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WBB • कक्षा X • Mathematics • अध्याय 16

अनुमानित समय: 75 minutes

प्रगति: अध्ययनरत

Right Circular Cone

Chapter 16 of WBBSE Class 10 Mathematics Ganit Prakash provides a comprehensive study of the mensuration of the Right Circular Cone (লম্ব বৃত্তাকার শঙ্কু). A right circular cone is a solid generated by the complete rotation of a right-angled triangle about one of its sides containing the right angle as an axis of revolution. The fixed side acts as the vertical height (h), the other perpendicular side traces out a circular base with radius (r), and the hypotenuse generates the curved lateral surface, becoming the slant height (l). By Pythagoras theorem, these three dimensions are bound by the fundamental relation l = sqrt(r^2 + h^2). When the curved surface of a cone is sliced along its slant height and flattened onto a plane, it forms a circular sector of radius l with arc length equal to the circumference of the circular base (2*pi*r). This geometric transformation proves that the curved surface area equals pi*r*l. Adding the area of the flat circular base (pi*r^2) gives the total surface area formula TSA = pi*r*(l + r). The volume of a right circular cone is exactly one-third the volume of a right circular cylinder having the same base radius and vertical height, given by V = (1/3)*pi*r^2*h. The chapter extensively covers high-frequency Madhyamik board problems, including calculations of canvas required for conical tents, budget estimations for canvas cloth, calculating the capacity and height of conical heaps of agricultural produce, and solving problems where dimensional parameters are varied in proportional ratios.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें

प्रस्तावना अध्याय अवधारणाएँ सूत्र एवं सर्वसमिकाएँ हल उदाहरण सामान्य गलतियाँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

क्या आपने कभी सोचा है?

Whether looking at the conical spire of a cathedral, a party clown hat, a traffic cone on the highway, or a massive conical heap of harvested golden paddy in rural Bengal, the right circular cone is one of the most mathematically elegant 3D solids in nature. How can rotating a flat, two-dimensional right-angled triangle through space instantly carve out a three-dimensional curved cone? And why does a cone hold exactly one-third the volume of a cylinder with identical base and height? In this chapter, we master the complete mensuration of right circular cones for the Madhyamik Board Examination.

यह अध्याय क्यों महत्वपूर्ण है

The right circular cone is an indispensable geometric form across architecture, industrial engineering, aerodynamics, and agricultural economics. In civil and architectural engineering, conical roofs, church steeples, cooling towers, and storage silos utilize the conical shape because it minimizes structural material while maximizing resistance against high wind loads and allowing rain and snow to slide off effortlessly. In aerospace engineering, the nose cones of rockets, supersonic fighter jets, and missiles are designed with conical and ogive geometry to minimize aerodynamic drag and disperse supersonic shock waves. In industry, funnels, hoppers, cyclone dust separators, and centrifugal concentrators rely on conical walls to guide granular materials or fluids downward under gravity. In agriculture, calculating the volume of conical grain heaps enables farmers and storage warehouses to estimate crop yields and storage capacities accurately. For Class 10 students preparing for the West Bengal Madhyamik Board Examination, Chapter 16 carries substantial weightage in the mensuration section, frequently featuring 4-mark word problems and compulsory objective questions testing the interplay between radius, vertical height, slant height, surface area, and volume.

अध्ययन से पूर्व (आवश्यक ज्ञान)

Concepts of circle mensuration: circumference C = 2πr and circular base area A = πr².
Pythagoras' Theorem for right-angled triangles: hypotenuse squared equals base squared plus perpendicular squared.
Mensuration of right circular cylinders from Chapter 8 (Curved surface area 2πrh and volume πr²h).
Elementary algebra: solving linear and quadratic equations and finding square roots.

इस अध्याय के लक्ष्य

Define a right circular cone as a solid of revolution generated by rotating a right-angled triangle about one of its perpendicular sides.
Identify and relate the fundamental geometric triad: base radius (r), vertical height (h), and slant height (l = √(r² + h²)).
Derive and apply the formula for the Curved (Lateral) Surface Area: CSA = πrl by analyzing the unfolded circular sector.
Derive and apply the formula for the Total Surface Area: TSA = πrl + πr² = πr(l + r).
Derive and apply the formula for the Volume: V = (1/3)πr²h, establishing the 1:3 volume ratio with a cylinder of identical base and height.
Solve practical real-world engineering problems involving conical tents, determining floor area, enclosed air volume, canvas area, and total canvas cost.
Analyze conical heaps of sand, gravel, and agricultural grain, relating the angle of repose to cone radius and height.
Master multi-step Madhyamik word problems involving ratios of radii, heights, surface areas, and volumes.

