# Conversion of Solid from One Shape to Another

You must have seen in our daily life that some solids are melted and recast into another solid for various purposes. For example a logwood is cut to make small pencils, which are in the shape of a cylinder surmounted by a cone. Also, for your school activities you must have converted a cylindrical shaped candle to form a spherical ball or conical lamp etc. by melting them. Now, in this topic we are going to learn as to how to find the volume and surface area of the converted solid from the original solid.

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An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls isÂ Â of the radius of the original ball, how many such balls are made? Compare the surface area, of all smaller balls combined together with that of the original ball.

Radius of the spherical ball = R

Radius of the smaller spherical ball (r) =Â R

Volume of a spherical ball =Â

Volume of a smaller spherical ball =Â

Number of balls =Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ==Â = 64

=Â = 16

Therefore the surface area of the spherical ball is 16 times bigger than surface area of the smaller spherical balls.

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An inverted conical vessel of radius of 6 cm and height 8 cm is filled with water. A sphere is lowered into the vessel. Find

(i) the radius of the sphere, given that when it touches the sides, the highest point of the sphere is in level with the base of the cone.

(ii) volume of water that flows out of conical vessel consequent to lowering the sphere in it.

(i) InÂ Î”Â AOC andÂ Î”Â ABD,

âˆ Â OAC =Â âˆ Â DAB (Common)âˆ Â OCA =Â âˆ Â ABD = 90

^{o}

Î”Â AOCÂ âˆ¼Â Î”Â ADB (AA Similarity)

â€¦â€¦â€¦â€¦..(i)

InÂ Î”Â ADB,

AD^{2}Â = AB^{2}Â + BD^{2}

Â AD =Â

Â Â Â Â Â =

Â Â Â Â Â =Â Â = 10 cm

AO = AB â€“ OB = 8 - OC

Substituting in (i)

10 OC = 48 â€“ 6 OC

16 OC = 48

Â Â Â OC = 3 cm

(ii) Volume of a sphere =Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Volume of a sphere = 56. 571 cm^{3}

Volume of water that flows out of conical vessel consequent to lowering the sphere in it = Volume of the Sphere =56.571 cm^{3}.

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From a sphere of radius 10 cm, a right circular cylinder diameter of whose base is 12 cm, is carved out. Calculate the volume of the right circular cylinder correct to 2 decimal places.

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From the figure, OB = OC = 10 cm (radius of the sphere)

InÂ Î”Â AOB,

OB^{2}Â = OA^{2}Â + AB^{2}

10^{2}Â = OA^{2}Â + 6^{2}

Â OA =Â

Â Â Â Â Â Â =Â

Â Â Â Â Â Â == 8 cm

Height of the cylinder = 2Â Ã—Â OA

Volume of a right circular cylinder =

^{2}h

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =Â

Ã—Â 6^{2}Â Ã—Â 16

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =Â Ã—Â 36Â Ã—Â 16

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1810.29 cm^{3}

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Find the number of coins, 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

Radius of a coin r = 0.75 cm and height h = 0.2 cm.

Volume of a coin =Ï€r^{2}

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =Â Â cm^{3}

Radius of a cylinder r = 2.25 cm and height h = 10 cm.Â

Volume of the cylinder =Â Â cm^{3}

Therefore number of coins =Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 3Ã—Â 3Â Ã—Â 50

Number of coins = 450

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A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of theÂ cylinder.

Let the required thickness of the cylinder be x cm, the external and internal radius of the hollow cylinder are 5 cm and (5 - x) cm.

Radius of the sphere = 6 cm

Then, volume of sphere = volume of the solid in hollow cylinder

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â

Ï€Â Ã—Â 6Â Ã—Â 6Â Ã—Â 6 =Â Ï€Â Ã—Â 32[5^{2}Â - (5 - x)

^{2}]

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 4

Ã—Â 6Â Ã—Â 6Â Ã—Â 6 = 3Â Ã—Â 32[(5Â Ã—Â 5) - (5 â€“ x)(5 - x)]Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 864 = 96 [25 - (5 â€“ x)(5 - x)]

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 9 = [25 - (5 â€“ x)(5 - x)Â Â Â Â Â

Â Â Â Â Â Â Â Â -16 = - (5 â€“ x)(5 - x)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 25 â€“ 10x + x^{2}Â = 16

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x^{2}Â â€“ 10 x + 9 = 0

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (x â€“ 9)(x - 1) = 0

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x = 9 or x = 1

Therefore the uniform thickness of the cylinder is 9 or 1cm.

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The rainwater from a roof 22 mÂ Ã—Â 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the vessel is just full, find the rainfall in cm.

Length of the roof = 22 m

Breadth of the roof = 20 mÂ âˆ´Â Area of the roof = 22Â Ã—Â 20 = 440 m^{2}Â = 440Â Ã—Â 100Â Ã—Â 100 cm^{2}

Diameter of the base of the cylinder = 2 mÂ âˆ´Â Radius of the base of the cylinder = 1 m

Height of the cylinder = 3.5 mÂ âˆ´Â Volume of the cylinder =Â Ï€Â r^{2}Â h

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =Â Ã—Â 1Â Ã—Â 1Â Ã—Â 3.5 m^{2}

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 11 m^{3}

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 11Â Ã—Â 100Â Ã—Â 100Â Ã—Â 100cm^{3}

Let the rainfall be h cm.Â âˆ´Â Area of the roofÂ Ã—Â rainfall in cm = Volume of the cylinderÂ âˆ´Â 440Â Ã—Â 100Ã—Â 100Â Ã—Â h = 11Â Ã—Â 100Â Ã—Â 100Â Ã—Â 100

âˆ´Â h =Â = 2.5 cm

âˆ´Â Rainfall = 2.5 cm.

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