Weight of the box.

Given, 1 cm³ of iron weighs 7.5 g.

So, weight = \(3960 \times 7.5 = 29700\,\text{g}\).

Convert grams to kilograms: \(29700 \div 1000 = 29.7\,\text{kg}\).

Final Answer: Iron volume = 3960 cm³, Weight of the box = 29.7 kg.

Step 1: Convert diameter into radius in cm.

Diameter of barrel = 5 mm = 0.5 cm. Radius = Diameter ÷ 2 = 0.5 ÷ 2 = 0.25 cm.

Step 2: Write the formula for volume of a cylinder.

Volume of cylinder = \(\pi r^2 h\)

Step 3: Substitute values.

\(r = 0.25\,\text{cm},\; h = 7\,\text{cm}\)

Volume = \(3.1416 \times (0.25)^2 \times 7\)

= \(3.1416 \times 0.0625 \times 7 = 1.375\,\text{cm}^3\)

Step 4: Find words written per cm³ of ink.

A full barrel (1.375 cm³) writes 3300 words.

So, words per cm³ = \(3300 \div 1.375 = 2400\,\text{words/cm}^3\)

Step 5: Convert given ink volume to cm³.

\(\dfrac{1}{5}\) litre = 0.2 litre. 1 litre = 1000 cm³. So, 0.2 litre = 200 cm³.

Step 6: Find total words from 200 cm³.

Total words = (200 cm³) × (2400 words/cm³) = 480000 words.

Final Answer: 4,80,000 words

Step 1: Write down the given data.

• Speed of water = \(10\,\text{m/min} = 1000\,\text{cm/min}\) (since \(1\,\text{m} = 100\,\text{cm}\)).
• Pipe diameter = \(5\,\text{mm} = 0.5\,\text{cm}\). Radius of pipe = \(0.25\,\text{cm}\).
• Conical vessel diameter = \(40\,\text{cm}\). Radius of cone = \(20\,\text{cm}\).
• Height (depth) of cone = \(24\,\text{cm}\).

Step 2: Find the volume of water flowing through the pipe per minute (flow rate).

The volume of water coming out in one minute = cross-sectional area of pipe × speed of flow.

Cross-sectional area of pipe = \(\pi r^2 = \pi (0.25)^2 = 0.0625\pi\,\text{cm}^2\).

So, volume per minute = \(0.0625\pi \times 1000 = 62.5\pi\,\text{cm}^3/ ext{min}\).

Step 3: Find the volume of the conical vessel.

Formula: \(V = \tfrac{1}{3}\pi r^2 h\).

Here, \(r = 20\,\text{cm}, h = 24\,\text{cm}\).

So, \(V = \tfrac{1}{3} \pi (20)^2 (24) = \tfrac{1}{3} \pi (400)(24) = 3200\pi\,\text{cm}^3\).

Step 4: Find the time required.

Time = Total volume ÷ Volume per minute

\(t = \dfrac{3200\pi}{62.5\pi} = \dfrac{3200}{62.5} = 51.2\,\text{minutes}\).

Final Answer: The conical vessel will be filled in 51.2 minutes.

Step 1: Identify the given values

• Diameter of cone = 9 m
• Radius, \(r = \dfrac{9}{2} = 4.5\,\text{m}\)
• Height, \(h = 3.5\,\text{m}\)

Step 2: Formula for volume of a cone

\(V = \dfrac{1}{3}\pi r^2 h\)

Step 3: Substitute the values

\(V = \dfrac{1}{3}\pi (4.5)^2 (3.5)\)

\(= \dfrac{1}{3}\pi (20.25)(3.5)\)

\(= \dfrac{1}{3}\pi (70.875)\)

\(= 23.625\pi\,\text{m}^3\)

Approximate value: \(V \approx 74.2\,\text{m}^3\)

Step 4: Formula for curved surface area (canvas needed)

Curved surface area (CSA) = \(\pi r l\)

where \(l\) is the slant height, given by: \(l = \sqrt{r^2 + h^2}\)

Step 5: Find slant height

\(l = \sqrt{(4.5)^2 + (3.5)^2} = \sqrt{20.25 + 12.25} = \sqrt{32.5}\)

\(l \approx 5.7\,\text{m}\)

Step 6: Find CSA

CSA = \(\pi (4.5)(5.7)\)

≈ \(80.6\,\text{m}^2\)

Final Answer:

• Volume of rice = \(74.2\,\text{m}^3\)
• Canvas required = \(80.6\,\text{m}^2\)

Step 1: Recall formula
The curved surface area (CSA) of a cylinder is:
\( CSA = \text{circumference of base} \times \text{height} \).

