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NCERT Exemplar · Class 10 Mathematics Surface Areas and Volumes

62 questions · 62 still being checked

EXERCISE 12.4 11–20 (part 7 of 7)

  1. Exercise 11

    16\displaystyle 16 glass spheres each of radius 2\displaystyle 2 cm are packed into a cuboidal box of internal dimensions 16\displaystyle 16 cm × 8\displaystyle 8 cm × 8\displaystyle 8 cm and then the box is filled with water. Find the volume of water filled in the box.

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    NCERT’s answer
    \(\displaystyle 487.6 \mathrm{~cm}^3\)
    Volume of the box: \[V_{\text{box}} = 16 \times 8 \times 8 = 1024\ \text{cm}^3 \] Volume of one sphere \(\displaystyle r = 2\ \text{cm}\): \[V_{\text{sphere}} = \frac{4}{3}\pi r^3 = \frac{4}{3}\times\frac{22}{7}\times 8 = \frac{704}{21}\ \text{cm}^3 \] Volume of the $\displaystyle 16$ spheres: \[16 \times \frac{704}{21} = \frac{11264}{21}\ \text{cm}^3 \approx 536.38\ \text{cm}^3 \] Water fills the remaining space: \[V_{\text{water}} = 1024 - 536.38 = 487.62\ \text{cm}^3 \] Answer: \(\displaystyle \approx 487.62\ \text{cm}^3\).

  2. Exercise 12

    A milk container of height 16\displaystyle 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8\displaystyle 8 cm and 20\displaystyle 20 cm respectively. Find the cost of milk at the rate of Rs. 22\displaystyle 22 per litre which the container can hold.

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    Frustum: lower radius \(\displaystyle r = 8\ \text{cm}\), upper radius \(\displaystyle R = 20\ \text{cm}\), height \(\displaystyle h = 16\ \text{cm}\). \[V = \frac{1}{3}\pi h\left(R^2 + Rr + r^2\right) = \frac{1}{3}\times\frac{22}{7}\times 16\times(400+160+64) \] \[V = \frac{1}{3}\times\frac{22}{7}\times 16 \times 624 = \frac{73216}{7}\ \text{cm}^3 \approx 10459.43\ \text{cm}^3 \] Convert to litres and price at Rs. $\displaystyle 22$/litre: \[\text{Cost} = \frac{10459.43}{1000}\times 22 \approx \text{Rs. } 230.11 \] Answer: Rs. $\displaystyle 230.11$ (approx.).

  3. Exercise 13

    A cylindrical bucket of height 32\displaystyle 32 cm and base radius 18\displaystyle 18 cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24\displaystyle 24 cm, find the radius and slant height of the heap.

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    NCERT’s answer
    $\displaystyle 36$ cm, $\displaystyle 43.27$ cm
    Volume of sand equals the cylinder's volume, radius \(\displaystyle 18\ \text{cm}\), height \(\displaystyle 32\ \text{cm}\): \[V = \pi r^2 h = \pi \times 18^2 \times 32 = 10368\pi\ \text{cm}^3 \] This forms a cone of height \(\displaystyle 24\ \text{cm}\), radius \(\displaystyle R\): \[\frac{1}{3}\pi R^2 \times 24 = 10368\pi \] \[R^2 = \frac{10368 \times 3}{24} = 1296 \implies R = 36\ \text{cm} \] Slant height: \[l = \sqrt{R^2 + h^2} = \sqrt{1296 + 576} = \sqrt{1872} = 12\sqrt{13}\ \text{cm} \approx 43.27\ \text{cm} \] Answer: radius \(\displaystyle 36\ \text{cm}\), slant height \(\displaystyle 12\sqrt{13} \approx 43.27\ \text{cm}\).

  4. Exercise 14

    A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6\displaystyle 6 cm and 12\displaystyle 12 cm, respectively. If the the slant height of the conical portion is 5\displaystyle 5 cm, find the total surface area and volume of the rocket [Use π=3.14\displaystyle \pi=3.14].

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    Cylinder: radius \(\displaystyle r = 3\ \text{cm}\), height \(\displaystyle h = 12\ \text{cm}\). Cone: same radius, slant height \(\displaystyle l = 5\ \text{cm}\). \[h_{\text{cone}} = \sqrt{l^2 - r^2} = \sqrt{25 - 9} = 4\ \text{cm} \] Total surface = cylinder's curved surface + cone's curved surface + base circle (the top is covered by the cone): \[\text{TSA} = 2\pi r h + \pi r l + \pi r^2 = 2(3.14)(3)(12) + (3.14)(3)(5) + (3.14)(3)^2 \] \[\text{TSA} = 226.08 + 47.1 + 28.26 = 301.44\ \text{cm}^2 \] Volume = cylinder + cone: \[V = \pi r^2 h + \frac{1}{3}\pi r^2 h_{\text{cone}} = (3.14)(9)(12) + \frac{1}{3}(3.14)(9)(4) \] \[V = 339.12 + 37.68 = 376.8\ \text{cm}^3 \] Answer: TSA \(\displaystyle = 301.44\ \text{cm}^2\), Volume \(\displaystyle = 376.8\ \text{cm}^3\).

  5. Exercise 15

    A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains 411921 m3\displaystyle 41 \frac{19}{21} \mathrm{~m}^3 of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building?