अध्याय रूपरेखा एवं प्रगति

1 Module 1: Geometric Definition and...

2 Module 2: The Fundamental Triad (r,...

3 Module 3: Curved Surface Area and T...

4 Module 4: Volume of Right Circular...

5 Module 5: Practical Applications &...

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

Module 1: Geometric Definition and Solid of Revolution

1.1 Definition of Right Circular Cone (লম্ব বৃত্তাকার শঙ্কু)

A Right Circular Cone is a three-dimensional geometric solid formed by rotating a right-angled triangle through one complete revolution ($360^\circ$) about one of the sides containing the right angle, keeping that side fixed.

Apex or Vertex (শীর্ষবিন্দু): The fixed top point $A$ of the cone opposite to the circular base.
Axis (অক্ষ): The straight line segment joining the apex $A$ to the center of the base $O$. In a right circular cone, this axis is strictly perpendicular to the plane of the circular base.
Circular Base (বৃত্তাকার ভূমি): The flat circular bottom surface traced by the perpendicular leg of the rotating right triangle.
Base Radius (ভূমির ব্যাসার্ধ, $r$): The radius $OB$ of the circular base.
Vertical Height (উচ্চতা, $h$): The perpendicular distance from the apex $A$ to the center of the circular base $O$.
Slant Height (তির্যক উচ্চতা, $l$): The distance from the apex $A$ to any point $B$ on the circumference of the base circle, generated by the hypotenuse of the rotating right triangle.

1.2 Why is it called 'Right Circular'?

It is termed circular because its base is a perfect circle. It is termed right because the straight line joining the vertex to the center of the base meets the base at a right angle ($90^\circ$). If the apex is tilted such that the axis is not perpendicular to the base, it is an oblique cone (which is outside the WBBSE secondary syllabus).

Element	Symbol	Geometric Role in Right Triangle ΔAOB
Vertical Height	$h$	Fixed perpendicular side along axis of rotation ($AO$)
Base Radius	$r$	Perpendicular base side rotating in horizontal plane ($OB$)
Slant Height	$l$	Hypotenuse generating the curved surface ($AB$)

Module 2: The Fundamental Triad (r, h, l) and Pythagoras' Relation

2.1 The Fundamental Geometric Relation

In the right circular cone, triangle $ riangle AOB$ is a right-angled triangle with $ngle AOB = 90^\circ$. By Pythagoras' Theorem:

$$\mathbf{l^2 = r^2 + h^2}$$ $$\mathbf{l = \sqrt{r^2 + h^2}}$$

From this fundamental relationship, if any two of the three parameters are known, the third can be found immediately:

To find Slant Height: $\mathbf{l = \sqrt{r^2 + h^2}}$
To find Vertical Height: $\mathbf{h = \sqrt{l^2 - r^2}}$
To find Base Radius: $\mathbf{r = \sqrt{l^2 - h^2}}$

2.2 Common Pythagorean Triplets in Cone Problems

Madhyamik examination questions frequently use standard integer Pythagorean triplets $(r, h, l)$:

$(3, 4, 5)$ or multiples: $(6, 8, 10)$, $(9, 12, 15)$, $(15, 20, 25)$
$(5, 12, 13)$ or multiples: $(10, 24, 26)$
$(7, 24, 25)$
$(8, 15, 17)$

Module 3: Curved Surface Area and Total Surface Area

3.1 Derivation of Curved Surface Area (CSA - বক্রতলের ক্ষেত্রফল)

To find the area of the curved surface, imagine cutting the cone open along a slant edge $AB$ and unrolling it flat on a table. The resulting shape is a sector of a circle:

The radius of this large sector is the slant height $l$ of the cone.
The curved arc length of the sector is equal to the perimeter of the circular base of the cone: $ ext{Arc length} = 2\pi r$.

