Volume and Surface Area of Solid Figures

Volume and Surface Area of Solid Figures

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Aptitude miscellaneous
  1. Find the area of the iron sheet required to prepare a cone double the height of 12 cm with diameter of the base 14 cm. ?








  1. View Hint View Answer Discuss in Forum

    r = Radius of the cone = 14/2 = 7 cm
    h = Height of the cone = 12 x 2 = 24 cm
    ∴ Slant height (l) = √r2 + h2
    = √(7)2 + (24)2
    = √49 + 576
    = √625
    = 25 cm
    Area of the sheet = Total surface area
    = (πrl + πr2)

    Correct Option: D

    r = Radius of the cone = 14/2 = 7 cm
    h = Height of the cone = 12 x 2 = 24 cm
    ∴ Slant height (l) = √r2 + h2
    = √(7)2 + (24)2
    = √49 + 576
    = √625
    = 25 cm
    Area of the sheet = Total surface area
    = (πrl + πr2)
    = πr(l + r)
    = (22/7) x 7 x (25 + 7) = 704 sq cm

  1. What part of a ditch 48 m long, 16.5 m broad and 4 m deep can be filled by the earth got by digging a cylindrical tunnel of diameter 4 m and length 56 m ?








  1. View Hint View Answer Discuss in Forum

    Volume of the earth dugout as a tunnel = πr2h = (22/7) x 2 x 2 x 56 = 704 m3
    Volume of the ditch = (48 x 33)/(2 x 4) = 24 x 33 x 4 = 3168 m3

    Correct Option: B

    Volume of the earth dugout as a tunnel = πr2h = (22/7) x 2 x 2 x 56 = 704 m3
    Volume of the ditch = (48 x 33)/(2 x 4) = 24 x 33 x 4 = 3168 m3
    ∴ Part required = 704/3168 = 2/9

  1. Water flows into a tank 100 m x 150 m x 2 m through a rectangular pipe 1.5 m x 1.25 m at a speed of 20 km/h. In what time, will the water rise by 2 m ?








  1. View Hint View Answer Discuss in Forum

    Volume of water in the tank = 100 x 500 x 2 = 30000 m3
    Speed of water = 20 x 1000/60 = 1000/3 m/min
    water flow per minute = (15/10) x (125/100) x (1000/3) = 625 m3

    Correct Option: C

    Volume of water in the tank = 100 x 500 x 2 = 30000 m3
    Speed of water = 20 x 1000/60 = 1000/3 m/min
    water flow per minute = (15/10) x (125/100) x (1000/3) = 625 m3
    Time taken = 30000/625 = 48 min

  1. A hospital room is to accommodate 56 patients. It should be done in such a way that every patient gets 2.2 m2 of floor and 8.83 of space. If the length of the room is 14 m, then breadth and height of the room are respectively ?








  1. View Hint View Answer Discuss in Forum

    Let the breadth and height of room be m and h m, respectively.
    Then, according to the question, l x b = n x Area occupied by one patient
    ⇒ 14 x b = 56 x 22
    ∴ b = (56 x 22)/14 = 8.8 m3
    ∴ h x Area of occupied by one patient = 8.8

    Correct Option: A

    Let the breadth and height of room be m and h m, respectively.
    Then, according to the question, l x b = n x Area occupied by one patient
    ⇒ 14 x b = 56 x 22
    ∴ b = (56 x 22)/14 = 8.8 m3
    ∴ h x Area of occupied by one patient = 8.8
    h = 8.8/2.2
    h = 4 m

  1. A right circular metal cone (solid) is 8 cm high and the radius is 2 cm. It is metaled and recast into a sphere. What is the radius of the sphere ?








  1. View Hint View Answer Discuss in Forum

    Given that, the height and radius of a right circular ,etal cone (solid) are 8 cm and 2 cm, respectively.
    i.e., h = 8 cm and r = 2 cm
    Let the radius of the sphere is R
    Then, by condition,
    1/3 π r2 h = 4/3 π R3 h
    ⇒ 4 x 8 = 4 R3

    Correct Option: A

    Given that, the height and radius of a right circular ,etal cone (solid) are 8 cm and 2 cm, respectively.
    i.e., h = 8 cm and r = 2 cm
    Let the radius of the sphere is R
    Then, by condition,
    1/3 π r2 h = 4/3 π R3 h
    ⇒ 4 x 8 = 4 R3
    ⇒ R3 = (2)3 = ⇒ R = 2
    ∴ Radius odf the sphere = 2 cm