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NCERT Exemplar · Class 9 Mathematics Surface Areas and Volumes

38 questions · 38 still being checked

EXERCISE 13.1 1–10 (part 1 of 4)

  1. Write the correct answer in each of the following :

    Exercise 1

    The radius of a sphere is 2r\displaystyle 2 r, then its volume will be (A) 43πr3\displaystyle \frac{4}{3} \pi r^{3} (B) 4πr3\displaystyle 4 \pi r^{3} (C) 8πr33\displaystyle \frac{8 \pi r^{3}}{3} (D) 323πr3\displaystyle \frac{32}{3} \pi r^{3}

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    NCERT’s answer
    (D)
    (D) \(\displaystyle \frac{32}{3}\pi r^{3}\)Sphere of radius \(\displaystyle 2r\). \[V = \frac{4}{3}\pi (2r)^3 = \frac{4}{3}\pi\cdot 8r^3 \] \[V = \frac{32}{3}\pi r^3 \]

  2. Exercise 2

    The total surface area of a cube is 96 cm2\displaystyle 96 \mathrm{~cm}^{2}. The volume of the cube is: (A) 8 cm3\displaystyle 8 \mathrm{~cm}^{3} (B) 512 cm3\displaystyle 512 \mathrm{~cm}^{3} (C) 64 cm3\displaystyle 64 \mathrm{~cm}^{3} (D) 27 cm3\displaystyle 27 \mathrm{~cm}^{3}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 64\ \text{cm}^3\)Cube, total surface area \(\displaystyle 96\ \text{cm}^2\). \[6a^2 = 96 \quad\Rightarrow\quad a^2 = 16,\ a = 4\ \text{cm} \] \[V = a^3 = 64\ \text{cm}^3 \]

  3. Exercise 3

    A cone is 8.4\displaystyle 8.4 cm high and the radius of its base is 2.1\displaystyle 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is : (A) 4.2\displaystyle 4.2 cm (B) 2.1\displaystyle 2.1 cm (C) 2.4\displaystyle 2.4 cm (D) 1.6\displaystyle 1.6 cm

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 2.1\ \text{cm}\)Cone \(\displaystyle r=2.1\) cm, \(\displaystyle h=8.4\) cm, recast into a sphere of radius \(\displaystyle R\). \[\frac{1}{3}\pi r^2 h = \frac{4}{3}\pi R^3 \] \[R^3 = \frac{r^2 h}{4} = \frac{(2.1)^2(8.4)}{4} = 9.261 \] \[R = 2.1\ \text{cm} \]

  4. Exercise 4

    In a cylinder, radius is doubled and height is halved, curved surface area will be (A) halved (B) doubled (C) same (D) four times

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    NCERT’s answer
    (C)
    (C) sameCylinder, radius \(\displaystyle r\to 2r\), height \(\displaystyle h\to \frac{h}{2}\). \[\text{CSA} = 2\pi r h \] \[\text{CSA}' = 2\pi(2r)\left(\frac{h}{2}\right) = 2\pi r h \]

  5. Exercise 5

    The total surface area of a cone whose radius is r2\displaystyle \frac{r}{2} and slant height 2l\displaystyle 2 l is (A) 2πr(l+r)\displaystyle 2 \pi r(l+r) (B) πr(l+r4)\displaystyle \pi r\left(l+\frac{r}{4}\right) (C) πr(l+r)\displaystyle \pi r(l+r) (D) 2πrl\displaystyle 2 \pi r l

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    NCERT’s answer
    (B)
    (B) \(\displaystyle \pi r\left(l+\frac{r}{4}\right)\)Cone, radius \(\displaystyle \frac{r}{2}\), slant height \(\displaystyle 2l\). \[\text{TSA} = \pi r'(l'+r') = \pi\cdot\frac{r}{2}\left(2l+\frac{r}{2}\right) \] \[\text{TSA} = \pi\left(rl+\frac{r^2}{4}\right) = \pi r\left(l+\frac{r}{4}\right) \]