Now, from the geometry of a circular sector: $$ ext{Area of Sector} = rac{1}{2} imes ext{Arc Length} imes ext{Radius of Sector}$$ $$ ext{CSA} = rac{1}{2} imes (2\pi r) imes l = \mathbf{\pi r l}$$

3.2 Sector Central Angle (θ)

The central angle $ heta$ (in degrees) of the unfolded sector is given by: $$rac{ heta}{360^\circ} = rac{ ext{Arc Length}}{ ext{Total Circumference of Radius } l} = rac{2\pi r}{2\pi l} = rac{r}{l}$$ $$\mathbf{ heta = rac{r}{l} imes 360^\circ}$$

3.3 Total Surface Area (TSA - সমগ্রতলের ক্ষেত্রফল)

A closed right circular cone consists of two distinct surfaces: the curved lateral surface and the flat circular base.

$$ ext{Total Surface Area} = ext{Curved Surface Area} + ext{Base Area}$$ $$ ext{TSA} = \pi r l + \pi r^2 = \mathbf{\pi r (l + r)}$$

Module 4: Volume of Right Circular Cone (আয়তন)

4.1 Volume Formula and Derivation

The volume (or capacity) of a right circular cone of base radius $r$ and vertical height $h$ is given by:

$$\mathbf{V = rac{1}{3}\pi r^2 h}$$

4.2 Relationship Between Cone and Cylinder Volumes

Consider a hollow cylinder and a hollow cone having the exact same base radius $r$ and height $h$. If you fill the cone completely with fine sand or water and pour it into the cylinder, it takes exactly three full cones to fill the cylinder completely!

$$ ext{Volume of Cylinder} = \pi r^2 h$$ $$\mathbf{ ext{Volume of Cone} = rac{1}{3} imes ext{Volume of Cylinder}}$$

The ratio of volumes of a cone and a cylinder having equal base and equal height is strictly $\mathbf{1 : 3}$.

Property	Right Circular Cylinder (Ch 8)	Right Circular Cone (Ch 16)
Curved Surface Area	$2\pi rh$	$\pi rl$
Total Surface Area	$2\pi r(r + h)$	$\pi r(l + r)$
Volume	$\pi r^2 h$	$rac{1}{3}\pi r^2 h$

Module 5: Practical Applications & Madhyamik Exam Word Problems

5.1 Conical Tents (তাঁবু সংক্রান্ত সমস্যা)

When a conical tent is pitched on the ground:

The floor area occupied by people or equipment is the flat circular base: $\mathbf{ ext{Floor Area} = \pi r^2}$.
The canvas (তিরপল বা কাপড়ের ক্ষেত্রফল) required to cover the tent is strictly the Curved Surface Area ($\mathbf{ ext{Canvas} = \pi r l}$), because the ground/floor is NOT made of tent canvas!
The air volume inside the tent available for breathing is the total volume: $\mathbf{V = rac{1}{3}\pi r^2 h}$.
If $N$ persons sleep inside the tent, and each person requires $a ext{ m}^2$ of floor space and $v ext{ m}^3$ of air: $$\pi r^2 = N imes a \quad ext{and} \quad rac{1}{3}\pi r^2 h = N imes v$$ $$\implies rac{1}{3}(N imes a) imes h = N imes v \implies \mathbf{h = rac{3v}{a}}$$ The required height of the tent is independent of the number of persons $N$!

5.2 Conical Heap of Sand or Grain (বালির স্তূপ)

When a cylindrical bucket of sand or grain is emptied onto the ground, it naturally forms a conical heap due to the friction between particles. By the Principle of Conservation of Volume: $$ ext{Volume of Cylindrical Container} = ext{Volume of Conical Heap}$$ $$\pi R_{ ext{cyl}}^2 H_{ ext{cyl}} = rac{1}{3}\pi r_{ ext{cone}}^2 h_{ ext{cone}}$$

महत्वपूर्ण सूत्र, सर्वसमिकाएँ एवं प्रमेय

Slant Height Formula

$$l = \sqrt{r^2 + h^2}$$

Vertical Height Formula

$$h = \sqrt{l^2 - r^2}$$

Base Radius Formula

$$r = \sqrt{l^2 - h^2}$$

Curved Surface Area (CSA)

$$\text{CSA} = \pi r l$$

Total Surface Area (TSA)

$$\text{TSA} = \pi r (l + r)$$

Volume of Cone

$$V = \frac{1}{3}\pi r^2 h$$

Sector Central Angle

$$\theta = \frac{r}{l} \times 360^\circ$$

Tent Height from Air & Floor Space

$$h = \frac{3v}{a}$$

अवधारणात्मक हल उदाहरण एवं अनुप्रयोग (Solved Examples)

उदाहरण 1

The base radius of a right circular cone is 7 cm and its vertical height is 24 cm. Find its slant height, curved surface area, total surface area, and volume (taking π = 22/7).

