A container shaped like a right circular cylinder having diamete

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 Multiple Choice QuestionsShort Answer Type

421.

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see the given Fig.).



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422.

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

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423.

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14)

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424.

A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

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425. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements and π = 3.14.
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426.

A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

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427.

Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

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428. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
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429.

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. 

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430.

A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.


Let R cm be the radius and H cm be the height of a container, then

R space equals space 12 over 2 = 6 cm and H = 15 cm
Therefore, volume of cylindrical container

                 
equals space πR squared straight H
equals space left parenthesis straight pi space straight x space 6 space straight x space 6 space straight x space 15 right parenthesis space cm cubed
equals space 540 space straight pi space cm cubed
Let rcm be the radius and h cm be the height of a cone, then
 
            straight r subscript 1 equals 6 over 2 equals 3 space cm space and space straight h space equals space 12 space cm comma
Therefore, Volume of conical part

                 equals space 1 third space straight pi space straight r subscript 1 squared straight h
equals space open parentheses 1 third πx 3 space straight x space 3 space straight x space 12 close parentheses space cm cubed
equals space 36 space straight pi space space cm cubed
Let rcm  be the radius of hemispherical part then  straight r subscript 2 equals 3 space cm comma space space space left square bracket because space straight r subscript 1 equals straight r subscript 2 right square bracket
Therefore, Volume of Hemispherical part

               equals space 2 over 3 space straight pi space straight r subscript 1 squared straight h
equals space open parentheses 2 over 3 straight pi space straight x space 3 space straight x space 3 space straight x space 3 close parentheses space cm cubed
equals space 18 space straight pi space space cm cubed
Now, Volume of ice-creme cone with hemispherical top

                       =  Volume of cone + Volume of Hemisphere
                       = (36 straight pi + 18 straight pi) cm3
                       = 54 straight pi cm
Therefore,
  the required no. of such cones

                    equals space fraction numerator Volume space of space cylindrical space container over denominator Volume space of space cone space with space hemispherical space top end fraction
equals space fraction numerator 540 straight pi over denominator 54 straight pi end fraction space equals space 10

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