Probability

Important Questions

Here are some essential questions for Class 10 Mathematics Chapter 12, Surface Area and Volumes, thoughtfully chosen to aid students in their preparation for the CBSE Class 10 Mathematics Examination in 2024-25. By practicing these varied problems, students can enhance their comprehension of Surface Area and Volumes concepts and refine their problem-solving skills.

Table of Contents

In Chapter 12 of Class 10 Mathematics,Surface Area and Volumes, we will study the surface areas and volumes of combinations of any two of the following geometric shapes: cubes, cuboids, spheres, hemispheres, and right circular cylinders/cones. We will explore how to calculate the total surface area, lateral surface area, and volume of various combinations involving these shapes.

The surface area of a three-dimensional shape refers to the total measure of all its external surfaces. It encompasses the combined areas of every face, side, and curved surface that make up the object. In essence, it quantifies the extent of the outer covering of the shape, accounting for all its visible and touchable parts.

(b) 14

(c) 28

(d) 56

**Ans.**(b) 14

**Explanation:**

Surface area of sphere = 616 cm^2\\
⇒ 4r^2 = 616\\
⇒ 4 × \dfrac{22}{7}× r^2 = 616\\
⇒ r^2 = 616 × \dfrac{7}{22×4}\\
⇒ r = 7 cm\\
So, the diameter = 2r = 2 × 7 = 14 cm.

(b) 1:64

(c) 1:4

(d) 1:128

**Ans.** (b) 1:64

**Explanation:**

The diameter of Moon is approximately one fourth of the diameter of Earth.

Let, Radius of Moon = r,

Then, Radius of Earth = 4r

\text{Required ratio = }\dfrac{\text{Volume of Moon}}{\text{Volume of Earth}} \\= \dfrac{\dfrac{4}{3}\pi r^3}{\dfrac{4}{3}\pi (4r)^3}\\= \dfrac{r^3}{64r^3} = \dfrac{1}{64}\\
= 1:64

**Ans.** Rainfall = 2.5 cm

**Explanation:**

Given, Area of the roof = 22 m × 20 cm

Height of cylindrical vessel = 3.5 m

Diameter of the vessel = 2 m

∴ Radius of the vessel = 1 m

Let the total rainfall be h m.

∴ Volume of rain water collected on the roof

= volume of the cylindrical vessel

⇒ 22 × 20 × h = π (1)3.5

⇒ 22 × 20 × h = \dfrac{22}{7} (3.5)\\
⇒ 20h = 0.5 \\
⇒ h = \dfrac{1}{40}m = 2.5 cm.

**Ans.** Height of the cylinder =\dfrac{8}{3}cm\\
**Explanation:**

Given, internal diameter of sphere = 6 cm

External diameter of sphere = 10cm

Base diameter of cylinder = 14 cm

Thus, internal radius of sphere ,r_1 = 3 cm\\
External radius of sphere r_2 = 5 cm\\
Base radius of cylinder, r = 7 cm

Let the height of the cylinder be h cm.

Thus, volume of cylinder = Volume of spherical shell

πr^2h = \dfrac{4}{3}π(r_2^3 -_1^3) \\
⇒ π(7)^2h = \dfrac{4}{3}π[(5)^3 - (3)^3]\\[4.5 bp]
⇒ 3(7)^2h = 4 [(5)^3 - (3)^3]\\[4.5 bp]
⇒ 3(49)h = 4 [125 - 27]\\[4.5 bp]
⇒ 3(49)h = 4 [98]\\[4.5 bp]
⇒ 3h = 4 [2]\\[4.5 bp]
⇒ h = \dfrac{8}{3} cm. \\[4.5 bp]

**Explanation:**

**Ans.** Given, Diameter of the cylinder = 12 cm

∴ Radius = 6 cm

Height of the cylinder = 15 cm

Number of toys = 12

Let the radius of the hemisphere be r cm

Thus, height of the cone = 3r cm

Now, volume of cylinder = π(6)^2 15 cm^3\\
=\text{ (36) (15) π cm}^3 \\
\text{= 540 π cm}^3\\
Now, total volume of 12 toys

= 12 [volume of hemisphere + volume of cone]
Thus, total volume of 12 toys

= 12\bigg[\dfrac{1}{3} \pi 3r(r)^2 + \dfrac{2}{3} \pi r^3\bigg] cm^3 \\[4.5 bp]
= 4[3πr^3 + 2πr^3]cm^3\\[4.5 bp]
=20πr^3 cm^3\\[4.5 bp]
Now, 20πr^3 = 540π\\[4.5 bp]
⇒ r^3 = 27 \\[4.5 bp]
⇒ r^3 = (3)^3\\[4.5 bp]
⇒ r = 3 cm

Thus, radius of the hemisphere = 3 cm and height of the cone = (3)(3) cm =9 cm

∴ Total height of the toy

= (9 + 3) cm = 12 cm.

Chapter No. | Chapter Name |
---|---|

Chapter 1 | Real Numbers |

Chapter 2 | Polynomials |

Chapter 3 | Pair of Linear Equations in Two Variable |

Chapter 4 | Quadratic Equations |

Chapter 5 | Arithmetic Progressions |

Chapter 6 | Triangles |

Chapter 7 | Coordinate Geometry |

Chapter 8 | Introduction to Trigonometry |

Chapter 9 | Some Applications of Trigonometry |

Chapter 10 | Circles |

Chapter 11 | Areas Related to Circle |

Chapter 12 | Surface Areas and Volumes |

Chapter 13 | Statistics |

Chapter 14 | Probability |

To improve your understanding of the Surface Area and Volumes chapter, you can explore oswal.io. This platform offers a range of practice questions designed to make learning easier. By practicing these questions, you can strengthen your understanding of various shapes and their measurements, as well as sharpen your skills in solving related problems.

Ans: The lateral surface area of a cylinder is calculated by multiplying its height by its circumference.

Ans: A hemisphere is half of a sphere. It shares the same base and curved surface, but the hemisphere lacks the bottom half of the sphere.

Ans: The volume of a cylinder is calculated by multiplying the area of its base by its height: V = πr²h, where ‘r’ is the radius and ‘h’ is the height.

Ans: The formula for the volume of a cone is V = (1/3)πr²h, where ‘r’ is the radius of the base and ‘h’ is the height.

Ans: The total surface area of a sphere is calculated using the formula A = 4πr², where ‘r’ represents the radius of the sphere.

Chapter Wise Important Questions for CBSE Board Class 10 Maths |
---|

Real Numbers |

Polynomials |

Pair of Linear Equations in Two Variables |

Quadratic Equations |

Arithmetic Progressions |

Triangles |

Coordinate Geometry |

Introduction to Trigonometry |

Some Applications of Trigonometry |

Circles |

Areas Related to Circles |

Surface Areas and Volumes |

Statistics |

Probability |

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