  6. Exercise 6

    The radii of two cylinders are in the ratio of 2\displaystyle 2:3\displaystyle 3 and their heights are in the ratio of 5\displaystyle 5:3. The ratio of their volumes is: (A) 10\displaystyle 10 : 17\displaystyle 17 (B) 20\displaystyle 20 : 27\displaystyle 27 (C) 17\displaystyle 17 : 27\displaystyle 27 (D) 20\displaystyle 20 : 37\displaystyle 37

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 20:27\)Radii \(\displaystyle 2:3\), heights \(\displaystyle 5:3\). \[\frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} = \frac{2^2\times 5}{3^2\times 3} \] \[\frac{V_1}{V_2} = \frac{20}{27} \]

  7. Exercise 7

    The lateral surface area of a cube is 256 m2\displaystyle 256 \mathrm{~m}^{2}. The volume of the cube is (A) 512 m3\displaystyle 512 \mathrm{~m}^{3} (B) 64 m3\displaystyle 64 \mathrm{~m}^{3} (C) 216 m3\displaystyle 216 \mathrm{~m}^{3} (D) 256 m3\displaystyle 256 \mathrm{~m}^{3}

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 512\ \text{m}^3\)Cube, lateral surface area \(\displaystyle 256\ \text{m}^2\). \[4a^2 = 256 \quad\Rightarrow\quad a^2 = 64,\ a = 8\ \text{m} \] \[V = a^3 = 512\ \text{m}^3 \]

  8. Exercise 8

    The number of planks of dimensions ( 4 m×50 cm×20 cm\displaystyle 4 \mathrm{~m} \times 50 \mathrm{~cm} \times 20 \mathrm{~cm} ) that can be stored in a pit which is 16\displaystyle 16 m long, 12m wide and 4\displaystyle 4 m deep is (A) 1900\displaystyle 1900 (B) 1920\displaystyle 1920 (C) 1800\displaystyle 1800 (D) 1840\displaystyle 1840

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 1920\)Plank \(\displaystyle 4\ \text{m}\times 0.5\ \text{m}\times 0.2\ \text{m}\); pit \(\displaystyle 16\ \text{m}\times 12\ \text{m}\times 4\ \text{m}\). \[\text{Number} = \frac{16\times12\times4}{4\times0.5\times0.2} = \frac{768}{0.4} \] \[\text{Number} = 1920 \]

  9. Exercise 9

    The length of the longest pole that can be put in a room of dimensions ( 10 m×10 m×5 m\displaystyle 10 \mathrm{~m} \times 10 \mathrm{~m} \times 5 \mathrm{~m} ) is (A) 15\displaystyle 15 m (B) 16\displaystyle 16 m (C) 10\displaystyle 10 m (D) 12\displaystyle 12 m

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 15\ \text{m}\)Room \(\displaystyle 10\ \text{m}\times 10\ \text{m}\times 5\ \text{m}\); longest pole is the space diagonal. \[d = \sqrt{10^2+10^2+5^2} = \sqrt{225} \] \[d = 15\ \text{m} \]

  10. Exercise 10

    The radius of a hemispherical balloon increases from 6\displaystyle 6 cm to 12\displaystyle 12 cm as air is being pumped into it. The ratios of the surface areas of the balloon in the two cases is (A) 1\displaystyle 1 : 4\displaystyle 4 (B) 1\displaystyle 1 : 3\displaystyle 3 (C) 2\displaystyle 2 : 3\displaystyle 3 (D) 2\displaystyle 2 : 1\displaystyle 1

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 1:4\)Hemisphere, radius \(\displaystyle 6\to12\) cm; surface area \(\displaystyle \propto r^2\). \[\frac{S_1}{S_2} = \frac{r_1^2}{r_2^2} = \frac{6^2}{12^2} \] \[\frac{S_1}{S_2} = \frac{1}{4} \]