विस्तृत समाधान / उत्तर:

Given Data: Base radius $r = 7\text{ cm}$. Vertical height $h = 24\text{ cm}$. Step 1: Calculate Slant Height (l) Using the fundamental relation: $$l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}$$ Step 2: Calculate Curved Surface Area (CSA) $$\text{CSA} = \pi r l = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550\text{ cm}^2$$ Step 3: Calculate Total Surface Area (TSA) $$\text{TSA} = \pi r(l + r) = \frac{22}{7} \times 7 \times (25 + 7) = 22 \times 32 = 704\text{ cm}^2$$ Step 4: Calculate Volume (V) $$V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24 = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24$$ $$V = \frac{1}{3} \times 22 \times 7 \times 24 = 22 \times 7 \times 8 = 154 \times 8 = 1232\text{ cm}^3$$ Final Answer: Slant Height $= \mathbf{25\text{ cm}}$, CSA $= \mathbf{550\text{ cm}^2}$, TSA $= \mathbf{704\text{ cm}^2}$, Volume $= \mathbf{1232\text{ cm}^3}$.

उदाहरण 2

A conical tent is 10 m high and the diameter of its base is 48 m. Find: (i) the slant height of the tent, (ii) the cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs 70 (use π = 22/7).

विस्तृत समाधान / उत्तर:

Given Data: Height of the tent $h = 10\text{ m}$. Diameter $= 48\text{ m} \implies$ Base radius $r = \frac{48}{2} = 24\text{ m}$. Rate of canvas $= \text{Rs } 70\text{ per m}^2$. Step 1: Calculate Slant Height (l) $$l = \sqrt{r^2 + h^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26\text{ m}$$ Step 2: Calculate Canvas Required (Curved Surface Area) Since a tent does not have canvas on the floor: $$\text{Canvas Required} = \text{CSA} = \pi r l$$ $$\text{Canvas} = \frac{22}{7} \times 24 \times 26 = \frac{22 \times 624}{7} = \frac{13728}{7}\text{ m}^2$$ Step 3: Calculate Total Cost $$\text{Total Cost} = \text{Canvas Area} \times \text{Rate}$$ $$\text{Total Cost} = \frac{13728}{7} \times 70 = 13728 \times 10 = 1,37,280\text{ Rupees}$$ Final Answer: (i) Slant height $= \mathbf{26\text{ m}}$ (ii) Cost of canvas $= \mathbf{\text{Rs } 1,37,280}$.

उदाहरण 3

The ratio of the radius and height of a right circular cone is 3 : 4 and its volume is 301.44 cm³. Find the slant height of the cone (take π = 3.14).

विस्तृत समाधान / उत्तर:

Given Data: Let common ratio multiple be $x > 0$. Base radius $r = 3x\text{ cm}$. Vertical height $h = 4x\text{ cm}$. Volume $V = 301.44\text{ cm}^3$, $\pi = 3.14$. Step 1: Use the Volume Formula to Solve for x $$V = \frac{1}{3}\pi r^2 h$$ $$301.44 = \frac{1}{3} \times 3.14 \times (3x)^2 \times (4x)$$ $$301.44 = \frac{1}{3} \times 3.14 \times 9x^2 \times 4x$$ $$301.44 = 3.14 \times 3 \times 4x^3$$ $$301.44 = 37.68 \times x^3$$ $$x^3 = \frac{301.44}{37.68} = 8$$ $$x = \sqrt[3]{8} = 2$$ Step 2: Determine r, h, and Slant Height l $$r = 3x = 3 \times 2 = 6\text{ cm}$$ $$h = 4x = 4 \times 2 = 8\text{ cm}$$ $$l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$ Final Answer: The slant height of the cone is $\mathbf{10\text{ cm}}$.

उदाहरण 4

A conical tent is designed to accommodate 11 persons. Each person requires 4 m² of the floor space and 20 m³ of air for breathing. Find the vertical height and the base radius of the tent.

