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NCERT Exemplar · Class 10 Mathematics Surface Areas and Volumes

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EXERCISE 12.3 1–10 (part 4 of 7)

  1. Exercise 1

    Three metallic solid cubes whose edges are 3\displaystyle 3 cm, 4\displaystyle 4 cm and 5\displaystyle 5 cm are melted and formed into a single cube.Find the edge of the cube so formed.

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    $\displaystyle 6$ cm
    Melting conserves volume, so the new cube's volume equals the sum of the three. \[V = 3^3+4^3+5^3 = 27+64+125 = 216 \text{ cm}^3 \] \[a^3 = 216 \implies a = \sqrt[3]{216} = 6 \]Answer: \(\displaystyle 6\) cm.

  2. Exercise 2

    How many shots each having diameter 3\displaystyle 3 cm can be made from a cuboidal lead solid of dimensions 9cm × 11cm × 12cm?

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    $\displaystyle 84$
    Melting conserves volume, so the cuboid's volume equals the total volume of the shots. \[V_{\text{cuboid}} = 9\times11\times12 = 1188 \text{ cm}^3 \] \[V_{\text{shot}} = \frac{4}{3}\pi r^3 = \frac{4}{3}\cdot\frac{22}{7}\cdot(1.5)^3 = \frac{99}{7} \text{ cm}^3 \] \[n = 1188 \div \frac{99}{7} = 84 \]Answer: \(\displaystyle 84\) shots.

  3. Exercise 3

    A bucket is in the form of a frustum of a cone and holds 28.490\displaystyle 28.490 litres of water. The radii of the top and bottom are 28\displaystyle 28 cm and 21\displaystyle 21 cm, respectively. Find the height of the bucket.

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    $\displaystyle 15$ cm
    \[V = 28.490 \text{ litres} = 28490 \text{ cm}^3 \] \[V = \frac{1}{3}\pi h\left(R^2+Rr+r^2\right) \] \[28490 = \frac{1}{3}\times\frac{22}{7}\times h\times\left(28^2+28\times21+21^2\right) \] \[28490 = \frac{1}{3}\times\frac{22}{7}\times h\times 1813 \] \[h = 15 \]Answer: \(\displaystyle 15\) cm.

  4. Exercise 4

    A cone of radius 8\displaystyle 8 cm and height 12\displaystyle 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

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    $\displaystyle 7$:$\displaystyle 1$
    The cut leaves a smaller cone at the apex, similar to the original with linear scale \(\displaystyle \tfrac12\).\[V_{\text{cone}} = \frac13\pi(8)^2(12) = 256\pi \text{ cm}^3 \] \[V_{\text{small cone}} = \left(\frac12\right)^3 V_{\text{cone}} = 32\pi \text{ cm}^3 \] \[V_{\text{frustum}} = 256\pi-32\pi = 224\pi \text{ cm}^3 \] \[V_{\text{frustum}} : V_{\text{small cone}} = 224\pi:32\pi = 7:1 \]Answer: \(\displaystyle 7:1\).

  5. Exercise 5

    Two identical cubes each of volume 64 cm3\displaystyle 64 \mathrm{~cm}^3 are joined together end to end. What is the surface area of the resulting cuboid?

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    $\displaystyle 160$ \(\displaystyle cm^{2}\)
    Each cube has edge \(\displaystyle \sqrt[3]{64}=4\) cm; joining them end to end hides one face of each.\[TSA = 2(lb+bh+hl) = 2(4\times4+4\times8+8\times4) \] \[TSA = 2(16+32+32) = 160 \text{ cm}^2 \]Answer: \(\displaystyle 160\) cm\(\displaystyle ^2\).

  6. Exercise 6

    From a solid cube of side 7\displaystyle 7 cm, a conical cavity of height 7\displaystyle 7 cm and radius 3\displaystyle 3 cm is hollowed out. Find the volume of the remaining solid.

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    $\displaystyle 277$ \(\displaystyle cm^{3}\)
    The cone's height equals the cube's edge, so its apex just reaches the opposite face.\[V_{\text{cube}} = 7^3 = 343 \text{ cm}^3 \] \[V_{\text{cone}} = \frac13\pi r^2h = \frac13\times\frac{22}{7}\times3^2\times7 = 66 \text{ cm}^3 \] \[V_{\text{remaining}} = 343-66 = 277 \text{ cm}^3 \]Answer: \(\displaystyle 277\) cm\(\displaystyle ^3\).

  7. Exercise 7

    Two cones with same base radius 8\displaystyle 8 cm and height 15\displaystyle 15 cm are joined together along their bases. Find the surface area of the shape so formed.

