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Mensuration-3D Test 275
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Mensuration-3D Test 275

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  • Question 1/10
    1 / -0.25
    A solid cylinder has radius of base 14 cm and height 15 cm. 4 identical cylinders are cut from each base as shown in the given figure. The height of the small cylinder is 5 cm. what is the total surface area (in cm2) of the remaining part?
    Solutions
    Since 4 identical cylinders has been cut from each base. So total no. of bases = 8
    C.S.A. of larger cylinder + 8 × C.S.A. of smaller cylinder + Area of both bases

    = 3432 cm2
    Note :
    The area which we have to reduce as the surface area of one base of the smaller cylinder, will also be added as when the smaller cylinder is cut out from the base it will leave the similar base area inside the hollow part.
    And won't have to calculate that portion.
  • Question 2/10
    1 / -0.25
    If the volume of a cube is 1728 cm3, then what is the total surface area (in cm2) of the cube
    Solutions
    Let the side of cube be a
    Then a3=1728
    A=12
    Then total surface area of cube =6a2=6× 144=864cm3
  • Question 3/10
    1 / -0.25
    A cone of radius 3.5 cm and height 12 cm is completely filled with water. This water is emptied into an empty cylindrical vessel of radius 7 cm. What will be the height of water in this vessel?
    Solutions
    Volume of cone = π r2h/3 = (π 3.52 × 12 )/3
    Volume of cylinder = π r2h = π 72 × h
    ⇒ (π 3.52 × 12)/3 = π 72 × h
    ⇒ h = 1 cm
  • Question 4/10
    1 / -0.25
    A tank is filled with water upto certain height and its base is of 65 m× 16 m. If some men dip into the tank and increase the level of water by 25 cm and if one man displaces 5 m3 of water, then find how many men dipped in the tank.
    Solutions
    Short Trick:-
    Height increased = 25cm = 0.25m
    So, volume displaced = 65 m× 16 m × 0.25m 
    One man displace 5 m3 of water
    Total men = 65 m× 16 m × 0.25m / 5 m3
     = 52 men
    Basic Method:-
    Let ‘h’ m be the original height of water in tank and h’ m be the height when some N(let) men dipped into it.
    ⇒ h’ – h = 0.25 m
    Volume of water in tank originally = 65*16*h = 1040h m3
    Volume of water after dipping = 65*16*h’= 1040h’ m3
    ⇒ volume of water displaced = N*5 = 5N m3
    ⇒ ∴ 1040h’ – 1040h = 5N
    ⇒ 1040*(0.25) = 5N
    ⇒ N= 52 men
  • Question 5/10
    1 / -0.25
    If the height of a given cone becomes four times and the radius of the base becomes twice, then what is the ratio of the volume of the given cone and the volume of the new cone?
    Solutions
    Volume of cone =
    Volume of new cone =
    Therefore, the ratio of the volume of the given cone and the volume of the new cone:
    ⇒(/(
    ⇒(/(
    ⇒1:16
    Hence, the correct option is C
  • Question 6/10
    1 / -0.25
    Sum of lengths of all edges of a cube is 84 cm, find its volume?
    Solutions
    Given: sum of all edges of the cube, 12a = 84 cm
    ⇒ a = 7 cm
    Then, its volume = a3
    = 73
    = 343 cubic cm
  • Question 7/10
    1 / -0.25
    The height of a right prism is 7 cm. The base of the prism is a square having diagonal 6 cm. Find the volume of prism.
    Solutions
    Side of the sqaure = (6/√2)
    Volume of prism = Area of base     × Height
    = (Side)2×Height = = 126 cm3
  • Question 8/10
    1 / -0.25
    Find the volume (in cm3) of a hemisphere of diameter 21 cm.
    Solutions
    Given, diameter of hemisphere = 21 cm
    Radius of hemisphere = 21/2 cm
    Volume of hemisphere =
    =
    = 2425.5 cm3
    Hence, option B is the correct option
  • Question 9/10
    1 / -0.25
    Three solid metallic spheres of radius 6cm, 8cm and 10cm are melted and recast into a new solid sphere. The volume of this new sphere will be how much?
    Solutions
    We can equate the volumes of the melted spheres & the newly formed sphere out of those.
    Let us assume the radius of the newly formed sphere to be r.
    And volume of a sphere = 
     
    r3 = 1728
    r = 12
    Volume = 
  • Question 10/10
    1 / -0.25
    The total surface area of a hemisphere is 41.58 sq cm. Find its curved surface area.
    Solutions
    Total surface area = 3 × π × r2
    Curved surface area = 2 × π × r2 = 41.58 × (2/3) = 27.72 cm2
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