Step 2: Write given values
Height (length of pencil) = \( 25\,\text{cm} \).
Circumference of base = \( 1.5\,\text{cm} \).

Step 3: Find CSA of 1 pencil
\( CSA = 1.5 \times 25 = 37.5\,\text{cm}^2 \).

Step 4: Total CSA for all pencils
Number of pencils = \( 1,20,000 = 1.2 \times 10^5 \).
Total CSA = \( 37.5 \times 1.2 \times 10^5 = 4.5 \times 10^6\,\text{cm}^2 \).

Step 5: Convert into dm² (SI requirement)
Recall: \( 1\,\text{dm} = 10\,\text{cm} \Rightarrow 1\,\text{dm}^2 = 100\,\text{cm}^2 \).
So, \( 4.5 \times 10^6\,\text{cm}^2 = \dfrac{4.5 \times 10^6}{100} = 45{,}000\,\text{dm}^2 \).

Step 6: Find cost of colouring
Cost per dm² = Rs 0.05.
Total cost = \( 45{,}000 \times 0.05 = 2{,}250 \).

Final Answer: Rs 2,250

Step 1: Volume of water required to raise the pond level

The pond is a cuboid of length \(50\,m\), breadth \(44\,m\), and the required rise in water level is \(21\,cm = 0.21\,m\).

So, volume required = \(50 \times 44 \times 0.21 = 462\,m^3\).

Step 2: Volume of water flowing from the pipe in 1 hour

Diameter of pipe = \(14\,cm = 0.14\,m\). Radius = \(0.07\,m\).

Cross-sectional area of pipe = \(\pi r^2 = 3.1416 \times (0.07)^2 \approx 0.0154\,m^2\).

Speed of water = \(15\,km/h = 15000\,m/h\).

Discharge per hour = area × speed = \(0.0154 \times 15000 \approx 231\,m^3/h\).

Step 3: Time taken to fill required volume

We need \(462\,m^3\) and in 1 hour \(231\,m^3\) flows in.

So, time = \(462 / 231 = 2\,h\).

Final Answer: The water level will rise by 21 cm in 2 hours.

Step 1: Volume of the cuboid

The cuboid is given with dimensions: length = 4.4 m, breadth = 2.6 m, height = 1 m.

Volume of cuboid = length × breadth × height

= \(4.4 \times 2.6 \times 1 = 11.44 \, \text{m}^3\)

Step 2: Understand the pipe structure

The pipe is hollow. This means it has an outer cylinder and an inner cylinder.

• Internal radius (r) = 30 cm = 0.30 m
• Thickness = 5 cm = 0.05 m
• So, outer radius (R) = internal radius + thickness = 0.30 + 0.05 = 0.35 m

Step 3: Cross-sectional area of the hollow pipe

Cross-sectional area of hollow cylinder (annulus) = \(\pi (R^2 - r^2)\)

= \(\pi (0.35^2 - 0.30^2)\)

= \(\pi (0.1225 - 0.0900)\)

= \(\pi (0.0325)\)

= \(0.0325\pi \, \text{m}^2\)

Step 4: Volume of the pipe

Volume of hollow cylinder = cross-sectional area × length

= \(0.0325\pi \times L\)

Step 5: Equating volumes

The cuboid is recast into the pipe, so volumes are equal:

\(11.44 = 0.0325\pi L\)

Step 6: Solve for length

\(L = \dfrac{11.44}{0.0325\pi}\)

\(L \approx 112\, \text{m}\)

Final Answer: The length of the pipe is 112 m.

