Volume and Surface Area Problems and Solutions

  • 15. A cistern 6 m long and 4 m wide contains water up to a breadth of 1 m 25 cm. Find the total area of the wet surface.

    1. 42 m sqaure
    2. 49 m sqaure
    3. 52 m sqaure
    4. 64 m sqaure
    Answer :

    Option B

    Explanation:

    Area of the wet surface =
    2[lb+bh+hl] - lb = 2 [bh+hl] + lb
    = 2[(4*1.25+6*1.25)]+6*4 = 49 m square

  • 16. If a right circular cone of height 24 cm has a volume of 1232 cm cube, then the area of its curved surface is :

    1. \begin{aligned} 450 cm^2 \end{aligned}
    2. \begin{aligned} 550 cm^2 \end{aligned}
    3. \begin{aligned} 650 cm^2 \end{aligned}
    4. \begin{aligned} 750 cm^2 \end{aligned}
    Answer :

    Option B

    Explanation:

    Volume is given, we can calculate the radius from it, then by calculating slant height, we can get curved surface area.
    \begin{aligned}
    \frac{1}{3}*\pi *r^2*h = 1232 \\
    \frac{1}{3}*\frac{22}{7}*r^2*24 = 1232 \\
    r^2 = \frac{1232*7*3}{22*24} = 49 \\
    r = 7 \\
    \text{Now, r = 7cm and h = 24 cm } \\
    l = \sqrt{r^2+h^2} \\
    = \sqrt{7^2+24^2} = 25cm \\
    \text{Curved surface area =}\pi rl\\
    = \frac{22}{7}*7*25 = 550 cm^2
    \end{aligned}

  • 17. A swimming pool 9 m wide and 12 m long is 1 m deep on the shallow side and 4 m deep on the deeper side. Its volume is:

    1. 260
    2. 262
    3. 270
    4. 272
    Answer :

    Option C

    Explanation:

    Volume will be length * breadth * height, but in this case two heights are given so we will take average,

    \begin{aligned}
    Volume = \left(12*9*\left(\frac{1+4}{2}\right)\right)m^3 \\
    12*9*2.5 m^3 = 270 m^3
    \end{aligned}

  • 18. The volume of the largest right circular cone that can be cut out of a cube of edge 7 cm is:

    1. \begin{aligned} 79.8 cm^3 \end{aligned}
    2. \begin{aligned} 79.4 cm^3 \end{aligned}
    3. \begin{aligned} 89.8 cm^3 \end{aligned}
    4. \begin{aligned} 89.4 cm^3 \end{aligned}
    Answer :

    Option C

    Explanation:

    Volume of the largest cone = Volume of the cone with diameter of base 7 and height 7 cm

    \begin{aligned}
    \text{Volume of cone =}\frac{1}{3}\pi r^2h \\
    = \frac{1}{3}*\frac{22}{7}*3.5*3.5*7 \\
    = \frac{269.5}{3}cm^3 \\
    = 89.8 cm^3
    \end{aligned}

    Note: radius is taken as 3.5, as diameter is 7 cm

  • 19. A circular well with a diameter of 2 meters, is dug to a depth of 14 meters. What is the volume of the earth dug out.

    1. \begin{aligned} 40 m^3 \end{aligned}
    2. \begin{aligned} 42 m^3 \end{aligned}
    3. \begin{aligned} 44 m^3 \end{aligned}
    4. \begin{aligned} 46 m^3 \end{aligned}
    Answer :

    Option C

    Explanation:

    \begin{aligned}
    Volume = \pi r^2h \\
    Volume = \left(\frac{22}{7}*1*1*14\right)m^3 \\
    = 44 m^3
    \end{aligned}

  • 20. A hemisphere and a cone have equal bases. If their heights are also equal, then the ratio of their curved surface will be :

    1. \begin{aligned} 2:1 \end{aligned}
    2. \begin{aligned} 1:\sqrt{2} \end{aligned}
    3. \begin{aligned} \sqrt{2}:1 \end{aligned}
    4. \begin{aligned} \sqrt{3}:1 \end{aligned}
    Answer :

    Option C

    Explanation:

    Let the radius of hemisphere and cone be R,
    Height of hemisphere H = R.
    So the height of the cone = height of the hemisphere = R
    Slant height of the cone
    \begin{aligned}
    = \sqrt{R^2+R^2} \\
    = \sqrt{2}R \\
    \frac{\text{Hemisphere Curved surface area}}{\text{Cone Curved surface area}} = \\
    \frac{2\pi R^2}{\pi *R*\sqrt{2}R} \\
    = \sqrt{2}:1
    \end{aligned}

  • 21. The maximum length of a pencil that can he kept is a rectangular box of dimensions 8 cm x 6 cm x 2 cm, is

    1. \begin{aligned} 2\sqrt{17} \end{aligned}
    2. \begin{aligned} 2\sqrt{16} \end{aligned}
    3. \begin{aligned} 2\sqrt{26} \end{aligned}
    4. \begin{aligned} 2\sqrt{24} \end{aligned}
    Answer :

    Option C

    Explanation:

    In this question we need to calculate the diagonal of cuboid,
    which is =
    \begin{aligned}
    \sqrt{l^2+b^2+h^2} \\
    = \sqrt{8^2+6^2+2^2} \\
    = \sqrt{104} \\
    = 2\sqrt{26}
    \end{aligned}

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