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NCERT Exemplar · Class 9 Mathematics Surface Areas and Volumes

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EXERCISE 13.3 1–10 (part 3 of 4)

  1. Exercise 1

    Metal spheres, each of radius 2\displaystyle 2 cm, are packed into a rectangular box of internal dimensions 16\displaystyle 16 cm × 8\displaystyle 8 cm × 8\displaystyle 8 cm. When 16\displaystyle 16 spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. Give your answer to the nearest integer. [Use π=3.14\displaystyle \pi=3.14 ]

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    NCERT’s answer
    $\displaystyle 488 \mathrm{~cm}^{3}$
    \[V_{\text{box}} = 16\times 8\times 8 = 1024\ \text{cm}^3 \] \[V_{16\ \text{spheres}} = 16\left(\frac{4}{3}\pi r^3\right) = 16\cdot\frac{4}{3}(3.14)(2)^3 = 535.89\ \text{cm}^3 \] \[V_{\text{liquid}} = V_{\text{box}} - V_{16\ \text{spheres}} = 1024 - 535.89 \] Answer: \(\displaystyle V_{\text{liquid}} \approx 488\ \text{cm}^3\).

  2. Exercise 2

    A storage tank is in the form of a cube. When it is full of water, the volume of water is 15.625 m3\displaystyle 15.625 \mathrm{~m}^{3}.If the present depth of water is 1.3\displaystyle 1.3 m, find the volume of water already used from the tank.

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    NCERT’s answer
    $\displaystyle 7.5 \mathrm{~cm}^{3}$
    \[a^3 = 15.625 \quad\Rightarrow\quad a = 2.5\ \text{m} \quad \text{(cube, full volume)} \] \[\text{base area} = a^2 = 6.25\ \text{m}^2 \] \[V_{\text{used}} = (a - 1.3)\times 6.25 = 1.2\times 6.25 \] Answer: \(\displaystyle V_{\text{used}} = 7.5\ \text{m}^3\).

  3. Exercise 3

    Find the amount of water displaced by a solid spherical ball of diameter 4.2\displaystyle 4.2 cm, when it is completely immersed in water.

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    \[r = \frac{4.2}{2} = 2.1\ \text{cm} \] \[V_{\text{displaced}} = V_{\text{sphere}} = \frac{4}{3}\pi r^3 \quad \text{(Archimedes)} \] \[V_{\text{displaced}} = \frac{4}{3}\cdot\frac{22}{7}\cdot(2.1)^3 = \frac{4}{3}\cdot\frac{22}{7}\cdot 9.261 \] Answer: \(\displaystyle V_{\text{displaced}} = 38.808\ \text{cm}^3\).NCERT prints: $\displaystyle 14.8$ cm³ — a misprint: no reading of a $\displaystyle 4.2$ cm diameter gives it; r = $\displaystyle 2.1$ cm gives V = ($\displaystyle 4$/$\displaystyle 3$)($\displaystyle 22$/$\displaystyle 7$)($\displaystyle 2.1$)³ = $\displaystyle 38.808$ cm³.

  4. Exercise 4

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

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    NCERT’s answer
    $\displaystyle 471.42 \mathrm{~m}^{2}$
    Canvas needed is the cone's curved surface area. \[l^2 = r^2 + h^2 = 12^2 + 3.5^2 = 144 + 12.25 = 156.25 \quad\Rightarrow\quad l = 12.5\ \text{m} \] \[S = \pi r l = \frac{22}{7}\times 12\times 12.5 = \frac{3300}{7} \] Answer: \(\displaystyle S \approx 471.43\ \text{m}^2\) of canvas.

  5. Exercise 5

    Two solid spheres made of the same metal have weights 5920\displaystyle 5920 g and 740\displaystyle 740 g, respectively. Determine the radius of the larger sphere, if the diameter of the smaller one is 5\displaystyle 5 cm.

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    NCERT’s answer
    $\displaystyle 5$ cm
    \[\frac{V_{\text{large}}}{V_{\text{small}}} = \frac{5920}{740} = 8 \quad \text{(same metal} \Rightarrow \text{weight} \propto \text{volume)} \] \[r_{\text{small}} = \frac{5}{2} = 2.5\ \text{cm} \] \[\left(\frac{R}{r_{\text{small}}}\right)^3 = 8 \quad\Rightarrow\quad \frac{R}{r_{\text{small}}} = 2 \] \[R = 2(2.5) = 5\ \text{cm} \] Answer: \(\displaystyle R = 5\ \text{cm}\).