Step 1: Find the total volume of water displaced.

Each person displaces \(0.04\,\text{m}^3\) of water.

For 500 persons, total displaced volume:

\(500 \times 0.04 = 20\,\text{m}^3\).

Step 2: Relating displaced volume to rise in water level.

The pond has a rectangular (cuboidal) shape with length = \(80\,\text{m}\) and breadth = \(50\,\text{m}\).

Base area of the pond = length × breadth:

\(80 \times 50 = 4000\,\text{m}^2\).

If water rises by \(h\,\text{m}\), the extra volume added = base area × height rise = \(4000 \times h\).

Step 3: Equating volumes.

Total displaced volume = Extra volume of pond water.

So, \(4000h = 20\).

\(h = \dfrac{20}{4000} = 0.005\,\text{m}\).

Step 4: Convert to cm.

\(0.005\,\text{m} = 0.005 \times 100 = 0.5\,\text{cm}\).

Final Answer: The water level rises by 0.5 cm.

Step 1: Volume of the cuboidal box

A cuboid's volume is found by multiplying its length, breadth, and height:

\(V_{\text{box}} = 16 \times 8 \times 8 = 1024\,\text{cm}^3\).

Step 2: Volume of one glass sphere

The formula for the volume of a sphere is:

\(V_{\text{sphere}} = \dfrac{4}{3}\pi r^3\).

Here radius \(r = 2\,\text{cm}\).

So, \(V_{\text{sphere}} = \dfrac{4}{3}\pi (2)^3 = \dfrac{4}{3}\pi (8) = \dfrac{32}{3}\pi\,\text{cm}^3\).

Step 3: Volume of 16 spheres

Total volume occupied by 16 spheres:

\(V_{\text{16 spheres}} = 16 \times \dfrac{32}{3}\pi = \dfrac{512}{3}\pi\,\text{cm}^3\).

Step 4: Space left for water

The remaining volume inside the box (after placing the spheres) will be filled with water:

\(V_{\text{water}} = V_{\text{box}} - V_{\text{16 spheres}}\).

\(V_{\text{water}} = 1024 - \dfrac{512}{3}\pi\,\text{cm}^3\).

Step 5: Approximate value

Using \(\pi \approx 3.14\):

\(V_{\text{water}} \approx 1024 - \dfrac{512}{3} (3.14)\).

\(\dfrac{512}{3} \times 3.14 \approx 536\).

\(V_{\text{water}} \approx 1024 - 536 = 488\,\text{cm}^3\).

Final Answer: The box can be filled with about \(488\,\text{cm}^3\) of water.

Step 1: Recall the formula for the volume of a frustum of a cone.

\[ V = \dfrac{1}{3} \pi h \big(R^2 + r^2 + Rr\big) \]

Here:

• \(h = 16\,\text{cm}\) (height)
• \(R = 20\,\text{cm}\) (larger radius)
• \(r = 8\,\text{cm}\) (smaller radius)

Step 2: Substitute the values into the formula.

\[ V = \dfrac{1}{3} \pi (16) \big(20^2 + 8^2 + 20 \times 8\big) \]

Step 3: Simplify inside the brackets.

\(20^2 = 400\), \(8^2 = 64\), and \(20 \times 8 = 160\).

So, \(400 + 64 + 160 = 624\).

Step 4: Multiply with height.

\(16 \times 624 = 9984\).

Step 5: Multiply with \(\pi/3\).

\[ V = \dfrac{1}{3} \pi (9984) = 3328\pi\,\text{cm}^3 \]

Step 6: Approximate using \(\pi \approx 3.1416\).

\(3328 \times 3.1416 \approx 10452.65\,\text{cm}^3\).

Step 7: Convert to litres (SI unit for capacity).

Since \(1000\,\text{cm}^3 = 1\,\text{L}\),

\(10452.65\,\text{cm}^3 \approx 10.45\,\text{L}\).

Step 8: Find the cost.

Cost per litre = Rs 22.

Total cost = \(10.45 \times 22 \approx 229.9 \approx Rs 230\).

