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

NCERT Solutions for Class 9 Mathematics Chapter 13 Free PDF Download

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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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