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Madhyamik Mathematics Suggestion – Different Solid Figures বিভিন্ন ঘনবস্তু Chapter 19 | WBBSE Class 10

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Madhyamik Mathematics | Chapter 19
Different Problems Related to Solid Figures — বিভিন্ন ঘনবস্তু সংক্রান্ত সমস্যা

Madhyamik Mathematics Chapter 19 deals with different problems involving solid figures such as cubes, spheres, cylinders, cones, hemispheres and hollow spheres. The chapter is particularly useful for practising volume, surface area and relationships between different three-dimensional shapes.

This revision article presents the important question-answer areas from the chapter in a cleaner and more student-friendly format, covering MCQ, True or False, Fill in the Blanks, 2-mark questions and 5-mark problems.

04 MCQ Questions
03 True or False
05 Short Questions
02 5-Mark Problems

Chapter 19 – Important Topics at a Glance

The questions in this chapter mainly revolve around volume, surface area, curved surface area, total surface area and the comparison of different solid figures.

Question Type Marks Main Focus
MCQ 1 Volume and surface-area relationships
True or False 1 Basic properties of solid figures
Fill in the Blanks 1 Formula and concept recall
Short Answer 2 Formula-based calculations
Long Answer 5 Application of volume and surface-area formulas

Multiple Choice Questions | MCQ

Question 1

If a cube and a sphere have equal surface areas, what is the ratio of their volumes?

A. π : 6
B. √π : √6
C. √6 : √π
D. 6π
Answer: B. √π : √6

Using the equality of the surface areas of the cube and sphere and simplifying the corresponding volume ratio gives:

Vcube : Vsphere = √π : √6

Question 2

A hemisphere, a cylinder and a cone have the same base radius and the same height. Find the ratio of their volumes.

A. 2 : 3 : 1
B. 3 : 2 : 1
C. 1 : 2 : 3
D. 1 : 3 : 2
Answer: A. 2 : 3 : 1

For equal radius and height, compare the standard volume expressions of a hemisphere, cylinder and cone.

Hemisphere : Cylinder : Cone = 2 : 3 : 1

Question 3

A sphere and a circular cylinder have the same radius r. If their volumes are equal, find the height of the cylinder.

A. r³
B. r⁴
C. 3r⁴
D. 4r/3
Answer: D. 4r/3

Let the cylinder height be h. Equal volumes give:

4/3 πr³ = πr²h

Therefore, h = 4r/3.

Question 4

Find the ratio of the volume of a sphere to the volume of its circumscribed right circular cylinder.

A. 3 : 2
B. 2 : 1
C. 2 : 3
D. 1 : 2
Answer: C. 2 : 3

For a cylinder circumscribed about a sphere, the cylinder has radius r and height 2r. Comparing the two volumes gives:

Vsphere : Vcylinder = 2 : 3

True or False

If a cylinder and a hemisphere have equal volumes and equal bases, the hemisphere has the greater height. True
A pencil with one pointed end can be represented as a combination of a cone and a cylinder. True
The radius and height of a hemisphere can never be equal. True

Fill in the Blanks

1. The volume of a pencil pointed at both ends = volume of one cone + volume of a cylinder + volume of the other cone.
2. If the outer radius of a hollow sphere is r1 and the inner radius is r2, its total surface-area expression is represented by:
4π(r12 + r22)
3. If a solid cylinder is melted and recast into a solid cone, the volumes of the cylinder and cone are equal.

Short Answer Questions | 2 Marks

Question 1 — Sphere and Cylinder

A sphere and a cylinder have the same radius. If their volumes are equal, find the ratio of the radius of the sphere to the height of the cylinder.

Solution

Let the common radius be r and the height of the cylinder be h.

```
Sphere volume = 4/3 πr³
Cylinder volume = πr²h

Since their volumes are equal:

4/3 πr³ = πr²h

After cancelling the common factors:

h = 4r/3

Therefore:

r : h = 3 : 4
```

Question 2 — Cone and Cylinder Surface Areas

A cone has height h and radius r. A cylinder with the same height and the same radius has a curved surface area equal to the total surface area of the cone. Find the relation between h and r.

Solution

```

Let the slant height of the cone be √(h² + r²).

The total surface area of the cone is:

πr² + πr√(h² + r²)

The curved surface area of the cylinder is:

2πrh

According to the condition:

πr² + πr√(h² + r²) = 2πrh

On simplifying the relation and squaring the required expression, we obtain:

3h = 4r

Hence the required relation is:

h : r = 4 : 3
```

Question 3 — Sphere and Circumscribed Cube

Find the ratio of the volume of a sphere to the volume of the cube circumscribed about the sphere.

Solution

```

Let the radius of the sphere be r. The side of the circumscribed cube is therefore 2r.

Sphere volume = 4/3 πr³
Cube volume = (2r)³ = 8r³

Taking π = 22/7 as used in the source solution:

4/3 πr³ : 8r³ = 11 : 21

Therefore, the required ratio is:

11 : 21
```

Question 4 — Equal Volumes of Sphere and Cylinder

A sphere and a cylinder have equal volumes. The diameter of the sphere is equal to the radius of the cylinder. Find the ratio of the cylinder's height to its base radius.

Solution

```

Let the radius of the sphere be r. Its diameter is therefore 2r. According to the condition, the radius of the cylinder is also 2r.

Let the height of the cylinder be h.

Sphere volume = 4/3 πr³
Cylinder volume = π(2r)²h = 4πr²h

Since the volumes are equal:

4/3 πr³ = 4πr²h

Therefore:

h = r/3

The cylinder radius is 2r, so:

h : base radius = r/3 : 2r = 1 : 6

Hence, the required ratio is:

1 : 6
```

Question 5 — Equal Volumes of Cone and Sphere

A cone and a sphere have equal volumes. If both have the same radius, find the ratio of the height of the cone to its radius.