Final Answer: Capacity ≈ 10.45 L, Cost ≈ Rs 230.

Step 1: Write down the given dimensions.

• Cylinder: height \(h_c = 32\,\text{cm}\), radius \(r_c = 18\,\text{cm}\).
• Cone: height \(h_{cone} = 24\,\text{cm}\), radius \(R = ?\).

Step 2: Volume of the cylinder.

The formula for volume of a cylinder is:

\(V_{cyl} = \pi r_c^2 h_c\)

Substitute values: \(V_{cyl} = \pi \times (18)^2 \times 32\)

\(V_{cyl} = \pi \times 324 \times 32 = \pi \times 10368\)

So, \(V_{cyl} = 10368\pi\,\text{cm}^3\).

Step 3: Volume of the cone.

The formula for volume of a cone is:

\(V_{cone} = \dfrac{1}{3}\pi R^2 h_{cone}\)

Substitute height: \(V_{cone} = \dfrac{1}{3}\pi R^2 (24)\)

\(V_{cone} = 8\pi R^2\).

Step 4: Since the sand volume is the same, equate volumes.

\(V_{cyl} = V_{cone}\)

\(10368\pi = 8\pi R^2\)

Step 5: Simplify the equation.

Cancel \(\pi\) from both sides:

\(10368 = 8R^2\)

Divide both sides by 8:

\(R^2 = 1296\)

So, \(R = \sqrt{1296} = 36\,\text{cm}\).

Step 6: Find the slant height of the cone.

Formula: \(l = \sqrt{R^2 + h_{cone}^2}\)

\(l = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872}\)

\(l \approx 43.3\,\text{cm}\).

Final Answer: Radius = 36 cm, Slant height ≈ 43.3 cm.

Step 1: Write down the given dimensions

• Radius of both cylinder and cone, \(r = 3\,\text{cm}\)
• Height of cylinder, \(h_{\rm cyl} = 12\,\text{cm}\)
• Slant height of cone, \(l = 5\,\text{cm}\)

Step 2: Find the vertical height of the cone

The cone’s vertical height \(h_{\rm cone}\) can be found using Pythagoras theorem:

\[ h_{\rm cone} = \sqrt{l^2 - r^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\,\text{cm}. \]

Step 3: Total Surface Area (TSA)

TSA includes:

• Curved surface area (CSA) of cylinder = \(2\pi r h_{\rm cyl}\)
• Curved surface area of cone = \(\pi r l\)
• Base of cylinder (only one base, since top is attached to cone) = \(\pi r^2\)

So,

\[ TSA = 2\pi r h_{\rm cyl} + \pi r^2 + \pi r l \]

Substitute values:

\[ TSA = 2 \times 3.14 \times 3 \times 12 + 3.14 \times 3^2 + 3.14 \times 3 \times 5 \]

\[ = 226.08 + 28.26 + 47.1 = 301.44\,\text{cm}^2 \]

Step 4: Volume

Total volume = Volume of cylinder + Volume of cone

\[ V = \pi r^2 h_{\rm cyl} + \tfrac{1}{3}\pi r^2 h_{\rm cone} \]

Substitute values:

\[ V = 3.14 \times 3^2 \times 12 + \tfrac{1}{3} \times 3.14 \times 3^2 \times 4 \]

\[ = 339.12 + 37.68 = 376.8\,\text{cm}^3 \]

Final Answer:

Total Surface Area = 301.44 cm²

Volume = 376.8 cm³

Step 1: Convert the given volume into an improper fraction.

\(41\dfrac{19}{21} = \dfrac{41 \times 21 + 19}{21} = \dfrac{861 + 19}{21} = \dfrac{880}{21}\,\text{m}^3.\)

Step 2: Let the radius of the hemispherical dome be \(r\,\text{m}.\)

The internal diameter of the dome = total height of the building.

So, \(2r = h + r \;\Rightarrow\; h = r.\)

This means the cylindrical part and the radius of the dome are equal in height.

Step 3: Write the formula for total volume of the building.

