Table of Contents
Ans.(b) 14
Explanation:
Surface area of sphere = 616 cm^2\\
⇒ 4πr^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.
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
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)^2 (3.5)\\[4.5 bp]
⇒ 22 × 20 × h = \dfrac{22}{7} (3.5)\\[4.5 bp]
⇒ 20 h = 0.5 \\[4.5 bp]
⇒ h = \dfrac{1}{40}m = 2.5 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 -r_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 |
| 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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