Solution

```

Let the common radius be r and the height of the cone be h.

Sphere volume = 4/3 πr³
Cone volume = 1/3 πr²h

Since the volumes are equal:

4/3 πr³ = 1/3 πr²h

Therefore:

h = 4r

Hence:

h : r = 4 : 1
```

Long Answer Questions | 5 Marks

Question 1 — Water Level Rise After Immersing Solid Cones

A cylindrical vessel has a diameter of 24 cm and contains some water. Sixty solid iron pieces in the shape of cones are completely immersed in the water. Each cone has a base diameter of 6 cm and height 4 cm. Find the increase in the water level.

Raised water level Cylindrical vessel
Schematic representation of the cylindrical vessel and completely immersed solid cones.

Solution

```
Diameter of each cone's base = 6 cm, so its radius = 3 cm.
Height of each cone = 4 cm.
Volume of one cone = 1/3 × π × 3² × 4 = 12π cm³.
Volume of 60 cones = 60 × 12π = 720π cm³.
Diameter of the cylindrical vessel = 24 cm, so its radius = 12 cm.
If the increase in water level is h cm, the additional cylindrical volume is π × 12² × h.
Since the immersed cones displace a volume of water equal to their total volume, 144πh = 720π.
Therefore h = 5 cm.
Increase in water level = 5 cm
```

Question 2 — Hollow Lead Sphere Recast into a Cylinder

A hollow sphere is made from a lead sheet 1 cm thick. Its outer radius is 6 cm. The lead sphere is melted and recast into a cylinder whose radius is 2 cm. Find the height of the cylinder.

r₁ = 6 cm Recast cylinder h
Schematic representation of the hollow sphere and the cylinder formed from the same lead.

Solution

```

The outer radius of the hollow sphere is 6 cm. Since the sheet thickness is 1 cm, the inner radius is:

r₂ = 6 − 1 = 5 cm

The volume of lead contained in the hollow sphere is the difference between the volumes of the outer and inner spheres:

Volume of lead = 4/3 π(6³ − 5³)
= 4/3 π(216 − 125) = 4/3 × 91π

The lead is recast into a cylinder of radius 2 cm. Let its height be h.

Cylinder volume = π × 2² × h = 4πh

Since no lead is lost during melting and recasting:

4πh = 4/3 × 91π

Therefore:

h = 91/3 = 30.33 cm approximately

Hence, the height of the resulting cylinder is approximately 30.33 cm.

```
Important source clarification: The wording of the second 5-mark problem is poorly rendered in the original source around the cylinder dimension. The published solution uses a cylinder radius of 2 cm, and the calculation above follows that source-supported value.

Important Solid Geometry Formulas

Solid Volume Important Surface Area
Cube a³ 6a²
Cuboid lbh 2(lb + bh + hl)
Cylinder πr²h Curved surface area = 2πrh
Cone 1/3 πr²h Total surface area = πr(r + l)
Sphere 4/3 πr³ Surface area = 4πr²
Hemisphere 2/3 πr³ Curved surface area = 2πr²

Quick Revision: Key Relationships

Situation Result
Sphere and cylinder have equal volume and common radius r h = 4r/3
Sphere and circumscribed cylinder Volume ratio = 2 : 3
Sphere and cone have equal volume with common radius h : r = 4 : 1
Sphere and cylinder equal volume, sphere diameter = cylinder radius h : r = 1 : 6
Sphere inside a circumscribed cube Side of cube = 2r

How to Solve Chapter 19 Problems Efficiently

Identify the solid figures. First determine whether the problem involves a sphere, cylinder, cone, hemisphere, cube or a combination of solids.
Write the required formula. Do not begin substituting values until the correct volume or surface-area formula is clear.
Convert dimensions carefully. Diameter and radius must not be confused. If the diameter is given, divide it by 2 before using the volume formula.
Use equality of volume when solids are melted or recast. If one solid is transformed into another without loss of material, their volumes remain equal.
Keep units consistent. Use the same unit throughout the calculation and write the final unit clearly.
Present the final answer separately. For a 2-mark or 5-mark question, make the final ratio, length or height easy to identify.
Exam Tip: Chapter 19 questions often become straightforward once the correct formula and dimension are identified. Pay special attention to whether the question gives a radius, diameter, height, curved surface area, total surface area or volume.

Chapter 19 Revision Checklist

Topic Revision Status
Cube and sphere volume comparison Important
Hemisphere, cylinder and cone volume ratio Important
Sphere and cylinder with equal volume Very Important
Circumscribed sphere and cylinder Important
Hollow sphere Very Important
Melting and recasting solids Very Important
Water displacement by solid objects Very Important

Final Revision Note

বিভিন্ন ঘনবস্তু সংক্রান্ত সমস্যা or Different Problems Related to Solid Figures is a calculation-focused part of Madhyamik Mathematics. The most useful preparation is to practise the standard formulas and then apply them carefully to mixed-solid situations.

For the examination, give special attention to problems involving equal volumes, recasting of solids, hollow spheres, water displacement and ratios between different solid figures. These problems require both formula knowledge and careful interpretation of the given dimensions.

Madhyamik Mathematics Suggestion Resources

For broader WBBSE Class 10 preparation, explore Cademy's Madhyamik Suggestion resource, which includes subject-wise revision resources including Mathematics.

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Madhyamik Mathematics Chapter 19 Different Solid Figures বিভিন্ন ঘনবস্তু WBBSE Class 10 Mathematics Suggestion Solid Geometry

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