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Surface Areas and Volumes

Short Answer Type

471. Three equal cubes of side 4 cm are joined end to end. Find the surface area of the resulting cuboid.

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472.

Three cubes of metal whose edges are 6 cm, 8 cm and 10 cm are melted and a new cube is made (i) Find the length of the edge of the new cube.
(ii) Find the total surface area of the new cube.

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473. Two cubes each of 12 cm edge are joined end to end. Find the surface area of the resulting cuboid.

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474. A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.

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475. A right circular cone is 8 cm heigh and the radius of its base is 2 cm. The cone is melted and re-cast into a sphere. Determine the diameter of the sphere.

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476. A solid cylinder of metal 8 m high and 4 m diameter is melted and recast into a cone of diameter 3 m. Find the height of the cone.

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477. A cone is 8.4 cm high and the radius of its base is 2.1 cm. it is melted and recast into a sphere. Find the radius of the sphere.

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478. A well of diameter 14 m is dug 15 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 7 m to form an embankment. Find the height of the embankment.

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479. An agriculture field is in the form of a rectangle of length 20 m, and width 14 m. A 10 m deep well of diameter 7 m is dug in a comer of the field and the earth taken out of the well is spread evenly over the remaining part of the field. Find the rise in the level of the field.

Since shape of the well is just like the shape of cylinder, so we consider well as cylinder.
Let r m be the radius and h m be the height of the well, then
$straight r space equals fraction numerator 7 degree over denominator 2 end fraction straight m comma space straight h space left parenthesis depth right parenthesis space equals space 10 straight m$
Now,
Volume of earth dug out = $πr squared straight h$

$equals space open parentheses 22 over 7 x 7 over 2 x 7 over 2 x 10 close parentheses m cubed equals 385 space m cubed$
Also, we have,
Length of the rectangular field = 20 m
Breadth of the field = 14 m
Therefore, are of the field = (20 x 14) m²
= 280 m²
And
Area of the remaining part of the field
= Area of field - Area of the base

$equals space open parentheses 280 minus 77 over 2 close parentheses straight m cubed$
Since,
Volume of the raised field
= Volume of the earth dug out
$rightwards double arrow space space open parentheses 483 over 2 close parentheses straight h space equals space 358 rightwards double arrow space space space straight h space equals space fraction numerator 358 space straight x space 2 over denominator 483 end fraction rightwards double arrow space space space straight h space equals space 1.6 space left parenthesis approx right parenthesis$
Hence, rise in the level of the field = 1.6 m (apporx.)

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480. A metallic sphere of radius 10.5 cm is melted and recast into 5 small cones each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.

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