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NCERT Exemplar · Class 10 Mathematics Surface Areas and Volumes

62 questions · 62 still being checked

EXERCISE 12.1 11–20 (part 2 of 7)

  1. Choose the correct answer from the given four options:

    Exercise 11

    A mason constructs a wall of dimensions 270 cm×300 cm×350 cm\displaystyle 270 \mathrm{~cm} \times 300 \mathrm{~cm} \times 350 \mathrm{~cm} with the bricks each of size 22.5 cm×11.25 cm×8.75 cm\displaystyle 22.5 \mathrm{~cm} \times 11.25 \mathrm{~cm} \times 8.75 \mathrm{~cm} and it is assumed that 18\displaystyle \frac{1}{8} space is covered by the mortar. Then the number of bricks used to construct the wall is
    (A)
    11100\displaystyle 11100
    (B)
    11200\displaystyle 11200
    (C)
    11000\displaystyle 11000
    (D)
    11300\displaystyle 11300

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 11200\)Mortar takes up one-eighth of the wall, so bricks fill \(\displaystyle \tfrac{7}{8}\) of it.\[V_{\text{wall}} = 270\times300\times350 = 28{,}350{,}000 \text{ cm}^3 \] \[V_{\text{bricks}} = \frac{7}{8}\times 28{,}350{,}000 = 24{,}806{,}250 \text{ cm}^3 \] \[V_{\text{brick}} = 22.5\times11.25\times8.75 = 2214.84375 \text{ cm}^3 \] \[n = \frac{24{,}806{,}250}{2214.84375} = 11200 \]

  2. Exercise 12

    Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2\displaystyle 2 cm and height 16\displaystyle 16 cm. The diameter of each sphere is
    (A)
    4\displaystyle 4 cm
    (B)
    3\displaystyle 3 cm
    (C)
    2\displaystyle 2 cm
    (D)
    6\displaystyle 6 cm

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 2 \text{ cm}\)Melting conserves volume, so the cylinder's volume equals twelve sphere volumes.\[V_{\text{cylinder}} = \pi r^2 h = \pi(1)^2(16) = 16\pi \text{ cm}^3 \] \[12\times\frac{4}{3}\pi r_s^3 = 16\pi \] \[r_s^3 = 1 \implies r_s = 1 \implies d = 2 \]

  3. Exercise 13

    The radii of the top and bottom of a bucket of slant height 45\displaystyle 45 cm are 28\displaystyle 28 cm and 7\displaystyle 7 cm, respectively. The curved surface area of the bucket is
    (A)
    4950 cm2\displaystyle 4950 \mathrm{~cm}^2
    (B)
    4951 cm2\displaystyle 4951 \mathrm{~cm}^2
    (C)
    4952 cm2\displaystyle 4952 \mathrm{~cm}^2
    (D)
    4953 cm2\displaystyle 4953 \mathrm{~cm}^2

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 4950 \text{ cm}^2\)The curved surface of a frustum uses the slant height with both radii.\[CSA = \pi(R+r)l = \pi(28+7)(45) \] \[CSA = \frac{22}{7}\times35\times45 = 4950 \text{ cm}^2 \]

  4. Exercise 14

    A medicine-capsule is in the shape of a cylinder of diameter 0.5\displaystyle 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2\displaystyle 2 cm. The capacity of the capsule is
    (A)
    0.36 cm3\displaystyle 0.36 \mathrm{~cm}^3
    (B)
    0.35 cm3\displaystyle 0.35 \mathrm{~cm}^3
    (C)
    0.34 cm3\displaystyle 0.34 \mathrm{~cm}^3
    (D)
    0.33 cm3\displaystyle 0.33 \mathrm{~cm}^3

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 0.36 \text{ cm}^3\)The two end hemispheres form one sphere; together with the cylinder they span the length.\[h + 2r = 2, \quad r = 0.25 \implies h = 1.5 \text{ cm} \] \[V = \pi r^2 h + \frac{4}{3}\pi r^3 = \frac{22}{7}(0.0625)(1.5) + \frac{4}{3}\cdot\frac{22}{7}(0.015625) \] \[V \approx 0.2946 + 0.0655 = 0.36 \text{ cm}^3 \]