Total volume = Volume of cylinder + Volume of hemisphere

\(= \pi r^2h + \dfrac{2}{3}\pi r^3\)

Step 4: Substitute \(h = r\).

Volume \(= \pi r^2(r) + \dfrac{2}{3}\pi r^3\)

\(= \pi r^3 + \dfrac{2}{3}\pi r^3\)

\(= \dfrac{5}{3}\pi r^3.\)

Step 5: Equate with given volume.

\(\dfrac{5}{3}\pi r^3 = \dfrac{880}{21}\)

Step 6: Take \(\pi = \dfrac{22}{7}.\)

\(\dfrac{5}{3} \times \dfrac{22}{7} r^3 = \dfrac{880}{21}\)

\(\dfrac{110}{21} r^3 = \dfrac{880}{21}\)

Step 7: Simplify.

\(110 r^3 = 880\)

\(r^3 = 8\)

\(r = 2\,\text{m}.\)

Step 8: Find the height of the cylindrical part.

Since \(h = r = 2\,\text{m},\) the cylinder’s height = 2 m.

Step 9: Find the total height of the building.

Total height = Height of cylinder + Radius of dome

= \(2 + 2 = 4\,\text{m}.\)

Final Answer: The height of the building = 4 m.

Step 1: Write the formula for the volume of a hemisphere.

Volume of hemisphere = \( \dfrac{2}{3} \pi r^3 \)

Here, radius \( r = 9 \; \text{cm} \).

Step 2: Calculate the volume of the hemispherical bowl.

\( \dfrac{2}{3} \pi (9)^3 = \dfrac{2}{3} \pi (729) \)

\( = 486 \pi \; \text{cm}^3 \)

So, the hemispherical bowl can hold \( 486 \pi \; \text{cm}^3 \) of liquid.

Step 3: Write the formula for the volume of a cylinder.

Volume of cylinder = \( \pi r^2 h \)

Here, radius \( r = 1.5 \; \text{cm}, \; h = 4 \; \text{cm} \).

Step 4: Calculate the volume of one bottle (cylinder).

\( \pi (1.5)^2 (4) = \pi (2.25)(4) \)

\( = 9 \pi \; \text{cm}^3 \)

So, each bottle can hold \( 9 \pi \; \text{cm}^3 \) of liquid.

Step 5: Find the number of bottles required.

Total liquid in bowl ÷ liquid in one bottle

\( \dfrac{486 \pi}{9 \pi} = 54 \)

Final Answer: 54 bottles are needed.

Step 1: Write down the dimensions.

• Radius of both cone and cylinder: \(r = 60\,\text{cm}\)
• Height of cone: \(h_c = 120\,\text{cm}\)
• Height of cylinder: \(h_{cy} = 180\,\text{cm}\)

Step 2: Formula for volume of a cylinder.

\(V_{\text{cylinder}} = \pi r^2 h\)

Substitute: \(V_{\text{cylinder}} = \pi (60)^2 (180)\)

\(= \pi (3600)(180) = 648000\pi\,\text{cm}^3\)

Step 3: Formula for volume of a cone.

\(V_{\text{cone}} = \tfrac{1}{3} \pi r^2 h\)

Substitute: \(V_{\text{cone}} = \tfrac{1}{3}\pi (60)^2 (120)\)

\(= \tfrac{1}{3}\pi (3600)(120) = 144000\pi\,\text{cm}^3\)

Step 4: Water displaced by the cone.

The cone is placed inside the cylinder, so it pushes away (displaces) water equal to its own volume.

Displaced water = \(144000\pi\,\text{cm}^3\).

Step 5: Remaining water in the cylinder.

Initial water in cylinder − Volume displaced by cone

\(= 648000\pi - 144000\pi = 504000\pi\,\text{cm}^3\).

Final Answer: \(5.04 \times 10^5 \pi\,\text{cm}^3\).

Step 1: Write down given data.

• Radius of pipe = 1 cm = 0.01 m
• Speed of water in pipe = 80 cm/s = 0.8 m/s
• Time = half an hour = 30 min = 1800 s
• Radius of tank = 40 cm = 0.40 m

Step 2: Find volume of water flowing into the tank in 1800 s.