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    NCERT’s answer
    $\displaystyle 4$ m
    Let the common radius of the cylinder and the dome be \(\displaystyle r\) and the cylinder's height be \(\displaystyle h\).\[\text{Total height} = h+r \quad \text{(cylinder height + hemisphere radius)} \] \[2r = h+r \quad \text{(given: dome's diameter = total height)} \] \[\implies h=r \] \[\pi r^2 h + \frac{2}{3}\pi r^3 = 41\tfrac{19}{21} = \frac{880}{21} \] \[\pi r^3 + \frac{2}{3}\pi r^3 = \frac{880}{21} \] \[\frac{5}{3}\pi r^3 = \frac{880}{21} \] \[\frac{5}{3}\times\frac{22}{7}\times r^3 = \frac{880}{21} \] \[r^3 = 8 \implies r=2\text{ m} \] \[H = h+r = r+r = 4\text{ m} \] Answer: height of the building \(\displaystyle =4\) m.

  6. Exercise 16

    A hemispherical bowl of internal radius 9\displaystyle 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5\displaystyle 1.5 cm and height 4\displaystyle 4 cm. How many bottles are needed to empty the bowl?

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    NCERT’s answer
    $\displaystyle 54$
    \[V_{\text{bowl}} = \frac{2}{3}\pi (9)^3 = 486\pi \ \text{cm}^3 \] \[V_{\text{bottle}} = \pi (1.5)^2(4) = 9\pi \ \text{cm}^3 \] \[n = \frac{V_{\text{bowl}}}{V_{\text{bottle}}} = \frac{486\pi}{9\pi} = 54 \] Answer: $\displaystyle 54$ bottles.

  7. Exercise 17

    A solid right circular cone of height 120\displaystyle 120 cm and radius 60\displaystyle 60 cm is placed in a right circular cylinder full of water of height 180\displaystyle 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.

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    NCERT’s answer
    \(\displaystyle 1.584 \mathrm{~m}^3\)
    The cone is fully submerged in the brim-full cylinder, so water equal to the cone's volume overflows.\[V_{\text{cyl}} = \pi (60)^2(180) = 648000\pi \ \text{cm}^3 \] \[V_{\text{cone}} = \frac{1}{3}\pi (60)^2(120) = 144000\pi \ \text{cm}^3 \] \[V_{\text{left}} = V_{\text{cyl}} - V_{\text{cone}} = 504000\pi \ \text{cm}^3 \] \[V_{\text{left}} = 504000\times\frac{22}{7} = 1\,584\,000 \ \text{cm}^3 \] Answer: \(\displaystyle 504000\pi\) cm\(\displaystyle ^3\) \(\displaystyle = 1{,}584{,}000\) cm\(\displaystyle ^3\) (\(\displaystyle 1.584\) m\(\displaystyle ^3\)).

  8. Exercise 18

    Water flows through a cylindrical pipe, whose inner radius is 1\displaystyle 1 cm, at the rate of 80\displaystyle 80 cm/sec in an empty cylindrical tank, the radius of whose base is 40\displaystyle 40 cm. What is the rise of water level in tank in half an hour?

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    NCERT’s answer
    $\displaystyle 90$ cm
    \[Q = \pi (1)^2(80) = 80\pi \ \text{cm}^3/\text{s} \quad \text{(rate of water entering)} \] \[t = 30\times60 = 1800 \ \text{s} \] \[V = 80\pi\times1800 = 144000\pi \ \text{cm}^3 \] \[\pi (40)^2 h = 144000\pi \quad \text{(same volume fills the tank to height } h\text{)} \] \[1600h = 144000 \implies h = 90 \ \text{cm} \] Answer: the water rises by $\displaystyle 90$ cm.

  9. Exercise 19

    The rain water from a roof of dimensions 22 m×20 m\displaystyle 22 \mathrm{~m} \times 20 \mathrm{~m} drains into a cylindrical vessel having diameter of base 2\displaystyle 2 m and height 3.5\displaystyle 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.

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    NCERT’s answer
    2.$\displaystyle 5$ cm
    Let the rainfall be \(\displaystyle h\) cm, so \(\displaystyle \tfrac{h}{100}\) m. \[\text{Rainwater volume} = 22\times20\times\frac{h}{100} \ \text{m}^3 \] \[\text{Vessel volume} = \pi (1)^2(3.5) = \frac{22}{7}\times3.5 = 11 \ \text{m}^3 \] \[22\times20\times\frac{h}{100} = 11 \] \[4.4h = 11 \implies h = 2.5 \ \text{cm} \] Answer: rainfall \(\displaystyle =2.5\) cm.

  10. Exercise 20

    A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are 10\displaystyle 10 cm, 5\displaystyle 5 cm and 4\displaystyle 4 cm. The radius of each of the conical depressions is 0.5\displaystyle 0.5 cm and the depth is 2.1\displaystyle 2.1 cm. The edge of the cubical depression is 3\displaystyle 3 cm. Find the volume of the wood in the entire stand.

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    NCERT’s answer
    \(\displaystyle 170.8 \mathrm{~cm}^3\)
    \[V_{\text{cuboid}} = 10\times5\times4 = 200 \ \text{cm}^3 \] \[V_{4\text{ cones}} = 4\times\frac{1}{3}\pi (0.5)^2(2.1) = 0.7\pi \ \text{cm}^3 \] \[V_{4\text{ cones}} = 0.7\times\frac{22}{7} = 2.2 \ \text{cm}^3 \] \[V_{\text{cube}} = 3^3 = 27 \ \text{cm}^3 \] \[V_{\text{wood}} = 200 - 2.2 - 27 = 170.8 \ \text{cm}^3 \] Answer: \(\displaystyle 170.8\) cm\(\displaystyle ^3\).