विस्तृत समाधान / उत्तर:

Given Data: Number of persons $N = 11$. Floor space required per person $= 4\text{ m}^2$. Air volume required per person $= 20\text{ m}^3$. Step 1: Determine Total Floor Area and Base Radius Total floor area $= N \times 4 = 11 \times 4 = 44\text{ m}^2$. Since floor is a circle: $$\pi r^2 = 44$$ $$\frac{22}{7} r^2 = 44 \implies r^2 = 44 \times \frac{7}{22} = 2 \times 7 = 14$$ $$r = \sqrt{14}\text{ m} \approx 3.74\text{ m}$$ Step 2: Determine Total Air Volume and Height Total air volume $= N \times 20 = 11 \times 20 = 220\text{ m}^3$. Volume of cone: $$V = \frac{1}{3}\pi r^2 h = 220$$ Substitute $\pi r^2 = 44$: $$\frac{1}{3} \times 44 \times h = 220$$ $$h = \frac{220 \times 3}{44} = 5 \times 3 = 15\text{ m}$$ Final Answer: The vertical height of the tent is $\mathbf{15\text{ m}}$ and the base radius is $\mathbf{\sqrt{14}\text{ m}}$ (approx $3.74\text{ m}$).

उदाहरण 5

A semi-circular thin sheet of metal of diameter 28 cm is bent and its bounding radii are joined to form an open conical cup. Find the depth and capacity (volume) of the conical cup (take π = 22/7).

विस्तृत समाधान / उत्तर:

Given Data: Diameter of semi-circular sheet $= 28\text{ cm} \implies$ Radius of sheet $R = 14\text{ cm}$.

Step 1: Geometric Transformation of Bending When the semi-circular sheet is bent into a cone:

The radius of the sheet becomes the slant height of the cone:

$$l = R = 14\text{ cm}$$

The semi-circular arc length becomes the circumference of the cone's base:

$$\text{Arc length of semicircle} = \pi R = \frac{22}{7} \times 14 = 44\text{ cm}$$

$$\implies 2\pi r = 44 \implies 2 \times \frac{22}{7} \times r = 44 \implies r = 7\text{ cm}$$

Step 2: Find Depth (Vertical Height h) of the Conical Cup

$$h = \sqrt{l^2 - r^2} = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} = \sqrt{49 \times 3} = 7\sqrt{3}\text{ cm}$$

Using $\sqrt{3} \approx 1.732$:

$$h \approx 7 \times 1.732 \approx 12.12\text{ cm}$$

Step 3: Calculate Capacity (Volume) of the Cup

$$V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 7\sqrt{3}$$

$$V = \frac{1}{3} \times 22 \times 7 \times 7\sqrt{3} = \frac{1078\sqrt{3}}{3}\text{ cm}^3$$

Evaluating with $\sqrt{3} \approx 1.732$:

$$V \approx \frac{1078 \times 1.732}{3} = \frac{1867.1}{3} \approx 622.37\text{ cm}^3$$

Final Answer: The depth of the cup is $\mathbf{7\sqrt{3}\text{ cm}}$ (approx $12.12\text{ cm}$) and its capacity is $\mathbf{\frac{1078\sqrt{3}}{3}\text{ cm}^3}$ (approx $622.37\text{ cm}^3$).

सामान्य गलतियाँ एवं परीक्षक के जाल (Examiner Traps)

सामान्य भ्रम / गलत उत्तर

Using vertical height h instead of slant height l in the curved surface area formula (writing πrh instead of πrl).

सही वैज्ञानिक तथ्य

Curved surface area ALWAYS depends on slant height: CSA = πrl. Height h is only used for volume: V = (1/3)πr²h.

सामान्य भ्रम / गलत उत्तर

Adding the base area (πr²) when calculating canvas required for a conical tent.

सही वैज्ञानिक तथ्य

A tent sits on the bare earth or a separate ground sheet; tent canvas only covers the lateral curved surface: Canvas = πrl.

सामान्य भ्रम / गलत उत्तर

Confusing diameter with radius when given the base diameter.

सही वैज्ञानिक तथ्य

Always check if diameter is given; immediately divide by 2: r = d / 2.

सामान्य भ्रम / गलत उत्तर

Forgetting the factor 1/3 in the volume formula.

सही वैज्ञानिक तथ्य

Volume of a cone is ONE-THIRD the volume of a cylinder: V = (1/3)πr²h.

सामान्य भ्रम / गलत उत्तर

Taking l = r + h instead of using Pythagoras theorem l = √(r² + h²).