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    NCERT’s answer
    \(\displaystyle 855 \mathrm{~cm}^2\) (approx.)
    Joining the cones at their bases hides both circular bases, leaving only the two curved surfaces.\[l = \sqrt{r^2+h^2} = \sqrt{8^2+15^2} = \sqrt{289} = 17 \] \[CSA_{\text{one cone}} = \pi r l = \frac{22}{7}\times8\times17 = \frac{2992}{7} \text{ cm}^2 \] \[TSA = 2\times\frac{2992}{7} = \frac{5984}{7} \approx 854.86 \text{ cm}^2 \]Answer: \(\displaystyle \dfrac{5984}{7}\approx854.86\) cm\(\displaystyle ^2\).

  8. Exercise 8

    Two solid cones A and B are placed in a cylinderical tube as shown in the Fig.12.9. The ratio of their capacities are 2\displaystyle 2:1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder. NCERT_Question_Class10_Maths_Exemplar_Ch12_Ex12-3_Q8

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    $\displaystyle 14$ cm, $\displaystyle 7$ cm; $\displaystyle 132$ \(\displaystyle cm^{3}\), $\displaystyle 66$ \(\displaystyle cm^{3}\); $\displaystyle 396$ \(\displaystyle cm^{3}\)
    Both cones share the tube's radius, so capacity is proportional to height: \[\frac{V_A}{V_B} = \frac{\dfrac{1}{3}\pi r^2 h_A}{\dfrac{1}{3}\pi r^2 h_B} = \frac{h_A}{h_B} = \frac{2}{1} \] \[h_A + h_B = 21 \text{ cm} \] \[h_A = 14 \text{ cm}, \quad h_B = 7 \text{ cm} \] \[V_A = \frac{1}{3}\pi (3)^2(14) = 132 \text{ cm}^3 \] \[V_B = \frac{1}{3}\pi (3)^2(7) = 66 \text{ cm}^3 \] \[V_{\text{cylinder}} = \pi (3)^2(21) = 594 \text{ cm}^3 \] \[V_{\text{remaining}} = 594 - (132+66) = 396 \text{ cm}^3 \]Answer: heights $\displaystyle 14$ cm, $\displaystyle 7$ cm; capacities $\displaystyle 132$ cm³, $\displaystyle 66$ cm³; remaining volume $\displaystyle 396$ cm³.

  9. Exercise 9

    An ice cream cone full of ice cream having radius 5\displaystyle 5 cm and height 10\displaystyle 10 cm as shown in the Fig.12.10. Calculate the volume of ice cream, provided that its 16\displaystyle \frac{1}{6} part is left unfilled with ice cream. NCERT_Question_Class10_Maths_Exemplar_Ch12_Ex12-3_Q9

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    NCERT’s answer
    327.$\displaystyle 4$ \(\displaystyle cm^{3}\)
    The $\displaystyle 10$ cm runs from the apex to the top of the hemisphere, so the cone's height is \(\displaystyle 10-5 = 5\) cm: \[V_{\text{cone}} = \frac{1}{3}\pi (5)^2(5) = \frac{125\pi}{3} \text{ cm}^3 \] \[V_{\text{hemisphere}} = \frac{2}{3}\pi (5)^3 = \frac{250\pi}{3} \text{ cm}^3 \] \[V_{\text{total}} = \frac{125\pi}{3}+\frac{250\pi}{3} = 125\pi = 125\times\frac{22}{7} = \frac{2750}{7} \text{ cm}^3 \] One-sixth is left unfilled, so the ice cream fills \(\displaystyle \frac{5}{6}\) of this: \[V_{\text{ice cream}} = \frac{5}{6}\times\frac{2750}{7} = \frac{6875}{21} \approx 327.4 \text{ cm}^3 \]Answer: \(\displaystyle \dfrac{6875}{21} \approx 327.4\) cm³.

  10. Exercise 10

    Marbles of diameter 1.4\displaystyle 1.4 cm are dropped into a cylindrical beaker of diameter 7\displaystyle 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6\displaystyle 5.6 cm.

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    $\displaystyle 150$
    The rise in water level equals the marbles' total volume: \[V_{\text{rise}} = \pi (3.5)^2(5.6) = 68.6\pi \text{ cm}^3 \] Each marble has radius $\displaystyle 0.7$ cm: \[V_{\text{marble}} = \frac{4}{3}\pi (0.7)^3 = \frac{1.372\pi}{3} \text{ cm}^3 \] \[n = \frac{V_{\text{rise}}}{V_{\text{marble}}} = \frac{68.6\times 3}{1.372} = 150 \] Answer: $\displaystyle 150$ marbles.