  6. Exercise 6

    A school provides milk to the students daily in a cylindrical glasses of diameter 7\displaystyle 7 cm . If the glass is filled with milk upto an height of 12\displaystyle 12 cm, find how many litres of milk is needed to serve 1600\displaystyle 1600 students.

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    NCERT’s answer
    739.$\displaystyle 2$ litres
    \[r = \frac{7}{2} = 3.5\ \text{cm}, \qquad h = 12\ \text{cm} \] \[V_{\text{glass}} = \pi r^2 h = \frac{22}{7}(3.5)^2(12) = 462\ \text{cm}^3 \] \[V_{\text{total}} = 1600\times 462 = 739200\ \text{cm}^3 = 739.2\ \text{litres} \quad (1\ \text{litre} = 1000\ \text{cm}^3) \] Answer: \(\displaystyle 739.2\) litres.

  7. Exercise 7

    A cylindrical roller 2.5\displaystyle 2.5 m in length, 1.75\displaystyle 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2\displaystyle 5500 \mathrm{~m}^{2}. How many revolutions did it make?

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    NCERT’s answer
    $\displaystyle 200$ revolutions
    Area covered in one revolution is the roller's curved surface area. \[S = 2\pi r l = 2\times\frac{22}{7}\times 1.75\times 2.5 = 27.5\ \text{m}^2 \] \[n = \frac{5500}{27.5} = 200 \] Answer: \(\displaystyle 200\) revolutions.

  8. Exercise 8

    A small village, having a population of 5000\displaystyle 5000 , requires 75\displaystyle 75 litres of water per head per day. The village has got an overhead tank of measurement 40 m×25 m×15 m\displaystyle 40 \mathrm{~m} \times 25 \mathrm{~m} \times 15 \mathrm{~m}. For how many days will the water of this tank last?

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    NCERT’s answer
    $\displaystyle 40$ days
    \[V_{\text{tank}} = 40\times 25\times 15 = 15000\ \text{m}^3 = 15000000\ \text{litres} \quad (1\ \text{m}^3 = 1000\ \text{litres}) \] \[\text{daily need} = 5000\times 75 = 375000\ \text{litres} \] \[\text{days} = \frac{15000000}{375000} = 40 \] Answer: the water lasts \(\displaystyle 40\) days.

  9. Exercise 9

    A shopkeeper has one spherical laddoo of radius 5cm. With the same amount of material, how many laddoos of radius 2.5\displaystyle 2.5 cm can be made?

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    NCERT’s answer
    $\displaystyle 8$ laddoos
    Reshaping the laddoo keeps its volume unchanged.\[\frac{4}{3}\pi (5)^{3} = n\cdot \frac{4}{3}\pi (2.5)^{3} \] \[n = \left(\frac{5}{2.5}\right)^{3} = 2^{3} \] \[n = 8 \]Answer: $\displaystyle 8$ laddoos of radius \(\displaystyle 2.5\) cm.

  10. Exercise 10

    A right triangle with sides 6\displaystyle 6 cm, 8\displaystyle 8 cm and 10\displaystyle 10 cm is revolved about the side 8\displaystyle 8 cm. Find the volume and the curved surface of the solid so formed.

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    Revolving \(\displaystyle \triangle ABC\) about the \(\displaystyle 8\) cm leg gives a cone.NCERT_Solution_Class9_Maths_Exemplar_Ch13_Ex13-3_Q10\[6^{2}+8^{2}=10^{2} \quad \text{(right angle at \(\displaystyle B\))} \] \[r = BA = 6, \quad h = BC = 8, \quad l = CA = 10 \] \[V = \frac{1}{3}\pi r^{2} h = \frac{1}{3}\pi (6)^{2}(8) = 96\pi \approx 301.59 \text{ cm}^3 \] \[S = \pi r l = \pi (6)(10) = 60\pi \approx 188.50 \text{ cm}^2 \]Answer: Volume \(\displaystyle 96\pi \approx 301.59\ \text{cm}^3\), curved surface area \(\displaystyle 60\pi \approx 188.50\ \text{cm}^2\).NCERT prints: $\displaystyle 304$ cm³ — a misprint: no value of π gives it; r = $\displaystyle 6$ cm, h = $\displaystyle 8$ cm give V = $\displaystyle 96$π ≈ $\displaystyle 301.6$ cm³.