NCERT Solutions for Class 9 Maths Chapter 13: Surface Areas and Volumes

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes PDF Download helps students solve all textbook problems with clear steps. These solutions follow the latest CBSE guidelines and support effective exam preparation. Surface Areas and Volumes Class 9th introduces students to practical applications of geometry in real life.

This chapter explains how to calculate the surface area and volume of 3D shapes like cubes, cylinders, and spheres. The Class 9 Maths Chapter 13 question answer section covers each exercise in a simple format. With NCERT Solutions for Class 9 Maths Chapter 13, students can clear their doubts quickly and score better marks.

Download the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes PDF Download to access complete practice material. Surface Areas and Volumes Class 9 questions and answers are also helpful for school tests and final exams. Get Class 9 Maths Surface Areas and Volumes question answer for free here.

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    Q. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

    Ans. 

    $$\text{Given, radius of bowl (r)} =\frac{10.5}{2}\\ = 5.25 cm\\ \text{Volume of hemisphere (V)} =\frac{2}{3}\pi r^2\\ =\frac{2}{3}×\frac{22}{7}×{(5.25)^3}\\ =\frac{2}{3}×\frac{22}{7}×{(5.25)×(5.25)×(5.25)}\\ = 303.1875 cm^3\\ \text{Hence, the hemispherical bowl can hold of milk 0.303 litre.}$$

    Q. How many square metres of canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m?

    Ans. 

    $$\text{Given, height of conical tent, h = 3.5 m}\\ \text{Radius of the base of conical tent, r = 12 m}\\ \text{Slant height, l} =\sqrt[]{h^2+r^2}\\ =\sqrt[]{{(3.5)}^2+{(12)}^2}\\ =\sqrt[]{{12.25}+{144}}\\ =\sqrt[]{{156.25}}\\ = 12.5 m\\ \text{Canvas required = Curved surface area of cone}\\(conical tent)\\ = \pi rl=\frac{22}{7}12×12.5\\ = 471.42 m^2\\ \text{Hence, the canvas required to make a conical tent is 471.42 m2.}$$

    Q. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface area of the balloon in two cases.

    Ans. 

    $$\text{ When, radius} (r_1) = 7 cm\\ \text{Surface area} = 4\pi r_1^2 =4×\frac{22}{7}×7×7= 616 cm^2\\ \text{When, radius} (r_2) = 14 cm\\ New\space\text{Surface area} = 4\pi r_2^2 =4×\frac{22}{7}×14×14= 2464cm^2\\ \text{Required ratio} =\frac{616}{2466}=\frac{1}{4}\space or\space1 : 4$$

    Q. A hemispherical tank is made up of an iron sheet of1 cm thickness. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

    Ans. 

    $$\text{Given, inner radius of hemispherical tank (r) = 1m}\\ = 100 cm\\ \text{Thickness of the iron sheet} = 1 cm\\ \text{Outer radius of hemispherical tank (r′) = 100 + 1}\\ = 101 cm\\ \text{Volume of the hemispherical tank} =\frac{2}{3}\pi(r′^3-r^3)\\ =\frac{2}{3}×\frac{27}{7}[{(101)^3-(100)^3}]\\ =\frac{44}{21}×[1030301×1000000]\\ =\frac{44}{21}×[30301]\\ = 63487.81 cm^3\\ = 0.063 m^3\space [ 1 m^3 = (10)^6 cm^3]\\ \text{Hence, the volume of the iron used to make the tank is 0.063 m3}$$

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