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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.

Let h m be the height and r m be the radius of the well, then

h = 15 m, r =

14 over 2 equals space 7 space m.

Now,
Volume of earth dug out =

πr squared straight h

equals space open parentheses 22 over 7 straight x space 7 space straight x space 7 space straight x space 15 close parentheses space straight m cubed

[

because

shape of well is similar to cylinder]
= ( 22 x 7 x 15) m³

= 2310 m³

Let H m be the required height of the embankment.
Since the shape of the embankment will be like the shape of cylinder of internal radius (r₂) 7 m and external radius (R₁) = 7 + 7 = 14 m. Now,
the volume of embankment
= volume of the earth dug out
⇒ π(R₁² –r₁²) H = 2310
⇒ π(14² – 7²) H = 2310
⇒ π (196 – 49) H = 2310
⇒ π(147) H = 2310

$rightwards double arrow space space space space space space space space straight H space equals space fraction numerator 2310 space straight x space 7 over denominator 147 space straight x space 22 end fraction space space space space space space space space space space space space space space space equals space 16170 over 3234 space equals space 5 space straight m$

Hence, the height of the embakment = 5m
Alternative Method :
Height of the embankment

$equals fraction numerator volume space of space the space earth space dug space out over denominator Area space of space the space embankment end fraction$

Note :
Area of the embankment
= Area of ring
= π R² – π r²= π (R² – π r²)
= π (R + r) (R – r)

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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.

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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.

346 Views