Volume flow per second through pipe = Cross-sectional area × velocity

\( A_{pipe} = \pi r^2 = \pi (0.01)^2 = \pi (0.0001) = 0.0001\pi \, m^2 \)

Flow per second = \( A_{pipe} \times v = 0.0001\pi \times 0.8 = 0.00008\pi \, m^3/s \)

Total volume in 1800 s = \( 0.00008\pi \times 1800 = 0.144\pi \, m^3 \)

Step 3: Relate this volume to the tank.

Tank volume = Base area × height rise

Base area of tank = \( \pi R^2 = \pi (0.40)^2 = 0.16\pi \, m^2 \)

Step 4: Find rise in water level.

Rise = \( \dfrac{\text{Volume of inflow}}{\text{Base area}} = \dfrac{0.144\pi}{0.16\pi} = 0.90 \, m \)

Step 5: Convert to cm.

\(0.90 \, m = 90 \, cm\)

Final Answer: Rise in water level = 90 cm.

Step 1: Write the given data.

• Roof length = \(22\,\text{m}\)
• Roof breadth = \(20\,\text{m}\)
• Vessel diameter = \(2\,\text{m}\) ⇒ radius = \(1\,\text{m}\)
• Vessel height = \(3.5\,\text{m}\)

Step 2: Find the volume of water collected in the cylindrical vessel (this equals the volume of rainwater fallen on the roof).

Volume of cylinder = \(\pi r^2 h\)

\(= \dfrac{22}{7} \times (1)^2 \times 3.5\)

\(= 11\,\text{m}^3\)

Step 3: Let the depth of rainfall be \(x\,\text{m}\). Then, volume of rainwater on the roof = roof area × rainfall depth.

Roof area = \(22 \times 20 = 440\,\text{m}^2\)

So, volume of rainwater = \(440 \times x\,\text{m}^3\)

Step 4: Since all the rainwater fills the vessel, equate the two volumes:

\(440x = 11\)

\(x = \dfrac{11}{440} = 0.025\,\text{m}\)

Step 5: Convert to cm:

\(0.025\,\text{m} = 0.025 \times 100 = 2.5\,\text{cm}\)

Final Answer: Rainfall = 2.5 cm

Step 1: First, calculate the volume of the cuboid (the outer shape of the pen stand).

Formula: \( V = l \times b \times h \)

Here, \( l = 10\,\text{cm}, b = 5\,\text{cm}, h = 4\,\text{cm} \).

So, \( V = 10 \times 5 \times 4 = 200\,\text{cm}^3 \).

Step 2: Now calculate the volume of one conical depression (hole).

Formula for cone: \( V = \tfrac{1}{3}\pi r^2 h \)

Here, \( r = 0.5\,\text{cm}, h = 2.1\,\text{cm} \).

So, \( V = \tfrac{1}{3}\pi (0.5)^2 (2.1) \).

\( V = \tfrac{1}{3}\pi (0.25)(2.1) = \tfrac{1}{3}\pi (0.525) = 0.175\pi\,\text{cm}^3 \).

Step 3: Since there are 4 such conical depressions:

Total volume removed = \( 4 \times 0.175\pi = 0.7\pi\,\text{cm}^3 \).

Step 4: Calculate the volume of the cubical depression (hole).

Formula for cube: \( V = a^3 \)

Here, \( a = 3\,\text{cm} \).

So, \( V = 3^3 = 27\,\text{cm}^3 \).

Step 5: Subtract the volumes of all depressions (holes) from the cuboid volume.

Net volume = \( 200 - 27 - 0.7\pi \).

Step 6: Approximate value of \( \pi \approx 3.14 \).

So, \( 0.7\pi \approx 0.7 \times 3.14 = 2.198 \).

Net volume \( \approx 200 - 27 - 2.198 = 170.8\,\text{cm}^3 \).

Final Answer: The volume of wood in the pen stand is \( 173 - 0.7\pi\,\text{cm}^3 \approx 170.8\,\text{cm}^3 \).