सही वैज्ञानिक तथ्य

Slant height is the hypotenuse of right triangle ΔAOB: l² = r² + h², so l = √(r² + h²).

Geometric Architecture: Right Circular Cone & Sector Unfolding (WBBSE Class 10 Ganit Prakash)

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1

Right Circular Cone Definition: Solid of revolution formed by rotating a right-angled triangle 360° about one of its perpendicular sides.

मुख्य बिंदु 2

Fundamental Triad: Radius r, vertical height h, and slant height l.

मुख्य बिंदु 3

Pythagorean Relation: l² = r² + h², which gives l = √(r² + h²), h = √(l² - r²), and r = √(l² - h²).

मुख्य बिंदु 4

Unfolded Lateral Surface: Cutting along slant height yields a circular sector of radius l and arc length 2πr.

मुख्य बिंदु 5

Curved Surface Area: CSA = πrl.

मुख्य बिंदु 6

Total Surface Area: TSA = CSA + Base Area = πrl + πr² = πr(l + r).

मुख्य बिंदु 7

Volume of Cone: V = (1/3)πr²h.

मुख्य बिंदु 8

Cylinder vs Cone Volume Ratio: Volume of cone is exactly 1/3 of the volume of a cylinder of identical base and height (Ratio 1 : 3).

मुख्य बिंदु 9

Conical Tent Principle: Floor area = πr²; Canvas required = CSA = πrl; Air volume = V = (1/3)πr²h.

मुख्य बिंदु 10

Bending Semicircle into Cone: Semicircular sheet of radius R folds into a cone where slant height l = R and base circumference 2πr = πR (giving r = R / 2).

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

If the curved surface area of a right circular cone is 4070 cm² and its slant height is 37 cm, find its base diameter (use π = 22/7).

उत्तर एवं व्याख्या देखें

उत्तर: CSA = πrl => 4070 = (22/7) * r * 37. r = (4070 * 7) / (22 * 37). Notice 4070 / 37 = 110. So r = (110 * 7) / 22 = 5 * 7 = 35 cm. Therefore, base diameter = 2r = 2 * 35 = 70 cm.
Set up πrl = 4070, substitute l = 37, solve for r, and then multiply by 2 to get the diameter.

The base radius and vertical height of two cones are in the ratio 3 : 5 and 2 : 3 respectively. Find the ratio of their volumes.

उत्तर एवं व्याख्या देखें

उत्तर: Let radii be r1 = 3x, r2 = 5x and heights be h1 = 2y, h2 = 3y. Ratio of volumes V1 / V2 = [(1/3)π * r1² * h1] / [(1/3)π * r2² * h2] = (r1 / r2)² * (h1 / h2) = (3/5)² * (2/3) = (9/25) * (2/3) = 6/25. The ratio is 6 : 25.
Express the ratio of volumes as (r1/r2)² * (h1/h2).

If the height of a cone is doubled and its radius is halved, how does its volume change?

उत्तर एवं व्याख्या देखें

उत्तर: Original volume V1 = (1/3)π r² h. New radius r' = r/2 and new height h' = 2h. New volume V2 = (1/3)π (r/2)² (2h) = (1/3)π (r²/4) (2h) = (1/2) * (1/3)π r² h = V1 / 2. The volume is halved (decreases by 50%).
Radius is squared in the volume formula, so halving radius quarters the r² term, while doubling height only multiplies by 2.

A right triangle with perpendicular sides 3 cm and 4 cm is rotated about the side of 4 cm. What is the volume of the cone generated?

उत्तर एवं व्याख्या देखें

उत्तर: The side of 4 cm is fixed as the axis, so vertical height h = 4 cm. The side of 3 cm rotates, so base radius r = 3 cm. Volume V = (1/3)π r² h = (1/3) * (22/7) * 3² * 4 = (1/3) * (22/7) * 9 * 4 = (22/7) * 12 = 264/7 = 37.71 cm³.
The side about which the triangle rotates becomes the vertical height h.

Find the sector angle of the unrolled curved surface of a cone whose base radius is 5 cm and slant height is 15 cm.

उत्तर एवं व्याख्या देखें

उत्तर: Sector angle θ = (r / l) * 360° = (5 / 15) * 360° = (1 / 3) * 360° = 120°.
Use the formula θ = (r / l) * 360°.

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