  5. Exercise 15

    If two solid hemispheres of same base radius r\displaystyle r are joined together along their bases, then curved surface area of this new solid is
    (A)
    4πr2\displaystyle 4 \pi r^2
    (B)
    6πr2\displaystyle 6 \pi r^2
    (C)
    3πr2\displaystyle 3 \pi r^2
    (D)
    8πr2\displaystyle 8 \pi r^2

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 4 \pi r^2\)The two flat faces are glued together and hidden inside; only the two curved surfaces remain exposed.\[\text{CSA of one hemisphere} = 2\pi r^2 \] \[\text{CSA of solid} = 2\pi r^2 + 2\pi r^2 = 4\pi r^2 \]

  6. Exercise 16

    A right circular cylinder of radius r cm\displaystyle r \mathrm{~cm} and height h cm(h>2r)\displaystyle h \mathrm{~cm}(h>2 r) just encloses a sphere of diameter
    (A)
    r cm\displaystyle r \mathrm{~cm}
    (B)
    2r cm\displaystyle 2 r \mathrm{~cm}
    (C)
    h\displaystyle h cm
    (D)
    2h cm\displaystyle 2 h \mathrm{~cm}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 2r\) cmThe sphere's width is capped by the cylinder's circular cross-section, not by its height.\[\text{cylinder radius} = r \implies \text{largest enclosed sphere radius} = r \] \[\text{diameter} = 2r \]

  7. Exercise 17

    During conversion of a solid from one shape to another, the volume of the new shape will
    (A)
    increase
    (B)
    decrease
    (C)
    remain unaltered
    (D)
    be doubled

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    NCERT’s answer
    (C)
    (C) remain unalteredRecasting or reshaping a solid neither adds nor removes material.\[V_{\text{new shape}} = V_{\text{original shape}} \]

  8. Exercise 18

    The diameters of the two circular ends of the bucket are 44\displaystyle 44 cm and 24\displaystyle 24 cm. The height of the bucket is 35\displaystyle 35 cm. The capacity of the bucket is
    (A)
    32.7\displaystyle 7 litres
    (B)
    33.7\displaystyle 7 litres
    (C)
    34.7\displaystyle 7 litres
    (D)
    31.7\displaystyle 7 litres

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    NCERT’s answer
    (A)
    (A) $\displaystyle 32.7$ litresThe bucket is a frustum; its volume converts to litres by dividing by 1000.\[R = 22 \text{ cm}, \quad r = 12 \text{ cm}, \quad h = 35 \text{ cm} \] \[V = \frac{\pi h}{3}\left(R^2+r^2+Rr\right) \] \[V = \frac{22}{7}\times\frac{35}{3}\times(484+144+264) \] \[V = \frac{110}{3}\times892 = 32706.67 \text{ cm}^3 \] \[V = 32.7 \text{ litres} \]

  9. Exercise 19

    In a right circular cone, the cross-section made by a plane parallel to the base is a
    (A)
    circle
    (B)
    frustum of a cone
    (C)
    sphere
    (D)
    hemisphere

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    NCERT’s answer
    (A)
    (A) a circleA plane parallel to the base meets the cone's slanted surface evenly all the way round, tracing a smaller circle.

  10. Exercise 20

    Volumes of two spheres are in the ratio 64\displaystyle 64:27. The ratio of their surface areas is
    (A)
    3:4\displaystyle 3: 4
    (B)
    4\displaystyle 4 : 3\displaystyle 3
    (C)
    9:16\displaystyle 9: 16
    (D)
    16\displaystyle 16 : 9\displaystyle 9

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 16:9\)Volume scales as the cube of the radius; surface area scales as its square.\[\frac{V_1}{V_2}=\frac{64}{27}=\left(\frac{r_1}{r_2}\right)^3 \implies \frac{r_1}{r_2}=\frac{4}{3} \] \[\frac{S_1}{S_2}=\left(\frac{r_1}{r_2}\right)^2=\frac{16}{9} \]